\(\int \frac {(e \cos (c+d x))^{7/2}}{(a+b \sin (c+d x))^2} \, dx\) [586]

   Optimal result
   Rubi [A] (verified)
   Mathematica [C] (warning: unable to verify)
   Maple [C] (warning: unable to verify)
   Fricas [F(-1)]
   Sympy [F(-1)]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 25, antiderivative size = 473 \[ \int \frac {(e \cos (c+d x))^{7/2}}{(a+b \sin (c+d x))^2} \, dx=-\frac {5 a \sqrt [4]{-a^2+b^2} e^{7/2} \arctan \left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{2 b^{7/2} d}-\frac {5 a \sqrt [4]{-a^2+b^2} e^{7/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{2 b^{7/2} d}+\frac {5 \left (3 a^2-b^2\right ) e^4 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 b^4 d \sqrt {e \cos (c+d x)}}-\frac {5 a^2 \left (a^2-b^2\right ) e^4 \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (c+d x),2\right )}{2 b^4 \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \cos (c+d x)}}-\frac {5 a^2 \left (a^2-b^2\right ) e^4 \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (c+d x),2\right )}{2 b^4 \left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \cos (c+d x)}}+\frac {5 e^3 \sqrt {e \cos (c+d x)} (3 a-b \sin (c+d x))}{3 b^3 d}-\frac {e (e \cos (c+d x))^{5/2}}{b d (a+b \sin (c+d x))} \]

[Out]

-5/2*a*(-a^2+b^2)^(1/4)*e^(7/2)*arctan(b^(1/2)*(e*cos(d*x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2))/b^(7/2)/d-5/2*a*
(-a^2+b^2)^(1/4)*e^(7/2)*arctanh(b^(1/2)*(e*cos(d*x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2))/b^(7/2)/d-e*(e*cos(d*x
+c))^(5/2)/b/d/(a+b*sin(d*x+c))+5/3*(3*a^2-b^2)*e^4*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(
sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)/b^4/d/(e*cos(d*x+c))^(1/2)-5/2*a^2*(a^2-b^2)*e^4*(cos(1/2*d*x+1/2
*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2*b/(b-(-a^2+b^2)^(1/2)),2^(1/2))*cos(d*x+c)^(1/
2)/b^4/d/(a^2-b*(b-(-a^2+b^2)^(1/2)))/(e*cos(d*x+c))^(1/2)-5/2*a^2*(a^2-b^2)*e^4*(cos(1/2*d*x+1/2*c)^2)^(1/2)/
cos(1/2*d*x+1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2*b/(b+(-a^2+b^2)^(1/2)),2^(1/2))*cos(d*x+c)^(1/2)/b^4/d/(a^2
-b*(b+(-a^2+b^2)^(1/2)))/(e*cos(d*x+c))^(1/2)+5/3*e^3*(3*a-b*sin(d*x+c))*(e*cos(d*x+c))^(1/2)/b^3/d

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 473, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {2772, 2944, 2946, 2721, 2720, 2781, 2886, 2884, 335, 218, 214, 211} \[ \int \frac {(e \cos (c+d x))^{7/2}}{(a+b \sin (c+d x))^2} \, dx=-\frac {5 a e^{7/2} \sqrt [4]{b^2-a^2} \arctan \left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{2 b^{7/2} d}-\frac {5 a e^{7/2} \sqrt [4]{b^2-a^2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{2 b^{7/2} d}+\frac {5 e^4 \left (3 a^2-b^2\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 b^4 d \sqrt {e \cos (c+d x)}}-\frac {5 a^2 e^4 \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} (c+d x),2\right )}{2 b^4 d \left (a^2-b \left (b-\sqrt {b^2-a^2}\right )\right ) \sqrt {e \cos (c+d x)}}-\frac {5 a^2 e^4 \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} (c+d x),2\right )}{2 b^4 d \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {e \cos (c+d x)}}+\frac {5 e^3 \sqrt {e \cos (c+d x)} (3 a-b \sin (c+d x))}{3 b^3 d}-\frac {e (e \cos (c+d x))^{5/2}}{b d (a+b \sin (c+d x))} \]

[In]

Int[(e*Cos[c + d*x])^(7/2)/(a + b*Sin[c + d*x])^2,x]

[Out]

