3.1155 \(\int \cos (c+d x) \cot ^3(c+d x) (a+b \sin (c+d x))^{3/2} \, dx\)

Optimal. Leaf size=383 \[ -\frac{\left (8 a^2-5 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{20 a^2 d}-\frac{\left (8 a^2-15 b^2\right ) \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{20 a d}-\frac{a \left (8 a^2+37 b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{20 b d \sqrt{a+b \sin (c+d x)}}+\frac{\left (8 a^2-81 b^2\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{20 b d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}-\frac{3 \left (4 a^2-b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{4 d \sqrt{a+b \sin (c+d x)}}-\frac{b \cot (c+d x) (a+b \sin (c+d x))^{5/2}}{4 a^2 d}-\frac{\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^{5/2}}{2 a d} \]

[Out]

-((8*a^2 - 15*b^2)*Cos[c + d*x]*Sqrt[a + b*Sin[c + d*x]])/(20*a*d) - ((8*a^2 - 5*b^2)*Cos[c + d*x]*(a + b*Sin[
c + d*x])^(3/2))/(20*a^2*d) - (b*Cot[c + d*x]*(a + b*Sin[c + d*x])^(5/2))/(4*a^2*d) - (Cot[c + d*x]*Csc[c + d*
x]*(a + b*Sin[c + d*x])^(5/2))/(2*a*d) + ((8*a^2 - 81*b^2)*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a
+ b*Sin[c + d*x]])/(20*b*d*Sqrt[(a + b*Sin[c + d*x])/(a + b)]) - (a*(8*a^2 + 37*b^2)*EllipticF[(c - Pi/2 + d*
x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(20*b*d*Sqrt[a + b*Sin[c + d*x]]) - (3*(4*a^2 - b^2)*
EllipticPi[2, (c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(4*d*Sqrt[a + b*Sin[c + d
*x]])

________________________________________________________________________________________

Rubi [A]  time = 1.16471, antiderivative size = 383, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 10, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.345, Rules used = {2893, 3049, 3059, 2655, 2653, 3002, 2663, 2661, 2807, 2805} \[ -\frac{\left (8 a^2-5 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{20 a^2 d}-\frac{\left (8 a^2-15 b^2\right ) \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{20 a d}-\frac{a \left (8 a^2+37 b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{20 b d \sqrt{a+b \sin (c+d x)}}+\frac{\left (8 a^2-81 b^2\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{20 b d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}-\frac{3 \left (4 a^2-b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{4 d \sqrt{a+b \sin (c+d x)}}-\frac{b \cot (c+d x) (a+b \sin (c+d x))^{5/2}}{4 a^2 d}-\frac{\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^{5/2}}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((8*a^2 - 15*b^2)*Cos[c + d*x]*Sqrt[a + b*Sin[c + d*x]])/(20*a*d) - ((8*a^2 - 5*b^2)*Cos[c + d*x]*(a + b*Sin[
c + d*x])^(3/2))/(20*a^2*d) - (b*Cot[c + d*x]*(a + b*Sin[c + d*x])^(5/2))/(4*a^2*d) - (Cot[c + d*x]*Csc[c + d*
x]*(a + b*Sin[c + d*x])^(5/2))/(2*a*d) + ((8*a^2 - 81*b^2)*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a
+ b*Sin[c + d*x]])/(20*b*d*Sqrt[(a + b*Sin[c + d*x])/(a + b)]) - (a*(8*a^2 + 37*b^2)*EllipticF[(c - Pi/2 + d*
x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(20*b*d*Sqrt[a + b*Sin[c + d*x]]) - (3*(4*a^2 - b^2)*
EllipticPi[2, (c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(4*d*Sqrt[a + b*Sin[c + d
*x]])

