3.1.40 \(\int x^m \sin ^3(a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log (c x^n)) \, dx\) [40]

Optimal. Leaf size=226 \[ -\frac {4 \sqrt {-\frac {(1+m)^2}{n^2}} n x^{1+m} \cos \left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right )}{5 (1+m)^2}+\frac {8 x^{1+m} \sin \left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right )}{5 (1+m)}+\frac {6 \sqrt {-\frac {(1+m)^2}{n^2}} n x^{1+m} \cos \left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \sin ^2\left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right )}{5 (1+m)^2}-\frac {4 x^{1+m} \sin ^3\left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right )}{5 (1+m)} \]

[Out]

8/5*x^(1+m)*sin(a+1/2*ln(c*x^n)*(-(1+m)^2/n^2)^(1/2))/(1+m)-4/5*x^(1+m)*sin(a+1/2*ln(c*x^n)*(-(1+m)^2/n^2)^(1/
2))^3/(1+m)-4/5*n*x^(1+m)*cos(a+1/2*ln(c*x^n)*(-(1+m)^2/n^2)^(1/2))*(-(1+m)^2/n^2)^(1/2)/(1+m)^2+6/5*n*x^(1+m)
*cos(a+1/2*ln(c*x^n)*(-(1+m)^2/n^2)^(1/2))*sin(a+1/2*ln(c*x^n)*(-(1+m)^2/n^2)^(1/2))^2*(-(1+m)^2/n^2)^(1/2)/(1
+m)^2

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Rubi [A]
time = 0.05, antiderivative size = 226, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {4575, 4573} \begin {gather*} -\frac {4 x^{m+1} \sin ^3\left (a+\frac {1}{2} \sqrt {-\frac {(m+1)^2}{n^2}} \log \left (c x^n\right )\right )}{5 (m+1)}+\frac {8 x^{m+1} \sin \left (a+\frac {1}{2} \sqrt {-\frac {(m+1)^2}{n^2}} \log \left (c x^n\right )\right )}{5 (m+1)}-\frac {4 n \sqrt {-\frac {(m+1)^2}{n^2}} x^{m+1} \cos \left (a+\frac {1}{2} \sqrt {-\frac {(m+1)^2}{n^2}} \log \left (c x^n\right )\right )}{5 (m+1)^2}+\frac {6 n \sqrt {-\frac {(m+1)^2}{n^2}} x^{m+1} \sin ^2\left (a+\frac {1}{2} \sqrt {-\frac {(m+1)^2}{n^2}} \log \left (c x^n\right )\right ) \cos \left (a+\frac {1}{2} \sqrt {-\frac {(m+1)^2}{n^2}} \log \left (c x^n\right )\right )}{5 (m+1)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^m*Sin[a + (Sqrt[-((1 + m)^2/n^2)]*Log[c*x^n])/2]^3,x]

[Out]

(-4*Sqrt[-((1 + m)^2/n^2)]*n*x^(1 + m)*Cos[a + (Sqrt[-((1 + m)^2/n^2)]*Log[c*x^n])/2])/(5*(1 + m)^2) + (8*x^(1
+ m)*Sin[a + (Sqrt[-((1 + m)^2/n^2)]*Log[c*x^n])/2])/(5*(1 + m)) + (6*Sqrt[-((1 + m)^2/n^2)]*n*x^(1 + m)*Cos[
a + (Sqrt[-((1 + m)^2/n^2)]*Log[c*x^n])/2]*Sin[a + (Sqrt[-((1 + m)^2/n^2)]*Log[c*x^n])/2]^2)/(5*(1 + m)^2) - (
4*x^(1 + m)*Sin[a + (Sqrt[-((1 + m)^2/n^2)]*Log[c*x^n])/2]^3)/(5*(1 + m))

Rule 4573

Int[((e_.)*(x_))^(m_.)*Sin[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)], x_Symbol] :> Simp[(m + 1)*(e*x)^(m +
1)*(Sin[d*(a + b*Log[c*x^n])]/(b^2*d^2*e*n^2 + e*(m + 1)^2)), x] - Simp[b*d*n*(e*x)^(m + 1)*(Cos[d*(a + b*Log[
c*x^n])]/(b^2*d^2*e*n^2 + e*(m + 1)^2)), x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b^2*d^2*n^2 + (m + 1)^2,
0]