(-5*a*(-a^2 + b^2)^(1/4)*e^(7/2)*ArcTan[(Sqrt[b]*Sqrt[e*Cos[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/(2*b^(7/
2)*d) - (5*a*(-a^2 + b^2)^(1/4)*e^(7/2)*ArcTanh[(Sqrt[b]*Sqrt[e*Cos[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/
(2*b^(7/2)*d) + (5*(3*a^2 - b^2)*e^4*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(3*b^4*d*Sqrt[e*Cos[c + d*x
]]) - (5*a^2*(a^2 - b^2)*e^4*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (c + d*x)/2, 2])/(2*b
^4*(a^2 - b*(b - Sqrt[-a^2 + b^2]))*d*Sqrt[e*Cos[c + d*x]]) - (5*a^2*(a^2 - b^2)*e^4*Sqrt[Cos[c + d*x]]*Ellipt
icPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (c + d*x)/2, 2])/(2*b^4*(a^2 - b*(b + Sqrt[-a^2 + b^2]))*d*Sqrt[e*Cos[c + d
*x]]) + (5*e^3*Sqrt[e*Cos[c + d*x]]*(3*a - b*Sin[c + d*x]))/(3*b^3*d) - (e*(e*Cos[c + d*x])^(5/2))/(b*d*(a + b
*Sin[c + d*x]))

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 218

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]},
Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !Gt
Q[a/b, 0]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
 && FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 2721

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*Sin[c + d*x])^n/Sin[c + d*x]^n, Int[Sin[c + d*x]
^n, x], x] /; FreeQ[{b, c, d}, x] && LtQ[-1, n, 1] && IntegerQ[2*n]

Rule 2772

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[g*(g*C
os[e + f*x])^(p - 1)*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Dist[g^2*((p - 1)/(b*(m + 1))), Int[(g
*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1)*Sin[e + f*x], x], x] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a
^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && IntegersQ[2*m, 2*p]

Rule 2781

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> With[{q = Rt[
-a^2 + b^2, 2]}, Dist[-a/(2*q), Int[1/(Sqrt[g*Cos[e + f*x]]*(q + b*Cos[e + f*x])), x], x] + (Dist[b*(g/f), Sub
st[Int[1/(Sqrt[x]*(g^2*(a^2 - b^2) + b^2*x^2)), x], x, g*Cos[e + f*x]], x] - Dist[a/(2*q), Int[1/(Sqrt[g*Cos[e
 + f*x]]*(q - b*Cos[e + f*x])), x], x])] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0]

Rule 2884

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a,
 b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]

Rule 2886

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist
[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt[c + d*Sin[e + f*x]], Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d/
(c + d))*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && N
eQ[c^2 - d^2, 0] &&  !GtQ[c + d, 0]

Rule 2944

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
 + (f_.)*(x_)]), x_Symbol] :> Simp[g*(g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1)*((b*c*(m + p + 1) -
 a*d*p + b*d*(m + p)*Sin[e + f*x])/(b^2*f*(m + p)*(m + p + 1))), x] + Dist[g^2*((p - 1)/(b^2*(m + p)*(m + p +
1))), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^m*Simp[b*(a*d*m + b*c*(m + p + 1)) + (a*b*c*(m + p + 1
) - d*(a^2*p - b^2*(m + p)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2,
0] && GtQ[p, 1] && NeQ[m + p, 0] && NeQ[m + p + 1, 0] && IntegerQ[2*m]

