3.1 \(\int (b \tan ^2(e+f x))^{5/2} \, dx\)

Optimal. Leaf size=98 \[ \frac{b^2 \tan ^3(e+f x) \sqrt{b \tan ^2(e+f x)}}{4 f}-\frac{b^2 \tan (e+f x) \sqrt{b \tan ^2(e+f x)}}{2 f}-\frac{b^2 \cot (e+f x) \sqrt{b \tan ^2(e+f x)} \log (\cos (e+f x))}{f} \]

[Out]

-((b^2*Cot[e + f*x]*Log[Cos[e + f*x]]*Sqrt[b*Tan[e + f*x]^2])/f) - (b^2*Tan[e + f*x]*Sqrt[b*Tan[e + f*x]^2])/(
2*f) + (b^2*Tan[e + f*x]^3*Sqrt[b*Tan[e + f*x]^2])/(4*f)

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Rubi [A]  time = 0.0390226, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {3658, 3473, 3475} \[ \frac{b^2 \tan ^3(e+f x) \sqrt{b \tan ^2(e+f x)}}{4 f}-\frac{b^2 \tan (e+f x) \sqrt{b \tan ^2(e+f x)}}{2 f}-\frac{b^2 \cot (e+f x) \sqrt{b \tan ^2(e+f x)} \log (\cos (e+f x))}{f} \]

Antiderivative was successfully verified.

[In]

Int[(b*Tan[e + f*x]^2)^(5/2),x]

[Out]

-((b^2*Cot[e + f*x]*Log[Cos[e + f*x]]*Sqrt[b*Tan[e + f*x]^2])/f) - (b^2*Tan[e + f*x]*Sqrt[b*Tan[e + f*x]^2])/(
2*f) + (b^2*Tan[e + f*x]^3*Sqrt[b*Tan[e + f*x]^2])/(4*f)

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
]])

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]

Rubi steps

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

Mathematica [A]  time = 0.372675, size = 56, normalized size = 0.57 \[ -\frac{\cot (e+f x) \left (b \tan ^2(e+f x)\right )^{5/2} \left (2 \cot ^2(e+f x)+4 \cot ^4(e+f x) \log (\cos (e+f x))-1\right )}{4 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Tan[e + f*x]^2)^(5/2),x]

[Out]

-(Cot[e + f*x]*(-1 + 2*Cot[e + f*x]^2 + 4*Cot[e + f*x]^4*Log[Cos[e + f*x]])*(b*Tan[e + f*x]^2)^(5/2))/(4*f)

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Maple [A]  time = 0.146, size = 58, normalized size = 0.6 \begin{align*}{\frac{ \left ( \tan \left ( fx+e \right ) \right ) ^{4}-2\, \left ( \tan \left ( fx+e \right ) \right ) ^{2}+2\,\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }{4\,f \left ( \tan \left ( fx+e \right ) \right ) ^{5}} \left ( b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(f*x+e)^2)^(5/2),x)

[Out]

1/4/f*(b*tan(f*x+e)^2)^(5/2)*(tan(f*x+e)^4-2*tan(f*x+e)^2+2*ln(1+tan(f*x+e)^2))/tan(f*x+e)^5

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Maxima [A]  time = 1.6261, size = 63, normalized size = 0.64 \begin{align*} \frac{b^{\frac{5}{2}} \tan \left (f x + e\right )^{4} - 2 \, b^{\frac{5}{2}} \tan \left (f x + e\right )^{2} + 2 \, b^{\frac{5}{2}} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{4 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(b^(5/2)*tan(f*x + e)^4 - 2*b^(5/2)*tan(f*x + e)^2 + 2*b^(5/2)*log(tan(f*x + e)^2 + 1))/f

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Fricas [A]  time = 1.93873, size = 180, normalized size = 1.84 \begin{align*} \frac{{\left (b^{2} \tan \left (f x + e\right )^{4} - 2 \, b^{2} \tan \left (f x + e\right )^{2} - 2 \, b^{2} \log \left (\frac{1}{\tan \left (f x + e\right )^{2} + 1}\right ) - 3 \, b^{2}\right )} \sqrt{b \tan \left (f x + e\right )^{2}}}{4 \, f \tan \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(b^2*tan(f*x + e)^4 - 2*b^2*tan(f*x + e)^2 - 2*b^2*log(1/(tan(f*x + e)^2 + 1)) - 3*b^2)*sqrt(b*tan(f*x + e
)^2)/(f*tan(f*x + e))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.4196, size = 80, normalized size = 0.82 \begin{align*} \frac{1}{4} \, b^{\frac{5}{2}}{\left (\frac{2 \, \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{f} + \frac{f \tan \left (f x + e\right )^{4} - 2 \, f \tan \left (f x + e\right )^{2}}{f^{2}}\right )} \mathrm{sgn}\left (\tan \left (f x + e\right )\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b^(5/2)*(2*log(tan(f*x + e)^2 + 1)/f + (f*tan(f*x + e)^4 - 2*f*tan(f*x + e)^2)/f^2)*sgn(tan(f*x + e))