3.1153 \(\int \cos ^3(c+d x) \cot (c+d x) (a+b \sin (c+d x))^{3/2} \, dx\)

Optimal. Leaf size=390 \[ -\frac{2 \left (8 a^2-77 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{315 b^2 d}-\frac{2 a \left (8 a^2-87 b^2\right ) \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{315 b^2 d}-\frac{2 a \left (-95 a^2 b^2+8 a^4-228 b^4\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{315 b^3 d \sqrt{a+b \sin (c+d x)}}+\frac{2 \left (-93 a^2 b^2+8 a^4+84 b^4\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{315 b^3 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{2 a^2 \sqrt{\frac{a+b \sin (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{d \sqrt{a+b \sin (c+d x)}}+\frac{8 a \cos (c+d x) (a+b \sin (c+d x))^{5/2}}{63 b^2 d}-\frac{2 \sin (c+d x) \cos (c+d x) (a+b \sin (c+d x))^{5/2}}{9 b d} \]

[Out]

(-2*a*(8*a^2 - 87*b^2)*Cos[c + d*x]*Sqrt[a + b*Sin[c + d*x]])/(315*b^2*d) - (2*(8*a^2 - 77*b^2)*Cos[c + d*x]*(
a + b*Sin[c + d*x])^(3/2))/(315*b^2*d) + (8*a*Cos[c + d*x]*(a + b*Sin[c + d*x])^(5/2))/(63*b^2*d) - (2*Cos[c +
d*x]*Sin[c + d*x]*(a + b*Sin[c + d*x])^(5/2))/(9*b*d) + (2*(8*a^4 - 93*a^2*b^2 + 84*b^4)*EllipticE[(c - Pi/2
+ d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(315*b^3*d*Sqrt[(a + b*Sin[c + d*x])/(a + b)]) - (2*a*(8*a^
4 - 95*a^2*b^2 - 228*b^4)*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(31
5*b^3*d*Sqrt[a + b*Sin[c + d*x]]) + (2*a^2*EllipticPi[2, (c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c
+ d*x])/(a + b)])/(d*Sqrt[a + b*Sin[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 1.15932, antiderivative size = 390, 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 = {2895, 3049, 3059, 2655, 2653, 3002, 2663, 2661, 2807, 2805} \[ -\frac{2 \left (8 a^2-77 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{315 b^2 d}-\frac{2 a \left (8 a^2-87 b^2\right ) \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{315 b^2 d}-\frac{2 a \left (-95 a^2 b^2+8 a^4-228 b^4\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{315 b^3 d \sqrt{a+b \sin (c+d x)}}+\frac{2 \left (-93 a^2 b^2+8 a^4+84 b^4\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{315 b^3 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{2 a^2 \sqrt{\frac{a+b \sin (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{d \sqrt{a+b \sin (c+d x)}}+\frac{8 a \cos (c+d x) (a+b \sin (c+d x))^{5/2}}{63 b^2 d}-\frac{2 \sin (c+d x) \cos (c+d x) (a+b \sin (c+d x))^{5/2}}{9 b d} \]

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*(8*a^2 - 87*b^2)*Cos[c + d*x]*Sqrt[a + b*Sin[c + d*x]])/(315*b^2*d) - (2*(8*a^2 - 77*b^2)*Cos[c + d*x]*(
a + b*Sin[c + d*x])^(3/2))/(315*b^2*d) + (8*a*Cos[c + d*x]*(a + b*Sin[c + d*x])^(5/2))/(63*b^2*d) - (2*Cos[c +
d*x]*Sin[c + d*x]*(a + b*Sin[c + d*x])^(5/2))/(9*b*d) + (2*(8*a^4 - 93*a^2*b^2 + 84*b^4)*EllipticE[(c - Pi/2
+ d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(315*b^3*d*Sqrt[(a + b*Sin[c + d*x])/(a + b)]) - (2*a*(8*a^
4 - 95*a^2*b^2 - 228*b^4)*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(31
5*b^3*d*Sqrt[a + b*Sin[c + d*x]]) + (2*a^2*EllipticPi[2, (c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c
+ d*x])/(a + b)])/(d*Sqrt[a + b*Sin[c + d*x]])

