3.139 \(\int \frac{\csc ^6(e+f x)}{(a+b \tan ^2(e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=171 \[ -\frac{2 b \left (15 a^2-40 a b+24 b^2\right ) \tan (e+f x)}{15 a^4 f \sqrt{a+b \tan ^2(e+f x)}}-\frac{\left (15 a^2-40 a b+24 b^2\right ) \cot (e+f x)}{15 a^3 f \sqrt{a+b \tan ^2(e+f x)}}-\frac{2 (5 a-3 b) \cot ^3(e+f x)}{15 a^2 f \sqrt{a+b \tan ^2(e+f x)}}-\frac{\cot ^5(e+f x)}{5 a f \sqrt{a+b \tan ^2(e+f x)}} \]

[Out]

-((15*a^2 - 40*a*b + 24*b^2)*Cot[e + f*x])/(15*a^3*f*Sqrt[a + b*Tan[e + f*x]^2]) - (2*(5*a - 3*b)*Cot[e + f*x]
^3)/(15*a^2*f*Sqrt[a + b*Tan[e + f*x]^2]) - Cot[e + f*x]^5/(5*a*f*Sqrt[a + b*Tan[e + f*x]^2]) - (2*b*(15*a^2 -
40*a*b + 24*b^2)*Tan[e + f*x])/(15*a^4*f*Sqrt[a + b*Tan[e + f*x]^2])

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Rubi [A]  time = 0.178864, antiderivative size = 171, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3663, 462, 453, 271, 191} \[ -\frac{2 b \left (15 a^2-40 a b+24 b^2\right ) \tan (e+f x)}{15 a^4 f \sqrt{a+b \tan ^2(e+f x)}}-\frac{\left (15 a^2-40 a b+24 b^2\right ) \cot (e+f x)}{15 a^3 f \sqrt{a+b \tan ^2(e+f x)}}-\frac{2 (5 a-3 b) \cot ^3(e+f x)}{15 a^2 f \sqrt{a+b \tan ^2(e+f x)}}-\frac{\cot ^5(e+f x)}{5 a f \sqrt{a+b \tan ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^6/(a + b*Tan[e + f*x]^2)^(3/2),x]

[Out]

-((15*a^2 - 40*a*b + 24*b^2)*Cot[e + f*x])/(15*a^3*f*Sqrt[a + b*Tan[e + f*x]^2]) - (2*(5*a - 3*b)*Cot[e + f*x]
^3)/(15*a^2*f*Sqrt[a + b*Tan[e + f*x]^2]) - Cot[e + f*x]^5/(5*a*f*Sqrt[a + b*Tan[e + f*x]^2]) - (2*b*(15*a^2 -
40*a*b + 24*b^2)*Tan[e + f*x])/(15*a^4*f*Sqrt[a + b*Tan[e + f*x]^2])

Rule 3663

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

Rule 462

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> Simp[(c^2*(e*x)^(
m + 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)*(a + b*x^n)^p*Simp[b
*c^2*n*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*(m + 1)*d^2*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && Ne
Q[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && GtQ[n, 0]

Rule 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 271

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x^(m + 1)*(a + b*x^n)^(p + 1))/(a*(m + 1)), x]
- Dist[(b*(m + n*(p + 1) + 1))/(a*(m + 1)), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rubi steps

