3.1187 \(\int \frac {\cos ^2(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=346 \[ -\frac {5 b \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{a^3 d \sqrt {a+b \sin (c+d x)}}+\frac {\left (2 a^2-5 b^2\right ) \cos (c+d x)}{3 a^2 b d (a+b \sin (c+d x))^{3/2}}+\frac {\left (4 a^2+5 b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{3 a^2 b^2 d \sqrt {a+b \sin (c+d x)}}-\frac {\left (4 a^2+15 b^2\right ) \cos (c+d x)}{3 a^3 b d \sqrt {a+b \sin (c+d x)}}-\frac {\left (4 a^2+15 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{3 a^3 b^2 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))^{3/2}} \]

[Out]

1/3*(2*a^2-5*b^2)*cos(d*x+c)/a^2/b/d/(a+b*sin(d*x+c))^(3/2)-cot(d*x+c)/a/d/(a+b*sin(d*x+c))^(3/2)-1/3*(4*a^2+1
5*b^2)*cos(d*x+c)/a^3/b/d/(a+b*sin(d*x+c))^(1/2)+1/3*(4*a^2+15*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^3/b^
2/d/((a+b*sin(d*x+c))/(a+b))^(1/2)-1/3*(4*a^2+5*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^2/b^2/d/(a+
b*sin(d*x+c))^(1/2)+5*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)/a^3/d/(a+b*sin(d*x+c))^(1/2)

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Rubi [A]  time = 0.99, antiderivative size = 346, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.323, Rules used = {2890, 3055, 3059, 2655, 2653, 3002, 2663, 2661, 2807, 2805} \[ -\frac {\left (4 a^2+15 b^2\right ) \cos (c+d x)}{3 a^3 b d \sqrt {a+b \sin (c+d x)}}+\frac {\left (2 a^2-5 b^2\right ) \cos (c+d x)}{3 a^2 b d (a+b \sin (c+d x))^{3/2}}+\frac {\left (4 a^2+5 b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{3 a^2 b^2 d \sqrt {a+b \sin (c+d x)}}-\frac {\left (4 a^2+15 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{3 a^3 b^2 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {5 b \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{a^3 d \sqrt {a+b \sin (c+d x)}}-\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a^2 - 5*b^2)*Cos[c + d*x])/(3*a^2*b*d*(a + b*Sin[c + d*x])^(3/2)) - Cot[c + d*x]/(a*d*(a + b*Sin[c + d*x])
^(3/2)) - ((4*a^2 + 15*b^2)*Cos[c + d*x])/(3*a^3*b*d*Sqrt[a + b*Sin[c + d*x]]) - ((4*a^2 + 15*b^2)*EllipticE[(
c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(3*a^3*b^2*d*Sqrt[(a + b*Sin[c + d*x])/(a + b)]) +
((4*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*a^2*b^2*
d*Sqrt[a + b*Sin[c + d*x]]) - (5*b*EllipticPi[2, (c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/
(a + b)])/(a^3*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 2890