Rule 2895

Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Simp[(a*(n + 3)*Cos[e + f*x]*(d*Sin[e + f*x])^(n + 1)*(a + b*Sin[e + f*x])^(m + 1))/(b^2*d*f*(m
+ n + 3)*(m + n + 4)), x] + (-Dist[1/(b^2*(m + n + 3)*(m + n + 4)), Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x
])^m*Simp[a^2*(n + 1)*(n + 3) - b^2*(m + n + 3)*(m + n + 4) + a*b*m*Sin[e + f*x] - (a^2*(n + 2)*(n + 3) - b^2*
(m + n + 3)*(m + n + 5))*Sin[e + f*x]^2, x], x], x] - Simp[(Cos[e + f*x]*(d*Sin[e + f*x])^(n + 2)*(a + b*Sin[e
+ f*x])^(m + 1))/(b*d^2*f*(m + n + 4)), x]) /; FreeQ[{a, b, d, e, f, m, n}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[
m, 0] || IntegersQ[2*m, 2*n]) &&  !m < -1 &&  !LtQ[n, -1] && NeQ[m + n + 3, 0] && NeQ[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 ^3(c+d x) \cot (c+d x) (a+b \sin (c+d x))^{3/2} \, dx &=\frac{8 a \cos (c+d x) (a+b \sin (c+d x))^{5/2}}{63 b^2 d}-\frac{2 \cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^{5/2}}{9 b d}-\frac{4 \int \csc (c+d x) (a+b \sin (c+d x))^{3/2} \left (-\frac{63 b^2}{4}+\frac{3}{2} a b \sin (c+d x)-\frac{1}{4} \left (8 a^2-77 b^2\right ) \sin ^2(c+d x)\right ) \, dx}{63 b^2}\\ &=-\frac{2 \left (8 a^2-77 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{315 b^2 d}+\frac{8 a \cos (c+d x) (a+b \sin (c+d x))^{5/2}}{63 b^2 d}-\frac{2 \cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^{5/2}}{9 b d}-\frac{8 \int \csc (c+d x) \sqrt{a+b \sin (c+d x)} \left (-\frac{315 a b^2}{8}+\frac{3}{4} b \left (a^2-14 b^2\right ) \sin (c+d x)-\frac{3}{8} a \left (8 a^2-87 b^2\right ) \sin ^2(c+d x)\right ) \, dx}{315 b^2}\\ &=-\frac{2 a \left (8 a^2-87 b^2\right ) \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{315 b^2 d}-\frac{2 \left (8 a^2-77 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{315 b^2 d}+\frac{8 a \cos (c+d x) (a+b \sin (c+d x))^{5/2}}{63 b^2 d}-\frac{2 \cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^{5/2}}{9 b d}-\frac{16 \int \frac{\csc (c+d x) \left (-\frac{945}{16} a^2 b^2-\frac{3}{8} a b \left (a^2+156 b^2\right ) \sin (c+d x)-\frac{3}{16} \left (8 a^4-93 a^2 b^2+84 b^4\right ) \sin ^2(c+d x)\right )}{\sqrt{a+b \sin (c+d x)}} \, dx}{945 b^2}\\ &=-\frac{2 a \left (8 a^2-87 b^2\right ) \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{315 b^2 d}-\frac{2 \left (8 a^2-77 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{315 b^2 d}+\frac{8 a \cos (c+d x) (a+b \sin (c+d x))^{5/2}}{63 b^2 d}-\frac{2 \cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^{5/2}}{9 b d}+\frac{16 \int \frac{\csc (c+d x) \left (\frac{945 a^2 b^3}{16}-\frac{3}{16} a \left (8 a^4-95 a^2 b^2-228 b^4\right ) \sin (c+d x)\right )}{\sqrt{a+b \sin (c+d x)}} \, dx}{945 b^3}+\frac{\left (8 a^4-93 a^2 b^2+84 b^4\right ) \int \sqrt{a+b \sin (c+d x)} \, dx}{315 b^3}\\ &=-\frac{2 a \left (8 a^2-87 b^2\right ) \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{315 b^2 d}-\frac{2 \left (8 a^2-77 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{315 b^2 d}+\frac{8 a \cos (c+d x) (a+b \sin (c+d x))^{5/2}}{63 b^2 d}-\frac{2 \cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^{5/2}}{9 b d}+a^2 \int \frac{\csc (c+d x)}{\sqrt{a+b \sin (c+d x)}} \, dx-\frac{\left (a \left (8 a^4-95 a^2 b^2-228 b^4\right )\right ) \int \frac{1}{\sqrt{a+b \sin (c+d x)}} \, dx}{315 b^3}+\frac{\left (\left (8 a^4-93 a^2 b^2+84 b^4\right ) \sqrt{a+b \sin (c+d x)}\right ) \int \sqrt{\frac{a}{a+b}+\frac{b \sin (c+d x)}{a+b}} \, dx}{315 b^3 \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}\\ &=-\frac{2 a \left (8 a^2-87 b^2\right ) \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{315 b^2 d}-\frac{2 \left (8 a^2-77 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{315 b^2 d}+\frac{8 a \cos (c+d x) (a+b \sin (c+d x))^{5/2}}{63 b^2 d}-\frac{2 \cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^{5/2}}{9 b d}+\frac{2 \left (8 a^4-93 a^2 b^2+84 b^4\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)}}{315 b^3 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{\left (a^2 \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}{\sqrt{a+b \sin (c+d x)}}-\frac{\left (a \left (8 a^4-95 a^2 b^2-228 b^4\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}{315 b^3 \sqrt{a+b \sin (c+d x)}}\\ &=-\frac{2 a \left (8 a^2-87 b^2\right ) \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{315 b^2 d}-\frac{2 \left (8 a^2-77 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{315 b^2 d}+\frac{8 a \cos (c+d x) (a+b \sin (c+d x))^{5/2}}{63 b^2 d}-\frac{2 \cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^{5/2}}{9 b d}+\frac{2 \left (8 a^4-93 a^2 b^2+84 b^4\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)}}{315 b^3 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}-\frac{2 a \left (8 a^4-95 a^2 b^2-228 b^4\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}}}{315 b^3 d \sqrt{a+b \sin (c+d x)}}+\frac{2 a^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}}}{d \sqrt{a+b \sin (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 3.50949, size = 477, normalized size = 1.22 \[ \frac{\cos (c+d x) \sqrt{a+b \sin (c+d x)} \left (\left (203 b^3-12 a^2 b\right ) \sin (c+d x)+16 a^3+100 a b^2 \cos (2 (c+d x))+556 a b^2+35 b^3 \sin (3 (c+d x))\right )-\frac{8 a b \left (a^2+156 b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )}{\sqrt{a+b \sin (c+d x)}}-\frac{2 \left (537 a^2 b^2+8 a^4+84 b^4\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} \Pi \left (2;\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )}{\sqrt{a+b \sin (c+d x)}}-\frac{2 i \left (-93 a^2 b^2+8 a^4+84 b^4\right ) \sec (c+d x) \sqrt{-\frac{b (\sin (c+d x)-1)}{a+b}} \sqrt{\frac{b (\sin (c+d x)+1)}{b-a}} \left (b \left (b \Pi \left (\frac{a+b}{a};i \sinh ^{-1}\left (\sqrt{-\frac{1}{a+b}} \sqrt{a+b \sin (c+d x)}\right )|\frac{a+b}{a-b}\right )-2 a F\left (i \sinh ^{-1}\left (\sqrt{-\frac{1}{a+b}} \sqrt{a+b \sin (c+d x)}\right )|\frac{a+b}{a-b}\right )\right )-2 a (a-b) E\left (i \sinh ^{-1}\left (\sqrt{-\frac{1}{a+b}} \sqrt{a+b \sin (c+d x)}\right )|\frac{a+b}{a-b}\right )\right )}{a b^2 \sqrt{-\frac{1}{a+b}}}}{630 b^2 d} \]

