3.1.18 \(\int \frac {1}{(b \tan ^4(e+f x))^{5/2}} \, dx\) [18]

Optimal. Leaf size=183 \[ \frac {\cot (e+f x)}{3 b^2 f \sqrt {b \tan ^4(e+f x)}}-\frac {\cot ^3(e+f x)}{5 b^2 f \sqrt {b \tan ^4(e+f x)}}+\frac {\cot ^5(e+f x)}{7 b^2 f \sqrt {b \tan ^4(e+f x)}}-\frac {\cot ^7(e+f x)}{9 b^2 f \sqrt {b \tan ^4(e+f x)}}-\frac {\tan (e+f x)}{b^2 f \sqrt {b \tan ^4(e+f x)}}-\frac {x \tan ^2(e+f x)}{b^2 \sqrt {b \tan ^4(e+f x)}} \]

[Out]

1/3*cot(f*x+e)/b^2/f/(b*tan(f*x+e)^4)^(1/2)-1/5*cot(f*x+e)^3/b^2/f/(b*tan(f*x+e)^4)^(1/2)+1/7*cot(f*x+e)^5/b^2
/f/(b*tan(f*x+e)^4)^(1/2)-1/9*cot(f*x+e)^7/b^2/f/(b*tan(f*x+e)^4)^(1/2)-tan(f*x+e)/b^2/f/(b*tan(f*x+e)^4)^(1/2
)-x*tan(f*x+e)^2/b^2/(b*tan(f*x+e)^4)^(1/2)

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Rubi [A]
time = 0.05, antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3739, 3554, 8} \begin {gather*} -\frac {\tan (e+f x)}{b^2 f \sqrt {b \tan ^4(e+f x)}}-\frac {x \tan ^2(e+f x)}{b^2 \sqrt {b \tan ^4(e+f x)}}-\frac {\cot ^7(e+f x)}{9 b^2 f \sqrt {b \tan ^4(e+f x)}}+\frac {\cot ^5(e+f x)}{7 b^2 f \sqrt {b \tan ^4(e+f x)}}-\frac {\cot ^3(e+f x)}{5 b^2 f \sqrt {b \tan ^4(e+f x)}}+\frac {\cot (e+f x)}{3 b^2 f \sqrt {b \tan ^4(e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*Tan[e + f*x]^4)^(-5/2),x]

[Out]

Cot[e + f*x]/(3*b^2*f*Sqrt[b*Tan[e + f*x]^4]) - Cot[e + f*x]^3/(5*b^2*f*Sqrt[b*Tan[e + f*x]^4]) + Cot[e + f*x]
^5/(7*b^2*f*Sqrt[b*Tan[e + f*x]^4]) - Cot[e + f*x]^7/(9*b^2*f*Sqrt[b*Tan[e + f*x]^4]) - Tan[e + f*x]/(b^2*f*Sq
rt[b*Tan[e + f*x]^4]) - (x*Tan[e + f*x]^2)/(b^2*Sqrt[b*Tan[e + f*x]^4])

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3554

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3739

Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Di
st[(b*ff^n)^IntPart[p]*((b*Tan[e + f*x]^n)^FracPart[p]/(Tan[e + f*x]/ff)^(n*FracPart[p])), Int[ActivateTrig[u]
*(Tan[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])