\begin{align*} \int \frac{\csc ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^2}{x^6 \left (a+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{\cot ^5(e+f x)}{5 a f \sqrt{a+b \tan ^2(e+f x)}}+\frac{\operatorname{Subst}\left (\int \frac{2 (5 a-3 b)+5 a x^2}{x^4 \left (a+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{5 a f}\\ &=-\frac{2 (5 a-3 b) \cot ^3(e+f x)}{15 a^2 f \sqrt{a+b \tan ^2(e+f x)}}-\frac{\cot ^5(e+f x)}{5 a f \sqrt{a+b \tan ^2(e+f x)}}-\frac{\left (-15 a^2+8 (5 a-3 b) b\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \left (a+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{15 a^2 f}\\ &=-\frac{\left (15 a^2-8 (5 a-3 b) b\right ) \cot (e+f x)}{15 a^3 f \sqrt{a+b \tan ^2(e+f x)}}-\frac{2 (5 a-3 b) \cot ^3(e+f x)}{15 a^2 f \sqrt{a+b \tan ^2(e+f x)}}-\frac{\cot ^5(e+f x)}{5 a f \sqrt{a+b \tan ^2(e+f x)}}+\frac{\left (2 b \left (-15 a^2+8 (5 a-3 b) b\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\left (a+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{15 a^3 f}\\ &=-\frac{\left (15 a^2-8 (5 a-3 b) b\right ) \cot (e+f x)}{15 a^3 f \sqrt{a+b \tan ^2(e+f x)}}-\frac{2 (5 a-3 b) \cot ^3(e+f x)}{15 a^2 f \sqrt{a+b \tan ^2(e+f x)}}-\frac{\cot ^5(e+f x)}{5 a f \sqrt{a+b \tan ^2(e+f x)}}-\frac{2 b \left (15 a^2-8 (5 a-3 b) b\right ) \tan (e+f x)}{15 a^4 f \sqrt{a+b \tan ^2(e+f x)}}\\ \end{align*}

Mathematica [A]  time = 1.28677, size = 135, normalized size = 0.79 \[ -\frac{\sqrt{\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)} \left (\cot (e+f x) \left (3 a^2 \csc ^4(e+f x)+8 a^2+a (4 a-9 b) \csc ^2(e+f x)-41 a b+33 b^2\right )+\frac{15 b (a-b)^2 \sin (2 (e+f x))}{(a-b) \cos (2 (e+f x))+a+b}\right )}{15 \sqrt{2} a^4 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^6/(a + b*Tan[e + f*x]^2)^(3/2),x]

[Out]

-(Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2]*(Cot[e + f*x]*(8*a^2 - 41*a*b + 33*b^2 + a*(4*a - 9*
b)*Csc[e + f*x]^2 + 3*a^2*Csc[e + f*x]^4) + (15*(a - b)^2*b*Sin[2*(e + f*x)])/(a + b + (a - b)*Cos[2*(e + f*x)
])))/(15*Sqrt[2]*a^4*f)

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Maple [A]  time = 0.239, size = 264, normalized size = 1.5 \begin{align*} -{\frac{ \left ( 8\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}{a}^{3}-64\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}{a}^{2}b+104\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}a{b}^{2}-48\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}{b}^{3}-20\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}{a}^{3}+164\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}{a}^{2}b-288\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}a{b}^{2}+144\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}{b}^{3}+15\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}{a}^{3}-130\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}{a}^{2}b+264\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}a{b}^{2}-144\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}{b}^{3}+30\,{a}^{2}b-80\,a{b}^{2}+48\,{b}^{3} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{3}}{15\,f{a}^{4} \left ( a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b \right ) ^{2} \left ( \sin \left ( fx+e \right ) \right ) ^{5}} \left ({\frac{a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b}{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^6/(a+b*tan(f*x+e)^2)^(3/2),x)

[Out]

-1/15/f/a^4/(a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)^2*(8*cos(f*x+e)^6*a^3-64*cos(f*x+e)^6*a^2*b+104*cos(f*x+e)^6*a*b
^2-48*cos(f*x+e)^6*b^3-20*cos(f*x+e)^4*a^3+164*cos(f*x+e)^4*a^2*b-288*cos(f*x+e)^4*a*b^2+144*cos(f*x+e)^4*b^3+
15*cos(f*x+e)^2*a^3-130*cos(f*x+e)^2*a^2*b+264*cos(f*x+e)^2*a*b^2-144*cos(f*x+e)^2*b^3+30*a^2*b-80*a*b^2+48*b^
3)*cos(f*x+e)^3*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/cos(f*x+e)^2)^(3/2)/sin(f*x+e)^5

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^6/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [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(f*x+e)^6/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

Timed out

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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(f*x+e)**6/(a+b*tan(f*x+e)**2)**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc \left (f x + e\right )^{6}}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^6/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

integrate(csc(f*x + e)^6/(b*tan(f*x + e)^2 + a)^(3/2), x)