Rule 2946

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]))/((a_) + (b_.)*sin[(e_.) + (
f_.)*(x_)]), x_Symbol] :> Dist[d/b, Int[(g*Cos[e + f*x])^p, x], x] + Dist[(b*c - a*d)/b, Int[(g*Cos[e + f*x])^
p/(a + b*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {e (e \cos (c+d x))^{5/2}}{b d (a+b \sin (c+d x))}-\frac {\left (5 e^2\right ) \int \frac {(e \cos (c+d x))^{3/2} \sin (c+d x)}{a+b \sin (c+d x)} \, dx}{2 b} \\ & = \frac {5 e^3 \sqrt {e \cos (c+d x)} (3 a-b \sin (c+d x))}{3 b^3 d}-\frac {e (e \cos (c+d x))^{5/2}}{b d (a+b \sin (c+d x))}-\frac {\left (5 e^4\right ) \int \frac {-a b-\frac {1}{2} \left (3 a^2-b^2\right ) \sin (c+d x)}{\sqrt {e \cos (c+d x)} (a+b \sin (c+d x))} \, dx}{3 b^3} \\ & = \frac {5 e^3 \sqrt {e \cos (c+d x)} (3 a-b \sin (c+d x))}{3 b^3 d}-\frac {e (e \cos (c+d x))^{5/2}}{b d (a+b \sin (c+d x))}-\frac {\left (5 a \left (a^2-b^2\right ) e^4\right ) \int \frac {1}{\sqrt {e \cos (c+d x)} (a+b \sin (c+d x))} \, dx}{2 b^4}+\frac {\left (5 \left (3 a^2-b^2\right ) e^4\right ) \int \frac {1}{\sqrt {e \cos (c+d x)}} \, dx}{6 b^4} \\ & = \frac {5 e^3 \sqrt {e \cos (c+d x)} (3 a-b \sin (c+d x))}{3 b^3 d}-\frac {e (e \cos (c+d x))^{5/2}}{b d (a+b \sin (c+d x))}-\frac {\left (5 a^2 \sqrt {-a^2+b^2} e^4\right ) \int \frac {1}{\sqrt {e \cos (c+d x)} \left (\sqrt {-a^2+b^2}-b \cos (c+d x)\right )} \, dx}{4 b^4}-\frac {\left (5 a^2 \sqrt {-a^2+b^2} e^4\right ) \int \frac {1}{\sqrt {e \cos (c+d x)} \left (\sqrt {-a^2+b^2}+b \cos (c+d x)\right )} \, dx}{4 b^4}-\frac {\left (5 a \left (a^2-b^2\right ) e^5\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} \left (\left (a^2-b^2\right ) e^2+b^2 x^2\right )} \, dx,x,e \cos (c+d x)\right )}{2 b^3 d}+\frac {\left (5 \left (3 a^2-b^2\right ) e^4 \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 b^4 \sqrt {e \cos (c+d x)}} \\ & = \frac {5 \left (3 a^2-b^2\right ) e^4 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 b^4 d \sqrt {e \cos (c+d x)}}+\frac {5 e^3 \sqrt {e \cos (c+d x)} (3 a-b \sin (c+d x))}{3 b^3 d}-\frac {e (e \cos (c+d x))^{5/2}}{b d (a+b \sin (c+d x))}-\frac {\left (5 a \left (a^2-b^2\right ) e^5\right ) \text {Subst}\left (\int \frac {1}{\left (a^2-b^2\right ) e^2+b^2 x^4} \, dx,x,\sqrt {e \cos (c+d x)}\right )}{b^3 d}-\frac {\left (5 a^2 \sqrt {-a^2+b^2} e^4 \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \left (\sqrt {-a^2+b^2}-b \cos (c+d x)\right )} \, dx}{4 b^4 \sqrt {e \cos (c+d x)}}-\frac {\left (5 a^2 \sqrt {-a^2+b^2} e^4 \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \left (\sqrt {-a^2+b^2}+b \cos (c+d x)\right )} \, dx}{4 b^4 \sqrt {e \cos (c+d x)}} \\ & = \frac {5 \left (3 a^2-b^2\right ) e^4 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 b^4 d \sqrt {e \cos (c+d x)}}+\frac {5 a^2 \sqrt {-a^2+b^2} e^4 \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (c+d x),2\right )}{2 b^4 \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}-\frac {5 a^2 \sqrt {-a^2+b^2} e^4 \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (c+d x),2\right )}{2 b^4 \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}+\frac {5 e^3 \sqrt {e \cos (c+d x)} (3 a-b \sin (c+d x))}{3 b^3 d}-\frac {e (e \cos (c+d x))^{5/2}}{b d (a+b \sin (c+d x))}-\frac {\left (5 a \sqrt {-a^2+b^2} e^4\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} e-b x^2} \, dx,x,\sqrt {e \cos (c+d x)}\right )}{2 b^3 d}-\frac {\left (5 a \sqrt {-a^2+b^2} e^4\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} e+b x^2} \, dx,x,\sqrt {e \cos (c+d x)}\right )}{2 b^3 d} \\ & = -\frac {5 a \sqrt [4]{-a^2+b^2} e^{7/2} \arctan \left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{2 b^{7/2} d}-\frac {5 a \sqrt [4]{-a^2+b^2} e^{7/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{2 b^{7/2} d}+\frac {5 \left (3 a^2-b^2\right ) e^4 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 b^4 d \sqrt {e \cos (c+d x)}}+\frac {5 a^2 \sqrt {-a^2+b^2} e^4 \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (c+d x),2\right )}{2 b^4 \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}-\frac {5 a^2 \sqrt {-a^2+b^2} e^4 \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (c+d x),2\right )}{2 b^4 \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}+\frac {5 e^3 \sqrt {e \cos (c+d x)} (3 a-b \sin (c+d x))}{3 b^3 d}-\frac {e (e \cos (c+d x))^{5/2}}{b d (a+b \sin (c+d x))} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.

