3.40 \(\int \frac {1}{(b \tan ^4(c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=119 \[ -\frac {\tan (c+d x)}{b d \sqrt {b \tan ^4(c+d x)}}-\frac {x \tan ^2(c+d x)}{b \sqrt {b \tan ^4(c+d x)}}-\frac {\cot ^3(c+d x)}{5 b d \sqrt {b \tan ^4(c+d x)}}+\frac {\cot (c+d x)}{3 b d \sqrt {b \tan ^4(c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3658, 3473, 8} \[ -\frac {x \tan ^2(c+d x)}{b \sqrt {b \tan ^4(c+d x)}}-\frac {\tan (c+d x)}{b d \sqrt {b \tan ^4(c+d x)}}-\frac {\cot ^3(c+d x)}{5 b d \sqrt {b \tan ^4(c+d x)}}+\frac {\cot (c+d x)}{3 b d \sqrt {b \tan ^4(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Tan[c + d*x]^4)^(-3/2),x]

[Out]

Cot[c + d*x]/(3*b*d*Sqrt[b*Tan[c + d*x]^4]) - Cot[c + d*x]^3/(5*b*d*Sqrt[b*Tan[c + d*x]^4]) - Tan[c + d*x]/(b*
d*Sqrt[b*Tan[c + d*x]^4]) - (x*Tan[c + d*x]^2)/(b*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 )^{3/2}} \, dx &=\frac {\tan ^2(c+d x) \int \cot ^6(c+d x) \, dx}{b \sqrt {b \tan ^4(c+d x)}}\\ &=-\frac {\cot ^3(c+d x)}{5 b d \sqrt {b \tan ^4(c+d x)}}-\frac {\tan ^2(c+d x) \int \cot ^4(c+d x) \, dx}{b \sqrt {b \tan ^4(c+d x)}}\\ &=\frac {\cot (c+d x)}{3 b d \sqrt {b \tan ^4(c+d x)}}-\frac {\cot ^3(c+d x)}{5 b d \sqrt {b \tan ^4(c+d x)}}+\frac {\tan ^2(c+d x) \int \cot ^2(c+d x) \, dx}{b \sqrt {b \tan ^4(c+d x)}}\\ &=\frac {\cot (c+d x)}{3 b d \sqrt {b \tan ^4(c+d x)}}-\frac {\cot ^3(c+d x)}{5 b d \sqrt {b \tan ^4(c+d x)}}-\frac {\tan (c+d x)}{b d \sqrt {b \tan ^4(c+d x)}}-\frac {\tan ^2(c+d x) \int 1 \, dx}{b \sqrt {b \tan ^4(c+d x)}}\\ &=\frac {\cot (c+d x)}{3 b d \sqrt {b \tan ^4(c+d x)}}-\frac {\cot ^3(c+d x)}{5 b d \sqrt {b \tan ^4(c+d x)}}-\frac {\tan (c+d x)}{b d \sqrt {b \tan ^4(c+d x)}}-\frac {x \tan ^2(c+d x)}{b \sqrt {b \tan ^4(c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 0.05, size = 45, normalized size = 0.38 \[ -\frac {\tan (c+d x) \, _2F_1\left (-\frac {5}{2},1;-\frac {3}{2};-\tan ^2(c+d x)\right )}{5 d \left (b \tan ^4(c+d x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Tan[c + d*x]^4)^(-3/2),x]

[Out]

-1/5*(Hypergeometric2F1[-5/2, 1, -3/2, -Tan[c + d*x]^2]*Tan[c + d*x])/(d*(b*Tan[c + d*x]^4)^(3/2))

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fricas [A]  time = 0.78, size = 62, normalized size = 0.52 \[ -\frac {{\left (15 \, d x \tan \left (d x + c\right )^{5} + 15 \, \tan \left (d x + c\right )^{4} - 5 \, \tan \left (d x + c\right )^{2} + 3\right )} \sqrt {b \tan \left (d x + c\right )^{4}}}{15 \, b^{2} d \tan \left (d x + c\right )^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(tan(d*x+c)^4*b)^(3/2),x, algorithm="fricas")

[Out]

-1/15*(15*d*x*tan(d*x + c)^5 + 15*tan(d*x + c)^4 - 5*tan(d*x + c)^2 + 3)*sqrt(b*tan(d*x + c)^4)/(b^2*d*tan(d*x
+ c)^7)

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giac [A]  time = 4.34, size = 124, normalized size = 1.04 \[ -\frac {\frac {480 \, {\left (d x + c\right )}}{\sqrt {b}} - \frac {3 \, b^{\frac {9}{2}} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 35 \, b^{\frac {9}{2}} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 330 \, b^{\frac {9}{2}} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{b^{5}} + \frac {330 \, \sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 35 \, \sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, \sqrt {b}}{b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{480 \, b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(tan(d*x+c)^4*b)^(3/2),x, algorithm="giac")

[Out]

-1/480*(480*(d*x + c)/sqrt(b) - (3*b^(9/2)*tan(1/2*d*x + 1/2*c)^5 - 35*b^(9/2)*tan(1/2*d*x + 1/2*c)^3 + 330*b^
(9/2)*tan(1/2*d*x + 1/2*c))/b^5 + (330*sqrt(b)*tan(1/2*d*x + 1/2*c)^4 - 35*sqrt(b)*tan(1/2*d*x + 1/2*c)^2 + 3*
sqrt(b))/(b*tan(1/2*d*x + 1/2*c)^5))/(b*d)

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maple [A]  time = 0.10, size = 63, normalized size = 0.53 \[ -\frac {\tan \left (d x +c \right ) \left (15 \arctan \left (\tan \left (d x +c \right )\right ) \left (\tan ^{5}\left (d x +c \right )\right )+15 \left (\tan ^{4}\left (d x +c \right )\right )-5 \left (\tan ^{2}\left (d x +c \right )\right )+3\right )}{15 d \left (b \left (\tan ^{4}\left (d x +c \right )\right )\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*tan(d*x+c)^4)^(3/2),x)

[Out]

-1/15/d*tan(d*x+c)*(15*arctan(tan(d*x+c))*tan(d*x+c)^5+15*tan(d*x+c)^4-5*tan(d*x+c)^2+3)/(b*tan(d*x+c)^4)^(3/2
)

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maxima [A]  time = 0.51, size = 50, normalized size = 0.42 \[ -\frac {\frac {15 \, {\left (d x + c\right )}}{b^{\frac {3}{2}}} + \frac {15 \, \tan \left (d x + c\right )^{4} - 5 \, \tan \left (d x + c\right )^{2} + 3}{b^{\frac {3}{2}} \tan \left (d x + c\right )^{5}}}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(tan(d*x+c)^4*b)^(3/2),x, algorithm="maxima")

[Out]

-1/15*(15*(d*x + c)/b^(3/2) + (15*tan(d*x + c)^4 - 5*tan(d*x + c)^2 + 3)/(b^(3/2)*tan(d*x + c)^5))/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (b\,{\mathrm {tan}\left (c+d\,x\right )}^4\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*tan(c + d*x)^4)^(3/2),x)

[Out]

int(1/(b*tan(c + d*x)^4)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \tan ^{4}{\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(tan(d*x+c)**4*b)**(3/2),x)

[Out]

Integral((b*tan(c + d*x)**4)**(-3/2), x)

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