3.1186 \(\int \frac {\cos ^3(c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=313 \[ \frac {2 \left (5 a^2+3 b^2\right ) \cos (c+d x)}{3 a^2 b^2 d \sqrt {a+b \sin (c+d x)}}-\frac {2 \left (a^2-b^2\right ) \cos (c+d x)}{3 a b^2 d (a+b \sin (c+d x))^{3/2}}-\frac {2 \left (8 a^2+b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{3 a b^3 d \sqrt {a+b \sin (c+d x)}}+\frac {2 \left (8 a^2+3 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{3 a^2 b^3 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {2 \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{a^2 d \sqrt {a+b \sin (c+d x)}} \]

[Out]

-2/3*(a^2-b^2)*cos(d*x+c)/a/b^2/d/(a+b*sin(d*x+c))^(3/2)+2/3*(5*a^2+3*b^2)*cos(d*x+c)/a^2/b^2/d/(a+b*sin(d*x+c
))^(1/2)-2/3*(8*a^2+3*b^2)*(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)*EllipticE(cos(1/2*c+1
/4*Pi+1/2*d*x),2^(1/2)*(b/(a+b))^(1/2))*(a+b*sin(d*x+c))^(1/2)/a^2/b^3/d/((a+b*sin(d*x+c))/(a+b))^(1/2)+2/3*(8
*a^2+b^2)*(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)*(b/(a+b))^(1/2))*((a+b*sin(d*x+c))/(a+b))^(1/2)/a/b^3/d/(a+b*sin(d*x+c))^(1/2)-2*(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),2,2^(1/2)*(b/(a+b))^(1/2))*((a+b*s
in(d*x+c))/(a+b))^(1/2)/a^2/d/(a+b*sin(d*x+c))^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.68, antiderivative size = 313, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {2891, 3059, 2655, 2653, 3002, 2663, 2661, 2807, 2805} \[ \frac {2 \left (5 a^2+3 b^2\right ) \cos (c+d x)}{3 a^2 b^2 d \sqrt {a+b \sin (c+d x)}}-\frac {2 \left (a^2-b^2\right ) \cos (c+d x)}{3 a b^2 d (a+b \sin (c+d x))^{3/2}}-\frac {2 \left (8 a^2+b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{3 a b^3 d \sqrt {a+b \sin (c+d x)}}+\frac {2 \left (8 a^2+3 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{3 a^2 b^3 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {2 \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{a^2 d \sqrt {a+b \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^3*Cot[c + d*x])/(a + b*Sin[c + d*x])^(5/2),x]

[Out]

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

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 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 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 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 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]

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 2891

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^2*b^2*(m + 1)*(m + 2)), Int[(a + b*Sin[e + f*x])^(m + 2)*(d*Sin[e + f*x])^n*Simp[
a^2*(n + 1)*(n + 3) - b^2*(m + n + 2)*(m + n + 3) + a*b*(m + 2)*Sin[e + f*x] - (a^2*(n + 2)*(n + 3) - b^2*(m +
n + 2)*(m + n + 4))*Sin[e + f*x]^2, x], x], x] + Simp[((a^2*(n - m + 1) - b^2*(m + n + 2))*Cos[e + f*x]*(a +
b*Sin[e + f*x])^(m + 2)*(d*Sin[e + f*x])^(n + 1))/(a^2*b^2*d*f*(m + 1)*(m + 2)), 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] && (LtQ[m, -2] || EqQ[m + n +
4, 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 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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [C]  time = 5.18, size = 443, normalized size = 1.42 \[ -\frac {\frac {2 a^2 \left (a^2-b^2\right ) \cos (c+d x)}{(a+b \sin (c+d x))^{3/2}}-\frac {2 a \left (5 a^2+3 b^2\right ) \cos (c+d x)}{\sqrt {a+b \sin (c+d x)}}+\frac {a \left (8 a^2+9 b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \Pi \left (2;\frac {1}{4} (-2 c-2 d x+\pi )|\frac {2 b}{a+b}\right )}{\sqrt {a+b \sin (c+d x)}}+\frac {i \left (8 a^2+3 b^2\right ) \sec (c+d x) \sqrt {-\frac {b (\sin (c+d x)-1)}{a+b}} \sqrt {\frac {b (\sin (c+d x)+1)}{b-a}} \left (b \left (b \Pi \left (\frac {a+b}{a};i \sinh ^{-1}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \sin (c+d x)}\right )|\frac {a+b}{a-b}\right )-2 a F\left (i \sinh ^{-1}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \sin (c+d x)}\right )|\frac {a+b}{a-b}\right )\right )-2 a (a-b) E\left (i \sinh ^{-1}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \sin (c+d x)}\right )|\frac {a+b}{a-b}\right )\right )}{b^2 \sqrt {-\frac {1}{a+b}}}+\frac {4 a^2 b \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{4} (-2 c-2 d x+\pi )|\frac {2 b}{a+b}\right )}{\sqrt {a+b \sin (c+d x)}}}{3 a^3 b^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^3*Cot[c + d*x])/(a + b*Sin[c + d*x])^(5/2),x]

[Out]

-1/3*((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)])/(b^2*Sqrt[-(a + b)^(-1)]) +
(2*a^2*(a^2 - b^2)*Cos[c + d*x])/(a + b*Sin[c + d*x])^(3/2) - (2*a*(5*a^2 + 3*b^2)*Cos[c + d*x])/Sqrt[a + b*Si
n[c + d*x]] + (4*a^2*b*EllipticF[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/Sqr
t[a + b*Sin[c + d*x]] + (a*(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]])/(a^3*b^2*d)

