3.1154 \(\int \cos ^2(c+d x) \cot ^2(c+d x) (a+b \sin (c+d x))^{3/2} \, dx\)

Optimal. Leaf size=374 \[ \frac {\left (4 a^2+35 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{35 a b d}+\frac {\left (4 a^2+65 b^2\right ) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{35 b d}-\frac {a \left (4 a^2+167 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{35 b^2 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {\left (4 a^4+61 a^2 b^2+40 b^4\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{35 b^2 d \sqrt {a+b \sin (c+d x)}}-\frac {2 \cos (c+d x) (a+b \sin (c+d x))^{5/2}}{7 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^{5/2}}{a d}+\frac {3 a b \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{d \sqrt {a+b \sin (c+d x)}} \]

[Out]

1/35*(4*a^2+35*b^2)*cos(d*x+c)*(a+b*sin(d*x+c))^(3/2)/a/b/d-2/7*cos(d*x+c)*(a+b*sin(d*x+c))^(5/2)/b/d-cot(d*x+
c)*(a+b*sin(d*x+c))^(5/2)/a/d+1/35*(4*a^2+65*b^2)*cos(d*x+c)*(a+b*sin(d*x+c))^(1/2)/b/d+1/35*a*(4*a^2+167*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)/b^2/d/((a+b*sin(d*x+c))/(a+b))^(1/2)-1/35*(4*a^4+61*a^2*b^2+40*b^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)*(b/(a+b))
^(1/2))*((a+b*sin(d*x+c))/(a+b))^(1/2)/b^2/d/(a+b*sin(d*x+c))^(1/2)-3*a*b*(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*sin(d*x+c))/(a
+b))^(1/2)/d/(a+b*sin(d*x+c))^(1/2)

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Rubi [A]  time = 1.17, antiderivative size = 374, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.323, Rules used = {2894, 3049, 3059, 2655, 2653, 3002, 2663, 2661, 2807, 2805} \[ \frac {\left (4 a^2+35 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{35 a b d}+\frac {\left (4 a^2+65 b^2\right ) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{35 b d}+\frac {\left (61 a^2 b^2+4 a^4+40 b^4\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{35 b^2 d \sqrt {a+b \sin (c+d x)}}-\frac {a \left (4 a^2+167 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{35 b^2 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {2 \cos (c+d x) (a+b \sin (c+d x))^{5/2}}{7 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^{5/2}}{a d}+\frac {3 a b \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{d \sqrt {a+b \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*a^2 + 65*b^2)*Cos[c + d*x]*Sqrt[a + b*Sin[c + d*x]])/(35*b*d) + ((4*a^2 + 35*b^2)*Cos[c + d*x]*(a + b*Sin[
c + d*x])^(3/2))/(35*a*b*d) - (2*Cos[c + d*x]*(a + b*Sin[c + d*x])^(5/2))/(7*b*d) - (Cot[c + d*x]*(a + b*Sin[c
+ d*x])^(5/2))/(a*d) - (a*(4*a^2 + 167*b^2)*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d
*x]])/(35*b^2*d*Sqrt[(a + b*Sin[c + d*x])/(a + b)]) + ((4*a^4 + 61*a^2*b^2 + 40*b^4)*EllipticF[(c - Pi/2 + d*x
)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(35*b^2*d*Sqrt[a + b*Sin[c + d*x]]) + (3*a*b*EllipticP
i[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 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 2894

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*b*d*(n + 1)*(m + n + 4)), Int[(a + b*Sin[e + f*x])^m*(d*Sin[e + f*x])^(n + 1)*Simp[a^2*(n + 1)*(n
+ 2) - b^2*(m + n + 2)*(m + n + 4) + a*b*(m + 3)*Sin[e + f*x] - (a^2*(n + 1)*(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 + 1)*(d*Sin[e + f*x])^(n + 2))/(
b*d^2*f*(m + n + 4)), x]) /; FreeQ[{a, b, d, e, f, m}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || IntegersQ[2*m
, 2*n]) &&  !m < -1 && LtQ[n, -1] && NeQ[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 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]

