3.51 \(\int x^m \sin ^3(a+\frac {1}{6} \sqrt {-(1+m)^2} \log (c x^2)) \, dx\)

Optimal. Leaf size=218 \[ \frac {9 e^{\frac {a \sqrt {-(m+1)^2}}{m+1}} x^{m+1} \left (c x^2\right )^{\frac {1}{6} (-m-1)}}{16 \sqrt {-(m+1)^2}}-\frac {9 e^{\frac {a (m+1)}{\sqrt {-(m+1)^2}}} x^{m+1} \left (c x^2\right )^{\frac {m+1}{6}}}{32 \sqrt {-(m+1)^2}}+\frac {e^{\frac {3 a (m+1)}{\sqrt {-(m+1)^2}}} x^{m+1} \left (c x^2\right )^{\frac {m+1}{2}}}{16 \sqrt {-(m+1)^2}}-\frac {(m+1) e^{-\frac {3 a (m+1)}{\sqrt {-(m+1)^2}}} x^{m+1} \log (x) \left (c x^2\right )^{\frac {1}{2} (-m-1)}}{8 \sqrt {-(m+1)^2}} \]

[Out]

9/16*exp(a*(-(1+m)^2)^(1/2)/(1+m))*x^(1+m)*(c*x^2)^(-1/6-1/6*m)/(-(1+m)^2)^(1/2)-9/32*exp(a*(1+m)/(-(1+m)^2)^(
1/2))*x^(1+m)*(c*x^2)^(1/6+1/6*m)/(-(1+m)^2)^(1/2)+1/16*exp(3*a*(1+m)/(-(1+m)^2)^(1/2))*x^(1+m)*(c*x^2)^(1/2+1
/2*m)/(-(1+m)^2)^(1/2)-1/8*(1+m)*x^(1+m)*(c*x^2)^(-1/2-1/2*m)*ln(x)/exp(3*a*(1+m)/(-(1+m)^2)^(1/2))/(-(1+m)^2)
^(1/2)

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Rubi [A]  time = 0.30, antiderivative size = 218, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {4493, 4489} \[ \frac {9 e^{\frac {a \sqrt {-(m+1)^2}}{m+1}} x^{m+1} \left (c x^2\right )^{\frac {1}{6} (-m-1)}}{16 \sqrt {-(m+1)^2}}-\frac {9 e^{\frac {a (m+1)}{\sqrt {-(m+1)^2}}} x^{m+1} \left (c x^2\right )^{\frac {m+1}{6}}}{32 \sqrt {-(m+1)^2}}+\frac {e^{\frac {3 a (m+1)}{\sqrt {-(m+1)^2}}} x^{m+1} \left (c x^2\right )^{\frac {m+1}{2}}}{16 \sqrt {-(m+1)^2}}-\frac {(m+1) e^{-\frac {3 a (m+1)}{\sqrt {-(m+1)^2}}} x^{m+1} \log (x) \left (c x^2\right )^{\frac {1}{2} (-m-1)}}{8 \sqrt {-(m+1)^2}} \]

Antiderivative was successfully verified.

[In]

Int[x^m*Sin[a + (Sqrt[-(1 + m)^2]*Log[c*x^2])/6]^3,x]

[Out]

(9*E^((a*Sqrt[-(1 + m)^2])/(1 + m))*x^(1 + m)*(c*x^2)^((-1 - m)/6))/(16*Sqrt[-(1 + m)^2]) - (9*E^((a*(1 + m))/
Sqrt[-(1 + m)^2])*x^(1 + m)*(c*x^2)^((1 + m)/6))/(32*Sqrt[-(1 + m)^2]) + (E^((3*a*(1 + m))/Sqrt[-(1 + m)^2])*x
^(1 + m)*(c*x^2)^((1 + m)/2))/(16*Sqrt[-(1 + m)^2]) - ((1 + m)*x^(1 + m)*(c*x^2)^((-1 - m)/2)*Log[x])/(8*E^((3
*a*(1 + m))/Sqrt[-(1 + m)^2])*Sqrt[-(1 + m)^2])

