3.1.25 \(\int (b \tan ^2(c+d x))^{3/2} \, dx\) [25]

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

[Out]

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

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Rubi [A]
time = 0.02, 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 = {3739, 3554, 3556} \begin {gather*} \frac {b \tan (c+d x) \sqrt {b \tan ^2(c+d x)}}{2 d}+\frac {b \cot (c+d x) \sqrt {b \tan ^2(c+d x)} \log (\cos (c+d x))}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*Tan[c + d*x]^2)^(3/2),x]

[Out]

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

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

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Mathematica [A]
time = 0.12, size = 47, normalized size = 0.77 \begin {gather*} \frac {\cot ^3(c+d x) \left (b \tan ^2(c+d x)\right )^{3/2} \left (2 \log (\cos (c+d x))+\tan ^2(c+d x)\right )}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*Tan[c + d*x]^2)^(3/2),x]

[Out]

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

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Maple [A]
time = 0.08, size = 48, normalized size = 0.79

method	result	size
derivativedivides	\(-\frac {\left (b \left (\tan ^{2}\left (d x +c \right )\right )\right )^{\frac {3}{2}} \left (-\left (\tan ^{2}\left (d x +c \right )\right )+\ln \left (1+\tan ^{2}\left (d x +c \right )\right )\right )}{2 d \tan \left (d x +c \right )^{3}}\)	\(48\)
default	\(-\frac {\left (b \left (\tan ^{2}\left (d x +c \right )\right )\right )^{\frac {3}{2}} \left (-\left (\tan ^{2}\left (d x +c \right )\right )+\ln \left (1+\tan ^{2}\left (d x +c \right )\right )\right )}{2 d \tan \left (d x +c \right )^{3}}\)	\(48\)
risch	\(-\frac {b \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {-\frac {b \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, x}{{\mathrm e}^{2 i \left (d x +c \right )}-1}+\frac {2 b \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {-\frac {b \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, \left (d x +c \right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) d}+\frac {2 i b \sqrt {-\frac {b \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, {\mathrm e}^{2 i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {i b \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {-\frac {b \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) d}\)	\(274\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.50, size = 34, normalized size = 0.56 \begin {gather*} \frac {b^{\frac {3}{2}} \tan \left (d x + c\right )^{2} - b^{\frac {3}{2}} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.37, size = 52, normalized size = 0.85 \begin {gather*} \frac {{\left (b \tan \left (d x + c\right )^{2} + b \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) + b\right )} \sqrt {b \tan \left (d x + c\right )^{2}}}{2 \, d \tan \left (d x + c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \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((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. 256 vs. \(2 (55) = 110\).
time = 0.74, size = 256, normalized size = 4.20 \begin {gather*} \frac {{\left (\log \left (\frac {4 \, {\left (\tan \left (d x\right )^{4} \tan \left (c\right )^{2} - 2 \, \tan \left (d x\right )^{3} \tan \left (c\right ) + \tan \left (d x\right )^{2} \tan \left (c\right )^{2} + \tan \left (d x\right )^{2} - 2 \, \tan \left (d x\right ) \tan \left (c\right ) + 1\right )}}{\tan \left (c\right )^{2} + 1}\right ) \tan \left (d x\right )^{2} \tan \left (c\right )^{2} + \tan \left (d x\right )^{2} \tan \left (c\right )^{2} - 2 \, \log \left (\frac {4 \, {\left (\tan \left (d x\right )^{4} \tan \left (c\right )^{2} - 2 \, \tan \left (d x\right )^{3} \tan \left (c\right ) + \tan \left (d x\right )^{2} \tan \left (c\right )^{2} + \tan \left (d x\right )^{2} - 2 \, \tan \left (d x\right ) \tan \left (c\right ) + 1\right )}}{\tan \left (c\right )^{2} + 1}\right ) \tan \left (d x\right ) \tan \left (c\right ) + \tan \left (d x\right )^{2} + \tan \left (c\right )^{2} + \log \left (\frac {4 \, {\left (\tan \left (d x\right )^{4} \tan \left (c\right )^{2} - 2 \, \tan \left (d x\right )^{3} \tan \left (c\right ) + \tan \left (d x\right )^{2} \tan \left (c\right )^{2} + \tan \left (d x\right )^{2} - 2 \, \tan \left (d x\right ) \tan \left (c\right ) + 1\right )}}{\tan \left (c\right )^{2} + 1}\right ) + 1\right )} b^{\frac {3}{2}} \mathrm {sgn}\left (\tan \left (d x + c\right )\right )}{2 \, {\left (d \tan \left (d x\right )^{2} \tan \left (c\right )^{2} - 2 \, d \tan \left (d x\right ) \tan \left (c\right ) + d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

int((b*tan(c + d*x)^2)^(3/2), x)

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