Rubi steps

\begin {align*} \int \frac {1}{\left (b \tan ^4(e+f x)\right )^{5/2}} \, dx &=\frac {\tan ^2(e+f x) \int \cot ^{10}(e+f x) \, dx}{b^2 \sqrt {b \tan ^4(e+f x)}}\\ &=-\frac {\cot ^7(e+f x)}{9 b^2 f \sqrt {b \tan ^4(e+f x)}}-\frac {\tan ^2(e+f x) \int \cot ^8(e+f x) \, dx}{b^2 \sqrt {b \tan ^4(e+f x)}}\\ &=\frac {\cot ^5(e+f x)}{7 b^2 f \sqrt {b \tan ^4(e+f x)}}-\frac {\cot ^7(e+f x)}{9 b^2 f \sqrt {b \tan ^4(e+f x)}}+\frac {\tan ^2(e+f x) \int \cot ^6(e+f x) \, dx}{b^2 \sqrt {b \tan ^4(e+f x)}}\\ &=-\frac {\cot ^3(e+f x)}{5 b^2 f \sqrt {b \tan ^4(e+f x)}}+\frac {\cot ^5(e+f x)}{7 b^2 f \sqrt {b \tan ^4(e+f x)}}-\frac {\cot ^7(e+f x)}{9 b^2 f \sqrt {b \tan ^4(e+f x)}}-\frac {\tan ^2(e+f x) \int \cot ^4(e+f x) \, dx}{b^2 \sqrt {b \tan ^4(e+f x)}}\\ &=\frac {\cot (e+f x)}{3 b^2 f \sqrt {b \tan ^4(e+f x)}}-\frac {\cot ^3(e+f x)}{5 b^2 f \sqrt {b \tan ^4(e+f x)}}+\frac {\cot ^5(e+f x)}{7 b^2 f \sqrt {b \tan ^4(e+f x)}}-\frac {\cot ^7(e+f x)}{9 b^2 f \sqrt {b \tan ^4(e+f x)}}+\frac {\tan ^2(e+f x) \int \cot ^2(e+f x) \, dx}{b^2 \sqrt {b \tan ^4(e+f x)}}\\ &=\frac {\cot (e+f x)}{3 b^2 f \sqrt {b \tan ^4(e+f x)}}-\frac {\cot ^3(e+f x)}{5 b^2 f \sqrt {b \tan ^4(e+f x)}}+\frac {\cot ^5(e+f x)}{7 b^2 f \sqrt {b \tan ^4(e+f x)}}-\frac {\cot ^7(e+f x)}{9 b^2 f \sqrt {b \tan ^4(e+f x)}}-\frac {\tan (e+f x)}{b^2 f \sqrt {b \tan ^4(e+f x)}}-\frac {\tan ^2(e+f x) \int 1 \, dx}{b^2 \sqrt {b \tan ^4(e+f x)}}\\ &=\frac {\cot (e+f x)}{3 b^2 f \sqrt {b \tan ^4(e+f x)}}-\frac {\cot ^3(e+f x)}{5 b^2 f \sqrt {b \tan ^4(e+f x)}}+\frac {\cot ^5(e+f x)}{7 b^2 f \sqrt {b \tan ^4(e+f x)}}-\frac {\cot ^7(e+f x)}{9 b^2 f \sqrt {b \tan ^4(e+f x)}}-\frac {\tan (e+f x)}{b^2 f \sqrt {b \tan ^4(e+f x)}}-\frac {x \tan ^2(e+f x)}{b^2 \sqrt {b \tan ^4(e+f x)}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.02, size = 45, normalized size = 0.25 \begin {gather*} -\frac {\, _2F_1\left (-\frac {9}{2},1;-\frac {7}{2};-\tan ^2(e+f x)\right ) \tan (e+f x)}{9 f \left (b \tan ^4(e+f x)\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*Tan[e + f*x]^4)^(-5/2),x]

[Out]

-1/9*(Hypergeometric2F1[-9/2, 1, -7/2, -Tan[e + f*x]^2]*Tan[e + f*x])/(f*(b*Tan[e + f*x]^4)^(5/2))

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Maple [A]
time = 0.05, size = 83, normalized size = 0.45

method	result	size
derivativedivides	\(-\frac {\tan \left (f x +e \right ) \left (315 \arctan \left (\tan \left (f x +e \right )\right ) \left (\tan ^{9}\left (f x +e \right )\right )+315 \left (\tan ^{8}\left (f x +e \right )\right )-105 \left (\tan ^{6}\left (f x +e \right )\right )+63 \left (\tan ^{4}\left (f x +e \right )\right )-45 \left (\tan ^{2}\left (f x +e \right )\right )+35\right )}{315 f \left (b \left (\tan ^{4}\left (f x +e \right )\right )\right )^{\frac {5}{2}}}\)	\(83\)
default	\(-\frac {\tan \left (f x +e \right ) \left (315 \arctan \left (\tan \left (f x +e \right )\right ) \left (\tan ^{9}\left (f x +e \right )\right )+315 \left (\tan ^{8}\left (f x +e \right )\right )-105 \left (\tan ^{6}\left (f x +e \right )\right )+63 \left (\tan ^{4}\left (f x +e \right )\right )-45 \left (\tan ^{2}\left (f x +e \right )\right )+35\right )}{315 f \left (b \left (\tan ^{4}\left (f x +e \right )\right )\right )^{\frac {5}{2}}}\)	\(83\)
risch	\(\frac {\left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{2} x}{b^{2} \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} \sqrt {\frac {b \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{4}}{\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{4}}}}+\frac {2 i \left (1575 \,{\mathrm e}^{16 i \left (f x +e \right )}-6300 \,{\mathrm e}^{14 i \left (f x +e \right )}+21000 \,{\mathrm e}^{12 i \left (f x +e \right )}-31500 \,{\mathrm e}^{10 i \left (f x +e \right )}+39438 \,{\mathrm e}^{8 i \left (f x +e \right )}-26292 \,{\mathrm e}^{6 i \left (f x +e \right )}+13968 \,{\mathrm e}^{4 i \left (f x +e \right )}-3492 \,{\mathrm e}^{2 i \left (f x +e \right )}+563\right )}{315 b^{2} \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{7} \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} \sqrt {\frac {b \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{4}}{\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{4}}}\, f}\)	\(218\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*tan(f*x+e)^4)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/315/f*tan(f*x+e)*(315*arctan(tan(f*x+e))*tan(f*x+e)^9+315*tan(f*x+e)^8-105*tan(f*x+e)^6+63*tan(f*x+e)^4-45*
tan(f*x+e)^2+35)/(b*tan(f*x+e)^4)^(5/2)

