3.306 \(\int \frac{(b \tan (e+f x))^{3/2}}{(d \sec (e+f x))^{9/2}} \, dx\)

Optimal. Leaf size=103 \[ \frac{8 b \sqrt{b \tan (e+f x)}}{45 d^4 f \sqrt{d \sec (e+f x)}}+\frac{2 b \sqrt{b \tan (e+f x)}}{45 d^2 f (d \sec (e+f x))^{5/2}}-\frac{2 b \sqrt{b \tan (e+f x)}}{9 f (d \sec (e+f x))^{9/2}} \]

[Out]

(-2*b*Sqrt[b*Tan[e + f*x]])/(9*f*(d*Sec[e + f*x])^(9/2)) + (2*b*Sqrt[b*Tan[e + f*x]])/(45*d^2*f*(d*Sec[e + f*x
])^(5/2)) + (8*b*Sqrt[b*Tan[e + f*x]])/(45*d^4*f*Sqrt[d*Sec[e + f*x]])

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Rubi [A]  time = 0.162904, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {2610, 2612, 2605} \[ \frac{8 b \sqrt{b \tan (e+f x)}}{45 d^4 f \sqrt{d \sec (e+f x)}}+\frac{2 b \sqrt{b \tan (e+f x)}}{45 d^2 f (d \sec (e+f x))^{5/2}}-\frac{2 b \sqrt{b \tan (e+f x)}}{9 f (d \sec (e+f x))^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Tan[e + f*x])^(3/2)/(d*Sec[e + f*x])^(9/2),x]

[Out]

(-2*b*Sqrt[b*Tan[e + f*x]])/(9*f*(d*Sec[e + f*x])^(9/2)) + (2*b*Sqrt[b*Tan[e + f*x]])/(45*d^2*f*(d*Sec[e + f*x
])^(5/2)) + (8*b*Sqrt[b*Tan[e + f*x]])/(45*d^4*f*Sqrt[d*Sec[e + f*x]])

Rule 2610

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e +
f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*m), x] - Dist[(b^2*(n - 1))/(a^2*m), Int[(a*Sec[e + f*x])^(m + 2)*(b*Tan
[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ[n, 1] && (LtQ[m, -1] || (EqQ[m, -1] && EqQ[n, 3/2]
)) && IntegersQ[2*m, 2*n]

Rule 2612

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[((a*Sec[e +
f*x])^m*(b*Tan[e + f*x])^(n + 1))/(b*f*m), x] + Dist[(m + n + 1)/(a^2*m), Int[(a*Sec[e + f*x])^(m + 2)*(b*Tan[
e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (LtQ[m, -1] || (EqQ[m, -1] && EqQ[n, -2^(-1)])) && Integer
sQ[2*m, 2*n]

Rule 2605

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> -Simp[((a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n + 1))/(b*f*m), x] /; FreeQ[{a, b, e, f, m, n}, x] && EqQ[m + n + 1, 0]

