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

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

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 98, normalized size of antiderivative = 1.00, 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 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(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*}

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Mathematica [A]  time = 0.38, 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]

-1/4*(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))/d

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fricas [A]  time = 0.63, size = 74, normalized size = 0.76 \[ \frac {{\left (b^{2} \tan \left (d x + c\right )^{4} - 2 \, b^{2} \tan \left (d x + c\right )^{2} - 2 \, b^{2} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) - 3 \, b^{2}\right )} \sqrt {b \tan \left (d x + c\right )^{2}}}{4 \, d \tan \left (d x + c\right )} \]

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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giac [B]  time = 5.71, size = 696, normalized size = 7.10 \[ \text {result too large to display} \]

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*(2*b^2*log(4*(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*t
an(c) + 1)/(tan(c)^2 + 1))*sgn(tan(d*x + c))*tan(d*x)^4*tan(c)^4 + 3*b^2*sgn(tan(d*x + c))*tan(d*x)^4*tan(c)^4
- 8*b^2*log(4*(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(
c) + 1)/(tan(c)^2 + 1))*sgn(tan(d*x + c))*tan(d*x)^3*tan(c)^3 + 2*b^2*sgn(tan(d*x + c))*tan(d*x)^4*tan(c)^2 -
8*b^2*sgn(tan(d*x + c))*tan(d*x)^3*tan(c)^3 + 2*b^2*sgn(tan(d*x + c))*tan(d*x)^2*tan(c)^4 + 12*b^2*log(4*(tan(
d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(c)^2 +
1))*sgn(tan(d*x + c))*tan(d*x)^2*tan(c)^2 - b^2*sgn(tan(d*x + c))*tan(d*x)^4 - 8*b^2*sgn(tan(d*x + c))*tan(d*x
)^3*tan(c) + 4*b^2*sgn(tan(d*x + c))*tan(d*x)^2*tan(c)^2 - 8*b^2*sgn(tan(d*x + c))*tan(d*x)*tan(c)^3 - b^2*sgn
(tan(d*x + c))*tan(c)^4 - 8*b^2*log(4*(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d
*x)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(c)^2 + 1))*sgn(tan(d*x + c))*tan(d*x)*tan(c) + 2*b^2*sgn(tan(d*x + c))*tan
(d*x)^2 - 8*b^2*sgn(tan(d*x + c))*tan(d*x)*tan(c) + 2*b^2*sgn(tan(d*x + c))*tan(c)^2 + 2*b^2*log(4*(tan(d*x)^4
*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(c)^2 + 1))*sg
n(tan(d*x + c)) + 3*b^2*sgn(tan(d*x + c)))*sqrt(b)/(d*tan(d*x)^4*tan(c)^4 - 4*d*tan(d*x)^3*tan(c)^3 + 6*d*tan(
d*x)^2*tan(c)^2 - 4*d*tan(d*x)*tan(c) + d)

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maple [A]  time = 0.15, size = 58, normalized size = 0.59 \[ \frac {\left (b \left (\tan ^{2}\left (d x +c \right )\right )\right )^{\frac {5}{2}} \left (\tan ^{4}\left (d x +c \right )-2 \left (\tan ^{2}\left (d x +c \right )\right )+2 \ln \left (1+\tan ^{2}\left (d x +c \right )\right )\right )}{4 d \tan \left (d x +c \right )^{5}} \]

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 = 0.78, size = 47, normalized size = 0.48 \[ \frac {b^{\frac {5}{2}} \tan \left (d x + c\right )^{4} - 2 \, b^{\frac {5}{2}} \tan \left (d x + c\right )^{2} + 2 \, b^{\frac {5}{2}} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{4 \, d} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (b\,{\mathrm {tan}\left (c+d\,x\right )}^2\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(c + d*x)^2)^(5/2),x)

[Out]

int((b*tan(c + d*x)^2)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \tan ^{2}{\left (c + d x \right )}\right )^{\frac {5}{2}}\, dx \]

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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