3.3.82 \(\int x^m \sec ^{\frac {3}{2}}(a+b \log (c x^n)) \, dx\) [282]

Optimal. Leaf size=130 \[ \frac {2 x^{1+m} \left (1+e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{3/2} \, _2F_1\left (\frac {3}{2},-\frac {2 i+2 i m-3 b n}{4 b n};-\frac {2 i+2 i m-7 b n}{4 b n};-e^{2 i a} \left (c x^n\right )^{2 i b}\right ) \sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}{2+2 m+3 i b n} \]

[Out]

2*x^(1+m)*(1+exp(2*I*a)*(c*x^n)^(2*I*b))^(3/2)*hypergeom([3/2, 1/4*(-2*I-2*I*m+3*b*n)/b/n],[1/4*(-2*I-2*I*m+7*
b*n)/b/n],-exp(2*I*a)*(c*x^n)^(2*I*b))*sec(a+b*ln(c*x^n))^(3/2)/(2+2*m+3*I*b*n)

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Rubi [A]
time = 0.07, 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 = {4605, 4603, 371} \begin {gather*} \frac {2 x^{m+1} \left (1+e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{3/2} \, _2F_1\left (\frac {3}{2},\frac {1}{4} \left (3-\frac {2 i (m+1)}{b n}\right );-\frac {2 i m-7 b n+2 i}{4 b n};-e^{2 i a} \left (c x^n\right )^{2 i b}\right ) \sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}{3 i b n+2 m+2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^m*Sec[a + b*Log[c*x^n]]^(3/2),x]

[Out]

(2*x^(1 + m)*(1 + E^((2*I)*a)*(c*x^n)^((2*I)*b))^(3/2)*Hypergeometric2F1[3/2, (3 - ((2*I)*(1 + m))/(b*n))/4, -
1/4*(2*I + (2*I)*m - 7*b*n)/(b*n), -(E^((2*I)*a)*(c*x^n)^((2*I)*b))]*Sec[a + b*Log[c*x^n]]^(3/2))/(2 + 2*m + (
3*I)*b*n)

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 4603

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 4605

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 {3}{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 ) \text {Subst}\left (\int x^{-1+\frac {1+m}{n}} \sec ^{\frac {3}{2}}(a+b \log (x)) \, dx,x,c x^n\right )}{n}\\ &=\frac {\left (x^{1+m} \left (c x^n\right )^{-\frac {3 i b}{2}-\frac {1+m}{n}} \left (1+e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{3/2} \sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )\right ) \text {Subst}\left (\int \frac {x^{-1+\frac {3 i b}{2}+\frac {1+m}{n}}}{\left (1+e^{2 i a} x^{2 i b}\right )^{3/2}} \, 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 )^{3/2} \, _2F_1\left (\frac {3}{2},\frac {1}{4} \left (3-\frac {2 i (1+m)}{b n}\right );-\frac {2 i+2 i m-7 b n}{4 b n};-e^{2 i a} \left (c x^n\right )^{2 i b}\right ) \sec ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}{2+2 m+3 i b n}\\ \end {align*}