Rubi steps

\begin {align*} \int \cos ^2(c+d x) \cot ^2(c+d x) (a+b \sin (c+d x))^{3/2} \, dx &=-\frac {2 \cos (c+d x) (a+b \sin (c+d x))^{5/2}}{7 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^{5/2}}{a d}-\frac {2 \int \csc (c+d x) (a+b \sin (c+d x))^{3/2} \left (-\frac {21 b^2}{4}+\frac {9}{2} a b \sin (c+d x)+\frac {1}{4} \left (4 a^2+35 b^2\right ) \sin ^2(c+d x)\right ) \, dx}{7 a b}\\ &=\frac {\left (4 a^2+35 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{35 a b d}-\frac {2 \cos (c+d x) (a+b \sin (c+d x))^{5/2}}{7 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^{5/2}}{a d}-\frac {4 \int \csc (c+d x) \sqrt {a+b \sin (c+d x)} \left (-\frac {105 a b^2}{8}+\frac {51}{4} a^2 b \sin (c+d x)+\frac {3}{8} a \left (4 a^2+65 b^2\right ) \sin ^2(c+d x)\right ) \, dx}{35 a b}\\ &=\frac {\left (4 a^2+65 b^2\right ) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{35 b d}+\frac {\left (4 a^2+35 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{35 a b d}-\frac {2 \cos (c+d x) (a+b \sin (c+d x))^{5/2}}{7 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^{5/2}}{a d}-\frac {8 \int \frac {\csc (c+d x) \left (-\frac {315}{16} a^2 b^2+\frac {3}{8} a b \left (53 a^2-20 b^2\right ) \sin (c+d x)+\frac {3}{16} a^2 \left (4 a^2+167 b^2\right ) \sin ^2(c+d x)\right )}{\sqrt {a+b \sin (c+d x)}} \, dx}{105 a b}\\ &=\frac {\left (4 a^2+65 b^2\right ) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{35 b d}+\frac {\left (4 a^2+35 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{35 a b d}-\frac {2 \cos (c+d x) (a+b \sin (c+d x))^{5/2}}{7 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^{5/2}}{a d}-\frac {1}{70} \left (a \left (167+\frac {4 a^2}{b^2}\right )\right ) \int \sqrt {a+b \sin (c+d x)} \, dx+\frac {8 \int \frac {\csc (c+d x) \left (\frac {315 a^2 b^3}{16}+\frac {3}{16} a \left (4 a^4+61 a^2 b^2+40 b^4\right ) \sin (c+d x)\right )}{\sqrt {a+b \sin (c+d x)}} \, dx}{105 a b^2}\\ &=\frac {\left (4 a^2+65 b^2\right ) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{35 b d}+\frac {\left (4 a^2+35 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{35 a b d}-\frac {2 \cos (c+d x) (a+b \sin (c+d x))^{5/2}}{7 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^{5/2}}{a d}+\frac {1}{2} (3 a b) \int \frac {\csc (c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx+\frac {\left (4 a^4+61 a^2 b^2+40 b^4\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}} \, dx}{70 b^2}-\frac {\left (a \left (167+\frac {4 a^2}{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}{70 \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}\\ &=\frac {\left (4 a^2+65 b^2\right ) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{35 b d}+\frac {\left (4 a^2+35 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{35 a b d}-\frac {2 \cos (c+d x) (a+b \sin (c+d x))^{5/2}}{7 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^{5/2}}{a d}-\frac {a \left (167+\frac {4 a^2}{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)}}{35 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {\left (3 a b \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}{2 \sqrt {a+b \sin (c+d x)}}+\frac {\left (\left (4 a^4+61 a^2 b^2+40 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}{70 b^2 \sqrt {a+b \sin (c+d x)}}\\ &=\frac {\left (4 a^2+65 b^2\right ) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{35 b d}+\frac {\left (4 a^2+35 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{35 a b d}-\frac {2 \cos (c+d x) (a+b \sin (c+d x))^{5/2}}{7 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^{5/2}}{a d}-\frac {a \left (167+\frac {4 a^2}{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)}}{35 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {\left (4 a^4+61 a^2 b^2+40 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}}}{35 b^2 d \sqrt {a+b \sin (c+d x)}}+\frac {3 a b \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*}