Time = 20.71 (sec) , antiderivative size = 1956, normalized size of antiderivative = 4.14 \[ \int \frac {(e \cos (c+d x))^{7/2}}{(a+b \sin (c+d x))^2} \, dx=\frac {(e \cos (c+d x))^{7/2} \sec ^3(c+d x) \left (-\frac {2 \sin (c+d x)}{3 b^2}+\frac {a^2-b^2}{b^3 (a+b \sin (c+d x))}\right )}{d}+\frac {(e \cos (c+d x))^{7/2} \left (-\frac {8 a b \left (a+b \sqrt {1-\cos ^2(c+d x)}\right ) \left (\frac {5 a \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},1,\frac {5}{4},\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{-a^2+b^2}\right ) \sqrt {\cos (c+d x)}}{\sqrt {1-\cos ^2(c+d x)} \left (5 \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},1,\frac {5}{4},\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{-a^2+b^2}\right )-2 \left (2 b^2 \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},2,\frac {9}{4},\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{-a^2+b^2}\right )+\left (-a^2+b^2\right ) \operatorname {AppellF1}\left (\frac {5}{4},\frac {3}{2},1,\frac {9}{4},\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{-a^2+b^2}\right )\right ) \cos ^2(c+d x)\right ) \left (a^2+b^2 \left (-1+\cos ^2(c+d x)\right )\right )}-\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \sqrt {b} \left (2 \arctan \left (1-\frac {(1+i) \sqrt {b} \sqrt {\cos (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )-2 \arctan \left (1+\frac {(1+i) \sqrt {b} \sqrt {\cos (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )+\log \left (\sqrt {-a^2+b^2}-(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\cos (c+d x)}+i b \cos (c+d x)\right )-\log \left (\sqrt {-a^2+b^2}+(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\cos (c+d x)}+i b \cos (c+d x)\right )\right )}{\left (-a^2+b^2\right )^{3/4}}\right ) \sin (c+d x)}{\sqrt {1-\cos ^2(c+d x)} (a+b \sin (c+d x))}+\frac {6 a b \left (a+b \sqrt {1-\cos ^2(c+d x)}\right ) \cos (2 (c+d x)) \left (\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \left (-2 a^2+b^2\right ) \arctan \left (1-\frac {(1+i) \sqrt {b} \sqrt {\cos (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )}{b^{3/2} \left (-a^2+b^2\right )^{3/4}}-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \left (-2 a^2+b^2\right ) \arctan \left (1+\frac {(1+i) \sqrt {b} \sqrt {\cos (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )}{b^{3/2} \left (-a^2+b^2\right )^{3/4}}+\frac {4 \sqrt {\cos (c+d x)}}{b}-\frac {4 a \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},1,\frac {9}{4},\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{-a^2+b^2}\right ) \cos ^{\frac {5}{2}}(c+d x)}{5 \left (a^2-b^2\right )}+\frac {10 a \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},1,\frac {5}{4},\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{-a^2+b^2}\right ) \sqrt {\cos (c+d x)}}{\sqrt {1-\cos ^2(c+d x)} \left (5 \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},1,\frac {5}{4},\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{-a^2+b^2}\right )-2 \left (2 b^2 \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},2,\frac {9}{4},\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{-a^2+b^2}\right )+\left (-a^2+b^2\right ) \operatorname {AppellF1}\left (\frac {5}{4},\frac {3}{2},1,\frac {9}{4},\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{-a^2+b^2}\right )\right ) \cos ^2(c+d x)\right ) \left (a^2+b^2 \left (-1+\cos ^2(c+d x)\right )\right )}+\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) \left (-2 a^2+b^2\right ) \log \left (\sqrt {-a^2+b^2}-(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\cos (c+d x)}+i b \cos (c+d x)\right )}{b^{3/2} \left (-a^2+b^2\right )^{3/4}}-\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) \left (-2 a^2+b^2\right ) \log \left (\sqrt {-a^2+b^2}+(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\cos (c+d x)}+i b \cos (c+d x)\right )}{b^{3/2} \left (-a^2+b^2\right )^{3/4}}\right ) \sin (c+d x)}{\sqrt {1-\cos ^2(c+d x)} \left (-1+2 \cos ^2(c+d x)\right ) (a+b \sin (c+d x))}-\frac {2 \left (3 a^2-5 b^2\right ) \left (a+b \sqrt {1-\cos ^2(c+d x)}\right ) \left (\frac {5 b \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},-\frac {1}{2},1,\frac {5}{4},\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{-a^2+b^2}\right ) \sqrt {\cos (c+d x)} \sqrt {1-\cos ^2(c+d x)}}{\left (-5 \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},-\frac {1}{2},1,\frac {5}{4},\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{-a^2+b^2}\right )+2 \left (2 b^2 \operatorname {AppellF1}\left (\frac {5}{4},-\frac {1}{2},2,\frac {9}{4},\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{-a^2+b^2}\right )+\left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},1,\frac {9}{4},\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{-a^2+b^2}\right )\right ) \cos ^2(c+d x)\right ) \left (a^2+b^2 \left (-1+\cos ^2(c+d x)\right )\right )}+\frac {a \left (-2 \arctan \left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )+2 \arctan \left (1+\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-\log \left (\sqrt {a^2-b^2}-\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (c+d x)}+b \cos (c+d x)\right )+\log \left (\sqrt {a^2-b^2}+\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (c+d x)}+b \cos (c+d x)\right )\right )}{4 \sqrt {2} \sqrt {b} \left (a^2-b^2\right )^{3/4}}\right ) \sin ^2(c+d x)}{\left (1-\cos ^2(c+d x)\right ) (a+b \sin (c+d x))}\right )}{6 b^3 d \cos ^{\frac {7}{2}}(c+d x)} \]

