3.24 \(\int (b \tan ^2(c+d x))^{5/2} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0416749, 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} \[ \frac{b^2 \tan ^3(c+d x) \sqrt{b \tan ^2(c+d x)}}{4 d}-\frac{b^2 \tan (c+d x) \sqrt{b \tan ^2(c+d x)}}{2 d}-\frac{b^2 \cot (c+d x) \sqrt{b \tan ^2(c+d x)} \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[(b*Tan[c + d*x]^2)^(5/2),x]

[Out]

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

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

Mathematica [A]  time = 0.376226, size = 56, normalized size = 0.57 \[ -\frac{\cot (c+d x) \left (b \tan ^2(c+d x)\right )^{5/2} \left (2 \cot ^2(c+d x)+4 \cot ^4(c+d x) \log (\cos (c+d x))-1\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Tan[c + d*x]^2)^(5/2),x]

[Out]

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

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Maple [A]  time = 0.033, size = 58, normalized size = 0.6 \begin{align*}{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{4}-2\, \left ( \tan \left ( dx+c \right ) \right ) ^{2}+2\,\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }{4\,d \left ( \tan \left ( dx+c \right ) \right ) ^{5}} \left ( b \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.56432, size = 63, normalized size = 0.64 \begin{align*} \frac{b^{\frac{5}{2}} \tan \left (d x + c\right )^{4} - 2 \, b^{\frac{5}{2}} \tan \left (d x + c\right )^{2} + 2 \, b^{\frac{5}{2}} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.29161, size = 180, normalized size = 1.84 \begin{align*} \frac{{\left (b^{2} \tan \left (d x + c\right )^{4} - 2 \, b^{2} \tan \left (d x + c\right )^{2} - 2 \, b^{2} \log \left (\frac{1}{\tan \left (d x + c\right )^{2} + 1}\right ) - 3 \, b^{2}\right )} \sqrt{b \tan \left (d x + c\right )^{2}}}{4 \, d \tan \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.58912, size = 74, normalized size = 0.76 \begin{align*} \frac{1}{4} \, b^{\frac{5}{2}}{\left (\frac{2 \, \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{d} + \frac{d \tan \left (d x + c\right )^{4} - 2 \, d \tan \left (d x + c\right )^{2}}{d^{2}}\right )} \mathrm{sgn}\left (\tan \left (d x + c\right )\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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