3.2.75 \(\int (e x)^m \tan (d (a+b \log (c x^n))) \, dx\) [175]

Optimal. Leaf size=101 \[ -\frac {i (e x)^{1+m}}{e (1+m)}+\frac {2 i (e x)^{1+m} \, _2F_1\left (1,-\frac {i (1+m)}{2 b d n};1-\frac {i (1+m)}{2 b d n};-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{e (1+m)} \]

[Out]

-I*(e*x)^(1+m)/e/(1+m)+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))/e/(1+m)

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Rubi [A]
time = 0.05, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {4593, 4591, 470, 371} \begin {gather*} \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 )}{e (m+1)}-\frac {i (e x)^{m+1}}{e (m+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*(e*x)^(1 + m))/(e*(1 + m)) + ((2*I)*(e*x)^(1 + m)*Hypergeometric2F1[1, ((-1/2*I)*(1 + m))/(b*d*n), 1 - (
(I/2)*(1 + m))/(b*d*n), -(E^((2*I)*a*d)*(c*x^n)^((2*I)*b*d))])/(e*(1 + m))

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 470

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(b*e*(m + n*(p + 1) + 1))), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]

Rule 4591

Int[((e_.)*(x_))^(m_.)*Tan[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Int[(e*x)^m*((I - I*E^(2*I*a*d)*
x^(2*I*b*d))/(1 + E^(2*I*a*d)*x^(2*I*b*d)))^p, x] /; FreeQ[{a, b, d, e, m, p}, x]

Rule 4593

Int[((e_.)*(x_))^(m_.)*Tan[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Dist[(e*x)^(m + 1)
/(e*n*(c*x^n)^((m + 1)/n)), Subst[Int[x^((m + 1)/n - 1)*Tan[d*(a + b*Log[x])]^p, x], x, c*x^n], x] /; FreeQ[{a
, b, c, d, e, m, n, p}, x] && (NeQ[c, 1] || NeQ[n, 1])

Rubi steps

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

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Mathematica [A]
time = 15.29, size = 186, normalized size = 1.84 \begin {gather*} \frac {i x (e x)^m \left (\, _2F_1\left (1,-\frac {i (1+m)}{2 b d n};1-\frac {i (1+m)}{2 b d n};-e^{2 i d \left (a+b \log \left (c x^n\right )\right )}\right )-\frac {e^{2 i a d} (1+m) \left (c x^n\right )^{2 i b d} \, _2F_1\left (1,-\frac {i (1+m+2 i b d n)}{2 b d n};-\frac {i (1+m+4 i b d n)}{2 b d n};-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{1+m+2 i b d n}\right )}{1+m} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(I*x*(e*x)^m*(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^((2*I)*a*d)*(1 + m)*(c*x^n)^((2*I)*b*d)*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)*a*d)*(c*x^n)^((2*I)*b*d))])/(1 + m + (2*I)*b
*d*n)))/(1 + m)

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \left (e x \right )^{m} \tan \left (d \left (a +b \ln \left (c \,x^{n}\right )\right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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