3.286 \(\int (e x)^m \sec ^p(d (a+b \log (c x^n))) \, dx\)

Optimal. Leaf size=139 \[ \frac{(e x)^{m+1} \left (1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )^p \sec ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \text{Hypergeometric2F1}\left (p,-\frac{-b d n p+i m+i}{2 b d n},\frac{1}{2} \left (-\frac{i (m+1)}{b d n}+p+2\right ),-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{e (i b d n p+m+1)} \]

[Out]

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

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Rubi [A]  time = 0.121491, antiderivative size = 133, normalized size of antiderivative = 0.96, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {4509, 4507, 364} \[ \frac{(e x)^{m+1} \left (1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )^p \, _2F_1\left (p,\frac{1}{2} \left (p-\frac{i (m+1)}{b d n}\right );\frac{1}{2} \left (-\frac{i (m+1)}{b d n}+p+2\right );-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right ) \sec ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right )}{e (i b d n p+m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4509

Int[((e_.)*(x_))^(m_.)*Sec[((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)*Sec[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])

Rule 4507

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

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int (e x)^m \sec ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx &=\frac{\left ((e x)^{1+m} \left (c x^n\right )^{-\frac{1+m}{n}}\right ) \operatorname{Subst}\left (\int x^{-1+\frac{1+m}{n}} \sec ^p(d (a+b \log (x))) \, dx,x,c x^n\right )}{e n}\\ &=\frac{\left ((e x)^{1+m} \left (c x^n\right )^{-\frac{1+m}{n}-i b d p} \left (1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )^p \sec ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right )\right ) \operatorname{Subst}\left (\int x^{-1+\frac{1+m}{n}+i b d p} \left (1+e^{2 i a d} x^{2 i b d}\right )^{-p} \, dx,x,c x^n\right )}{e n}\\ &=\frac{(e x)^{1+m} \left (1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )^p \, _2F_1\left (p,\frac{1}{2} \left (-\frac{i (1+m)}{b d n}+p\right );\frac{1}{2} \left (2-\frac{i (1+m)}{b d n}+p\right );-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right ) \sec ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right )}{e (1+m+i b d n p)}\\ \end{align*}

Mathematica [A]  time = 1.55888, size = 169, normalized size = 1.22 \[ \frac{2^p x (e x)^m \left (\frac{e^{i a d} \left (c x^n\right )^{i b d}}{1+e^{2 i a d} \left (c x^n\right )^{2 i b d}}\right )^p \left (1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )^p \text{Hypergeometric2F1}\left (p,-\frac{i (i b d n p+m+1)}{2 b d n},\frac{1}{2} \left (-\frac{i (m+1)}{b d n}+p+2\right ),-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{i b d n p+m+1} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [F]  time = 0.27, size = 0, normalized size = 0. \begin{align*} \int \left ( ex \right ) ^{m} \left ( \sec \left ( d \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) \right ) \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (e x\right )^{m} \sec \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right )^{p}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\left (e x\right )^{m} \sec \left (b d \log \left (c x^{n}\right ) + a d\right )^{p}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (e x\right )^{m} \sec \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right )^{p}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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