Rule 4575

Int[((e_.)*(x_))^(m_.)*Sin[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_), x_Symbol] :> Simp[(m + 1)*(e*x)^
(m + 1)*(Sin[d*(a + b*Log[c*x^n])]^p/(b^2*d^2*e*n^2*p^2 + e*(m + 1)^2)), x] + (Dist[b^2*d^2*n^2*p*((p - 1)/(b^
2*d^2*n^2*p^2 + (m + 1)^2)), Int[(e*x)^m*Sin[d*(a + b*Log[c*x^n])]^(p - 2), x], x] - Simp[b*d*n*p*(e*x)^(m + 1
)*Cos[d*(a + b*Log[c*x^n])]*(Sin[d*(a + b*Log[c*x^n])]^(p - 1)/(b^2*d^2*e*n^2*p^2 + e*(m + 1)^2)), x]) /; Free
Q[{a, b, c, d, e, m, n}, x] && IGtQ[p, 1] && NeQ[b^2*d^2*n^2*p^2 + (m + 1)^2, 0]

Rubi steps

\begin {align*} \int x^m \sin ^3\left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx &=\frac {6 \sqrt {-\frac {(1+m)^2}{n^2}} n x^{1+m} \cos \left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \sin ^2\left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right )}{5 (1+m)^2}-\frac {4 x^{1+m} \sin ^3\left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right )}{5 (1+m)}+\frac {6}{5} \int x^m \sin \left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx\\ &=-\frac {4 \sqrt {-\frac {(1+m)^2}{n^2}} n x^{1+m} \cos \left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right )}{5 (1+m)^2}+\frac {8 x^{1+m} \sin \left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right )}{5 (1+m)}+\frac {6 \sqrt {-\frac {(1+m)^2}{n^2}} n x^{1+m} \cos \left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \sin ^2\left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right )}{5 (1+m)^2}-\frac {4 x^{1+m} \sin ^3\left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right )}{5 (1+m)}\\ \end {align*}

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Mathematica [A]
time = 1.39, size = 169, normalized size = 0.75 \begin {gather*} \frac {x^{1+m} \left (-5 \sqrt {-\frac {(1+m)^2}{n^2}} n \cos \left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right )-3 \sqrt {-\frac {(1+m)^2}{n^2}} n \cos \left (3 a+\frac {3}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right )+2 (1+m) \left (5 \sin \left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right )+\sin \left (3 a+\frac {3}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right )\right )\right )}{10 (1+m)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^m*Sin[a + (Sqrt[-((1 + m)^2/n^2)]*Log[c*x^n])/2]^3,x]

[Out]

(x^(1 + m)*(-5*Sqrt[-((1 + m)^2/n^2)]*n*Cos[a + (Sqrt[-((1 + m)^2/n^2)]*Log[c*x^n])/2] - 3*Sqrt[-((1 + m)^2/n^
2)]*n*Cos[3*a + (3*Sqrt[-((1 + m)^2/n^2)]*Log[c*x^n])/2] + 2*(1 + m)*(5*Sin[a + (Sqrt[-((1 + m)^2/n^2)]*Log[c*
x^n])/2] + Sin[3*a + (3*Sqrt[-((1 + m)^2/n^2)]*Log[c*x^n])/2])))/(10*(1 + m)^2)

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Maple [F]
time = 0.07, size = 0, normalized size = 0.00 \[\int x^{m} \left (\sin ^{3}\left (a +\frac {\ln \left (c \,x^{n}\right ) \sqrt {-\frac {\left (1+m \right )^{2}}{n^{2}}}}{2}\right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*sin(a+1/2*ln(c*x^n)*(-(1+m)^2/n^2)^(1/2))^3,x)

[Out]

int(x^m*sin(a+1/2*ln(c*x^n)*(-(1+m)^2/n^2)^(1/2))^3,x)

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Maxima [A]
time = 0.32, size = 195, normalized size = 0.86 \begin {gather*} -\frac {{\left (c^{\frac {3 \, m}{n} + \frac {3}{n}} x e^{\left (m \log \left (x\right ) + \frac {3 \, m \log \left (x^{n}\right )}{n} + \frac {3 \, \log \left (x^{n}\right )}{n}\right )} \sin \left (3 \, a\right ) - 5 \, c^{\frac {2 \, m}{n} + \frac {2}{n}} x e^{\left (m \log \left (x\right ) + \frac {2 \, m \log \left (x^{n}\right )}{n} + \frac {2 \, \log \left (x^{n}\right )}{n}\right )} \sin \left (a\right ) - 15 \, c^{\frac {m}{n} + \frac {1}{n}} x e^{\left (m \log \left (x\right ) + \frac {m \log \left (x^{n}\right )}{n} + \frac {\log \left (x^{n}\right )}{n}\right )} \sin \left (a\right ) - 5 \, x x^{m} \sin \left (3 \, a\right )\right )} e^{\left (-\frac {3 \, m \log \left (x^{n}\right )}{2 \, n} - \frac {3 \, \log \left (x^{n}\right )}{2 \, n}\right )}}{20 \, {\left (c^{\frac {3 \, m}{2 \, n} + \frac {3}{2 \, n}} m + c^{\frac {3 \, m}{2 \, n} + \frac {3}{2 \, n}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*sin(a+1/2*log(c*x^n)*(-(1+m)^2/n^2)^(1/2))^3,x, algorithm="maxima")