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(470\) vs. \(2(130)=260\).
time = 10.10, size = 470, normalized size = 3.62 \begin {gather*} \frac {\sqrt {2} x^{1+m-i b n} \left (-\left (\left (4+8 m+4 m^2+b^2 n^2\right ) x^{2 i b n} \sqrt {\frac {e^{i a} \left (c x^n\right )^{i b}}{1+e^{2 i a} \left (c x^n\right )^{2 i b}}} \sqrt {1+e^{2 i a} \left (c x^n\right )^{2 i b}} \, _2F_1\left (\frac {1}{2},-\frac {2 i+2 i m-3 b n}{4 b n};-\frac {2 i+2 i m-7 b n}{4 b n};-e^{2 i a} \left (c x^n\right )^{2 i b}\right )\right )+(2+2 m+3 i b n) \left ((2+2 m+i b n) \sqrt {\frac {e^{i a} \left (c x^n\right )^{i b}}{1+e^{2 i a} \left (c x^n\right )^{2 i b}}} \sqrt {1+e^{2 i a} \left (c x^n\right )^{2 i b}} \, _2F_1\left (\frac {1}{2},-\frac {2 i+2 i m+b n}{4 b n};-\frac {2 i+2 i m-3 b n}{4 b n};-e^{2 i a} \left (c x^n\right )^{2 i b}\right )-i \sqrt {2} x^{i b n} \sqrt {\sec \left (a+b \log \left (c x^n\right )\right )} (b n \cos (b n \log (x))-2 (1+m) \sin (b n \log (x)))\right )\right )}{b n (-2 i-2 i m+3 b n) \left (-2 (1+m) \cos \left (a-b n \log (x)+b \log \left (c x^n\right )\right )+b n \sin \left (a-b n \log (x)+b \log \left (c x^n\right )\right )\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[x^m*Sec[a + b*Log[c*x^n]]^(3/2),x]

[Out]

(Sqrt[2]*x^(1 + m - I*b*n)*(-((4 + 8*m + 4*m^2 + b^2*n^2)*x^((2*I)*b*n)*Sqrt[(E^(I*a)*(c*x^n)^(I*b))/(1 + E^((
2*I)*a)*(c*x^n)^((2*I)*b))]*Sqrt[1 + E^((2*I)*a)*(c*x^n)^((2*I)*b)]*Hypergeometric2F1[1/2, -1/4*(2*I + (2*I)*m
- 3*b*n)/(b*n), -1/4*(2*I + (2*I)*m - 7*b*n)/(b*n), -(E^((2*I)*a)*(c*x^n)^((2*I)*b))]) + (2 + 2*m + (3*I)*b*n
)*((2 + 2*m + I*b*n)*Sqrt[(E^(I*a)*(c*x^n)^(I*b))/(1 + E^((2*I)*a)*(c*x^n)^((2*I)*b))]*Sqrt[1 + E^((2*I)*a)*(c
*x^n)^((2*I)*b)]*Hypergeometric2F1[1/2, -1/4*(2*I + (2*I)*m + b*n)/(b*n), -1/4*(2*I + (2*I)*m - 3*b*n)/(b*n),
-(E^((2*I)*a)*(c*x^n)^((2*I)*b))] - I*Sqrt[2]*x^(I*b*n)*Sqrt[Sec[a + b*Log[c*x^n]]]*(b*n*Cos[b*n*Log[x]] - 2*(
1 + m)*Sin[b*n*Log[x]]))))/(b*n*(-2*I - (2*I)*m + 3*b*n)*(-2*(1 + m)*Cos[a - b*n*Log[x] + b*Log[c*x^n]] + b*n*
Sin[a - b*n*Log[x] + b*Log[c*x^n]]))

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Maple [F]
time = 0.08, size = 0, normalized size = 0.00 \[\int x^{m} \left (\sec ^{\frac {3}{2}}\left (a +b \ln \left (c \,x^{n}\right )\right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*sec(a+b*ln(c*x^n))^(3/2),x)

[Out]

int(x^m*sec(a+b*ln(c*x^n))^(3/2),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(x^m*sec(a+b*log(c*x^n))^(3/2),x, algorithm="maxima")

[Out]

integrate(x^m*sec(b*log(c*x^n) + a)^(3/2), x)

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*sec(a+b*log(c*x^n))^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m*sec(a+b*ln(c*x**n))**(3/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 6437 deep

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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(x^m*sec(a+b*log(c*x^n))^(3/2),x, algorithm="giac")

[Out]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^m\,{\left (\frac {1}{\cos \left (a+b\,\ln \left (c\,x^n\right )\right )}\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*(1/cos(a + b*log(c*x^n)))^(3/2),x)

[Out]

int(x^m*(1/cos(a + b*log(c*x^n)))^(3/2), x)

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