3.41 \(\int \frac {1}{(b \tan ^4(c+d x))^{5/2}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.06, 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 = {3658, 3473, 8} \[ -\frac {x \tan ^2(c+d x)}{b^2 \sqrt {b \tan ^4(c+d x)}}-\frac {\tan (c+d x)}{b^2 d \sqrt {b \tan ^4(c+d x)}}-\frac {\cot ^7(c+d x)}{9 b^2 d \sqrt {b \tan ^4(c+d x)}}+\frac {\cot ^5(c+d x)}{7 b^2 d \sqrt {b \tan ^4(c+d x)}}-\frac {\cot ^3(c+d x)}{5 b^2 d \sqrt {b \tan ^4(c+d x)}}+\frac {\cot (c+d x)}{3 b^2 d \sqrt {b \tan ^4(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Tan[c + d*x]^4)^(-5/2),x]

[Out]

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

Rule 8

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

Rule 3473

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 3658

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

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Mathematica [C]  time = 0.04, size = 45, normalized size = 0.25 \[ -\frac {\tan (c+d x) \, _2F_1\left (-\frac {9}{2},1;-\frac {7}{2};-\tan ^2(c+d x)\right )}{9 d \left (b \tan ^4(c+d x)\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.80, size = 82, normalized size = 0.45 \[ -\frac {{\left (315 \, d x \tan \left (d x + c\right )^{9} + 315 \, \tan \left (d x + c\right )^{8} - 105 \, \tan \left (d x + c\right )^{6} + 63 \, \tan \left (d x + c\right )^{4} - 45 \, \tan \left (d x + c\right )^{2} + 35\right )} \sqrt {b \tan \left (d x + c\right )^{4}}}{315 \, b^{3} d \tan \left (d x + c\right )^{11}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 8.36, size = 185, normalized size = 1.01 \[ -\frac {\frac {161280 \, {\left (d x + c\right )}}{b^{\frac {5}{2}}} + \frac {121590 \, \sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 18480 \, \sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 3528 \, \sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 495 \, \sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 35 \, \sqrt {b}}{b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9}} - \frac {35 \, b^{\frac {49}{2}} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 495 \, b^{\frac {49}{2}} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 3528 \, b^{\frac {49}{2}} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 18480 \, b^{\frac {49}{2}} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 121590 \, b^{\frac {49}{2}} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{b^{27}}}{161280 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.10, size = 83, normalized size = 0.45 \[ -\frac {\tan \left (d x +c \right ) \left (315 \arctan \left (\tan \left (d x +c \right )\right ) \left (\tan ^{9}\left (d x +c \right )\right )+315 \left (\tan ^{8}\left (d x +c \right )\right )-105 \left (\tan ^{6}\left (d x +c \right )\right )+63 \left (\tan ^{4}\left (d x +c \right )\right )-45 \left (\tan ^{2}\left (d x +c \right )\right )+35\right )}{315 d \left (b \left (\tan ^{4}\left (d x +c \right )\right )\right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*tan(d*x+c)^4)^(5/2),x)

[Out]

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

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maxima [A]  time = 0.95, size = 70, normalized size = 0.38 \[ -\frac {\frac {315 \, {\left (d x + c\right )}}{b^{\frac {5}{2}}} + \frac {315 \, \tan \left (d x + c\right )^{8} - 105 \, \tan \left (d x + c\right )^{6} + 63 \, \tan \left (d x + c\right )^{4} - 45 \, \tan \left (d x + c\right )^{2} + 35}{b^{\frac {5}{2}} \tan \left (d x + c\right )^{9}}}{315 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (b\,{\mathrm {tan}\left (c+d\,x\right )}^4\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*tan(c + d*x)^4)^(5/2),x)

[Out]

int(1/(b*tan(c + d*x)^4)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \tan ^{4}{\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(tan(d*x+c)**4*b)**(5/2),x)

[Out]

Integral((b*tan(c + d*x)**4)**(-5/2), x)

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