3.2 \(\int (b \tan ^2(e+f x))^{3/2} \, dx\)

Optimal. Leaf size=61 \[ \frac {b \tan (e+f x) \sqrt {b \tan ^2(e+f x)}}{2 f}+\frac {b \cot (e+f x) \sqrt {b \tan ^2(e+f x)} \log (\cos (e+f x))}{f} \]

[Out]

b*cot(f*x+e)*ln(cos(f*x+e))*(b*tan(f*x+e)^2)^(1/2)/f+1/2*b*(b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)/f

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Rubi [A]  time = 0.03, antiderivative size = 61, 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 = {3658, 3473, 3475} \[ \frac {b \tan (e+f x) \sqrt {b \tan ^2(e+f x)}}{2 f}+\frac {b \cot (e+f x) \sqrt {b \tan ^2(e+f x)} \log (\cos (e+f x))}{f} \]

Antiderivative was successfully verified.

[In]

Int[(b*Tan[e + f*x]^2)^(3/2),x]

[Out]

(b*Cot[e + f*x]*Log[Cos[e + f*x]]*Sqrt[b*Tan[e + f*x]^2])/f + (b*Tan[e + f*x]*Sqrt[b*Tan[e + f*x]^2])/(2*f)

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 3475

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

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 \left (b \tan ^2(e+f x)\right )^{3/2} \, dx &=\left (b \cot (e+f x) \sqrt {b \tan ^2(e+f x)}\right ) \int \tan ^3(e+f x) \, dx\\ &=\frac {b \tan (e+f x) \sqrt {b \tan ^2(e+f x)}}{2 f}-\left (b \cot (e+f x) \sqrt {b \tan ^2(e+f x)}\right ) \int \tan (e+f x) \, dx\\ &=\frac {b \cot (e+f x) \log (\cos (e+f x)) \sqrt {b \tan ^2(e+f x)}}{f}+\frac {b \tan (e+f x) \sqrt {b \tan ^2(e+f x)}}{2 f}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 47, normalized size = 0.77 \[ \frac {\cot ^3(e+f x) \left (b \tan ^2(e+f x)\right )^{3/2} \left (\tan ^2(e+f x)+2 \log (\cos (e+f x))\right )}{2 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Tan[e + f*x]^2)^(3/2),x]

[Out]

(Cot[e + f*x]^3*(b*Tan[e + f*x]^2)^(3/2)*(2*Log[Cos[e + f*x]] + Tan[e + f*x]^2))/(2*f)

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fricas [A]  time = 0.40, size = 52, normalized size = 0.85 \[ \frac {{\left (b \tan \left (f x + e\right )^{2} + b \log \left (\frac {1}{\tan \left (f x + e\right )^{2} + 1}\right ) + b\right )} \sqrt {b \tan \left (f x + e\right )^{2}}}{2 \, f \tan \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

1/2*(b*tan(f*x + e)^2 + b*log(1/(tan(f*x + e)^2 + 1)) + b)*sqrt(b*tan(f*x + e)^2)/(f*tan(f*x + e))

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (2*pi/x/2)>(-2*pi/
x/2)sqrt(b)*b*(tan(f*x)^2*tan(exp(1))^2+tan(f*x)^2+tan(exp(1))^2+tan(f*x)^2*tan(exp(1))^2*ln((4*tan(f*x)^2*tan
(exp(1))^2-8*tan(f*x)^3*tan(exp(1))+4*tan(f*x)^4*tan(exp(1))^2+4*tan(f*x)^2-8*tan(f*x)*tan(exp(1))+4)/(tan(exp
(1))^2+1))-2*tan(f*x)*tan(exp(1))*ln((4*tan(f*x)^2*tan(exp(1))^2-8*tan(f*x)^3*tan(exp(1))+4*tan(f*x)^4*tan(exp
(1))^2+4*tan(f*x)^2-8*tan(f*x)*tan(exp(1))+4)/(tan(exp(1))^2+1))+ln((-8*tan(f*x)^3*tan(exp(1))+4*tan(f*x)^4*ta
n(exp(1))^2+4*tan(f*x)^2*tan(exp(1))^2+4*tan(f*x)^2-8*tan(f*x)*tan(exp(1))+4)/(tan(exp(1))^2+1))+1)*sign(tan(f
*x+exp(1)))/(2*f*tan(f*x)^2*tan(exp(1))^2-4*f*tan(f*x)*tan(exp(1))+2*f)

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maple [A]  time = 0.26, size = 48, normalized size = 0.79 \[ -\frac {\left (b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{\frac {3}{2}} \left (-\left (\tan ^{2}\left (f x +e \right )\right )+\ln \left (1+\tan ^{2}\left (f x +e \right )\right )\right )}{2 f \tan \left (f x +e \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(f*x+e)^2)^(3/2),x)

[Out]

-1/2/f*(b*tan(f*x+e)^2)^(3/2)*(-tan(f*x+e)^2+ln(1+tan(f*x+e)^2))/tan(f*x+e)^3

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maxima [A]  time = 0.83, size = 34, normalized size = 0.56 \[ \frac {b^{\frac {3}{2}} \tan \left (f x + e\right )^{2} - b^{\frac {3}{2}} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

1/2*(b^(3/2)*tan(f*x + e)^2 - b^(3/2)*log(tan(f*x + e)^2 + 1))/f

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int {\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(e + f*x)^2)^(3/2),x)

[Out]

int((b*tan(e + f*x)^2)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e)**2)**(3/2),x)

[Out]

Integral((b*tan(e + f*x)**2)**(3/2), x)

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