3.7.9 \(\int \frac {(e \cos (c+d x))^{7/2}}{(a+b \sin (c+d x))^4} \, dx\) [609]

Optimal. Leaf size=597 \[ -\frac {5 a \left (a^2-2 b^2\right ) e^{7/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{16 b^{7/2} \left (-a^2+b^2\right )^{7/4} d}-\frac {5 a \left (a^2-2 b^2\right ) e^{7/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{16 b^{7/2} \left (-a^2+b^2\right )^{7/4} d}+\frac {5 \left (3 a^2-4 b^2\right ) e^4 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{24 b^4 \left (a^2-b^2\right ) d \sqrt {e \cos (c+d x)}}-\frac {5 a^2 \left (a^2-2 b^2\right ) e^4 \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b-\sqrt {-a^2+b^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{16 b^4 \left (a^2-b^2\right ) \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \cos (c+d x)}}-\frac {5 a^2 \left (a^2-2 b^2\right ) e^4 \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b+\sqrt {-a^2+b^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{16 b^4 \left (a^2-b^2\right ) \left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \cos (c+d x)}}-\frac {e (e \cos (c+d x))^{5/2}}{3 b d (a+b \sin (c+d x))^3}-\frac {5 \left (3 a^2-4 b^2\right ) e^3 \sqrt {e \cos (c+d x)}}{24 b^3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac {5 e^3 \sqrt {e \cos (c+d x)} (3 a+4 b \sin (c+d x))}{12 b^3 d (a+b \sin (c+d x))^2} \]

[Out]

-5/16*a*(a^2-2*b^2)*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)/(-a^2+b^2)^(
7/4)/d-5/16*a*(a^2-2*b^2)*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)/(-a^2
+b^2)^(7/4)/d-1/3*e*(e*cos(d*x+c))^(5/2)/b/d/(a+b*sin(d*x+c))^3+5/24*(3*a^2-4*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/(a^2-b^2)/d/(e*cos(d*x+c))
^(1/2)-5/16*a^2*(a^2-2*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/(a^2-b^2)/d/(a^2-b*(b-(-a^2+b^2)^(1/2)))/(e*cos(d*x+c))
^(1/2)-5/16*a^2*(a^2-2*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/(a^2-b^2)/d/(a^2-b*(b+(-a^2+b^2)^(1/2)))/(e*cos(d*x+c))
^(1/2)-5/24*(3*a^2-4*b^2)*e^3*(e*cos(d*x+c))^(1/2)/b^3/(a^2-b^2)/d/(a+b*sin(d*x+c))+5/12*e^3*(3*a+4*b*sin(d*x+
c))*(e*cos(d*x+c))^(1/2)/b^3/d/(a+b*sin(d*x+c))^2

________________________________________________________________________________________

Rubi [A]
time = 1.02, antiderivative size = 597, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 13, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.520, Rules used = {2772, 2942, 2943, 2946, 2721, 2720, 2781, 2886, 2884, 335, 218, 214, 211} \begin {gather*} -\frac {5 a e^{7/2} \left (a^2-2 b^2\right ) \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{16 b^{7/2} d \left (b^2-a^2\right )^{7/4}}-\frac {5 a e^{7/2} \left (a^2-2 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{16 b^{7/2} d \left (b^2-a^2\right )^{7/4}}+\frac {5 e^4 \left (3 a^2-4 b^2\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{24 b^4 d \left (a^2-b^2\right ) \sqrt {e \cos (c+d x)}}-\frac {5 a^2 e^4 \left (a^2-2 b^2\right ) \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b-\sqrt {b^2-a^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{16 b^4 d \left (a^2-b^2\right ) \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^2-2 b^2\right ) \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b+\sqrt {b^2-a^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{16 b^4 d \left (a^2-b^2\right ) \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {e \cos (c+d x)}}-\frac {5 e^3 \left (3 a^2-4 b^2\right ) \sqrt {e \cos (c+d x)}}{24 b^3 d \left (a^2-b^2\right ) (a+b \sin (c+d x))}+\frac {5 e^3 \sqrt {e \cos (c+d x)} (3 a+4 b \sin (c+d x))}{12 b^3 d (a+b \sin (c+d x))^2}-\frac {e (e \cos (c+d x))^{5/2}}{3 b d (a+b \sin (c+d x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(e*Cos[c + d*x])^(7/2)/(a + b*Sin[c + d*x])^4,x]

[Out]