[In]

Integrate[(e*Cos[c + d*x])^(7/2)/(a + b*Sin[c + d*x])^2,x]

[Out]

((e*Cos[c + d*x])^(7/2)*Sec[c + d*x]^3*((-2*Sin[c + d*x])/(3*b^2) + (a^2 - b^2)/(b^3*(a + b*Sin[c + d*x]))))/d
 + ((e*Cos[c + d*x])^(7/2)*((-8*a*b*(a + b*Sqrt[1 - Cos[c + d*x]^2])*((5*a*(a^2 - b^2)*AppellF1[1/4, 1/2, 1, 5
/4, Cos[c + d*x]^2, (b^2*Cos[c + d*x]^2)/(-a^2 + b^2)]*Sqrt[Cos[c + d*x]])/(Sqrt[1 - Cos[c + d*x]^2]*(5*(a^2 -
 b^2)*AppellF1[1/4, 1/2, 1, 5/4, Cos[c + d*x]^2, (b^2*Cos[c + d*x]^2)/(-a^2 + b^2)] - 2*(2*b^2*AppellF1[5/4, 1
/2, 2, 9/4, Cos[c + d*x]^2, (b^2*Cos[c + d*x]^2)/(-a^2 + b^2)] + (-a^2 + b^2)*AppellF1[5/4, 3/2, 1, 9/4, Cos[c
 + d*x]^2, (b^2*Cos[c + d*x]^2)/(-a^2 + b^2)])*Cos[c + d*x]^2)*(a^2 + b^2*(-1 + Cos[c + d*x]^2))) - ((1/8 - I/
8)*Sqrt[b]*(2*ArcTan[1 - ((1 + I)*Sqrt[b]*Sqrt[Cos[c + d*x]])/(-a^2 + b^2)^(1/4)] - 2*ArcTan[1 + ((1 + I)*Sqrt
[b]*Sqrt[Cos[c + d*x]])/(-a^2 + b^2)^(1/4)] + Log[Sqrt[-a^2 + b^2] - (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[C
os[c + d*x]] + I*b*Cos[c + d*x]] - Log[Sqrt[-a^2 + b^2] + (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Cos[c + d*x]
] + I*b*Cos[c + d*x]]))/(-a^2 + b^2)^(3/4))*Sin[c + d*x])/(Sqrt[1 - Cos[c + d*x]^2]*(a + b*Sin[c + d*x])) + (6
*a*b*(a + b*Sqrt[1 - Cos[c + d*x]^2])*Cos[2*(c + d*x)]*(((1/2 - I/2)*(-2*a^2 + b^2)*ArcTan[1 - ((1 + I)*Sqrt[b
]*Sqrt[Cos[c + d*x]])/(-a^2 + b^2)^(1/4)])/(b^(3/2)*(-a^2 + b^2)^(3/4)) - ((1/2 - I/2)*(-2*a^2 + b^2)*ArcTan[1
 + ((1 + I)*Sqrt[b]*Sqrt[Cos[c + d*x]])/(-a^2 + b^2)^(1/4)])/(b^(3/2)*(-a^2 + b^2)^(3/4)) + (4*Sqrt[Cos[c + d*
x]])/b - (4*a*AppellF1[5/4, 1/2, 1, 9/4, Cos[c + d*x]^2, (b^2*Cos[c + d*x]^2)/(-a^2 + b^2)]*Cos[c + d*x]^(5/2)
)/(5*(a^2 - b^2)) + (10*a*(a^2 - b^2)*AppellF1[1/4, 1/2, 1, 5/4, Cos[c + d*x]^2, (b^2*Cos[c + d*x]^2)/(-a^2 +
