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

Optimal. Leaf size=63 \[ -\frac{2 d^5}{5 b (d \tan (a+b x))^{5/2}}-\frac{4 d^3}{b \sqrt{d \tan (a+b x)}}+\frac{2 d (d \tan (a+b x))^{3/2}}{3 b} \]

[Out]

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

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Rubi [A]  time = 0.0539081, antiderivative size = 63, 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} \[ -\frac{2 d^5}{5 b (d \tan (a+b x))^{5/2}}-\frac{4 d^3}{b \sqrt{d \tan (a+b x)}}+\frac{2 d (d \tan (a+b x))^{3/2}}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 \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^{7/2}} \, dx,x,d \tan (a+b x)\right )}{b}\\ &=\frac{d \operatorname{Subst}\left (\int \left (\frac{d^4}{x^{7/2}}+\frac{2 d^2}{x^{3/2}}+\sqrt{x}\right ) \, dx,x,d \tan (a+b x)\right )}{b}\\ &=-\frac{2 d^5}{5 b (d \tan (a+b x))^{5/2}}-\frac{4 d^3}{b \sqrt{d \tan (a+b x)}}+\frac{2 d (d \tan (a+b x))^{3/2}}{3 b}\\ \end{align*}

Mathematica [A]  time = 0.223922, size = 42, normalized size = 0.67 \[ -\frac{2 d (d \tan (a+b x))^{3/2} \left (3 \cot ^2(a+b x) \left (\csc ^2(a+b x)+9\right )-5\right )}{15 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*d*(-5 + 3*Cot[a + b*x]^2*(9 + Csc[a + b*x]^2))*(d*Tan[a + b*x])^(3/2))/(15*b)

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Maple [A]  time = 0.154, size = 60, normalized size = 1. \begin{align*}{\frac{ \left ( 64\, \left ( \cos \left ( bx+a \right ) \right ) ^{4}-80\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}+10 \right ) \cos \left ( bx+a \right ) }{15\,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]

2/15/b*(32*cos(b*x+a)^4-40*cos(b*x+a)^2+5)*cos(b*x+a)*(d*sin(b*x+a)/cos(b*x+a))^(5/2)/sin(b*x+a)^5

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Maxima [A]  time = 1.51859, size = 76, normalized size = 1.21 \begin{align*} \frac{2 \, d^{5}{\left (\frac{5 \, \left (d \tan \left (b x + a\right )\right )^{\frac{3}{2}}}{d^{4}} - \frac{3 \,{\left (10 \, d^{2} \tan \left (b x + a\right )^{2} + d^{2}\right )}}{\left (d \tan \left (b x + a\right )\right )^{\frac{5}{2}} d^{2}}\right )}}{15 \, 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]

2/15*d^5*(5*(d*tan(b*x + a))^(3/2)/d^4 - 3*(10*d^2*tan(b*x + a)^2 + d^2)/((d*tan(b*x + a))^(5/2)*d^2))/b

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Fricas [A]  time = 2.24406, size = 200, normalized size = 3.17 \begin{align*} -\frac{2 \,{\left (32 \, d^{2} \cos \left (b x + a\right )^{4} - 40 \, d^{2} \cos \left (b x + a\right )^{2} + 5 \, d^{2}\right )} \sqrt{\frac{d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}}}{15 \,{\left (b \cos \left (b x + a\right )^{3} - b \cos \left (b x + a\right )\right )} \sin \left (b x + a\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]

-2/15*(32*d^2*cos(b*x + a)^4 - 40*d^2*cos(b*x + a)^2 + 5*d^2)*sqrt(d*sin(b*x + a)/cos(b*x + a))/((b*cos(b*x +
a)^3 - b*cos(b*x + a))*sin(b*x + a))

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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 [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d \tan \left (b x + a\right )\right )^{\frac{5}{2}} \csc \left (b x + a\right )^{6}\,{d x} \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]

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