3.1.60 \(\int \frac {(e \sin (c+d x))^{7/2}}{a+b \cos (c+d x)} \, dx\) [60]

Optimal. Leaf size=474 \[ \frac {\left (-a^2+b^2\right )^{5/4} e^{7/2} \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{b^{7/2} d}+\frac {\left (-a^2+b^2\right )^{5/4} e^{7/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{b^{7/2} d}-\frac {2 a \left (3 a^2-4 b^2\right ) e^4 F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{3 b^4 d \sqrt {e \sin (c+d x)}}+\frac {a \left (a^2-b^2\right )^2 e^4 \Pi \left (\frac {2 b}{b-\sqrt {-a^2+b^2}};\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{b^4 \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \sin (c+d x)}}+\frac {a \left (a^2-b^2\right )^2 e^4 \Pi \left (\frac {2 b}{b+\sqrt {-a^2+b^2}};\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{b^4 \left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \sin (c+d x)}}+\frac {2 e^3 \left (3 \left (a^2-b^2\right )-a b \cos (c+d x)\right ) \sqrt {e \sin (c+d x)}}{3 b^3 d}-\frac {2 e (e \sin (c+d x))^{5/2}}{5 b d} \]

[Out]

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

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Rubi [A]
time = 0.86, antiderivative size = 474, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 12, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {2774, 2944, 2946, 2721, 2720, 2781, 2886, 2884, 335, 218, 214, 211} \begin {gather*} \frac {e^{7/2} \left (b^2-a^2\right )^{5/4} \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{b^{7/2} d}+\frac {e^{7/2} \left (b^2-a^2\right )^{5/4} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{b^{7/2} d}-\frac {2 a e^4 \left (3 a^2-4 b^2\right ) \sqrt {\sin (c+d x)} F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{3 b^4 d \sqrt {e \sin (c+d x)}}+\frac {a e^4 \left (a^2-b^2\right )^2 \sqrt {\sin (c+d x)} \Pi \left (\frac {2 b}{b-\sqrt {b^2-a^2}};\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{b^4 d \left (a^2-b \left (b-\sqrt {b^2-a^2}\right )\right ) \sqrt {e \sin (c+d x)}}+\frac {a e^4 \left (a^2-b^2\right )^2 \sqrt {\sin (c+d x)} \Pi \left (\frac {2 b}{b+\sqrt {b^2-a^2}};\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{b^4 d \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {e \sin (c+d x)}}+\frac {2 e^3 \sqrt {e \sin (c+d x)} \left (3 \left (a^2-b^2\right )-a b \cos (c+d x)\right )}{3 b^3 d}-\frac {2 e (e \sin (c+d x))^{5/2}}{5 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2774

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 + p))), x] + Dist[g^2*((p - 1)/(b*(m + p))), Int[(g
*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^m*(b + a*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, m}, x] &&
NeQ[a^2 - b^2, 0] && GtQ[p, 1] && NeQ[m + p, 0] && 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*} \int \frac {(e \sin (c+d x))^{7/2}}{a+b \cos (c+d x)} \, dx &=-\frac {2 e (e \sin (c+d x))^{5/2}}{5 b d}-\frac {e^2 \int \frac {(-b-a \cos (c+d x)) (e \sin (c+d x))^{3/2}}{a+b \cos (c+d x)} \, dx}{b}\\ &=\frac {2 e^3 \left (3 \left (a^2-b^2\right )-a b \cos (c+d x)\right ) \sqrt {e \sin (c+d x)}}{3 b^3 d}-\frac {2 e (e \sin (c+d x))^{5/2}}{5 b d}-\frac {\left (2 e^4\right ) \int \frac {\frac {1}{2} b \left (2 a^2-3 b^2\right )+\frac {1}{2} a \left (3 a^2-4 b^2\right ) \cos (c+d x)}{(a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}} \, dx}{3 b^3}\\ &=\frac {2 e^3 \left (3 \left (a^2-b^2\right )-a b \cos (c+d x)\right ) \sqrt {e \sin (c+d x)}}{3 b^3 d}-\frac {2 e (e \sin (c+d x))^{5/2}}{5 b d}-\frac {\left (a \left (3 a^2-4 b^2\right ) e^4\right ) \int \frac {1}{\sqrt {e \sin (c+d x)}} \, dx}{3 b^4}+\frac {\left (\left (a^2-b^2\right )^2 e^4\right ) \int \frac {1}{(a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}} \, dx}{b^4}\\ &=\frac {2 e^3 \left (3 \left (a^2-b^2\right )-a b \cos (c+d x)\right ) \sqrt {e \sin (c+d