3.1179 \(\int \frac{\cos ^3(c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=296 \[ -\frac{2 \left (a^2-b^2\right ) \cos (c+d x)}{a b^2 d \sqrt{a+b \sin (c+d x)}}+\frac{2 \left (8 a^2-5 b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{3 b^3 d \sqrt{a+b \sin (c+d x)}}-\frac{2 \left (8 a^2-3 b^2\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{3 a b^3 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}-\frac{2 \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{3 b^2 d}+\frac{2 \sqrt{\frac{a+b \sin (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{a d \sqrt{a+b \sin (c+d x)}} \]

[Out]

(-2*(a^2 - b^2)*Cos[c + d*x])/(a*b^2*d*Sqrt[a + b*Sin[c + d*x]]) - (2*Cos[c + d*x]*Sqrt[a + b*Sin[c + d*x]])/(
3*b^2*d) - (2*(8*a^2 - 3*b^2)*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(3*a*b^3*
d*Sqrt[(a + b*Sin[c + d*x])/(a + b)]) + (2*(8*a^2 - 5*b^2)*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(
a + b*Sin[c + d*x])/(a + b)])/(3*b^3*d*Sqrt[a + b*Sin[c + d*x]]) + (2*EllipticPi[2, (c - Pi/2 + d*x)/2, (2*b)/
(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(a*d*Sqrt[a + b*Sin[c + d*x]])

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Rubi [A]  time = 0.678251, antiderivative size = 296, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.31, Rules used = {2892, 3059, 2655, 2653, 3002, 2663, 2661, 2807, 2805} \[ -\frac{2 \left (a^2-b^2\right ) \cos (c+d x)}{a b^2 d \sqrt{a+b \sin (c+d x)}}+\frac{2 \left (8 a^2-5 b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{3 b^3 d \sqrt{a+b \sin (c+d x)}}-\frac{2 \left (8 a^2-3 b^2\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{3 a b^3 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}-\frac{2 \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{3 b^2 d}+\frac{2 \sqrt{\frac{a+b \sin (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{a d \sqrt{a+b \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^3*Cot[c + d*x])/(a + b*Sin[c + d*x])^(3/2),x]

[Out]

(-2*(a^2 - b^2)*Cos[c + d*x])/(a*b^2*d*Sqrt[a + b*Sin[c + d*x]]) - (2*Cos[c + d*x]*Sqrt[a + b*Sin[c + d*x]])/(
3*b^2*d) - (2*(8*a^2 - 3*b^2)*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(3*a*b^3*
d*Sqrt[(a + b*Sin[c + d*x])/(a + b)]) + (2*(8*a^2 - 5*b^2)*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(
a + b*Sin[c + d*x])/(a + b)])/(3*b^3*d*Sqrt[a + b*Sin[c + d*x]]) + (2*EllipticPi[2, (c - Pi/2 + d*x)/2, (2*b)/
(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(a*d*Sqrt[a + b*Sin[c + d*x]])

Rule 2892

Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Simp[((a^2 - b^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(d*Sin[e + f*x])^(n + 1))/(a*b^2*d*
f*(m + 1)), x] + (-Dist[1/(a*b^2*(m + 1)*(m + n + 4)), Int[(a + b*Sin[e + f*x])^(m + 1)*(d*Sin[e + f*x])^n*Sim
p[a^2*(n + 1)*(n + 3) - b^2*(m + n + 2)*(m + n + 4) + a*b*(m + 1)*Sin[e + f*x] - (a^2*(n + 2)*(n + 3) - b^2*(m
+ n + 3)*(m + n + 4))*Sin[e + f*x]^2, x], x], x] - Simp[(Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 2)*(d*Sin[e +
f*x])^(n + 1))/(b^2*d*f*(m + n + 4)), x]) /; FreeQ[{a, b, d, e, f, n}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[2
*m, 2*n] && LtQ[m, -1] &&  !LtQ[n, -1] && NeQ[m + n + 4, 0]

Rule 3059

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) +
(f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Dist[C/(b*d), Int[Sqrt[a + b*Sin[e + f*x]]
, x], x] - Dist[1/(b*d), Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[e + f*x], x]/(Sqrt[a + b*Sin[e +
f*x]]*(c + d*Sin[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 2655

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

Rule 2653

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

Rule 3002

Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[
(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[B/d, Int[(a + b*Sin[e + f*x])^m, x], x] - Dist[(B*c - A*d)/d, Int[(a +
b*Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 2663