Rule 4489

Int[((e_.)*(x_))^(m_.)*Sin[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Dist[(m + 1)^p/(2^p*b^p*d^p*p^p)
, Int[ExpandIntegrand[(e*x)^m*(E^((a*b*d^2*p)/(m + 1))/x^((m + 1)/p) - x^((m + 1)/p)/E^((a*b*d^2*p)/(m + 1)))^
p, x], x], x] /; FreeQ[{a, b, d, e, m}, x] && IGtQ[p, 0] && EqQ[b^2*d^2*p^2 + (m + 1)^2, 0]

Rule 4493

Int[((e_.)*(x_))^(m_.)*Sin[((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)*Sin[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 \sin ^3\left (a+\frac {1}{6} \sqrt {-(1+m)^2} \log \left (c x^2\right )\right ) \, dx &=\frac {1}{2} \left (x^{1+m} \left (c x^2\right )^{\frac {1}{2} (-1-m)}\right ) \operatorname {Subst}\left (\int x^{-1+\frac {1+m}{2}} \sin ^3\left (a+\frac {1}{6} \sqrt {-(1+m)^2} \log (x)\right ) \, dx,x,c x^2\right )\\ &=\frac {\left (\sqrt {-(1+m)^2} x^{1+m} \left (c x^2\right )^{\frac {1}{2} (-1-m)}\right ) \operatorname {Subst}\left (\int \left (\frac {e^{-\frac {3 a (1+m)}{\sqrt {-(1+m)^2}}}}{x}-3 e^{\frac {a \sqrt {-(1+m)^2}}{1+m}} x^{\frac {1}{3} (-2+m)}-e^{\frac {3 a (1+m)}{\sqrt {-(1+m)^2}}} x^m+3 e^{\frac {a (1+m)}{\sqrt {-(1+m)^2}}} x^{\frac {1}{3} (-1+2 m)}\right ) \, dx,x,c x^2\right )}{16 (1+m)}\\ &=\frac {9 e^{\frac {a \sqrt {-(1+m)^2}}{1+m}} x^{1+m} \left (c x^2\right )^{\frac {1}{6} (-1-m)}}{16 \sqrt {-(1+m)^2}}-\frac {9 e^{\frac {a (1+m)}{\sqrt {-(1+m)^2}}} x^{1+m} \left (c x^2\right )^{\frac {1+m}{6}}}{32 \sqrt {-(1+m)^2}}+\frac {e^{\frac {3 a (1+m)}{\sqrt {-(1+m)^2}}} x^{1+m} \left (c x^2\right )^{\frac {1+m}{2}}}{16 \sqrt {-(1+m)^2}}+\frac {e^{-\frac {3 a (1+m)}{\sqrt {-(1+m)^2}}} \sqrt {-(1+m)^2} x^{1+m} \left (c x^2\right )^{\frac {1}{2} (-1-m)} \log (x)}{8 (1+m)}\\ \end {align*}

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Mathematica [F]  time = 0.52, size = 0, normalized size = 0.00 \[ \int x^m \sin ^3\left (a+\frac {1}{6} \sqrt {-(1+m)^2} \log \left (c x^2\right )\right ) \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[x^m*Sin[a + (Sqrt[-(1 + m)^2]*Log[c*x^2])/6]^3,x]

[Out]

Integrate[x^m*Sin[a + (Sqrt[-(1 + m)^2]*Log[c*x^2])/6]^3, x]

