3.334 \(\int \frac{1}{(d \sec (e+f x))^{5/2} (b \tan (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=106 \[ -\frac{64 \sqrt{b \tan (e+f x)}}{15 b^3 d^2 f \sqrt{d \sec (e+f x)}}-\frac{16 \sqrt{b \tan (e+f x)}}{15 b^3 f (d \sec (e+f x))^{5/2}}-\frac{2}{3 b f (b \tan (e+f x))^{3/2} (d \sec (e+f x))^{5/2}} \]

[Out]

-2/(3*b*f*(d*Sec[e + f*x])^(5/2)*(b*Tan[e + f*x])^(3/2)) - (16*Sqrt[b*Tan[e + f*x]])/(15*b^3*f*(d*Sec[e + f*x]
)^(5/2)) - (64*Sqrt[b*Tan[e + f*x]])/(15*b^3*d^2*f*Sqrt[d*Sec[e + f*x]])

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Rubi [A]  time = 0.165777, antiderivative size = 106, 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 = {2609, 2612, 2605} \[ -\frac{64 \sqrt{b \tan (e+f x)}}{15 b^3 d^2 f \sqrt{d \sec (e+f x)}}-\frac{16 \sqrt{b \tan (e+f x)}}{15 b^3 f (d \sec (e+f x))^{5/2}}-\frac{2}{3 b f (b \tan (e+f x))^{3/2} (d \sec (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((d*Sec[e + f*x])^(5/2)*(b*Tan[e + f*x])^(5/2)),x]

[Out]

-2/(3*b*f*(d*Sec[e + f*x])^(5/2)*(b*Tan[e + f*x])^(3/2)) - (16*Sqrt[b*Tan[e + f*x]])/(15*b^3*f*(d*Sec[e + f*x]
)^(5/2)) - (64*Sqrt[b*Tan[e + f*x]])/(15*b^3*d^2*f*Sqrt[d*Sec[e + f*x]])

Rule 2609

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*(n + 1)), x] - Dist[(m + n + 1)/(b^2*(n + 1)), Int[(a*Sec[e + f*x])^m*(
b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && LtQ[n, -1] && 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{1}{(d \sec (e+f x))^{5/2} (b \tan (e+f x))^{5/2}} \, dx &=-\frac{2}{3 b f (d \sec (e+f x))^{5/2} (b \tan (e+f x))^{3/2}}-\frac{8 \int \frac{1}{(d \sec (e+f x))^{5/2} \sqrt{b \tan (e+f x)}} \, dx}{3 b^2}\\ &=-\frac{2}{3 b f (d \sec (e+f x))^{5/2} (b \tan (e+f x))^{3/2}}-\frac{16 \sqrt{b \tan (e+f x)}}{15 b^3 f (d \sec (e+f x))^{5/2}}-\frac{32 \int \frac{1}{\sqrt{d \sec (e+f x)} \sqrt{b \tan (e+f x)}} \, dx}{15 b^2 d^2}\\ &=-\frac{2}{3 b f (d \sec (e+f x))^{5/2} (b \tan (e+f x))^{3/2}}-\frac{16 \sqrt{b \tan (e+f x)}}{15 b^3 f (d \sec (e+f x))^{5/2}}-\frac{64 \sqrt{b \tan (e+f x)}}{15 b^3 d^2 f \sqrt{d \sec (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 3.05235, size = 159, normalized size = 1.5 \[ \frac{-6 \sqrt{\frac{1}{\cos (e+f x)+1}} (2 \cos (2 (e+f x))-1) \tan (e+f x)-228 \tan \left (\frac{1}{2} (e+f x)\right ) \sqrt{\sec (e+f x)+1} \sqrt{\sec (e+f x)}+\sqrt{\frac{1}{\cos (e+f x)+1}} (3 \cos (2 (e+f x))-43) \csc (e+f x) \sec (e+f x)}{60 b^2 d^2 f \sqrt{\frac{1}{\cos (e+f x)+1}} \sqrt{b \tan (e+f x)} \sqrt{d \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((d*Sec[e + f*x])^(5/2)*(b*Tan[e + f*x])^(5/2)),x]

[Out]

(Sqrt[(1 + Cos[e + f*x])^(-1)]*(-43 + 3*Cos[2*(e + f*x)])*Csc[e + f*x]*Sec[e + f*x] - 228*Sqrt[Sec[e + f*x]]*S
qrt[1 + Sec[e + f*x]]*Tan[(e + f*x)/2] - 6*Sqrt[(1 + Cos[e + f*x])^(-1)]*(-1 + 2*Cos[2*(e + f*x)])*Tan[e + f*x
])/(60*b^2*d^2*f*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[d*Sec[e + f*x]]*Sqrt[b*Tan[e + f*x]])

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Maple [A]  time = 0.156, size = 72, normalized size = 0.7 \begin{align*}{\frac{2\,\sin \left ( fx+e \right ) \left ( 3\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}+24\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}-32 \right ) }{15\,f \left ( \cos \left ( fx+e \right ) \right ) ^{5}} \left ({\frac{d}{\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{5}{2}}} \left ({\frac{b\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(d*sec(f*x+e))^(5/2)/(b*tan(f*x+e))^(5/2),x)

[Out]

2/15/f*sin(f*x+e)*(3*cos(f*x+e)^4+24*cos(f*x+e)^2-32)/(d/cos(f*x+e))^(5/2)/(b*sin(f*x+e)/cos(f*x+e))^(5/2)/cos
(f*x+e)^5

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (d \sec \left (f x + e\right )\right )^{\frac{5}{2}} \left (b \tan \left (f x + e\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*sec(f*x+e))^(5/2)/(b*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate(1/((d*sec(f*x + e))^(5/2)*(b*tan(f*x + e))^(5/2)), x)

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Fricas [A]  time = 1.93458, size = 213, normalized size = 2.01 \begin{align*} -\frac{2 \,{\left (3 \, \cos \left (f x + e\right )^{5} + 24 \, \cos \left (f x + e\right )^{3} - 32 \, \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 )}}}{15 \,{\left (b^{3} d^{3} f \cos \left (f x + e\right )^{2} - b^{3} d^{3} f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*sec(f*x+e))^(5/2)/(b*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

-2/15*(3*cos(f*x + e)^5 + 24*cos(f*x + e)^3 - 32*cos(f*x + e))*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*
x + e))/(b^3*d^3*f*cos(f*x + e)^2 - b^3*d^3*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(1/(d*sec(f*x+e))**(5/2)/(b*tan(f*x+e))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (d \sec \left (f x + e\right )\right )^{\frac{5}{2}} \left (b \tan \left (f x + e\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*sec(f*x+e))^(5/2)/(b*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate(1/((d*sec(f*x + e))^(5/2)*(b*tan(f*x + e))^(5/2)), x)