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

Rule 2661

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

Rule 2807

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*
Sin[e + f*x])/(c + d)]), 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 2805

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[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]

Rubi steps

\begin{align*} \int \frac{\cos ^3(c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx &=-\frac{2 \left (a^2-b^2\right ) \cos (c+d x)}{a b^2 d \sqrt{a+b \sin (c+d x)}}-\frac{2 \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{3 b^2 d}+\frac{4 \int \frac{\csc (c+d x) \left (\frac{3 b^2}{4}-\frac{1}{2} a b \sin (c+d x)-\frac{1}{4} \left (8 a^2-3 b^2\right ) \sin ^2(c+d x)\right )}{\sqrt{a+b \sin (c+d x)}} \, dx}{3 a b^2}\\ &=-\frac{2 \left (a^2-b^2\right ) \cos (c+d x)}{a b^2 d \sqrt{a+b \sin (c+d x)}}-\frac{2 \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{3 b^2 d}-\frac{4 \int \frac{\csc (c+d x) \left (-\frac{3 b^3}{4}-\frac{1}{4} a \left (8 a^2-5 b^2\right ) \sin (c+d x)\right )}{\sqrt{a+b \sin (c+d x)}} \, dx}{3 a b^3}+\frac{\left (-8 a^2+3 b^2\right ) \int \sqrt{a+b \sin (c+d x)} \, dx}{3 a b^3}\\ &=-\frac{2 \left (a^2-b^2\right ) \cos (c+d x)}{a b^2 d \sqrt{a+b \sin (c+d x)}}-\frac{2 \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{3 b^2 d}+\frac{\int \frac{\csc (c+d x)}{\sqrt{a+b \sin (c+d x)}} \, dx}{a}+\frac{\left (8 a^2-5 b^2\right ) \int \frac{1}{\sqrt{a+b \sin (c+d x)}} \, dx}{3 b^3}+\frac{\left (\left (-8 a^2+3 b^2\right ) \sqrt{a+b \sin (c+d x)}\right ) \int \sqrt{\frac{a}{a+b}+\frac{b \sin (c+d x)}{a+b}} \, dx}{3 a b^3 \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}\\ &=-\frac{2 \left (a^2-b^2\right ) \cos (c+d x)}{a b^2 d \sqrt{a+b \sin (c+d x)}}-\frac{2 \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{3 b^2 d}-\frac{2 \left (8 a^2-3 b^2\right ) E\left (\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )|\frac{2 b}{a+b}\right ) \sqrt{a+b \sin (c+d x)}}{3 a b^3 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{\sqrt{\frac{a+b \sin (c+d x)}{a+b}} \int \frac{\csc (c+d x)}{\sqrt{\frac{a}{a+b}+\frac{b \sin (c+d x)}{a+b}}} \, dx}{a \sqrt{a+b \sin (c+d x)}}+\frac{\left (\left (8 a^2-5 b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}}\right ) \int \frac{1}{\sqrt{\frac{a}{a+b}+\frac{b \sin (c+d x)}{a+b}}} \, dx}{3 b^3 \sqrt{a+b \sin (c+d x)}}\\ &=-\frac{2 \left (a^2-b^2\right ) \cos (c+d x)}{a b^2 d \sqrt{a+b \sin (c+d x)}}-\frac{2 \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{3 b^2 d}-\frac{2 \left (8 a^2-3 b^2\right ) E\left (\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )|\frac{2 b}{a+b}\right ) \sqrt{a+b \sin (c+d x)}}{3 a b^3 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{2 \left (8 a^2-5 b^2\right ) F\left (\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )|\frac{2 b}{a+b}\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}{3 b^3 d \sqrt{a+b \sin (c+d x)}}+\frac{2 \Pi \left (2;\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )|\frac{2 b}{a+b}\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}{a d \sqrt{a+b \sin (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 3.67926, size = 419, normalized size = 1.42 \[ \frac{-\frac{4 \cos (c+d x) \left (4 a^2+a b \sin (c+d x)-3 b^2\right )}{\sqrt{a+b \sin (c+d x)}}+\frac{2 \left (8 a^2-9 b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} \Pi \left (2;\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )}{\sqrt{a+b \sin (c+d x)}}-\frac{2 i \left (3 b^2-8 a^2\right ) \sec (c+d x) \sqrt{-\frac{b (\sin (c+d x)-1)}{a+b}} \sqrt{-\frac{b (\sin (c+d x)+1)}{a-b}} \left (b \left (b \Pi \left (\frac{a+b}{a};i \sinh ^{-1}\left (\sqrt{-\frac{1}{a+b}} \sqrt{a+b \sin (c+d x)}\right )|\frac{a+b}{a-b}\right )-2 a F\left (i \sinh ^{-1}\left (\sqrt{-\frac{1}{a+b}} \sqrt{a+b \sin (c+d x)}\right )|\frac{a+b}{a-b}\right )\right )-2 a (a-b) E\left (i \sinh ^{-1}\left (\sqrt{-\frac{1}{a+b}} \sqrt{a+b \sin (c+d x)}\right )|\frac{a+b}{a-b}\right )\right )}{a b^2 \sqrt{-\frac{1}{a+b}}}+\frac{8 a b \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )}{\sqrt{a+b \sin (c+d x)}}}{6 a b^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^3*Cot[c + d*x])/(a + b*Sin[c + d*x])^(3/2),x]