Rubi steps

\begin{align*} \int \frac{(b \tan (e+f x))^{3/2}}{(d \sec (e+f x))^{9/2}} \, dx &=-\frac{2 b \sqrt{b \tan (e+f x)}}{9 f (d \sec (e+f x))^{9/2}}+\frac{b^2 \int \frac{1}{(d \sec (e+f x))^{5/2} \sqrt{b \tan (e+f x)}} \, dx}{9 d^2}\\ &=-\frac{2 b \sqrt{b \tan (e+f x)}}{9 f (d \sec (e+f x))^{9/2}}+\frac{2 b \sqrt{b \tan (e+f x)}}{45 d^2 f (d \sec (e+f x))^{5/2}}+\frac{\left (4 b^2\right ) \int \frac{1}{\sqrt{d \sec (e+f x)} \sqrt{b \tan (e+f x)}} \, dx}{45 d^4}\\ &=-\frac{2 b \sqrt{b \tan (e+f x)}}{9 f (d \sec (e+f x))^{9/2}}+\frac{2 b \sqrt{b \tan (e+f x)}}{45 d^2 f (d \sec (e+f x))^{5/2}}+\frac{8 b \sqrt{b \tan (e+f x)}}{45 d^4 f \sqrt{d \sec (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 3.3127, size = 158, normalized size = 1.53 \[ -\frac{b \sqrt{\sec (e+f x)} \sqrt{b \tan (e+f x)} \left (-21 \sqrt{\sec (e+f x)+1} \sec ^2\left (\frac{1}{2} (e+f x)\right )+\sqrt{\frac{1}{\cos (e+f x)+1}} (21 \cos (3 (e+f x))+5 \cos (5 (e+f x))) \sec ^{\frac{3}{2}}(e+f x)+16 \sqrt{\frac{1}{\cos (e+f x)+1}} \sqrt{\sec (e+f x)}\right )}{360 d^3 f \sqrt{\frac{1}{\cos (e+f x)+1}} (d \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Tan[e + f*x])^(3/2)/(d*Sec[e + f*x])^(9/2),x]

[Out]

-(b*Sqrt[Sec[e + f*x]]*(16*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[Sec[e + f*x]] + Sqrt[(1 + Cos[e + f*x])^(-1)]*(2
1*Cos[3*(e + f*x)] + 5*Cos[5*(e + f*x)])*Sec[e + f*x]^(3/2) - 21*Sec[(e + f*x)/2]^2*Sqrt[1 + Sec[e + f*x]])*Sq
rt[b*Tan[e + f*x]])/(360*d^3*f*Sqrt[(1 + Cos[e + f*x])^(-1)]*(d*Sec[e + f*x])^(3/2))

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Maple [A]  time = 0.162, size = 62, normalized size = 0.6 \begin{align*}{\frac{ \left ( 10\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+8 \right ) \sin \left ( fx+e \right ) }{45\,f \left ( \cos \left ( fx+e \right ) \right ) ^{3}} \left ({\frac{b\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }} \right ) ^{{\frac{3}{2}}} \left ({\frac{d}{\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{9}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(f*x+e))^(3/2)/(d*sec(f*x+e))^(9/2),x)

[Out]

2/45/f*(5*cos(f*x+e)^2+4)*(b*sin(f*x+e)/cos(f*x+e))^(3/2)*sin(f*x+e)/cos(f*x+e)^3/(d/cos(f*x+e))^(9/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (b \tan \left (f x + e\right )\right )^{\frac{3}{2}}}{\left (d \sec \left (f x + e\right )\right )^{\frac{9}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e))^(3/2)/(d*sec(f*x+e))^(9/2),x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e))^(3/2)/(d*sec(f*x + e))^(9/2), x)

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Fricas [A]  time = 1.81583, size = 174, normalized size = 1.69 \begin{align*} -\frac{2 \,{\left (5 \, b \cos \left (f x + e\right )^{5} - b \cos \left (f x + e\right )^{3} - 4 \, b \cos \left (f x + e\right )\right )} \sqrt{\frac{b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt{\frac{d}{\cos \left (f x + e\right )}}}{45 \, d^{5} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e))^(3/2)/(d*sec(f*x+e))^(9/2),x, algorithm="fricas")

[Out]

-2/45*(5*b*cos(f*x + e)^5 - b*cos(f*x + e)^3 - 4*b*cos(f*x + e))*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(
f*x + e))/(d^5*f)

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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((b*tan(f*x+e))**(3/2)/(d*sec(f*x+e))**(9/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (b \tan \left (f x + e\right )\right )^{\frac{3}{2}}}{\left (d \sec \left (f x + e\right )\right )^{\frac{9}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e))^(3/2)/(d*sec(f*x+e))^(9/2),x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e))^(3/2)/(d*sec(f*x + e))^(9/2), x)