b^2)]*Sqrt[Cos[c + d*x]])/(Sqrt[1 - Cos[c + d*x]^2]*(5*(a^2 - b^2)*AppellF1[1/4, 1/2, 1, 5/4, Cos[c + d*x]^2,
(b^2*Cos[c + d*x]^2)/(-a^2 + b^2)] - 2*(2*b^2*AppellF1[5/4, 1/2, 2, 9/4, Cos[c + d*x]^2, (b^2*Cos[c + d*x]^2)/
(-a^2 + b^2)] + (-a^2 + b^2)*AppellF1[5/4, 3/2, 1, 9/4, Cos[c + d*x]^2, (b^2*Cos[c + d*x]^2)/(-a^2 + b^2)])*Co
s[c + d*x]^2)*(a^2 + b^2*(-1 + Cos[c + d*x]^2))) + ((1/4 - I/4)*(-2*a^2 + b^2)*Log[Sqrt[-a^2 + b^2] - (1 + I)*
Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Cos[c + d*x]] + I*b*Cos[c + d*x]])/(b^(3/2)*(-a^2 + b^2)^(3/4)) - ((1/4 - I/4)
*(-2*a^2 + b^2)*Log[Sqrt[-a^2 + b^2] + (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Cos[c + d*x]] + I*b*Cos[c + d*x
]])/(b^(3/2)*(-a^2 + b^2)^(3/4)))*Sin[c + d*x])/(Sqrt[1 - Cos[c + d*x]^2]*(-1 + 2*Cos[c + d*x]^2)*(a + b*Sin[c
 + d*x])) - (2*(3*a^2 - 5*b^2)*(a + b*Sqrt[1 - Cos[c + d*x]^2])*((5*b*(a^2 - b^2)*AppellF1[1/4, -1/2, 1, 5/4,
Cos[c + d*x]^2, (b^2*Cos[c + d*x]^2)/(-a^2 + b^2)]*Sqrt[Cos[c + d*x]]*Sqrt[1 - Cos[c + d*x]^2])/((-5*(a^2 - b^
2)*AppellF1[1/4, -1/2, 1, 5/4, Cos[c + d*x]^2, (b^2*Cos[c + d*x]^2)/(-a^2 + b^2)] + 2*(2*b^2*AppellF1[5/4, -1/
2, 2, 9/4, Cos[c + d*x]^2, (b^2*Cos[c + d*x]^2)/(-a^2 + b^2)] + (a^2 - b^2)*AppellF1[5/4, 1/2, 1, 9/4, Cos[c +
 d*x]^2, (b^2*Cos[c + d*x]^2)/(-a^2 + b^2)])*Cos[c + d*x]^2)*(a^2 + b^2*(-1 + Cos[c + d*x]^2))) + (a*(-2*ArcTa
n[1 - (Sqrt[2]*Sqrt[b]*Sqrt[Cos[c + d*x]])/(a^2 - b^2)^(1/4)] + 2*ArcTan[1 + (Sqrt[2]*Sqrt[b]*Sqrt[Cos[c + d*x
]])/(a^2 - b^2)^(1/4)] - Log[Sqrt[a^2 - b^2] - Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Cos[c + d*x]] + b*Cos[c
+ d*x]] + Log[Sqrt[a^2 - b^2] + Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Cos[c + d*x]] + b*Cos[c + d*x]]))/(4*Sq
rt[2]*Sqrt[b]*(a^2 - b^2)^(3/4)))*Sin[c + d*x]^2)/((1 - Cos[c + d*x]^2)*(a + b*Sin[c + d*x]))))/(6*b^3*d*Cos[c
 + d*x]^(7/2))