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Mathematica [C]  time = 4.23, size = 452, normalized size = 1.21 \[ \frac {\frac {8 \left (53 a^2-20 b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{4} (-2 c-2 d x+\pi )|\frac {2 b}{a+b}\right )}{\sqrt {a+b \sin (c+d x)}}+\frac {2 a \left (4 a^2-43 b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \Pi \left (2;\frac {1}{4} (-2 c-2 d x+\pi )|\frac {2 b}{a+b}\right )}{b \sqrt {a+b \sin (c+d x)}}-\frac {2 \sqrt {a+b \sin (c+d x)} \left (\left (4 a^2-55 b^2\right ) \cos (c+d x)+b (16 a \sin (2 (c+d x))+70 a \cot (c+d x)-5 b \cos (3 (c+d x)))\right )}{b}+\frac {2 i \left (4 a^2+167 b^2\right ) \sec (c+d x) \sqrt {-\frac {b (\sin (c+d x)-1)}{a+b}} \sqrt {\frac {b (\sin (c+d x)+1)}{b-a}} \left (b \left (b \Pi \left (\frac {a+b}{a};i \sinh ^{-1}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \sin (c+d x)}\right )|\frac {a+b}{a-b}\right )-2 a F\left (i \sinh ^{-1}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \sin (c+d x)}\right )|\frac {a+b}{a-b}\right )\right )-2 a (a-b) E\left (i \sinh ^{-1}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \sin (c+d x)}\right )|\frac {a+b}{a-b}\right )\right )}{b^3 \sqrt {-\frac {1}{a+b}}}}{140 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((2*I)*(4*a^2 + 167*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^3*Sqrt[-(a + b)^(-1)]) +
(8*(53*a^2 - 20*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*a*(4*a^2 - 43*b^2)*EllipticPi[2, (-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*Sqrt[(a + b*S
in[c + d*x])/(a + b)])/(b*Sqrt[a + b*Sin[c + d*x]]) - (2*Sqrt[a + b*Sin[c + d*x]]*((4*a^2 - 55*b^2)*Cos[c + d*
x] + b*(-5*b*Cos[3*(c + d*x)] + 70*a*Cot[c + d*x] + 16*a*Sin[2*(c + d*x)])))/b)/(140*d)

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

[Out]

Timed out

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

[Out]

Timed out

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maple [A]  time = 1.77, size = 726, normalized size = 1.94 \[ -\frac {-26 a \,b^{4} \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )+\left (2 a^{3} b^{2}+31 a \,b^{4}\right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+\sqrt {-\frac {b \sin \left (d x +c \right )}{a +b}+\frac {b}{a +b}}\, \sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}\, \sqrt {-\frac {b \sin \left (d x +c \right )}{a -b}-\frac {b}{a -b}}\, \left (105 \EllipticPi \left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \frac {a -b}{a}, \sqrt {\frac {a -b}{a +b}}\right ) a \,b^{4}-105 \EllipticPi \left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \frac {a -b}{a}, \sqrt {\frac {a -b}{a +b}}\right ) b^{5}-4 \EllipticE \left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{5}-163 \EllipticE \left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{3} b^{2}+167 \EllipticE \left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a \,b^{4}+4 \EllipticF \left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{4} b +102 \EllipticF \left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{3} b^{2}+61 \EllipticF \left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{2} b^{3}-207 \EllipticF \left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a \,b^{4}+40 \EllipticF \left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) b^{5}\right ) \sin \left (d x +c \right )+10 b^{5} \left (\cos ^{6}\left (d x +c \right )\right )+\left (-18 a^{2} b^{3}+10 b^{5}\right ) \left (\cos ^{4}\left (d x +c \right )\right )+\left (53 a^{2} b^{3}-20 b^{5}\right ) \left (\cos ^{2}\left (d x +c \right )\right )}{35 \sin \left (d x +c \right ) b^{3} \cos \left (d x +c \right ) \sqrt {a +b \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/35*(-26*a*b^4*sin(d*x+c)*cos(d*x+c)^4+(2*a^3*b^2+31*a*b^4)*cos(d*x+c)^2*sin(d*x+c)+(-b/(a+b)*sin(d*x+c)+b/(
a+b))^(1/2)*(b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2)*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*(105*EllipticPi((b/(a-b)*si
n(d*x+c)+a/(a-b))^(1/2),(a-b)/a,((a-b)/(a+b))^(1/2))*a*b^4-105*EllipticPi((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),(
a-b)/a,((a-b)/(a+b))^(1/2))*b^5-4*EllipticE((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^5-163*El
lipticE((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^3*b^2+167*EllipticE((b/(a-b)*sin(d*x+c)+a/(a
-b))^(1/2),((a-b)/(a+b))^(1/2))*a*b^4+4*EllipticF((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^4*
b+102*EllipticF((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^3*b^2+61*EllipticF((b/(a-b)*sin(d*x+
c)+a/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^2*b^3-207*EllipticF((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),((a-b)/(a+b))^
(1/2))*a*b^4+40*EllipticF((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*b^5)*sin(d*x+c)+10*b^5*cos(d
*x+c)^6+(-18*a^2*b^3+10*b^5)*cos(d*x+c)^4+(53*a^2*b^3-20*b^5)*cos(d*x+c)^2)/sin(d*x+c)/b^3/cos(d*x+c)/(a+b*sin
(d*x+c))^(1/2)/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*cot(d*x+c)^2*(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)^2*cot(d*x + c)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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