3.107 \(\int \frac{\csc ^6(a+b x)}{(d \tan (a+b x))^{5/2}} \, dx\)

Optimal. Leaf size=65 \[ -\frac{2 d^5}{15 b (d \tan (a+b x))^{15/2}}-\frac{4 d^3}{11 b (d \tan (a+b x))^{11/2}}-\frac{2 d}{7 b (d \tan (a+b x))^{7/2}} \]

[Out]

(-2*d^5)/(15*b*(d*Tan[a + b*x])^(15/2)) - (4*d^3)/(11*b*(d*Tan[a + b*x])^(11/2)) - (2*d)/(7*b*(d*Tan[a + b*x])
^(7/2))

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Rubi [A]  time = 0.0562098, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2591, 270} \[ -\frac{2 d^5}{15 b (d \tan (a+b x))^{15/2}}-\frac{4 d^3}{11 b (d \tan (a+b x))^{11/2}}-\frac{2 d}{7 b (d \tan (a+b x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^6/(d*Tan[a + b*x])^(5/2),x]

[Out]

(-2*d^5)/(15*b*(d*Tan[a + b*x])^(15/2)) - (4*d^3)/(11*b*(d*Tan[a + b*x])^(11/2)) - (2*d)/(7*b*(d*Tan[a + b*x])
^(7/2))

Rule 2591

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> With[{ff = FreeFactors[Ta
n[e + f*x], x]}, Dist[(b*ff)/f, Subst[Int[(ff*x)^(m + n)/(b^2 + ff^2*x^2)^(m/2 + 1), x], x, (b*Tan[e + f*x])/f
f], x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2]

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \frac{\csc ^6(a+b x)}{(d \tan (a+b x))^{5/2}} \, dx &=\frac{d \operatorname{Subst}\left (\int \frac{\left (d^2+x^2\right )^2}{x^{17/2}} \, dx,x,d \tan (a+b x)\right )}{b}\\ &=\frac{d \operatorname{Subst}\left (\int \left (\frac{d^4}{x^{17/2}}+\frac{2 d^2}{x^{13/2}}+\frac{1}{x^{9/2}}\right ) \, dx,x,d \tan (a+b x)\right )}{b}\\ &=-\frac{2 d^5}{15 b (d \tan (a+b x))^{15/2}}-\frac{4 d^3}{11 b (d \tan (a+b x))^{11/2}}-\frac{2 d}{7 b (d \tan (a+b x))^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.232077, size = 60, normalized size = 0.92 \[ \frac{2 (44 \cos (2 (a+b x))-4 \cos (4 (a+b x))-117) \cot ^4(a+b x) \csc ^4(a+b x) \sqrt{d \tan (a+b x)}}{1155 b d^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]^6/(d*Tan[a + b*x])^(5/2),x]

[Out]

(2*(-117 + 44*Cos[2*(a + b*x)] - 4*Cos[4*(a + b*x)])*Cot[a + b*x]^4*Csc[a + b*x]^4*Sqrt[d*Tan[a + b*x]])/(1155
*b*d^3)

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Maple [A]  time = 0.172, size = 60, normalized size = 0.9 \begin{align*} -{\frac{ \left ( 64\, \left ( \cos \left ( bx+a \right ) \right ) ^{4}-240\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}+330 \right ) \cos \left ( bx+a \right ) }{1155\,b \left ( \sin \left ( bx+a \right ) \right ) ^{5}} \left ({\frac{d\sin \left ( bx+a \right ) }{\cos \left ( bx+a \right ) }} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(b*x+a)^6/(d*tan(b*x+a))^(5/2),x)

[Out]

-2/1155/b*(32*cos(b*x+a)^4-120*cos(b*x+a)^2+165)*cos(b*x+a)/sin(b*x+a)^5/(d*sin(b*x+a)/cos(b*x+a))^(5/2)

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Maxima [A]  time = 1.0186, size = 65, normalized size = 1. \begin{align*} -\frac{2 \,{\left (165 \, d^{4} \tan \left (b x + a\right )^{4} + 210 \, d^{4} \tan \left (b x + a\right )^{2} + 77 \, d^{4}\right )} d}{1155 \, \left (d \tan \left (b x + a\right )\right )^{\frac{15}{2}} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^6/(d*tan(b*x+a))^(5/2),x, algorithm="maxima")

[Out]

-2/1155*(165*d^4*tan(b*x + a)^4 + 210*d^4*tan(b*x + a)^2 + 77*d^4)*d/((d*tan(b*x + a))^(15/2)*b)

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Fricas [B]  time = 3.20698, size = 285, normalized size = 4.38 \begin{align*} -\frac{2 \,{\left (32 \, \cos \left (b x + a\right )^{8} - 120 \, \cos \left (b x + a\right )^{6} + 165 \, \cos \left (b x + a\right )^{4}\right )} \sqrt{\frac{d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}}}{1155 \,{\left (b d^{3} \cos \left (b x + a\right )^{8} - 4 \, b d^{3} \cos \left (b x + a\right )^{6} + 6 \, b d^{3} \cos \left (b x + a\right )^{4} - 4 \, b d^{3} \cos \left (b x + a\right )^{2} + b d^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^6/(d*tan(b*x+a))^(5/2),x, algorithm="fricas")

[Out]

-2/1155*(32*cos(b*x + a)^8 - 120*cos(b*x + a)^6 + 165*cos(b*x + a)^4)*sqrt(d*sin(b*x + a)/cos(b*x + a))/(b*d^3
*cos(b*x + a)^8 - 4*b*d^3*cos(b*x + a)^6 + 6*b*d^3*cos(b*x + a)^4 - 4*b*d^3*cos(b*x + a)^2 + b*d^3)

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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(csc(b*x+a)**6/(d*tan(b*x+a))**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.14282, size = 78, normalized size = 1.2 \begin{align*} -\frac{2 \,{\left (165 \, d^{5} \tan \left (b x + a\right )^{4} + 210 \, d^{5} \tan \left (b x + a\right )^{2} + 77 \, d^{5}\right )}}{1155 \, \sqrt{d \tan \left (b x + a\right )} b d^{7} \tan \left (b x + a\right )^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^6/(d*tan(b*x+a))^(5/2),x, algorithm="giac")

[Out]

-2/1155*(165*d^5*tan(b*x + a)^4 + 210*d^5*tan(b*x + a)^2 + 77*d^5)/(sqrt(d*tan(b*x + a))*b*d^7*tan(b*x + a)^7)