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fricas [C]  time = 0.43, size = 97, normalized size = 0.44 \[ \frac {{\left ({\left (4 i \, m + 4 i\right )} e^{\left (-{\left (m + 1\right )} \log \relax (c) - 2 \, {\left (m + 1\right )} \log \relax (x) + 6 i \, a\right )} \log \relax (x) + 9 i \, e^{\left (-\frac {1}{3} \, {\left (m + 1\right )} \log \relax (c) - \frac {2}{3} \, {\left (m + 1\right )} \log \relax (x) + 2 i \, a\right )} - 18 i \, e^{\left (-\frac {2}{3} \, {\left (m + 1\right )} \log \relax (c) - \frac {4}{3} \, {\left (m + 1\right )} \log \relax (x) + 4 i \, a\right )} - 2 i\right )} e^{\left (\frac {1}{2} \, {\left (m + 1\right )} \log \relax (c) + 2 \, {\left (m + 1\right )} \log \relax (x) - 3 i \, a\right )}}{32 \, {\left (m + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*sin(a+1/6*log(c*x^2)*(-(1+m)^2)^(1/2))^3,x, algorithm="fricas")

[Out]

1/32*((4*I*m + 4*I)*e^(-(m + 1)*log(c) - 2*(m + 1)*log(x) + 6*I*a)*log(x) + 9*I*e^(-1/3*(m + 1)*log(c) - 2/3*(
m + 1)*log(x) + 2*I*a) - 18*I*e^(-2/3*(m + 1)*log(c) - 4/3*(m + 1)*log(x) + 4*I*a) - 2*I)*e^(1/2*(m + 1)*log(c
) + 2*(m + 1)*log(x) - 3*I*a)/(m + 1)

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giac [C]  time = 3.96, size = 1297, normalized size = 5.95 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*sin(a+1/6*log(c*x^2)*(-(1+m)^2)^(1/2))^3,x, algorithm="giac")

[Out]

