3.176 \(\int (e x)^m \tan ^2(d (a+b \log (c x^n))) \, dx\)

Optimal. Leaf size=196 \[ -\frac {2 i (e x)^{m+1} \, _2F_1\left (1,-\frac {i (m+1)}{2 b d n};1-\frac {i (m+1)}{2 b d n};-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{b d e n}+\frac {i (e x)^{m+1} \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{b d e n \left (1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}+\frac {(e x)^{m+1} (-b d n+i (m+1))}{b d e (m+1) n} \]

[Out]

(I*(1+m)-b*d*n)*(e*x)^(1+m)/b/d/e/(1+m)/n+I*(e*x)^(1+m)*(1-exp(2*I*a*d)*(c*x^n)^(2*I*b*d))/b/d/e/n/(1+exp(2*I*
a*d)*(c*x^n)^(2*I*b*d))-2*I*(e*x)^(1+m)*hypergeom([1, -1/2*I*(1+m)/b/d/n],[1-1/2*I*(1+m)/b/d/n],-exp(2*I*a*d)*
(c*x^n)^(2*I*b*d))/b/d/e/n

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Rubi [F]  time = 0.08, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int (e x)^m \tan ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx \]

Verification is Not applicable to the result.

[In]

Int[(e*x)^m*Tan[d*(a + b*Log[c*x^n])]^2,x]

[Out]

Defer[Int][(e*x)^m*Tan[d*(a + b*Log[c*x^n])]^2, x]

Rubi steps

\begin {align*} \int (e x)^m \tan ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx &=\int (e x)^m \tan ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx\\ \end {align*}

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Mathematica [B]  time = 17.55, size = 550, normalized size = 2.81 \[ -\frac {(m+1) x^{-m} (e x)^m \sec \left (d \left (a+b \left (\log \left (c x^n\right )-n \log (x)\right )\right )\right ) \left (\frac {x^{m+1} \sin (b d n \log (x)) \sec \left (d \left (a+b \log \left (c x^n\right )\right )\right )}{m+1}-\frac {i \cos \left (d \left (a+b \left (\log \left (c x^n\right )-n \log (x)\right )\right )\right ) \exp \left (-\frac {(2 m+1) \left (a+b \left (\log \left (c x^n\right )-n \log (x)\right )\right )}{b n}\right ) \left ((2 i b d n+m+1) \left (-\exp \left (\frac {2 a m+a+b (2 m+1) \left (\log \left (c x^n\right )-n \log (x)\right )+b (m+1) n \log (x)}{b n}\right )\right ) \, _2F_1\left (1,-\frac {i (m+1)}{2 b d n};1-\frac {i (m+1)}{2 b d n};-e^{2 i d \left (a+b \log \left (c x^n\right )\right )}\right )+(m+1) \exp \left (\frac {a (2 i b d n+2 m+1)}{b n}+\frac {(2 i b d n+2 m+1) \left (\log \left (c x^n\right )-n \log (x)\right )}{n}+\log (x) (2 i b d n+m+1)\right ) \, _2F_1\left (1,-\frac {i (m+2 i b d n+1)}{2 b d n};-\frac {i (m+4 i b d n+1)}{2 b d n};-e^{2 i d \left (a+b \log \left (c x^n\right )\right )}\right )-i (2 i b d n+m+1) \tan \left (d \left (a+b \log \left (c x^n\right )\right )\right ) \exp \left (\frac {2 a m+a+b (2 m+1) \left (\log \left (c x^n\right )-n \log (x)\right )+b (m+1) n \log (x)}{b n}\right )\right )}{(m+1) (2 i b d n+m+1)}\right )}{b d n}+\frac {x (e x)^m \sin (b d n \log (x)) \sec \left (d \left (a+b \left (\log \left (c x^n\right )-n \log (x)\right )\right )\right ) \sec \left (d \left (a+b \left (\log \left (c x^n\right )-n \log (x)\right )\right )+b d n \log (x)\right )}{b d n}-\frac {x (e x)^m}{m+1} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*x)^m*Tan[d*(a + b*Log[c*x^n])]^2,x]

[Out]

-((x*(e*x)^m)/(1 + m)) + (x*(e*x)^m*Sec[d*(a + b*(-(n*Log[x]) + Log[c*x^n]))]*Sec[b*d*n*Log[x] + d*(a + b*(-(n
*Log[x]) + Log[c*x^n]))]*Sin[b*d*n*Log[x]])/(b*d*n) - ((1 + m)*(e*x)^m*Sec[d*(a + b*(-(n*Log[x]) + Log[c*x^n])
)]*((x^(1 + m)*Sec[d*(a + b*Log[c*x^n])]*Sin[b*d*n*Log[x]])/(1 + m) - (I*Cos[d*(a + b*(-(n*Log[x]) + Log[c*x^n
]))]*(-(E^((a + 2*a*m + b*(1 + m)*n*Log[x] + b*(1 + 2*m)*(-(n*Log[x]) + Log[c*x^n]))/(b*n))*(1 + m + (2*I)*b*d
*n)*Hypergeometric2F1[1, ((-1/2*I)*(1 + m))/(b*d*n), 1 - ((I/2)*(1 + m))/(b*d*n), -E^((2*I)*d*(a + b*Log[c*x^n
]))]) + E^((a*(1 + 2*m + (2*I)*b*d*n))/(b*n) + (1 + m + (2*I)*b*d*n)*Log[x] + ((1 + 2*m + (2*I)*b*d*n)*(-(n*Lo
g[x]) + Log[c*x^n]))/n)*(1 + m)*Hypergeometric2F1[1, ((-1/2*I)*(1 + m + (2*I)*b*d*n))/(b*d*n), ((-1/2*I)*(1 +
m + (4*I)*b*d*n))/(b*d*n), -E^((2*I)*d*(a + b*Log[c*x^n]))] - I*E^((a + 2*a*m + b*(1 + m)*n*Log[x] + b*(1 + 2*
m)*(-(n*Log[x]) + Log[c*x^n]))/(b*n))*(1 + m + (2*I)*b*d*n)*Tan[d*(a + b*Log[c*x^n])]))/(E^(((1 + 2*m)*(a + b*
(-(n*Log[x]) + Log[c*x^n])))/(b*n))*(1 + m)*(1 + m + (2*I)*b*d*n))))/(b*d*n*x^m)

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (e x\right )^{m} \tan \left (b d \log \left (c x^{n}\right ) + a d\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*tan(d*(a+b*log(c*x^n)))^2,x, algorithm="fricas")

[Out]

integral((e*x)^m*tan(b*d*log(c*x^n) + a*d)^2, x)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*tan(d*(a+b*log(c*x^n)))^2,x, algorithm="giac")

[Out]

Timed out

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maple [F]  time = 0.35, size = 0, normalized size = 0.00 \[ \int \left (e x \right )^{m} \left (\tan ^{2}\left (d \left (a +b \ln \left (c \,x^{n}\right )\right )\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^m*tan(d*(a+b*ln(c*x^n)))^2,x)

[Out]

int((e*x)^m*tan(d*(a+b*ln(c*x^n)))^2,x)

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*tan(d*(a+b*log(c*x^n)))^2,x, algorithm="maxima")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*(a + b*log(c*x^n)))^2*(e*x)^m,x)

[Out]

int(tan(d*(a + b*log(c*x^n)))^2*(e*x)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**m*tan(d*(a+b*ln(c*x**n)))**2,x)

[Out]

Integral((e*x)**m*tan(a*d + b*d*log(c*x**n))**2, x)

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