(-5*a*(a^2 - 2*b^2)*e^(7/2)*ArcTan[(Sqrt[b]*Sqrt[e*Cos[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/(16*b^(7/2)*(
-a^2 + b^2)^(7/4)*d) - (5*a*(a^2 - 2*b^2)*e^(7/2)*ArcTanh[(Sqrt[b]*Sqrt[e*Cos[c + d*x]])/((-a^2 + b^2)^(1/4)*S
qrt[e])])/(16*b^(7/2)*(-a^2 + b^2)^(7/4)*d) + (5*(3*a^2 - 4*b^2)*e^4*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2,
2])/(24*b^4*(a^2 - b^2)*d*Sqrt[e*Cos[c + d*x]]) - (5*a^2*(a^2 - 2*b^2)*e^4*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b
)/(b - Sqrt[-a^2 + b^2]), (c + d*x)/2, 2])/(16*b^4*(a^2 - b^2)*(a^2 - b*(b - Sqrt[-a^2 + b^2]))*d*Sqrt[e*Cos[c
+ d*x]]) - (5*a^2*(a^2 - 2*b^2)*e^4*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (c + d*x)/2,
2])/(16*b^4*(a^2 - b^2)*(a^2 - b*(b + Sqrt[-a^2 + b^2]))*d*Sqrt[e*Cos[c + d*x]]) - (e*(e*Cos[c + d*x])^(5/2))/
(3*b*d*(a + b*Sin[c + d*x])^3) - (5*(3*a^2 - 4*b^2)*e^3*Sqrt[e*Cos[c + d*x]])/(24*b^3*(a^2 - b^2)*d*(a + b*Sin
[c + d*x])) + (5*e^3*Sqrt[e*Cos[c + d*x]]*(3*a + 4*b*Sin[c + d*x]))/(12*b^3*d*(a + b*Sin[c + d*x])^2)

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 2942

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 + 1)*Sin[e + f*x])/(b^2*f*(m + 1)*(m + p + 1))), x] + Dist[g^2*((p - 1)/(b^2*(m + 1)*(m + p + 1
))), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1)*Simp[b*d*(m + 1) + (b*c*(m + p + 1) - a*d*p)*Si
n[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && N
eQ[m + p + 1, 0] && IntegerQ[2*m]

