3.1.28 \(\int \frac {1}{(b \tan ^2(c+d x))^{3/2}} \, dx\) [28]

Optimal. Leaf size=66 \[ -\frac {\cot (c+d x)}{2 b d \sqrt {b \tan ^2(c+d x)}}-\frac {\log (\sin (c+d x)) \tan (c+d x)}{b d \sqrt {b \tan ^2(c+d x)}} \]

[Out]

-1/2*cot(d*x+c)/b/d/(b*tan(d*x+c)^2)^(1/2)-ln(sin(d*x+c))*tan(d*x+c)/b/d/(b*tan(d*x+c)^2)^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3739, 3554, 3556} \begin {gather*} -\frac {\cot (c+d x)}{2 b d \sqrt {b \tan ^2(c+d x)}}-\frac {\tan (c+d x) \log (\sin (c+d x))}{b d \sqrt {b \tan ^2(c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*Tan[c + d*x]^2)^(-3/2),x]

[Out]

-1/2*Cot[c + d*x]/(b*d*Sqrt[b*Tan[c + d*x]^2]) - (Log[Sin[c + d*x]]*Tan[c + d*x])/(b*d*Sqrt[b*Tan[c + d*x]^2])

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 3556

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

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 ^2(c+d x)\right )^{3/2}} \, dx &=\frac {\tan (c+d x) \int \cot ^3(c+d x) \, dx}{b \sqrt {b \tan ^2(c+d x)}}\\ &=-\frac {\cot (c+d x)}{2 b d \sqrt {b \tan ^2(c+d x)}}-\frac {\tan (c+d x) \int \cot (c+d x) \, dx}{b \sqrt {b \tan ^2(c+d x)}}\\ &=-\frac {\cot (c+d x)}{2 b d \sqrt {b \tan ^2(c+d x)}}-\frac {\log (\sin (c+d x)) \tan (c+d x)}{b d \sqrt {b \tan ^2(c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.41, size = 56, normalized size = 0.85 \begin {gather*} -\frac {\left (\cot ^2(c+d x)+2 \log (\cos (c+d x))+2 \log (\tan (c+d x))\right ) \tan ^3(c+d x)}{2 d \left (b \tan ^2(c+d x)\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((Cot[c + d*x]^2 + 2*Log[Cos[c + d*x]] + 2*Log[Tan[c + d*x]])*Tan[c + d*x]^3)/(d*(b*Tan[c + d*x]^2)^(3/2)
)

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Maple [A]
time = 0.10, size = 64, normalized size = 0.97

method	result	size
derivativedivides	\(-\frac {\tan \left (d x +c \right ) \left (2 \ln \left (\tan \left (d x +c \right )\right ) \left (\tan ^{2}\left (d x +c \right )\right )-\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) \left (\tan ^{2}\left (d x +c \right )\right )+1\right )}{2 d \left (b \left (\tan ^{2}\left (d x +c \right )\right )\right )^{\frac {3}{2}}}\)	\(64\)
default	\(-\frac {\tan \left (d x +c \right ) \left (2 \ln \left (\tan \left (d x +c \right )\right ) \left (\tan ^{2}\left (d x +c \right )\right )-\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) \left (\tan ^{2}\left (d x +c \right )\right )+1\right )}{2 d \left (b \left (\tan ^{2}\left (d x +c \right )\right )\right )^{\frac {3}{2}}}\)	\(64\)
risch	\(-\frac {\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) x}{b \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {-\frac {b \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}}+\frac {2 \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left (d x +c \right )}{b \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {-\frac {b \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, d}-\frac {2 i {\mathrm e}^{2 i \left (d x +c \right )}}{b \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {-\frac {b \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, d}+\frac {i \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{b \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {-\frac {b \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, d}\)	\(282\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/d*tan(d*x+c)*(2*ln(tan(d*x+c))*tan(d*x+c)^2-ln(1+tan(d*x+c)^2)*tan(d*x+c)^2+1)/(b*tan(d*x+c)^2)^(3/2)

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Maxima [A]
time = 0.49, size = 46, normalized size = 0.70 \begin {gather*} \frac {\frac {\log \left (\tan \left (d x + c\right )^{2} + 1\right )}{b^{\frac {3}{2}}} - \frac {2 \, \log \left (\tan \left (d x + c\right )\right )}{b^{\frac {3}{2}}} - \frac {1}{b^{\frac {3}{2}} \tan \left (d x + c\right )^{2}}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(log(tan(d*x + c)^2 + 1)/b^(3/2) - 2*log(tan(d*x + c))/b^(3/2) - 1/(b^(3/2)*tan(d*x + c)^2))/d

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Fricas [A]
time = 0.35, size = 69, normalized size = 1.05 \begin {gather*} -\frac {\sqrt {b \tan \left (d x + c\right )^{2}} {\left (\log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{2} + \tan \left (d x + c\right )^{2} + 1\right )}}{2 \, b^{2} d \tan \left (d x + c\right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(b*tan(d*x + c)^2)*(log(tan(d*x + c)^2/(tan(d*x + c)^2 + 1))*tan(d*x + c)^2 + tan(d*x + c)^2 + 1)/(b^
2*d*tan(d*x + c)^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 176 vs. \(2 (60) = 120\).
time = 0.52, size = 176, normalized size = 2.67 \begin {gather*} -\frac {\frac {4 \, \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{\sqrt {b} \mathrm {sgn}\left (\tan \left (d x + c\right )\right )} - \frac {8 \, \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right )}{\sqrt {b} \mathrm {sgn}\left (\tan \left (d x + c\right )\right )} - \frac {{\left (\sqrt {b} + \frac {4 \, \sqrt {b} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}{b {\left (\cos \left (d x + c\right ) - 1\right )} \mathrm {sgn}\left (\tan \left (d x + c\right )\right )} - \frac {\cos \left (d x + c\right ) - 1}{\sqrt {b} {\left (\cos \left (d x + c\right ) + 1\right )} \mathrm {sgn}\left (\tan \left (d x + c\right )\right )}}{8 \, b d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(4*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/(sqrt(b)*sgn(tan(d*x + c))) - 8*log(abs(-(cos(d*x +
c) - 1)/(cos(d*x + c) + 1) + 1))/(sqrt(b)*sgn(tan(d*x + c))) - (sqrt(b) + 4*sqrt(b)*(cos(d*x + c) - 1)/(cos(d*
x + c) + 1))*(cos(d*x + c) + 1)/(b*(cos(d*x + c) - 1)*sgn(tan(d*x + c))) - (cos(d*x + c) - 1)/(sqrt(b)*(cos(d*
x + c) + 1)*sgn(tan(d*x + c))))/(b*d)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{{\left (b\,{\mathrm {tan}\left (c+d\,x\right )}^2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*tan(c + d*x)^2)^(3/2),x)

[Out]

int(1/(b*tan(c + d*x)^2)^(3/2), x)

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