3.4.34 \(\int \frac {1}{(d \sec (e+f x))^{5/2} (b \tan (e+f x))^{5/2}} \, dx\) [334]

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

[Out]

-16/15*(b*tan(f*x+e))^(1/2)/b^3/f/(d*sec(f*x+e))^(5/2)-64/15*(b*tan(f*x+e))^(1/2)/b^3/d^2/f/(d*sec(f*x+e))^(1/
2)-2/3/b/f/(d*sec(f*x+e))^(5/2)/(b*tan(f*x+e))^(3/2)

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Rubi [A]
time = 0.11, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2689, 2692, 2685} \begin {gather*} -\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}} \end {gather*}

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 2685

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]

Rule 2689

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 2692

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*Ta
n[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (LtQ[m, -1] || (EqQ[m, -1] && EqQ[n, -2^(-1)])) && Integ
ersQ[2*m, 2*n]

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*}

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Mathematica [A]
time = 3.32, size = 159, normalized size = 1.50 \begin {gather*} \frac {\sqrt {\frac {1}{1+\cos (e+f x)}} (-43+3 \cos (2 (e+f x))) \csc (e+f x) \sec (e+f x)-228 \sqrt {\sec (e+f x)} \sqrt {1+\sec (e+f x)} \tan \left (\frac {1}{2} (e+f x)\right )-6 \sqrt {\frac {1}{1+\cos (e+f x)}} (-1+2 \cos (2 (e+f x))) \tan (e+f x)}{60 b^2 d^2 f \sqrt {\frac {1}{1+\cos (e+f x)}} \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}} \end {gather*}

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 = 0.32, size = 72, normalized size = 0.68

method	result	size
default	\(\frac {2 \sin \left (f x +e \right ) \left (3 \left (\cos ^{4}\left (f x +e \right )\right )+24 \left (\cos ^{2}\left (f x +e \right )\right )-32\right )}{15 f \left (\frac {d}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}} \left (\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}} \cos \left (f x +e \right )^{5}}\)	\(72\)

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,method=_RETURNVERBOSE)

[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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.43, size = 96, normalized size = 0.91 \begin {gather*} -\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 {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [B]
time = 4.35, size = 93, normalized size = 0.88 \begin {gather*} -\frac {\sqrt {\frac {d}{\cos \left (e+f\,x\right )}}\,\left (105\,\sin \left (3\,e+3\,f\,x\right )-410\,\sin \left (e+f\,x\right )+3\,\sin \left (5\,e+5\,f\,x\right )\right )}{60\,b^2\,d^3\,f\,\left (\cos \left (2\,e+2\,f\,x\right )-1\right )\,\sqrt {\frac {b\,\sin \left (2\,e+2\,f\,x\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((d/cos(e + f*x))^(1/2)*(105*sin(3*e + 3*f*x) - 410*sin(e + f*x) + 3*sin(5*e + 5*f*x)))/(60*b^2*d^3*f*(cos(2*
e + 2*f*x) - 1)*((b*sin(2*e + 2*f*x))/(cos(2*e + 2*f*x) + 1))^(1/2))

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