Rule 2893

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

Rule 3049

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e +
f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n + 2)), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x]
)^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2)
- C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2
, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 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 \cos (c+d x) \cot ^3(c+d x) (a+b \sin (c+d x))^{3/2} \, dx &=-\frac{b \cot (c+d x) (a+b \sin (c+d x))^{5/2}}{4 a^2 d}-\frac{\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^{5/2}}{2 a d}-\frac{\int \csc (c+d x) (a+b \sin (c+d x))^{3/2} \left (\frac{3}{4} \left (4 a^2-b^2\right )+\frac{3}{2} a b \sin (c+d x)-\frac{1}{4} \left (8 a^2-5 b^2\right ) \sin ^2(c+d x)\right ) \, dx}{2 a^2}\\ &=-\frac{\left (8 a^2-5 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{20 a^2 d}-\frac{b \cot (c+d x) (a+b \sin (c+d x))^{5/2}}{4 a^2 d}-\frac{\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^{5/2}}{2 a d}-\frac{\int \csc (c+d x) \sqrt{a+b \sin (c+d x)} \left (\frac{15}{8} a \left (4 a^2-b^2\right )+\frac{33}{4} a^2 b \sin (c+d x)-\frac{3}{8} a \left (8 a^2-15 b^2\right ) \sin ^2(c+d x)\right ) \, dx}{5 a^2}\\ &=-\frac{\left (8 a^2-15 b^2\right ) \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{20 a d}-\frac{\left (8 a^2-5 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{20 a^2 d}-\frac{b \cot (c+d x) (a+b \sin (c+d x))^{5/2}}{4 a^2 d}-\frac{\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^{5/2}}{2 a d}-\frac{2 \int \frac{\csc (c+d x) \left (\frac{45}{16} a^2 \left (4 a^2-b^2\right )+\frac{177}{8} a^3 b \sin (c+d x)-\frac{3}{16} a^2 \left (8 a^2-81 b^2\right ) \sin ^2(c+d x)\right )}{\sqrt{a+b \sin (c+d x)}} \, dx}{15 a^2}\\ &=-\frac{\left (8 a^2-15 b^2\right ) \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{20 a d}-\frac{\left (8 a^2-5 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{20 a^2 d}-\frac{b \cot (c+d x) (a+b \sin (c+d x))^{5/2}}{4 a^2 d}-\frac{\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^{5/2}}{2 a d}+\frac{2 \int \frac{\csc (c+d x) \left (-\frac{45}{16} a^2 b \left (4 a^2-b^2\right )-\frac{3}{16} a^3 \left (8 a^2+37 b^2\right ) \sin (c+d x)\right )}{\sqrt{a+b \sin (c+d x)}} \, dx}{15 a^2 b}+\frac{\left (8 a^2-81 b^2\right ) \int \sqrt{a+b \sin (c+d x)} \, dx}{40 b}\\ &=-\frac{\left (8 a^2-15 b^2\right ) \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{20 a d}-\frac{\left (8 a^2-5 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{20 a^2 d}-\frac{b \cot (c+d x) (a+b \sin (c+d x))^{5/2}}{4 a^2 d}-\frac{\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^{5/2}}{2 a d}-\frac{1}{8} \left (3 \left (4 a^2-b^2\right )\right ) \int \frac{\csc (c+d x)}{\sqrt{a+b \sin (c+d x)}} \, dx-\frac{\left (a \left (8 a^2+37 b^2\right )\right ) \int \frac{1}{\sqrt{a+b \sin (c+d x)}} \, dx}{40 b}+\frac{\left (\left (8 a^2-81 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}{40 b \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}\\ &=-\frac{\left (8 a^2-15 b^2\right ) \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{20 a d}-\frac{\left (8 a^2-5 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{20 a^2 d}-\frac{b \cot (c+d x) (a+b \sin (c+d x))^{5/2}}{4 a^2 d}-\frac{\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^{5/2}}{2 a d}+\frac{\left (8 a^2-81 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)}}{20 b d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}-\frac{\left (3 \left (4 a^2-b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}}\right ) \int \frac{\csc (c+d x)}{\sqrt{\frac{a}{a+b}+\frac{b \sin (c+d x)}{a+b}}} \, dx}{8 \sqrt{a+b \sin (c+d x)}}-\frac{\left (a \left (8 a^2+37 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}{40 b \sqrt{a+b \sin (c+d x)}}\\ &=-\frac{\left (8 a^2-15 b^2\right ) \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{20 a d}-\frac{\left (8 a^2-5 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{20 a^2 d}-\frac{b \cot (c+d x) (a+b \sin (c+d x))^{5/2}}{4 a^2 d}-\frac{\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^{5/2}}{2 a d}+\frac{\left (8 a^2-81 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)}}{20 b d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}-\frac{a \left (8 a^2+37 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}}}{20 b d \sqrt{a+b \sin (c+d x)}}-\frac{3 \left (4 a^2-b^2\right ) \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}}}{4 d \sqrt{a+b \sin (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 3.1366, size = 434, normalized size = 1.13 \[ \frac{\frac{2 \left (112 a^2+51 b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} \Pi \left (2;\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )}{\sqrt{a+b \sin (c+d x)}}+\frac{2 i \left (81 b^2-8 a^2\right ) \sec (c+d x) \sqrt{-\frac{b (\sin (c+d x)-1)}{a+b}} \sqrt{-\frac{b (\sin (c+d x)+1)}{a-b}} \left (b \left (b \Pi \left (\frac{a+b}{a};i \sinh ^{-1}\left (\sqrt{-\frac{1}{a+b}} \sqrt{a+b \sin (c+d x)}\right )|\frac{a+b}{a-b}\right )-2 a F\left (i \sinh ^{-1}\left (\sqrt{-\frac{1}{a+b}} \sqrt{a+b \sin (c+d x)}\right )|\frac{a+b}{a-b}\right )\right )-2 a (a-b) E\left (i \sinh ^{-1}\left (\sqrt{-\frac{1}{a+b}} \sqrt{a+b \sin (c+d x)}\right )|\frac{a+b}{a-b}\right )\right )}{a b^2 \sqrt{-\frac{1}{a+b}}}+\frac{472 a b \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )}{\sqrt{a+b \sin (c+d x)}}+4 \cot (c+d x) \csc (c+d x) \sqrt{a+b \sin (c+d x)} (8 a \cos (2 (c+d x))-18 a-31 b \sin (c+d x)+2 b \sin (3 (c+d x)))}{80 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((2*I)*(-8*a^2 + 81*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)
]) + (472*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*S
in[c + d*x]] + (2*(112*a^2 + 51*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*Cot[c + d*x]*Csc[c + d*x]*Sqrt[a + b*Sin[c + d*x]]*(-18*a + 8*a*Co
s[2*(c + d*x)] - 31*b*Sin[c + d*x] + 2*b*Sin[3*(c + d*x)]))/(80*d)