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Maxima [A]
time = 0.53, size = 76, normalized size = 0.42 \begin {gather*} -\frac {\frac {315 \, {\left (f x + e\right )}}{b^{\frac {5}{2}}} + \frac {315 \, \tan \left (f x + e\right )^{8} - 105 \, \tan \left (f x + e\right )^{6} + 63 \, \tan \left (f x + e\right )^{4} - 45 \, \tan \left (f x + e\right )^{2} + 35}{b^{\frac {5}{2}} \tan \left (f x + e\right )^{9}}}{315 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*tan(f*x+e)^4)^(5/2),x, algorithm="maxima")

[Out]

-1/315*(315*(f*x + e)/b^(5/2) + (315*tan(f*x + e)^8 - 105*tan(f*x + e)^6 + 63*tan(f*x + e)^4 - 45*tan(f*x + e)
^2 + 35)/(b^(5/2)*tan(f*x + e)^9))/f

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Fricas [A]
time = 1.45, size = 89, normalized size = 0.49 \begin {gather*} -\frac {{\left (315 \, f x \tan \left (f x + e\right )^{9} + 315 \, \tan \left (f x + e\right )^{8} - 105 \, \tan \left (f x + e\right )^{6} + 63 \, \tan \left (f x + e\right )^{4} - 45 \, \tan \left (f x + e\right )^{2} + 35\right )} \sqrt {b \tan \left (f x + e\right )^{4}}}{315 \, b^{3} f \tan \left (f x + e\right )^{11}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*tan(f*x+e)^4)^(5/2),x, algorithm="fricas")

[Out]

-1/315*(315*f*x*tan(f*x + e)^9 + 315*tan(f*x + e)^8 - 105*tan(f*x + e)^6 + 63*tan(f*x + e)^4 - 45*tan(f*x + e)
^2 + 35)*sqrt(b*tan(f*x + e)^4)/(b^3*f*tan(f*x + e)^11)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (b \tan ^{4}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*tan(f*x+e)**4)**(5/2),x)

[Out]

Integral((b*tan(e + f*x)**4)**(-5/2), x)

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Giac [A]
time = 1.22, size = 196, normalized size = 1.07 \begin {gather*} -\frac {\frac {161280 \, {\left (f x + e\right )}}{b^{\frac {5}{2}}} + \frac {121590 \, \sqrt {b} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} - 18480 \, \sqrt {b} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} + 3528 \, \sqrt {b} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 495 \, \sqrt {b} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 35 \, \sqrt {b}}{b^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{9}} - \frac {35 \, b^{\frac {49}{2}} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{9} - 495 \, b^{\frac {49}{2}} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} + 3528 \, b^{\frac {49}{2}} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 18480 \, b^{\frac {49}{2}} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 121590 \, b^{\frac {49}{2}} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{b^{27}}}{161280 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*tan(f*x+e)^4)^(5/2),x, algorithm="giac")

[Out]

-1/161280*(161280*(f*x + e)/b^(5/2) + (121590*sqrt(b)*tan(1/2*f*x + 1/2*e)^8 - 18480*sqrt(b)*tan(1/2*f*x + 1/2
*e)^6 + 3528*sqrt(b)*tan(1/2*f*x + 1/2*e)^4 - 495*sqrt(b)*tan(1/2*f*x + 1/2*e)^2 + 35*sqrt(b))/(b^3*tan(1/2*f*
x + 1/2*e)^9) - (35*b^(49/2)*tan(1/2*f*x + 1/2*e)^9 - 495*b^(49/2)*tan(1/2*f*x + 1/2*e)^7 + 3528*b^(49/2)*tan(
1/2*f*x + 1/2*e)^5 - 18480*b^(49/2)*tan(1/2*f*x + 1/2*e)^3 + 121590*b^(49/2)*tan(1/2*f*x + 1/2*e))/b^27)/f

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^4\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*tan(e + f*x)^4)^(5/2),x)

[Out]

int(1/(b*tan(e + f*x)^4)^(5/2), x)

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