x)}}{3 b^3 d}-\frac {2 e (e \sin (c+d x))^{5/2}}{5 b d}-\frac {\left (a \left (-a^2+b^2\right )^{3/2} e^4\right ) \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {-a^2+b^2}-b \sin (c+d x)\right )} \, dx}{2 b^4}-\frac {\left (a \left (-a^2+b^2\right )^{3/2} e^4\right ) \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {-a^2+b^2}+b \sin (c+d x)\right )} \, dx}{2 b^4}-\frac {\left (\left (a^2-b^2\right )^2 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 \sin (c+d x)\right )}{b^3 d}-\frac {\left (a \left (3 a^2-4 b^2\right ) e^4 \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)}} \, dx}{3 b^4 \sqrt {e \sin (c+d x)}}\\ &=-\frac {2 a \left (3 a^2-4 b^2\right ) e^4 F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{3 b^4 d \sqrt {e \sin (c+d x)}}+\frac {2 e^3 \left (3 \left (a^2-b^2\right )-a b \cos (c+d x)\right ) \sqrt {e \sin (c+d x)}}{3 b^3 d}-\frac {2 e (e \sin (c+d x))^{5/2}}{5 b d}-\frac {\left (2 \left (a^2-b^2\right )^2 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 \sin (c+d x)}\right )}{b^3 d}-\frac {\left (a \left (-a^2+b^2\right )^{3/2} e^4 \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)} \left (\sqrt {-a^2+b^2}-b \sin (c+d x)\right )} \, dx}{2 b^4 \sqrt {e \sin (c+d x)}}-\frac {\left (a \left (-a^2+b^2\right )^{3/2} e^4 \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)} \left (\sqrt {-a^2+b^2}+b \sin (c+d x)\right )} \, dx}{2 b^4 \sqrt {e \sin (c+d x)}}\\ &=-\frac {2 a \left (3 a^2-4 b^2\right ) e^4 F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{3 b^4 d \sqrt {e \sin (c+d x)}}+\frac {a \left (-a^2+b^2\right )^{3/2} e^4 \Pi \left (\frac {2 b}{b-\sqrt {-a^2+b^2}};\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{b^4 \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}-\frac {a \left (-a^2+b^2\right )^{3/2} e^4 \Pi \left (\frac {2 b}{b+\sqrt {-a^2+b^2}};\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{b^4 \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {2 e^3 \left (3 \left (a^2-b^2\right )-a b \cos (c+d x)\right ) \sqrt {e \sin (c+d x)}}{3 b^3 d}-\frac {2 e (e \sin (c+d x))^{5/2}}{5 b d}+\frac {\left (\left (-a^2+b^2\right )^{3/2} e^4\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} e-b x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{b^3 d}+\frac {\left (\left (-a^2+b^2\right )^{3/2} e^4\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} e+b x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{b^3 d}\\ &=\frac {\left (-a^2+b^2\right )^{5/4} e^{7/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{b^{7/2} d}+\frac {\left (-a^2+b^2\right )^{5/4} e^{7/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{b^{7/2} d}-\frac {2 a \left (3 a^2-4 b^2\right ) e^4 F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{3 b^4 d \sqrt {e \sin (c+d x)}}+\frac {a \left (-a^2+b^2\right )^{3/2} e^4 \Pi \left (\frac {2 b}{b-\sqrt {-a^2+b^2}};\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{b^4 \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}-\frac {a \left (-a^2+b^2\right )^{3/2} e^4 \Pi \left (\frac {2 b}{b+\sqrt {-a^2+b^2}};\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{b^4 \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {2 e^3 \left (3 \left (a^2-b^2\right )-a b \cos (c+d x)\right ) \sqrt {e \sin (c+d x)}}{3 b^3 d}-\frac {2 e (e \sin (c+d x))^{5/2}}{5 b d}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