1/8*(I*(m + 1)^2*m*x*x^m*e^(1/2*abs(m + 1)*log(c) + abs(m + 1)*log(x) - 3*I*a) - 9*I*m^3*x*x^m*e^(1/2*abs(m +
1)*log(c) + abs(m + 1)*log(x) - 3*I*a) - I*(m + 1)^2*x*x^m*abs(m + 1)*e^(1/2*abs(m + 1)*log(c) + abs(m + 1)*lo
g(x) - 3*I*a) + 9*I*m^2*x*x^m*abs(m + 1)*e^(1/2*abs(m + 1)*log(c) + abs(m + 1)*log(x) - 3*I*a) - 27*I*(m + 1)^
2*m*x*x^m*e^(1/6*abs(m + 1)*log(c) + 1/3*abs(m + 1)*log(x) - I*a) + 27*I*m^3*x*x^m*e^(1/6*abs(m + 1)*log(c) +
1/3*abs(m + 1)*log(x) - I*a) + 9*I*(m + 1)^2*x*x^m*abs(m + 1)*e^(1/6*abs(m + 1)*log(c) + 1/3*abs(m + 1)*log(x)
- I*a) - 9*I*m^2*x*x^m*abs(m + 1)*e^(1/6*abs(m + 1)*log(c) + 1/3*abs(m + 1)*log(x) - I*a) + 27*I*(m + 1)^2*m*
x*x^m*e^(-1/6*abs(m + 1)*log(c) - 1/3*abs(m + 1)*log(x) + I*a) - 27*I*m^3*x*x^m*e^(-1/6*abs(m + 1)*log(c) - 1/
3*abs(m + 1)*log(x) + I*a) + 9*I*(m + 1)^2*x*x^m*abs(m + 1)*e^(-1/6*abs(m + 1)*log(c) - 1/3*abs(m + 1)*log(x)
+ I*a) - 9*I*m^2*x*x^m*abs(m + 1)*e^(-1/6*abs(m + 1)*log(c) - 1/3*abs(m + 1)*log(x) + I*a) - I*(m + 1)^2*m*x*x
^m*e^(-1/2*abs(m + 1)*log(c) - abs(m + 1)*log(x) + 3*I*a) + 9*I*m^3*x*x^m*e^(-1/2*abs(m + 1)*log(c) - abs(m +
1)*log(x) + 3*I*a) - I*(m + 1)^2*x*x^m*abs(m + 1)*e^(-1/2*abs(m + 1)*log(c) - abs(m + 1)*log(x) + 3*I*a) + 9*I
*m^2*x*x^m*abs(m + 1)*e^(-1/2*abs(m + 1)*log(c) - abs(m + 1)*log(x) + 3*I*a) + I*(m + 1)^2*x*x^m*e^(1/2*abs(m
+ 1)*log(c) + abs(m + 1)*log(x) - 3*I*a) - 27*I*m^2*x*x^m*e^(1/2*abs(m + 1)*log(c) + abs(m + 1)*log(x) - 3*I*a
) + 18*I*m*x*x^m*abs(m + 1)*e^(1/2*abs(m + 1)*log(c) + abs(m + 1)*log(x) - 3*I*a) - 27*I*(m + 1)^2*x*x^m*e^(1/
6*abs(m + 1)*log(c) + 1/3*abs(m + 1)*log(x) - I*a) + 81*I*m^2*x*x^m*e^(1/6*abs(m + 1)*log(c) + 1/3*abs(m + 1)*
log(x) - I*a) - 18*I*m*x*x^m*abs(m + 1)*e^(1/6*abs(m + 1)*log(c) + 1/3*abs(m + 1)*log(x) - I*a) + 27*I*(m + 1)
^2*x*x^m*e^(-1/6*abs(m + 1)*log(c) - 1/3*abs(m + 1)*log(x) + I*a) - 81*I*m^2*x*x^m*e^(-1/6*abs(m + 1)*log(c) -
1/3*abs(m + 1)*log(x) + I*a) - 18*I*m*x*x^m*abs(m + 1)*e^(-1/6*abs(m + 1)*log(c) - 1/3*abs(m + 1)*log(x) + I*
a) - I*(m + 1)^2*x*x^m*e^(-1/2*abs(m + 1)*log(c) - abs(m + 1)*log(x) + 3*I*a) + 27*I*m^2*x*x^m*e^(-1/2*abs(m +
1)*log(c) - abs(m + 1)*log(x) + 3*I*a) + 18*I*m*x*x^m*abs(m + 1)*e^(-1/2*abs(m + 1)*log(c) - abs(m + 1)*log(x
) + 3*I*a) - 27*I*m*x*x^m*e^(1/2*abs(m + 1)*log(c) + abs(m + 1)*log(x) - 3*I*a) + 9*I*x*x^m*abs(m + 1)*e^(1/2*
abs(m + 1)*log(c) + abs(m + 1)*log(x) - 3*I*a) + 81*I*m*x*x^m*e^(1/6*abs(m + 1)*log(c) + 1/3*abs(m + 1)*log(x)
- I*a) - 9*I*x*x^m*abs(m + 1)*e^(1/6*abs(m + 1)*log(c) + 1/3*abs(m + 1)*log(x) - I*a) - 81*I*m*x*x^m*e^(-1/6*
abs(m + 1)*log(c) - 1/3*abs(m + 1)*log(x) + I*a) - 9*I*x*x^m*abs(m + 1)*e^(-1/6*abs(m + 1)*log(c) - 1/3*abs(m
+ 1)*log(x) + I*a) + 27*I*m*x*x^m*e^(-1/2*abs(m + 1)*log(c) - abs(m + 1)*log(x) + 3*I*a) + 9*I*x*x^m*abs(m + 1
)*e^(-1/2*abs(m + 1)*log(c) - abs(m + 1)*log(x) + 3*I*a) - 9*I*x*x^m*e^(1/2*abs(m + 1)*log(c) + abs(m + 1)*log
(x) - 3*I*a) + 27*I*x*x^m*e^(1/6*abs(m + 1)*log(c) + 1/3*abs(m + 1)*log(x) - I*a) - 27*I*x*x^m*e^(-1/6*abs(m +
1)*log(c) - 1/3*abs(m + 1)*log(x) + I*a) + 9*I*x*x^m*e^(-1/2*abs(m + 1)*log(c) - abs(m + 1)*log(x) + 3*I*a))/
((m + 1)^4 - 10*(m + 1)^2*m^2 + 9*m^4 - 20*(m + 1)^2*m + 36*m^3 - 10*(m + 1)^2 + 54*m^2 + 36*m + 9)