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 16.28 (sec) , antiderivative size = 1962, normalized size of antiderivative = 4.15

method result size
default \(\text {Expression too large to display}\) \(1962\)

[In]

int((e*cos(d*x+c))^(7/2)/(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

(8*e^4*a*b*(1/2/b^4/e*(e*(2*cos(1/2*d*x+1/2*c)^2-1))^(1/2)-2*(a^2-b^2)/b^4*(e^2*(a^2-b^2)/b^2)^(1/4)*2^(1/2)*(
ln((2*e*cos(1/2*d*x+1/2*c)^2-e+(e^2*(a^2-b^2)/b^2)^(1/4)*(2*e*cos(1/2*d*x+1/2*c)^2-e)^(1/2)*2^(1/2)+(e^2*(a^2-
b^2)/b^2)^(1/2))/(2*e*cos(1/2*d*x+1/2*c)^2-e-(e^2*(a^2-b^2)/b^2)^(1/4)*(2*e*cos(1/2*d*x+1/2*c)^2-e)^(1/2)*2^(1
/2)+(e^2*(a^2-b^2)/b^2)^(1/2)))+2*arctan((2^(1/2)*(2*e*cos(1/2*d*x+1/2*c)^2-e)^(1/2)+(e^2*(a^2-b^2)/b^2)^(1/4)
)/(e^2*(a^2-b^2)/b^2)^(1/4))+2*arctan((2^(1/2)*(2*e*cos(1/2*d*x+1/2*c)^2-e)^(1/2)-(e^2*(a^2-b^2)/b^2)^(1/4))/(
e^2*(a^2-b^2)/b^2)^(1/4)))/(16*a^2-16*b^2)/e+1/64*(a^4-2*a^2*b^2+b^4)/b^4*(3*(ln((2*e*cos(1/2*d*x+1/2*c)^2-e+(
e^2*(a^2-b^2)/b^2)^(1/4)*(2*e*cos(1/2*d*x+1/2*c)^2-e)^(1/2)*2^(1/2)+(e^2*(a^2-b^2)/b^2)^(1/2))/(2*e*cos(1/2*d*
x+1/2*c)^2-e-(e^2*(a^2-b^2)/b^2)^(1/4)*(2*e*cos(1/2*d*x+1/2*c)^2-e)^(1/2)*2^(1/2)+(e^2*(a^2-b^2)/b^2)^(1/2)))+
2*arctan((2^(1/2)*(2*e*cos(1/2*d*x+1/2*c)^2-e)^(1/2)+(e^2*(a^2-b^2)/b^2)^(1/4))/(e^2*(a^2-b^2)/b^2)^(1/4))+2*a
rctan((2^(1/2)*(2*e*cos(1/2*d*x+1/2*c)^2-e)^(1/2)-(e^2*(a^2-b^2)/b^2)^(1/4))/(e^2*(a^2-b^2)/b^2)^(1/4)))*(4*co
s(1/2*d*x+1/2*c)^4*b^2-4*cos(1/2*d*x+1/2*c)^2*b^2+a^2)*2^(1/2)*(e^2*(a^2-b^2)/b^2)^(1/4)+(8*a^2-8*b^2)*(2*e*co
s(1/2*d*x+1/2*c)^2-e)^(1/2))/e/(a-b)^2/(a+b)^2/(4*cos(1/2*d*x+1/2*c)^4*b^2-4*cos(1/2*d*x+1/2*c)^2*b^2+a^2))-2*
(e*(2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*e^4*(-1/3/b^4/(-2*sin(1/2*d*x+1/2*c)^4*e+sin(1/2*d*x
+1/2*c)^2*e)^(1/2)*(4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4*b^2-2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*b^
2-9*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^2+4*
(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*b^2)-1/16/
b^6*(5*a^4-6*a^2*b^2+b^4)*sum(1/_alpha/(2*_alpha^2-1)*(2^(1/2)/(e*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2)*arctan
h(1/2*e*(4*_alpha^2-3)/(4*a^2-3*b^2)*(4*a^2*cos(1/2*d*x+1/2*c)^2-3*cos(1/2*d*x+1/2*c)^2*b^2+b^2*_alpha^2-3*a^2
+2*b^2)*2^(1/2)/(e*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2)/(-e*(2*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^(1
/2))+8*b^2/a^2*_alpha*(_alpha^2-1)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-e*sin(1/2*
d*x+1/2*c)^2*(2*sin(1/2*d*x+1/2*c)^2-1))^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-4*b^2/a^2*(_alpha^2-1),2^(1/2)))
,_alpha=RootOf(4*_Z^4*b^2-4*_Z^2*b^2+a^2))+2*a^2*(a^4-2*a^2*b^2+b^4)/b^4*(1/2*b^2/(a^2-b^2)/a^2/e*cos(1/2*d*x+
1/2*c)*(-e*(2*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^(1/2)/(4*cos(1/2*d*x+1/2*c)^4*b^2-4*cos(1/2*d*x+1/2*
c)^2*b^2+a^2)-1/4/(a^2-b^2)/a^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-e*(2*sin(1/2*
d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-1/64/a^2/b^2*sum((-5*a^2+2*b^2
)/(a-b)/(a+b)/(2*_alpha^2-1)/_alpha*(2^(1/2)/(e*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2)*arctanh(1/2*e*(4*_alpha^
2-3)/(4*a^2-3*b^2)*(4*a^2*cos(1/2*d*x+1/2*c)^2-3*cos(1/2*d*x+1/2*c)^2*b^2+b^2*_alpha^2-3*a^2+2*b^2)*2^(1/2)/(e
*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2)/(-e*(2*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^(1/2))+8*b^2/a^2*_al
pha*(_alpha^2-1)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-e*sin(1/2*d*x+1/2*c)^2*(2*si
n(1/2*d*x+1/2*c)^2-1))^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-4*b^2/a^2*(_alpha^2-1),2^(1/2))),_alpha=RootOf(4*_
Z^4*b^2-4*_Z^2*b^2+a^2))))/sin(1/2*d*x+1/2*c)/(e*(2*cos(1/2*d*x+1/2*c)^2-1))^(1/2))/d