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^4 - 93*a^2*b^2 + 84*b^4)*(-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*(a^2 + 156*b^2)*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^4 + 537*a^2*b^2 + 84*b^4)*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]] + Cos[c + d*x]*Sqrt[a + b*Sin[c +
d*x]]*(16*a^3 + 556*a*b^2 + 100*a*b^2*Cos[2*(c + d*x)] + (-12*a^2*b + 203*b^3)*Sin[c + d*x] + 35*b^3*Sin[3*(c
+ d*x)]))/(630*b^2*d)

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Maple [B]  time = 1.644, size = 1405, 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)^3*cot(d*x+c)*(a+b*sin(d*x+c))^(3/2),x)

[Out]

2/315*(164*a^2*b^4+85*a*b^5*sin(d*x+c)^5+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)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^5*b+35*b^6*sin(d*x
+c)^6-112*b^6*sin(d*x+c)^4+77*b^6*sin(d*x+c)^2+53*a^2*b^4*sin(d*x+c)^4-a^3*b^3*sin(d*x+c)^3-326*a*b^5*sin(d*x+
c)^3-4*a^4*b^2*sin(d*x+c)^2-217*a^2*b^4*sin(d*x+c)^2+a^3*b^3*sin(d*x+c)+241*a*b^5*sin(d*x+c)+4*a^4*b^2-84*((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))*b^6-8*((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^6+84*((a+b*s
in(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))*b^6-6*((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^4*b^2-95*((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^3-315*a^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)*b^4*EllipticPi(((a+b*sin(d*x+c))/(a-b))^(1/2),(a-b)/a,((a-b)/(a+b))^
(1/2))+315*a*((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))+405*((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*b^4-228*((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^5+101*((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^4*b^2-177*((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^2*b^4)/b^4/cos(d*x+c)/(a+b*si
n(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 )^{3} \cot \left (d x + c\right )\,{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="maxima")

[Out]

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

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

[Out]

Timed out