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[ \int x^{m} \left (\sin ^{3}\left (a +\frac {\ln \left (c \,x^{2}\right ) \sqrt {-\left (1+m \right )^{2}}}{6}\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*sin(a+1/6*ln(c*x^2)*(-(1+m)^2)^(1/2))^3,x)

[Out]

int(x^m*sin(a+1/6*ln(c*x^2)*(-(1+m)^2)^(1/2))^3,x)

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maxima [A]  time = 0.37, size = 206, normalized size = 0.94 \[ \frac {9 \, {\left (\cos \left (2 \, a\right ) \sin \left (3 \, a\right ) - \cos \left (3 \, a\right ) \sin \left (2 \, a\right )\right )} c^{\frac {5}{6} \, m + \frac {5}{6}} x^{\frac {5}{3}} x^{\frac {4}{3} \, m} + 18 \, {\left (\cos \left (3 \, a\right ) \sin \left (4 \, a\right ) - \cos \left (4 \, a\right ) \sin \left (3 \, a\right )\right )} c^{\frac {1}{2} \, m + \frac {1}{2}} x x^{\frac {2}{3} \, m} - 2 \, {\left (c^{\frac {7}{6} \, m + 1} x^{2} x^{2 \, m} \sin \left (3 \, a\right ) + 2 \, {\left ({\left (\cos \left (3 \, a\right )^{2} \sin \left (3 \, a\right ) + \sin \left (3 \, a\right )^{3}\right )} c^{\frac {1}{6} \, m} m + {\left (\cos \left (3 \, a\right )^{2} \sin \left (3 \, a\right ) + \sin \left (3 \, a\right )^{3}\right )} c^{\frac {1}{6} \, m}\right )} \log \relax (x)\right )} c^{\frac {1}{6}} x^{\frac {1}{3}}}{32 \, {\left ({\left (\cos \left (3 \, a\right )^{2} + \sin \left (3 \, a\right )^{2}\right )} c^{\frac {2}{3} \, m} m + {\left (\cos \left (3 \, a\right )^{2} + \sin \left (3 \, a\right )^{2}\right )} c^{\frac {2}{3} \, m}\right )} c^{\frac {2}{3}} x^{\frac {1}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*sin(a+1/6*log(c*x^2)*(-(1+m)^2)^(1/2))^3,x, algorithm="maxima")

[Out]

1/32*(9*(cos(2*a)*sin(3*a) - cos(3*a)*sin(2*a))*c^(5/6*m + 5/6)*x^(5/3)*x^(4/3*m) + 18*(cos(3*a)*sin(4*a) - co
s(4*a)*sin(3*a))*c^(1/2*m + 1/2)*x*x^(2/3*m) - 2*(c^(7/6*m + 1)*x^2*x^(2*m)*sin(3*a) + 2*((cos(3*a)^2*sin(3*a)
+ sin(3*a)^3)*c^(1/6*m)*m + (cos(3*a)^2*sin(3*a) + sin(3*a)^3)*c^(1/6*m))*log(x))*c^(1/6)*x^(1/3))/(((cos(3*a
)^2 + sin(3*a)^2)*c^(2/3*m)*m + (cos(3*a)^2 + sin(3*a)^2)*c^(2/3*m))*c^(2/3)*x^(1/3))