time = 45.91, size = 1955, normalized size = 4.12 \begin {gather*} \frac {\left (-\frac {2 a \cos (c+d x)}{3 b^2}+\frac {\cos (2 (c+d x))}{5 b}\right ) \csc ^3(c+d x) (e \sin (c+d x))^{7/2}}{d}+\frac {(e \sin (c+d x))^{7/2} \left (\frac {28 a b \cos ^2(c+d x) \left (a+b \sqrt {1-\sin ^2(c+d x)}\right ) \left (\frac {a \left (-2 \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )+2 \text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt {b} \sqrt {\sin (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 {\sin (c+d x)}+b \sin (c+d x)\right )+\log \left (\sqrt {a^2-b^2}+\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\sin (c+d x)}+b \sin (c+d x)\right )\right )}{4 \sqrt {2} \sqrt {b} \left (a^2-b^2\right )^{3/4}}+\frac {5 b \left (a^2-b^2\right ) F_1\left (\frac {1}{4};-\frac {1}{2},1;\frac {5}{4};\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{-a^2+b^2}\right ) \sqrt {\sin (c+d x)} \sqrt {1-\sin ^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};\sin ^2(c+d x),\frac {b^2 \sin ^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};\sin ^2(c+d x),\frac {b^2 \sin ^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};\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{-a^2+b^2}\right )\right ) \sin ^2(c+d x)\right ) \left (a^2+b^2 \left (-1+\sin ^2(c+d x)\right )\right )}\right )}{(a+b \cos (c+d x)) \left (1-\sin ^2(c+d x)\right )}+\frac {2 \left (-10 a^2+27 b^2\right ) \cos (c+d x) \left (a+b \sqrt {1-\sin ^2(c+d x)}\right ) \left (-\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \sqrt {b} \left (2 \text {ArcTan}\left (1-\frac {(1+i) \sqrt {b} \sqrt {\sin (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )-2 \text {ArcTan}\left (1+\frac {(1+i) \sqrt {b} \sqrt {\sin (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 {\sin (c+d x)}+i b \sin (c+d x)\right )-\log \left (\sqrt {-a^2+b^2}+(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\sin (c+d x)}+i b \sin (c+d x)\right )\right )}{\left (-a^2+b^2\right )^{3/4}}+\frac {5 a \left (a^2-b^2\right ) F_1\left (\frac {1}{4};\frac {1}{2},1;\frac {5}{4};\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{-a^2+b^2}\right ) \sqrt {\sin (c+d x)}}{\sqrt {1-\sin ^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};\sin ^2(c+d x),\frac {b^2 \sin ^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};\sin ^2(c+d x),\frac {b^2 \sin ^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};\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{-a^2+b^2}\right )\right ) \sin ^2(c+d x)\right ) \left (a^2+b^2 \left (-1+\sin ^2(c+d x)\right )\right )}\right )}{(a+b \cos (c+d x)) \sqrt {1-\sin ^2(c+d x)}}+\frac {\left (30 a^2-33 b^2\right ) \cos (c+d x) \cos (2 (c+d x)) \left (a+b \sqrt {1-\sin ^2(c+d x)}\right ) \left (\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \left (-2 a^2+b^2\right ) \text {ArcTan}\left (1-\frac {(1+i) \sqrt {b} \sqrt {\sin (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 ) \text {ArcTan}\left (1+\frac {(1+i) \sqrt {b} \sqrt {\sin (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}{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 {\sin (c+d x)}+i b \sin (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 {\sin (c+d x)}+i b \sin (c+d x)\right )}{b^{3/2} \left (-a^2+b^2\right )^{3/4}}+\frac {4 \sqrt {\sin (c+d x)}}{b}-\frac {4 a F_1\left (\frac {5}{4};\frac {1}{2},1;\frac {9}{4};\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{-a^2+b^2}\right ) \sin ^{\frac {5}{2}}(c+d x)}{5 \left (a^2-b^2\right )}+\frac {10 a \left (a^2-b^2\right ) F_1\left (\frac {1}{4};\frac {1}{2},1;\frac {5}{4};\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{-a^2+b^2}\right ) \sqrt {\sin (c+d x)}}{\sqrt {1-\sin ^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};\sin ^2(c+d x),\frac {b^2 \sin ^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};\sin ^2(c+d x),\frac {b^2 \sin ^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};\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{-a^2+b^2}\right )\right ) \sin ^2(c+d x)\right ) \left (a^2+b^2 \left (-1+\sin ^2(c+d x)\right )\right )}\right )}{(a+b \cos (c+d x)) \left (1-2 \sin ^2(c+d x)\right ) \sqrt {1-\sin ^2(c+d x)}}\right )}{60 b^2 d \sin ^{\frac {7}{2}}(c+d x)} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(((-2*a*Cos[c + d*x])/(3*b^2) + Cos[2*(c + d*x)]/(5*b))*Csc[c + d*x]^3*(e*Sin[c + d*x])^(7/2))/d + ((e*Sin[c +