[Out]

-1/20*(c^(3*m/n + 3/n)*x*e^(m*log(x) + 3*m*log(x^n)/n + 3*log(x^n)/n)*sin(3*a) - 5*c^(2*m/n + 2/n)*x*e^(m*log(
x) + 2*m*log(x^n)/n + 2*log(x^n)/n)*sin(a) - 15*c^(m/n + 1/n)*x*e^(m*log(x) + m*log(x^n)/n + log(x^n)/n)*sin(a
) - 5*x*x^m*sin(3*a))*e^(-3/2*m*log(x^n)/n - 3/2*log(x^n)/n)/(c^(3/2*m/n + 3/2/n)*m + c^(3/2*m/n + 3/2/n))

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Fricas [C] Result contains complex when optimal does not.
time = 2.35, size = 128, normalized size = 0.57 \begin {gather*} \frac {{\left (5 i \, e^{\left (-\frac {{\left (m + 1\right )} n \log \left (x\right ) - 2 i \, a n + {\left (m + 1\right )} \log \left (c\right )}{n}\right )} - 15 i \, e^{\left (-\frac {2 \, {\left ({\left (m + 1\right )} n \log \left (x\right ) - 2 i \, a n + {\left (m + 1\right )} \log \left (c\right )\right )}}{n}\right )} - 5 i \, e^{\left (-\frac {3 \, {\left ({\left (m + 1\right )} n \log \left (x\right ) - 2 i \, a n + {\left (m + 1\right )} \log \left (c\right )\right )}}{n}\right )} - i\right )} e^{\left (\frac {5 \, {\left ({\left (m + 1\right )} n \log \left (x\right ) - 2 i \, a n + {\left (m + 1\right )} \log \left (c\right )\right )}}{2 \, n} + \frac {2 i \, a n - {\left (m + 1\right )} \log \left (c\right )}{n}\right )}}{20 \, {\left (m + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*sin(a+1/2*log(c*x^n)*(-(1+m)^2/n^2)^(1/2))^3,x, algorithm="fricas")

[Out]

1/20*(5*I*e^(-((m + 1)*n*log(x) - 2*I*a*n + (m + 1)*log(c))/n) - 15*I*e^(-2*((m + 1)*n*log(x) - 2*I*a*n + (m +
1)*log(c))/n) - 5*I*e^(-3*((m + 1)*n*log(x) - 2*I*a*n + (m + 1)*log(c))/n) - I)*e^(5/2*((m + 1)*n*log(x) - 2*
I*a*n + (m + 1)*log(c))/n + (2*I*a*n - (m + 1)*log(c))/n)/(m + 1)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{m} \sin ^{3}{\left (a + \frac {\sqrt {- \frac {m^{2}}{n^{2}} - \frac {2 m}{n^{2}} - \frac {1}{n^{2}}} \log {\left (c x^{n} \right )}}{2} \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m*sin(a+1/2*ln(c*x**n)*(-(1+m)**2/n**2)**(1/2))**3,x)

[Out]

Integral(x**m*sin(a + sqrt(-m**2/n**2 - 2*m/n**2 - 1/n**2)*log(c*x**n)/2)**3, x)

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Giac [C] Result contains complex when optimal does not.
time = 5.78, size = 1870, normalized size = 8.27 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*sin(a+1/2*log(c*x^n)*(-(1+m)^2/n^2)^(1/2))^3,x, algorithm="giac")

[Out]