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

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 3055

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*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] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 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 ^2(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx &=\frac {\left (2 a^2-5 b^2\right ) \cos (c+d x)}{3 a^2 b d (a+b \sin (c+d x))^{3/2}}-\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))^{3/2}}+\frac {2 \int \frac {\csc (c+d x) \left (-\frac {15 b^2}{4}-\frac {3}{2} a b \sin (c+d x)+\frac {1}{4} \left (4 a^2+5 b^2\right ) \sin ^2(c+d x)\right )}{(a+b \sin (c+d x))^{3/2}} \, dx}{3 a^2 b}\\ &=\frac {\left (2 a^2-5 b^2\right ) \cos (c+d x)}{3 a^2 b d (a+b \sin (c+d x))^{3/2}}-\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))^{3/2}}-\frac {\left (4 a^2+15 b^2\right ) \cos (c+d x)}{3 a^3 b d \sqrt {a+b \sin (c+d x)}}+\frac {4 \int \frac {\csc (c+d x) \left (-\frac {15}{8} b^2 \left (a^2-b^2\right )-\frac {5}{4} a b \left (a^2-b^2\right ) \sin (c+d x)-\frac {1}{8} \left (4 a^4+11 a^2 b^2-15 b^4\right ) \sin ^2(c+d x)\right )}{\sqrt {a+b \sin (c+d x)}} \, dx}{3 a^3 b \left (a^2-b^2\right )}\\ &=\frac {\left (2 a^2-5 b^2\right ) \cos (c+d x)}{3 a^2 b d (a+b \sin (c+d x))^{3/2}}-\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))^{3/2}}-\frac {\left (4 a^2+15 b^2\right ) \cos (c+d x)}{3 a^3 b d \sqrt {a+b \sin (c+d x)}}-\frac {4 \int \frac {\csc (c+d x) \left (\frac {15}{8} b^3 \left (a^2-b^2\right )-\frac {1}{8} a \left (4 a^4+a^2 b^2-5 b^4\right ) \sin (c+d x)\right )}{\sqrt {a+b \sin (c+d x)}} \, dx}{3 a^3 b^2 \left (a^2-b^2\right )}-\frac {\left (4 a^4+11 a^2 b^2-15 b^4\right ) \int \sqrt {a+b \sin (c+d x)} \, dx}{6 a^3 b^2 \left (a^2-b^2\right )}\\ &=\frac {\left (2 a^2-5 b^2\right ) \cos (c+d x)}{3 a^2 b d (a+b \sin (c+d x))^{3/2}}-\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))^{3/2}}-\frac {\left (4 a^2+15 b^2\right ) \cos (c+d x)}{3 a^3 b d \sqrt {a+b \sin (c+d x)}}+\frac {1}{6} \left (\frac {5}{a^2}+\frac {4}{b^2}\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}} \, dx-\frac {(5 b) \int \frac {\csc (c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx}{2 a^3}-\frac {\left (\left (4 a^4+11 a^2 b^2-15 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}{6 a^3 b^2 \left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}\\ &=\frac {\left (2 a^2-5 b^2\right ) \cos (c+d x)}{3 a^2 b d (a+b \sin (c+d x))^{3/2}}-\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))^{3/2}}-\frac {\left (4 a^2+15 b^2\right ) \cos (c+d x)}{3 a^3 b d \sqrt {a+b \sin (c+d x)}}-\frac {\left (4 a^2+15 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^3 b^2 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {\left (\left (\frac {5}{a^2}+\frac {4}{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}{6 \sqrt {a+b \sin (c+d x)}}-\frac {\left (5 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 a^3 \sqrt {a+b \sin (c+d x)}}\\ &=\frac {\left (2 a^2-5 b^2\right ) \cos (c+d x)}{3 a^2 b d (a+b \sin (c+d x))^{3/2}}-\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))^{3/2}}-\frac {\left (4 a^2+15 b^2\right ) \cos (c+d x)}{3 a^3 b d \sqrt {a+b \sin (c+d x)}}-\frac {\left (4 a^2+15 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^3 b^2 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {\left (\frac {5}{a^2}+\frac {4}{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 d \sqrt {a+b \sin (c+d x)}}-\frac {5 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}}}{a^3 d \sqrt {a+b \sin (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 5.46, size = 445, normalized size = 1.29 \[ \frac {\frac {2 \left (4 a^2+45 b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \Pi \left (2;\frac {1}{4} (-2 c-2 d x+\pi )|\frac {2 b}{a+b}\right )}{\sqrt {a+b \sin (c+d x)}}-\frac {2 \left (b \left (4 a^2+15 b^2\right ) \sin (2 (c+d x))+4 a \left (a^2+10 b^2\right ) \cos (c+d x)+6 a^2 b \cot (c+d x)\right )}{(a+b \sin (c+d x))^{3/2}}+\frac {2 i \left (4 a^2+15 b^2\right ) \sec (c+d x) \sqrt {-\frac {b (\sin (c+d x)-1)}{a+b}} \sqrt {-\frac {b (\sin (c+d x)+1)}{a-b}} \left (b \left (b \Pi \left (\frac {a+b}{a};i \sinh ^{-1}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \sin (c+d x)}\right )|\frac {a+b}{a-b}\right )-2 a F\left (i \sinh ^{-1}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \sin (c+d x)}\right )|\frac {a+b}{a-b}\right )\right )-2 a (a-b) E\left (i \sinh ^{-1}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \sin (c+d x)}\right )|\frac {a+b}{a-b}\right )\right )}{a b^2 \sqrt {-\frac {1}{a+b}}}+\frac {40 a b \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{4} (-2 c-2 d x+\pi )|\frac {2 b}{a+b}\right )}{\sqrt {a+b \sin (c+d x)}}}{12 a^3 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((2*I)*(4*a^2 + 15*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)]
) + (40*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*(4*a^2 + 45*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]] - (2*(4*a*(a^2 + 10*b^2)*Cos[c + d*x] + 6*a^2*b*Cot[c + d*x] + b*(4*a^2 +
15*b^2)*Sin[2*(c + d*x)]))/(a + b*Sin[c + d*x])^(3/2))/(12*a^3*b*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))^(5/2),x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*cot(d*x+c)^2/(a+b*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 2.19, size = 2112, normalized size = 6.10 \[ \text {result too large to display} \]

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))^(5/2),x)

[Out]

-1/3*(15*a*b^5*sin(d*x+c)^2-4*a^3*b^3*sin(d*x+c)^4-15*a*b^5*sin(d*x+c)^4-2*a^4*b^2*sin(d*x+c)^3-20*a^2*b^4*sin
(d*x+c)^3+a^3*b^3*sin(d*x+c)^2+2*a^4*b^2*sin(d*x+c)+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^3*s
in(d*x+c)^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)*E
llipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^2*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)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2)
,((a-b)/(a+b))^(1/2))*a*b^5*sin(d*x+c)^2-4*((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*b*sin(d*x+c)^2-1
1*((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^3*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)*EllipticE(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b
))^(1/2))*a*b^5*sin(d*x+c)^2+20*a^2*b^4*sin(d*x+c)+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^6*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^5*sin(d*x+c)^2+4*((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*sin(d*x+c)+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*si
n(d*x+c)+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)*Elli
pticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^3*b^3*sin(d*x+c)-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)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-
b)/(a+b))^(1/2))*a^2*b^4*sin(d*x+c)-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^4*b^2*sin(d*x+c)+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)*EllipticE(((a+b*s
in(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^2*b^4*sin(d*x+c)-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^2*b^4*sin(d*x+c)+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^5*sin(d*x+c)+4*
((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*sin(d*x+c)^2+3*a^3*b^3-4*((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*sin(d*x+c))/a^4/sin(d*x+c)/(a+b*sin(d*x+c))^(3/2)/b^3/cos(d*x+c)/d

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maxima [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))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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

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))^(5/2),x)

[Out]

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

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

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

[Out]

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

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