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Maple [B]  time = 1.774, size = 1379, normalized size = 3.6 \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)*cot(d*x+c)^3*(a+b*sin(d*x+c))^(3/2),x)

[Out]

1/20*(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*sin(d*x+c)^2-126*b^2*((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*sin(d*x+c)^2+37*b^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^2*sin(d*x+c)^2+81*
((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*sin(d*x+c)^2-8*((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)*EllipticE(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1
/2))*a^5*sin(d*x+c)^2+89*((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*sin(d*x+c)^2-81*((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*b^4*sin(d*x+c)^2+60*((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^2*EllipticPi(((a+b*sin(d*x+c))/(a-b))^(1/2),(a-b)/a,((a-b)/(a+b))^(1
/2))*a^3*sin(d*x+c)^2-60*((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^3*EllipticPi(((a+b*sin(d*x+c))/(a-b))^(1/2),(a-b)/a,((a-b)/(a+b))^(1/2))*a^2*sin(d*x+c)^2-15*((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)*EllipticPi(((a+b*sin
(d*x+c))/(a-b))^(1/2),(a-b)/a,((a-b)/(a+b))^(1/2))*a*b^4*sin(d*x+c)^2+15*((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)*EllipticPi(((a+b*sin(d*x+c))/(a-b))^(1/2),(a-b)/a,((
a-b)/(a+b))^(1/2))*b^5*sin(d*x+c)^2+8*a*b^4*sin(d*x+c)^6+24*a^2*b^3*sin(d*x+c)^5+16*a^3*b^2*sin(d*x+c)^4+17*a*
b^4*sin(d*x+c)^4+11*a^2*b^3*sin(d*x+c)^3-6*a^3*b^2*sin(d*x+c)^2-25*a*b^4*sin(d*x+c)^2-35*a^2*b^3*sin(d*x+c)-10
*a^3*b^2)/a/b^2/sin(d*x+c)^2/cos(d*x+c)/(a+b*sin(d*x+c))^(1/2)/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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

[Out]

Timed out

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

[Out]

Timed out