1/4*(8*I*m^3*n^4*x*x^m*e^(3*I*a - 3/2*(n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c))/n^2) - 24*I*m^3*n^4*x*x^m*
e^(I*a - 1/2*(n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c))/n^2) + 24*I*m^3*n^4*x*x^m*e^(-I*a + 1/2*(n*abs(m*n
+ n)*log(x) + abs(m*n + n)*log(c))/n^2) - 8*I*m^3*n^4*x*x^m*e^(-3*I*a + 3/2*(n*abs(m*n + n)*log(x) + abs(m*n +
n)*log(c))/n^2) + 24*I*m^2*n^4*x*x^m*e^(3*I*a - 3/2*(n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c))/n^2) + 12*I
*m^2*n^3*x*x^m*abs(m*n + n)*e^(3*I*a - 3/2*(n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c))/n^2) - 72*I*m^2*n^4*x
*x^m*e^(I*a - 1/2*(n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c))/n^2) - 12*I*m^2*n^3*x*x^m*abs(m*n + n)*e^(I*a
- 1/2*(n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c))/n^2) + 72*I*m^2*n^4*x*x^m*e^(-I*a + 1/2*(n*abs(m*n + n)*lo
g(x) + abs(m*n + n)*log(c))/n^2) - 12*I*m^2*n^3*x*x^m*abs(m*n + n)*e^(-I*a + 1/2*(n*abs(m*n + n)*log(x) + abs(
m*n + n)*log(c))/n^2) - 24*I*m^2*n^4*x*x^m*e^(-3*I*a + 3/2*(n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c))/n^2)
+ 12*I*m^2*n^3*x*x^m*abs(m*n + n)*e^(-3*I*a + 3/2*(n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c))/n^2) - 2*I*(m*
n + n)^2*m*n^2*x*x^m*e^(3*I*a - 3/2*(n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c))/n^2) + 24*I*m*n^4*x*x^m*e^(3
*I*a - 3/2*(n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c))/n^2) + 24*I*m*n^3*x*x^m*abs(m*n + n)*e^(3*I*a - 3/2*(
n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c))/n^2) + 54*I*(m*n + n)^2*m*n^2*x*x^m*e^(I*a - 1/2*(n*abs(m*n + n)*
log(x) + abs(m*n + n)*log(c))/n^2) - 72*I*m*n^4*x*x^m*e^(I*a - 1/2*(n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c
))/n^2) - 24*I*m*n^3*x*x^m*abs(m*n + n)*e^(I*a - 1/2*(n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c))/n^2) - 54*I
*(m*n + n)^2*m*n^2*x*x^m*e^(-I*a + 1/2*(n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c))/n^2) + 72*I*m*n^4*x*x^m*e
^(-I*a + 1/2*(n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c))/n^2) - 24*I*m*n^3*x*x^m*abs(m*n + n)*e^(-I*a + 1/2*
(n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c))/n^2) + 2*I*(m*n + n)^2*m*n^2*x*x^m*e^(-3*I*a + 3/2*(n*abs(m*n +
n)*log(x) + abs(m*n + n)*log(c))/n^2) - 24*I*m*n^4*x*x^m*e^(-3*I*a + 3/2*(n*abs(m*n + n)*log(x) + abs(m*n + n)
*log(c))/n^2) + 24*I*m*n^3*x*x^m*abs(m*n + n)*e^(-3*I*a + 3/2*(n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c))/n^
2) - 2*I*(m*n + n)^2*n^2*x*x^m*e^(3*I*a - 3/2*(n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c))/n^2) + 8*I*n^4*x*x
^m*e^(3*I*a - 3/2*(n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c))/n^2) - 3*I*(m*n + n)^2*n*x*x^m*abs(m*n + n)*e^
(3*I*a - 3/2*(n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c))/n^2) + 12*I*n^3*x*x^m*abs(m*n + n)*e^(3*I*a - 3/2*(
n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c))/n^2) + 54*I*(m*n + n)^2*n^2*x*x^m*e^(I*a - 1/2*(n*abs(m*n + n)*lo
g(x) + abs(m*n + n)*log(c))/n^2) - 24*I*n^4*x*x^m*e^(I*a - 1/2*(n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c))/n
^2) + 27*I*(m*n + n)^2*n*x*x^m*abs(m*n + n)*e^(I*a - 1/2*(n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c))/n^2) -
12*I*n^3*x*x^m*abs(m*n + n)*e^(I*a - 1/2*(n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c))/n^2) - 54*I*(m*n + n)^2
*n^2*x*x^m*e^(-I*a + 1/2*(n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c))/n^2) + 24*I*n^4*x*x^m*e^(-I*a + 1/2*(n*
abs(m*n + n)*log(x) + abs(m*n + n)*log(c))/n^2) + 27*I*(m*n + n)^2*n*x*x^m*abs(m*n + n)*e^(-I*a + 1/2*(n*abs(m
*n + n)*log(x) + abs(m*n + n)*log(c))/n^2) - 12*I*n^3*x*x^m*abs(m*n + n)*e^(-I*a + 1/2*(n*abs(m*n + n)*log(x)
+ abs(m*n + n)*log(c))/n^2) + 2*I*(m*n + n)^2*n^2*x*x^m*e^(-3*I*a + 3/2*(n*abs(m*n + n)*log(x) + abs(m*n + n)*
log(c))/n^2) - 8*I*n^4*x*x^m*e^(-3*I*a + 3/2*(n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c))/n^2) - 3*I*(m*n + n
)^2*n*x*x^m*abs(m*n + n)*e^(-3*I*a + 3/2*(n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c))/n^2) + 12*I*n^3*x*x^m*a
bs(m*n + n)*e^(-3*I*a + 3/2*(n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c))/n^2))/(16*m^4*n^4 + 64*m^3*n^4 - 40*
(m*n + n)^2*m^2*n^2 + 96*m^2*n^4 - 80*(m*n + n)^2*m*n^2 + 64*m*n^4 + 9*(m*n + n)^4 - 40*(m*n + n)^2*n^2 + 16*n
^4)