[Out]

(((-2*I)*(-8*a^2 + 3*b^2)*(-2*a*(a - b)*EllipticE[I*ArcSinh[Sqrt[-(a + b)^(-1)]*Sqrt[a + b*Sin[c + d*x]]], (a
+ b)/(a - b)] + b*(-2*a*EllipticF[I*ArcSinh[Sqrt[-(a + b)^(-1)]*Sqrt[a + b*Sin[c + d*x]]], (a + b)/(a - b)] +
b*EllipticPi[(a + b)/a, I*ArcSinh[Sqrt[-(a + b)^(-1)]*Sqrt[a + b*Sin[c + d*x]]], (a + b)/(a - b)]))*Sec[c + d*
x]*Sqrt[-((b*(-1 + Sin[c + d*x]))/(a + b))]*Sqrt[-((b*(1 + Sin[c + d*x]))/(a - b))])/(a*b^2*Sqrt[-(a + b)^(-1)
]) + (8*a*b*EllipticF[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/Sqrt[a + b*Sin
[c + d*x]] + (2*(8*a^2 - 9*b^2)*EllipticPi[2, (-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/
(a + b)])/Sqrt[a + b*Sin[c + d*x]] - (4*Cos[c + d*x]*(4*a^2 - 3*b^2 + a*b*Sin[c + d*x]))/Sqrt[a + b*Sin[c + d*
x]])/(6*a*b^2*d)

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Maple [B]  time = 1.5, size = 1010, normalized size = 3.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*cot(d*x+c)/(a+b*sin(d*x+c))^(3/2),x)

[Out]

-2/3*(8*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*Ellipti
cF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^4*b-6*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)
*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*
a^3*b^2-5*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*Ellip
ticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^2*b^3+3*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c
)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/
2))*a*b^4-8*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*Ell
ipticE(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^5+11*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)
-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*EllipticE(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2
))*a^3*b^2-3*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*El
lipticE(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a*b^4+3*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+
c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*b^4*EllipticPi(((a+b*sin(d*x+c))/(a-b))^(1/2),(a-b)/a,((a
-b)/(a+b))^(1/2))*a-3*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))
^(1/2)*b^5*EllipticPi(((a+b*sin(d*x+c))/(a-b))^(1/2),(a-b)/a,((a-b)/(a+b))^(1/2))-a^2*b^3*sin(d*x+c)^3-4*a^3*b
^2*sin(d*x+c)^2+3*a*b^4*sin(d*x+c)^2+a^2*b^3*sin(d*x+c)+4*a^3*b^2-3*a*b^4)/a^2/b^4/cos(d*x+c)/(a+b*sin(d*x+c))
^(1/2)/d

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*cot(d*x+c)/(a+b*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{b \sin \left (d x + c\right ) + a} \cos \left (d x + c\right )^{3} \cot \left (d x + c\right )}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*cot(d*x+c)/(a+b*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

integral(-sqrt(b*sin(d*x + c) + a)*cos(d*x + c)^3*cot(d*x + c)/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2
- b^2), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3*cot(d*x+c)/(a+b*sin(d*x+c))**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (d x + c\right )^{3} \cot \left (d x + c\right )}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*cot(d*x+c)/(a+b*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(cos(d*x + c)^3*cot(d*x + c)/(b*sin(d*x + c) + a)^(3/2), x)