________________________________________________________________________________________

fricas [F]  time = 73.82, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {b \sin \left (d x + c\right ) + a} \cos \left (d x + c\right )^{3} \cot \left (d x + c\right )}{3 \, a b^{2} \cos \left (d x + c\right )^{2} - a^{3} - 3 \, a b^{2} + {\left (b^{3} \cos \left (d x + c\right )^{2} - 3 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )}, x\right ) \]

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))^(5/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (d x + c\right )^{3} \cot \left (d x + c\right )}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

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))^(5/2),x, algorithm="giac")

[Out]

integrate(cos(d*x + c)^3*cot(d*x + c)/(b*sin(d*x + c) + a)^(5/2), x)

________________________________________________________________________________________

maple [B]  time = 6.70, size = 1375, normalized size = 4.39 \[ \text {result too large to display} \]

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))^(5/2),x)

[Out]

(-(-a-b*sin(d*x+c))*cos(d*x+c)^2)^(1/2)*(1/b^3*(2*b*(a/b-1)*((a+b*sin(d*x+c))/(a-b))^(1/2)*(b*(1-sin(d*x+c))/(
a+b))^(1/2)*((-1-sin(d*x+c))*b/(a-b))^(1/2)/(-(-a-b*sin(d*x+c))*cos(d*x+c)^2)^(1/2)*((-a/b-1)*EllipticE(((a+b*
sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))+EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2)))-4
*a*(a/b-1)*((a+b*sin(d*x+c))/(a-b))^(1/2)*(b*(1-sin(d*x+c))/(a+b))^(1/2)*((-1-sin(d*x+c))*b/(a-b))^(1/2)/(-(-a
-b*sin(d*x+c))*cos(d*x+c)^2)^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2)))+1/b^3*(3*a^4
-2*a^2*b^2-b^4)/a^2*(2*b*cos(d*x+c)^2/(a^2-b^2)/(-(-a-b*sin(d*x+c))*cos(d*x+c)^2)^(1/2)+2*a/(a^2-b^2)*(a/b-1)*
((a+b*sin(d*x+c))/(a-b))^(1/2)*(b*(1-sin(d*x+c))/(a+b))^(1/2)*((-1-sin(d*x+c))*b/(a-b))^(1/2)/(-(-a-b*sin(d*x+
c))*cos(d*x+c)^2)^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))+2*b/(a^2-b^2)*(a/b-1)*((
a+b*sin(d*x+c))/(a-b))^(1/2)*(b*(1-sin(d*x+c))/(a+b))^(1/2)*((-1-sin(d*x+c))*b/(a-b))^(1/2)/(-(-a-b*sin(d*x+c)
)*cos(d*x+c)^2)^(1/2)*((-a/b-1)*EllipticE(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))+EllipticF(((a+b*
sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))))-2/a^3*(a/b-1)*((a+b*sin(d*x+c))/(a-b))^(1/2)*(b*(1-sin(d*x+c))
/(a+b))^(1/2)*((-1-sin(d*x+c))*b/(a-b))^(1/2)/(-(-a-b*sin(d*x+c))*cos(d*x+c)^2)^(1/2)*b*EllipticPi(((a+b*sin(d
*x+c))/(a-b))^(1/2),-(-a/b+1)/a*b,((a-b)/(a+b))^(1/2))+(-a^4+2*a^2*b^2-b^4)/a/b^3*(2/3/b/(a^2-b^2)*(-(-a-b*sin
(d*x+c))*cos(d*x+c)^2)^(1/2)/(sin(d*x+c)+a/b)^2+8/3*b*cos(d*x+c)^2/(a^2-b^2)^2*a/(-(-a-b*sin(d*x+c))*cos(d*x+c
)^2)^(1/2)+2*(3*a^2+b^2)/(3*a^4-6*a^2*b^2+3*b^4)*(a/b-1)*((a+b*sin(d*x+c))/(a-b))^(1/2)*(b*(1-sin(d*x+c))/(a+b
))^(1/2)*((-1-sin(d*x+c))*b/(a-b))^(1/2)/(-(-a-b*sin(d*x+c))*cos(d*x+c)^2)^(1/2)*EllipticF(((a+b*sin(d*x+c))/(
a-b))^(1/2),((a-b)/(a+b))^(1/2))+8/3*a*b/(a^2-b^2)^2*(a/b-1)*((a+b*sin(d*x+c))/(a-b))^(1/2)*(b*(1-sin(d*x+c))/
(a+b))^(1/2)*((-1-sin(d*x+c))*b/(a-b))^(1/2)/(-(-a-b*sin(d*x+c))*cos(d*x+c)^2)^(1/2)*((-a/b-1)*EllipticE(((a+b
*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))+EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))))
)/cos(d*x+c)/(a+b*sin(d*x+c))^(1/2)/d

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (d x + c\right )^{3} \cot \left (d x + c\right )}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

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))^(5/2),x, algorithm="maxima")

[Out]

integrate(cos(d*x + c)^3*cot(d*x + c)/(b*sin(d*x + c) + a)^(5/2), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\cos \left (c+d\,x\right )}^3\,\mathrm {cot}\left (c+d\,x\right )}{{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^3*cot(c + d*x))/(a + b*sin(c + d*x))^(5/2),x)

[Out]

int((cos(c + d*x)^3*cot(c + d*x))/(a + b*sin(c + d*x))^(5/2), x)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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))**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________