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Mupad [B]
time = 4.71, size = 297, normalized size = 1.31 \begin {gather*} -\frac {x\,x^m\,{\mathrm {e}}^{-a\,1{}\mathrm {i}}\,\frac {1}{{\left (c\,x^n\right )}^{\frac {\sqrt {-\frac {2\,m}{n^2}-\frac {1}{n^2}-\frac {m^2}{n^2}}\,1{}\mathrm {i}}{2}}}\,\left (2\,m+2+n\,\sqrt {-\frac {{\left (m+1\right )}^2}{n^2}}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4\,{\left (m\,1{}\mathrm {i}+1{}\mathrm {i}\right )}^2}+\frac {x\,x^m\,{\mathrm {e}}^{a\,1{}\mathrm {i}}\,{\left (c\,x^n\right )}^{\frac {\sqrt {-\frac {2\,m}{n^2}-\frac {1}{n^2}-\frac {m^2}{n^2}}\,1{}\mathrm {i}}{2}}\,\left (2\,m+2-n\,\sqrt {-\frac {{\left (m+1\right )}^2}{n^2}}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4\,{\left (m\,1{}\mathrm {i}+1{}\mathrm {i}\right )}^2}-\frac {x\,x^m\,{\mathrm {e}}^{-a\,3{}\mathrm {i}}\,\frac {1}{{\left (c\,x^n\right )}^{\frac {\sqrt {-\frac {2\,m}{n^2}-\frac {1}{n^2}-\frac {m^2}{n^2}}\,3{}\mathrm {i}}{2}}}\,\left (2\,m+2+n\,\sqrt {-\frac {{\left (m+1\right )}^2}{n^2}}\,3{}\mathrm {i}\right )\,1{}\mathrm {i}}{20\,{\left (m\,1{}\mathrm {i}+1{}\mathrm {i}\right )}^2}+\frac {x\,x^m\,{\mathrm {e}}^{a\,3{}\mathrm {i}}\,{\left (c\,x^n\right )}^{\frac {\sqrt {-\frac {2\,m}{n^2}-\frac {1}{n^2}-\frac {m^2}{n^2}}\,3{}\mathrm {i}}{2}}\,\left (2\,m+2-n\,\sqrt {-\frac {{\left (m+1\right )}^2}{n^2}}\,3{}\mathrm {i}\right )\,1{}\mathrm {i}}{20\,{\left (m\,1{}\mathrm {i}+1{}\mathrm {i}\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x*x^m*exp(a*1i)*(c*x^n)^(((- (2*m)/n^2 - 1/n^2 - m^2/n^2)^(1/2)*1i)/2)*(2*m - n*(-(m + 1)^2/n^2)^(1/2)*1i + 2
)*1i)/(4*(m*1i + 1i)^2) - (x*x^m*exp(-a*1i)/(c*x^n)^(((- (2*m)/n^2 - 1/n^2 - m^2/n^2)^(1/2)*1i)/2)*(2*m + n*(-
(m + 1)^2/n^2)^(1/2)*1i + 2)*1i)/(4*(m*1i + 1i)^2) - (x*x^m*exp(-a*3i)/(c*x^n)^(((- (2*m)/n^2 - 1/n^2 - m^2/n^
2)^(1/2)*3i)/2)*(2*m + n*(-(m + 1)^2/n^2)^(1/2)*3i + 2)*1i)/(20*(m*1i + 1i)^2) + (x*x^m*exp(a*3i)*(c*x^n)^(((-
(2*m)/n^2 - 1/n^2 - m^2/n^2)^(1/2)*3i)/2)*(2*m - n*(-(m + 1)^2/n^2)^(1/2)*3i + 2)*1i)/(20*(m*1i + 1i)^2)

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