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mupad [B]  time = 4.08, size = 291, normalized size = 1.33 \[ -\frac {\frac {1}{c^{\frac {\sqrt {-m^2-2\,m-1}\,1{}\mathrm {i}}{2}}}\,x\,x^m\,{\mathrm {e}}^{-a\,3{}\mathrm {i}}\,\frac {1}{{\left (x^2\right )}^{\frac {\sqrt {-m^2-2\,m-1}\,1{}\mathrm {i}}{2}}}\,1{}\mathrm {i}}{8\,m+8-\sqrt {-{\left (m+1\right )}^2}\,8{}\mathrm {i}}+\frac {c^{\frac {\sqrt {-m^2-2\,m-1}\,1{}\mathrm {i}}{2}}\,x\,x^m\,{\mathrm {e}}^{a\,3{}\mathrm {i}}\,{\left (x^2\right )}^{\frac {\sqrt {-m^2-2\,m-1}\,1{}\mathrm {i}}{2}}\,1{}\mathrm {i}}{8\,m+8+\sqrt {-{\left (m+1\right )}^2}\,8{}\mathrm {i}}-\frac {\frac {1}{c^{\frac {\sqrt {-m^2-2\,m-1}\,1{}\mathrm {i}}{6}}}\,x\,x^m\,{\mathrm {e}}^{-a\,1{}\mathrm {i}}\,\frac {1}{{\left (x^2\right )}^{\frac {\sqrt {-m^2-2\,m-1}\,1{}\mathrm {i}}{6}}}\,\left (27\,m+27+\sqrt {-{\left (m+1\right )}^2}\,9{}\mathrm {i}\right )\,1{}\mathrm {i}}{64\,{\left (m\,1{}\mathrm {i}+1{}\mathrm {i}\right )}^2}+\frac {c^{\frac {\sqrt {-m^2-2\,m-1}\,1{}\mathrm {i}}{6}}\,x\,x^m\,{\mathrm {e}}^{a\,1{}\mathrm {i}}\,{\left (x^2\right )}^{\frac {\sqrt {-m^2-2\,m-1}\,1{}\mathrm {i}}{6}}\,\left (27\,m+27-\sqrt {-{\left (m+1\right )}^2}\,9{}\mathrm {i}\right )\,1{}\mathrm {i}}{64\,{\left (m\,1{}\mathrm {i}+1{}\mathrm {i}\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*sin(a + (log(c*x^2)*(-(m + 1)^2)^(1/2))/6)^3,x)

[Out]

(c^(((- 2*m - m^2 - 1)^(1/2)*1i)/2)*x*x^m*exp(a*3i)*(x^2)^(((- 2*m - m^2 - 1)^(1/2)*1i)/2)*1i)/(8*m + (-(m + 1
)^2)^(1/2)*8i + 8) - (1/c^(((- 2*m - m^2 - 1)^(1/2)*1i)/2)*x*x^m*exp(-a*3i)/(x^2)^(((- 2*m - m^2 - 1)^(1/2)*1i
)/2)*1i)/(8*m - (-(m + 1)^2)^(1/2)*8i + 8) - (1/c^(((- 2*m - m^2 - 1)^(1/2)*1i)/6)*x*x^m*exp(-a*1i)/(x^2)^(((-
2*m - m^2 - 1)^(1/2)*1i)/6)*(27*m + (-(m + 1)^2)^(1/2)*9i + 27)*1i)/(64*(m*1i + 1i)^2) + (c^(((- 2*m - m^2 -
1)^(1/2)*1i)/6)*x*x^m*exp(a*1i)*(x^2)^(((- 2*m - m^2 - 1)^(1/2)*1i)/6)*(27*m - (-(m + 1)^2)^(1/2)*9i + 27)*1i)
/(64*(m*1i + 1i)^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{m} \sin ^{3}{\left (a + \frac {\sqrt {- m^{2} - 2 m - 1} \log {\left (c x^{2} \right )}}{6} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m*sin(a+1/6*ln(c*x**2)*(-(1+m)**2)**(1/2))**3,x)

[Out]

Integral(x**m*sin(a + sqrt(-m**2 - 2*m - 1)*log(c*x**2)/6)**3, x)

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