Rule 2943

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[(-(b*c - a*d))*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^(m + 1)/(f*g*(
a^2 - b^2)*(m + 1))), x] + Dist[1/((a^2 - b^2)*(m + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1)*S
imp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + p + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p},
x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && 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*} \int \frac {(e \cos (c+d x))^{7/2}}{(a+b \sin (c+d x))^4} \, dx &=-\frac {e (e \cos (c+d x))^{5/2}}{3 b d (a+b \sin (c+d x))^3}-\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))^3} \, dx}{6 b}\\ &=-\frac {e (e \cos (c+d x))^{5/2}}{3 b d (a+b \sin (c+d x))^3}+\frac {5 e^3 \sqrt {e \cos (c+d x)} (3 a+4 b \sin (c+d x))}{12 b^3 d (a+b \sin (c+d x))^2}-\frac {\left (5 e^4\right ) \int \frac {-2 b-\frac {3}{2} a \sin (c+d x)}{\sqrt {e \cos (c+d x)} (a+b \sin (c+d x))^2} \, dx}{12 b^3}\\ &=-\frac {e (e \cos (c+d x))^{5/2}}{3 b d (a+b \sin (c+d x))^3}-\frac {5 \left (3 a^2-4 b^2\right ) e^3 \sqrt {e \cos (c+d x)}}{24 b^3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac {5 e^3 \sqrt {e \cos (c+d x)} (3 a+4 b \sin (c+d x))}{12 b^3 d (a+b \sin (c+d x))^2}+\frac {\left (5 e^4\right ) \int \frac {\frac {a b}{2}+\frac {1}{4} \left (3 a^2-4 b^2\right ) \sin (c+d x)}{\sqrt {e \cos (c+d x)} (a+b \sin (c+d x))} \, dx}{12 b^3 \left (a^2-b^2\right )}\\ &=-\frac {e (e \cos (c+d x))^{5/2}}{3 b d (a+b \sin (c+d x))^3}-\frac {5 \left (3 a^2-4 b^2\right ) e^3 \sqrt {e \cos (c+d x)}}{24 b^3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac {5 e^3 \sqrt {e \cos (c+d x)} (3 a+4 b \sin (c+d x))}{12 b^3 d (a+b \sin (c+d x))^2}+\frac {\left (5 \left (3 a^2-4 b^2\right ) e^4\right ) \int \frac {1}{\sqrt {e \cos (c+d x)}} \, dx}{48 b^4 \left (a^2-b^2\right )}-\frac {\left (5 a \left (a^2-2 b^2\right ) e^4\right ) \int \frac {1}{\sqrt {e \cos (c+d x)} (a+b \sin (c+d x))} \, dx}{16 b^4 \left (a^2-b^2\right )}\\ &=-\frac {e (e \cos (c+d x))^{5/2}}{3 b d (a+b \sin (c+d x))^3}-\frac {5 \left (3 a^2-4 b^2\right ) e^3 \sqrt {e \cos (c+d x)}}{24 b^3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac {5 e^3 \sqrt {e \cos (c+d x)} (3 a+4 b \sin (c+d x))}{12 b^3 d (a+b \sin (c+d x))^2}-\frac {\left (5 a^2 \left (a^2-2 b^2\right ) 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}{32 b^4 \left (-a^2+b^2\right )^{3/2}}-\frac {\left (5 a^2 \left (a^2-2 b^2\right ) 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}{32 b^4 \left (-a^2+b^2\right )^{3/2}}-\frac {\left (5 a \left (a^2-2 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 )}{16 b^3 \left (a^2-b^2\right ) d}+\frac {\left (5 \left (3 a^2-4 b^2\right ) e^4 \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{48 b^4 \left (a^2-b^2\right ) \sqrt {e \cos (c+d x)}}\\ &=\frac {5 \left (3 a^2-4 b^2\right ) e^4 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{24 b^4 \left (a^2-b^2\right ) d \sqrt {e \cos (c+d x)}}-\frac {e (e \cos (c+d x))^{5/2}}{3 b d (a+b \sin (c+d x))^3}-\frac {5 \left (3 a^2-4 b^2\right ) e^3 \sqrt {e \cos (c+d x)}}{24 b^3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac {5 e^3 \sqrt {e \cos (c+d x)} (3 a+4 b \sin (c+d x))}{12 b^3 d (a+b \sin (c+d x))^2}-\frac {\left (5 a \left (a^2-2 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 )}{8 b^3 \left (a^2-b^2\right ) d}-\frac {\left (5 a^2 \left (a^2-2 b^2\right ) 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}{32 b^4 \left (-a^2+b^2\right )^{3/2} \sqrt {e \cos (c+d x)}}-\frac {\left (5 a^2 \left (a^2-2 b^2\right ) 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}{32 b^4 \left (-a^2+b^2\right )^{3/2} \sqrt {e \cos (c+d x)}}\\ &=\frac {5 \left (3 a^2-4 b^2\right ) e^4 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{24 b^4 \left (a^2-b^2\right ) d \sqrt {e \cos (c+d x)}}+\frac {5 a^2 \left (a^2-2 b^2\right ) e^4 \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b-\sqrt {-a^2+b^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{16 b^4 \left (-a^2+b^2\right )^{3/2} \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}-\frac {5 a^2 \left (a^2-2 b^2\right ) e^4 \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b+\sqrt {-a^2+b^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{16 b^4 \left (-a^2+b^2\right )^{3/2} \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}-\frac {e (e \cos (c+d x))^{5/2}}{3 b d (a+b \sin (c+d x))^3}-\frac {5 \left (3 a^2-4 b^2\right ) e^3 \sqrt {e \cos (c+d x)}}{24 b^3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac {5 e^3 \sqrt {e \cos (c+d x)} (3 a+4 b \sin (c+d x))}{12 b^3 d (a+b \sin (c+d x))^2}-\frac {\left (5 a \left (a^2-2 b^2\right ) 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 )}{16 b^3 \left (-a^2+b^2\right )^{3/2} d}-\frac {\left (5 a \left (a^2-2 b^2\right ) 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 )}{16 b^3 \left (-a^2+b^2\right )^{3/2} d}\\ &=-\frac {5 a \left (a^2-2 b^2\right ) e^{7/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{16 b^{7/2} \left (-a^2+b^2\right )^{7/4} d}-\frac {5 a \left (a^2-2 b^2\right ) e^{7/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{16 b^{7/2} \left (-a^2+b^2\right )^{7/4} d}+\frac {5 \left (3 a^2-4 b^2\right ) e^4 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{24 b^4 \left (a^2-b^2\right ) d \sqrt {e \cos (c+d x)}}+\frac {5 a^2 \left (a^2-2 b^2\right ) e^4 \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b-\sqrt {-a^2+b^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{16 b^4 \left (-a^2+b^2\right )^{3/2} \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}-\frac {5 a^2 \left (a^2-2 b^2\right ) e^4 \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b+\sqrt {-a^2+b^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{16 b^4 \left (-a^2+b^2\right )^{3/2} \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}-\frac {e (e \cos (c+d x))^{5/2}}{3 b d (a+b \sin (c+d x))^3}-\frac {5 \left (3 a^2-4 b^2\right ) e^3 \sqrt {e \cos (c+d x)}}{24 b^3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac {5 e^3 \sqrt {e \cos (c+d x)} (3 a+4 b \sin (c+d x))}{12 b^3 d (a+b \sin (c+d x))^2}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
time = 42.34, size = 1263, normalized size = 2.12 \begin {gather*} \frac {(e \cos (c+d x))^{7/2} \sec ^3(c+d x) \left (\frac {a^2-b^2}{3 b^3 (a+b \sin (c+d x))^3}-\frac {13 a}{12 b^3 (a+b \sin (c+d x))^2}+\frac {-33 a^2+28 b^2}{24 b^3 \left (-a^2+b^2\right ) (a+b \sin (c+d x))}\right )}{d}+\frac {5 (e \cos (c+d x))^{7/2} \left (-\frac {4 a b \left (a+b \sqrt {1-\cos ^2(c+d x)}\right ) \left (\frac {5 a \left (a^2-b^2\right ) F_1\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 ) F_1\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 F_1\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 ) F_1\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 \tan ^{-1}\left (1-\frac {(1+i) \sqrt {b} \sqrt {\cos (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )-2 \tan ^{-1}\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 {2 \left (3 a^2-4 b^2\right ) \left (a+b \sqrt {1-\cos ^2(c+d x)}\right ) \left (\frac {5 b \left (a^2-b^2\right ) F_1\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 ) F_1\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 F_1\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 ) F_1\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 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )+2 \tan ^{-1}\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 )}{48 (a-b) b^3 (a+b) d \cos ^{\frac {7}{2}}(c+d x)} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(e*Cos[c + d*x])^(7/2)/(a + b*Sin[c + d*x])^4,x]