d*x])^(7/2)*((28*a*b*Cos[c + d*x]^2*(a + b*Sqrt[1 - Sin[c + d*x]^2])*((a*(-2*ArcTan[1 - (Sqrt[2]*Sqrt[b]*Sqrt
[Sin[c + d*x]])/(a^2 - b^2)^(1/4)] + 2*ArcTan[1 + (Sqrt[2]*Sqrt[b]*Sqrt[Sin[c + d*x]])/(a^2 - b^2)^(1/4)] - Lo
g[Sqrt[a^2 - b^2] - Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Sin[c + d*x]] + b*Sin[c + d*x]] + Log[Sqrt[a^2 - b^
2] + Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Sin[c + d*x]] + b*Sin[c + d*x]]))/(4*Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(
3/4)) + (5*b*(a^2 - b^2)*AppellF1[1/4, -1/2, 1, 5/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + b^2)]*Sqrt[S
in[c + d*x]]*Sqrt[1 - Sin[c + d*x]^2])/((-5*(a^2 - b^2)*AppellF1[1/4, -1/2, 1, 5/4, Sin[c + d*x]^2, (b^2*Sin[c
+ d*x]^2)/(-a^2 + b^2)] + 2*(2*b^2*AppellF1[5/4, -1/2, 2, 9/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + b
^2)] + (a^2 - b^2)*AppellF1[5/4, 1/2, 1, 9/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + b^2)])*Sin[c + d*x]
^2)*(a^2 + b^2*(-1 + Sin[c + d*x]^2)))))/((a + b*Cos[c + d*x])*(1 - Sin[c + d*x]^2)) + (2*(-10*a^2 + 27*b^2)*C
os[c + d*x]*(a + b*Sqrt[1 - Sin[c + d*x]^2])*(((-1/8 + I/8)*Sqrt[b]*(2*ArcTan[1 - ((1 + I)*Sqrt[b]*Sqrt[Sin[c
+ d*x]])/(-a^2 + b^2)^(1/4)] - 2*ArcTan[1 + ((1 + I)*Sqrt[b]*Sqrt[Sin[c + d*x]])/(-a^2 + b^2)^(1/4)] + Log[Sqr
t[-a^2 + b^2] - (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Sin[c + d*x]] + I*b*Sin[c + d*x]] - Log[Sqrt[-a^2 + b^
2] + (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Sin[c + d*x]] + I*b*Sin[c + d*x]]))/(-a^2 + b^2)^(3/4) + (5*a*(a^
2 - b^2)*AppellF1[1/4, 1/2, 1, 5/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + b^2)]*Sqrt[Sin[c + d*x]])/(Sq
rt[1 - Sin[c + d*x]^2]*(5*(a^2 - b^2)*AppellF1[1/4, 1/2, 1, 5/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 +
b^2)] - 2*(2*b^2*AppellF1[5/4, 1/2, 2, 9/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + b^2)] + (-a^2 + b^2)*
AppellF1[5/4, 3/2, 1, 9/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + b^2)])*Sin[c + d*x]^2)*(a^2 + b^2*(-1
+ Sin[c + d*x]^2)))))/((a + b*Cos[c + d*x])*Sqrt[1 - Sin[c + d*x]^2]) + ((30*a^2 - 33*b^2)*Cos[c + d*x]*Cos[2*
(c + d*x)]*(a + b*Sqrt[1 - Sin[c + d*x]^2])*(((1/2 - I/2)*(-2*a^2 + b^2)*ArcTan[1 - ((1 + I)*Sqrt[b]*Sqrt[Sin[
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[Sin[c + d*x]])/(-a^2 + b^2)^(1/4)])/(b^(3/2)*(-a^2 + b^2)^(3/4)) + ((1/4 - I/4)*(-2*a^2 + b^2)*L
og[Sqrt[-a^2 + b^2] - (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Sin[c + d*x]] + I*b*Sin[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[S
in[c + d*x]] + I*b*Sin[c + d*x]])/(b^(3/2)*(-a^2 + b^2)^(3/4)) + (4*Sqrt[Sin[c + d*x]])/b - (4*a*AppellF1[5/4,
1/2, 1, 9/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + b^2)]*Sin[c + d*x]^(5/2))/(5*(a^2 - b^2)) + (10*a*(
a^2 - b^2)*AppellF1[1/4, 1/2, 1, 5/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + b^2)]*Sqrt[Sin[c + d*x]])/(
Sqrt[1 - Sin[c + d*x]^2]*(5*(a^2 - b^2)*AppellF1[1/4, 1/2, 1, 5/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2
+ b^2)] - 2*(2*b^2*AppellF1[5/4, 1/2, 2, 9/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + b^2)] + (-a^2 + b^2
)*AppellF1[5/4, 3/2, 1, 9/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + b^2)])*Sin[c + d*x]^2)*(a^2 + b^2*(-
1 + Sin[c + d*x]^2)))))/((a + b*Cos[c + d*x])*(1 - 2*Sin[c + d*x]^2)*Sqrt[1 - Sin[c + d*x]^2])))/(60*b^2*d*Sin
[c + d*x]^(7/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1533\) vs. \(2(507)=1014\).
time = 0.24, size = 1534, normalized size = 3.24

method	result	size
default	\(\text {Expression too large to display}\)	\(1534\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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