Fricas [F(-1)]

Timed out. \[ \int \frac {(e \cos (c+d x))^{7/2}}{(a+b \sin (c+d x))^2} \, dx=\text {Timed out} \]

[In]

integrate((e*cos(d*x+c))^(7/2)/(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

Timed out

Sympy [F(-1)]

Timed out. \[ \int \frac {(e \cos (c+d x))^{7/2}}{(a+b \sin (c+d x))^2} \, dx=\text {Timed out} \]

[In]

integrate((e*cos(d*x+c))**(7/2)/(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(e \cos (c+d x))^{7/2}}{(a+b \sin (c+d x))^2} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {7}{2}}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{2}} \,d x } \]

[In]

integrate((e*cos(d*x+c))^(7/2)/(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

integrate((e*cos(d*x + c))^(7/2)/(b*sin(d*x + c) + a)^2, x)

Giac [F]

\[ \int \frac {(e \cos (c+d x))^{7/2}}{(a+b \sin (c+d x))^2} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {7}{2}}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{2}} \,d x } \]

[In]

integrate((e*cos(d*x+c))^(7/2)/(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

integrate((e*cos(d*x + c))^(7/2)/(b*sin(d*x + c) + a)^2, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(e \cos (c+d x))^{7/2}}{(a+b \sin (c+d x))^2} \, dx=\int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{7/2}}{{\left (a+b\,\sin \left (c+d\,x\right )\right )}^2} \,d x \]

[In]

int((e*cos(c + d*x))^(7/2)/(a + b*sin(c + d*x))^2,x)

[Out]

int((e*cos(c + d*x))^(7/2)/(a + b*sin(c + d*x))^2, x)