[Out]

((e*Cos[c + d*x])^(7/2)*Sec[c + d*x]^3*((a^2 - b^2)/(3*b^3*(a + b*Sin[c + d*x])^3) - (13*a)/(12*b^3*(a + b*Sin
[c + d*x])^2) + (-33*a^2 + 28*b^2)/(24*b^3*(-a^2 + b^2)*(a + b*Sin[c + d*x]))))/d + (5*(e*Cos[c + d*x])^(7/2)*
((-4*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*Co
s[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[Cos[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])) - (2*(3*a^2 - 4*b^2)*(a + b*Sqr
t[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*ArcTan[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]*S
qrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Cos[c + d*x]] + b*Cos[c + d*x]]))/(4*Sqrt[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]))))/(48*(a - b)*b^3*(a + b)*d*Cos[c + d*x]^(7/2))

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {\left (e \cos \left (d x +c \right )\right )^{\frac {7}{2}}}{\left (a +b \sin \left (d x +c \right )\right )^{4}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(d*x+c))^(7/2)/(a+b*sin(d*x+c))^4,x)

[Out]

int((e*cos(d*x+c))^(7/2)/(a+b*sin(d*x+c))^4,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(7/2)/(a+b*sin(d*x+c))^4,x, algorithm="maxima")

[Out]

e^(7/2)*integrate(cos(d*x + c)^(7/2)/(b*sin(d*x + c) + a)^4, x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(7/2)/(a+b*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

integral(cos(d*x + c)^(7/2)*e^(7/2)/(b^4*cos(d*x + c)^4 + a^4 + 6*a^2*b^2 + b^4 - 2*(3*a^2*b^2 + b^4)*cos(d*x
+ c)^2 - 4*(a*b^3*cos(d*x + c)^2 - a^3*b - a*b^3)*sin(d*x + c)), x)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))**(7/2)/(a+b*sin(d*x+c))**4,x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3878 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(7/2)/(a+b*sin(d*x+c))^4,x, algorithm="giac")

[Out]

integrate(cos(d*x + c)^(7/2)*e^(7/2)/(b*sin(d*x + c) + a)^4, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{7/2}}{{\left (a+b\,\sin \left (c+d\,x\right )\right )}^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(c + d*x))^(7/2)/(a + b*sin(c + d*x))^4,x)

[Out]

int((e*cos(c + d*x))^(7/2)/(a + b*sin(c + d*x))^4, x)

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