3.281 \(\int x^m \sec ^{\frac {5}{2}}(a+b \log (c x^n)) \, dx\)

Optimal. Leaf size=130 \[ \frac {2 x^{m+1} \left (1+e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{5/2} \, _2F_1\left (\frac {5}{2},-\frac {2 i m-5 b n+2 i}{4 b n};-\frac {2 i m-9 b n+2 i}{4 b n};-e^{2 i a} \left (c x^n\right )^{2 i b}\right ) \sec ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )}{5 i b n+2 m+2} \]

[Out]

2*x^(1+m)*(1+exp(2*I*a)*(c*x^n)^(2*I*b))^(5/2)*hypergeom([5/2, 1/4*(-2*I-2*I*m+5*b*n)/b/n],[1/4*(-2*I-2*I*m+9*
b*n)/b/n],-exp(2*I*a)*(c*x^n)^(2*I*b))*sec(a+b*ln(c*x^n))^(5/2)/(2+2*m+5*I*b*n)

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Rubi [A]  time = 0.10, antiderivative size = 126, normalized size of antiderivative = 0.97, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {4509, 4507, 364} \[ \frac {2 x^{m+1} \left (1+e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{5/2} \, _2F_1\left (\frac {5}{2},\frac {1}{4} \left (5-\frac {2 i (m+1)}{b n}\right );-\frac {2 i m-9 b n+2 i}{4 b n};-e^{2 i a} \left (c x^n\right )^{2 i b}\right ) \sec ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )}{5 i b n+2 m+2} \]

Antiderivative was successfully verified.

[In]

Int[x^m*Sec[a + b*Log[c*x^n]]^(5/2),x]

[Out]

(2*x^(1 + m)*(1 + E^((2*I)*a)*(c*x^n)^((2*I)*b))^(5/2)*Hypergeometric2F1[5/2, (5 - ((2*I)*(1 + m))/(b*n))/4, -
(2*I + (2*I)*m - 9*b*n)/(4*b*n), -(E^((2*I)*a)*(c*x^n)^((2*I)*b))]*Sec[a + b*Log[c*x^n]]^(5/2))/(2 + 2*m + (5*
I)*b*n)

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])

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 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])

Rubi steps

\begin {align*} \int x^m \sec ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right ) \, dx &=\frac {\left (x^{1+m} \left (c x^n\right )^{-\frac {1+m}{n}}\right ) \operatorname {Subst}\left (\int x^{-1+\frac {1+m}{n}} \sec ^{\frac {5}{2}}(a+b \log (x)) \, dx,x,c x^n\right )}{n}\\ &=\frac {\left (x^{1+m} \left (c x^n\right )^{-\frac {5 i b}{2}-\frac {1+m}{n}} \left (1+e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{5/2} \sec ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )\right ) \operatorname {Subst}\left (\int \frac {x^{-1+\frac {5 i b}{2}+\frac {1+m}{n}}}{\left (1+e^{2 i a} x^{2 i b}\right )^{5/2}} \, dx,x,c x^n\right )}{n}\\ &=\frac {2 x^{1+m} \left (1+e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{5/2} \, _2F_1\left (\frac {5}{2},\frac {1}{4} \left (5-\frac {2 i (1+m)}{b n}\right );-\frac {2 i+2 i m-9 b n}{4 b n};-e^{2 i a} \left (c x^n\right )^{2 i b}\right ) \sec ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )}{2+2 m+5 i b n}\\ \end {align*}

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Mathematica [A]  time = 2.14, size = 182, normalized size = 1.40 \[ \frac {2 x^{m+1} \sqrt {\sec \left (a+b \log \left (c x^n\right )\right )} \left (\left (b^2 n^2+4 m^2+8 m+4\right ) \left (1+e^{2 i a} \left (c x^n\right )^{2 i b}\right ) \, _2F_1\left (1,-\frac {2 i m-3 b n+2 i}{4 b n};-\frac {2 i m-5 b n+2 i}{4 b n};-e^{2 i \left (a+b \log \left (c x^n\right )\right )}\right )-(i b n+2 m+2) \left (-b n \tan \left (a+b \log \left (c x^n\right )\right )+2 m+2\right )\right )}{3 b^2 n^2 (i b n+2 m+2)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^m*Sec[a + b*Log[c*x^n]]^(5/2),x]

[Out]

(2*x^(1 + m)*Sqrt[Sec[a + b*Log[c*x^n]]]*((4 + 8*m + 4*m^2 + b^2*n^2)*(1 + E^((2*I)*a)*(c*x^n)^((2*I)*b))*Hype
rgeometric2F1[1, -1/4*(2*I + (2*I)*m - 3*b*n)/(b*n), -1/4*(2*I + (2*I)*m - 5*b*n)/(b*n), -E^((2*I)*(a + b*Log[
c*x^n]))] - (2 + 2*m + I*b*n)*(2 + 2*m - b*n*Tan[a + b*Log[c*x^n]])))/(3*b^2*n^2*(2 + 2*m + I*b*n))

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*sec(a+b*log(c*x^n))^(5/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

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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(x^m*sec(a+b*log(c*x^n))^(5/2),x, algorithm="giac")

[Out]

Timed out

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maple [F]  time = 0.12, size = 0, normalized size = 0.00 \[ \int x^{m} \left (\sec ^{\frac {5}{2}}\left (a +b \ln \left (c \,x^{n}\right )\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*sec(a+b*ln(c*x^n))^(5/2),x)

[Out]

int(x^m*sec(a+b*ln(c*x^n))^(5/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{m} \sec \left (b \log \left (c x^{n}\right ) + a\right )^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*sec(a+b*log(c*x^n))^(5/2),x, algorithm="maxima")

[Out]

integrate(x^m*sec(b*log(c*x^n) + a)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^m\,{\left (\frac {1}{\cos \left (a+b\,\ln \left (c\,x^n\right )\right )}\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*(1/cos(a + b*log(c*x^n)))^(5/2),x)

[Out]

int(x^m*(1/cos(a + b*log(c*x^n)))^(5/2), x)

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sympy [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(x**m*sec(a+b*ln(c*x**n))**(5/2),x)

[Out]

Timed out

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