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

Optimal. Leaf size=133 \[ \frac {(m+1) x^{m+1} \log (x) e^{\frac {a n \sqrt {-\frac {(m+1)^2}{n^2}}}{m+1}} \left (c x^n\right )^{-\frac {m+1}{n}}}{2 n \sqrt {-\frac {(m+1)^2}{n^2}}}-\frac {x^{m+1} e^{\frac {a (m+1)}{n \sqrt {-\frac {(m+1)^2}{n^2}}}} \left (c x^n\right )^{\frac {m+1}{n}}}{4 n \sqrt {-\frac {(m+1)^2}{n^2}}} \]

[Out]

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

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Rubi [A]  time = 0.28, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {4493, 4489} \[ \frac {(m+1) x^{m+1} \log (x) e^{\frac {a n \sqrt {-\frac {(m+1)^2}{n^2}}}{m+1}} \left (c x^n\right )^{-\frac {m+1}{n}}}{2 n \sqrt {-\frac {(m+1)^2}{n^2}}}-\frac {x^{m+1} e^{\frac {a (m+1)}{n \sqrt {-\frac {(m+1)^2}{n^2}}}} \left (c x^n\right )^{\frac {m+1}{n}}}{4 n \sqrt {-\frac {(m+1)^2}{n^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(E^((a*(1 + m))/(Sqrt[-((1 + m)^2/n^2)]*n))*x^(1 + m)*(c*x^n)^((1 + m)/n))/(4*Sqrt[-((1 + m)^2/n^2)]*n) + (E^
((a*Sqrt[-((1 + m)^2/n^2)]*n)/(1 + m))*(1 + m)*x^(1 + m)*Log[x])/(2*Sqrt[-((1 + m)^2/n^2)]*n*(c*x^n)^((1 + m)/
n))

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 \left (a+\sqrt {-\frac {(1+m)^2}{n^2}} \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}} \sin \left (a+\sqrt {-\frac {(1+m)^2}{n^2}} \log (x)\right ) \, dx,x,c x^n\right )}{n}\\ &=\frac {\left ((1+m) x^{1+m} \left (c x^n\right )^{-\frac {1+m}{n}}\right ) \operatorname {Subst}\left (\int \left (\frac {e^{\frac {a \sqrt {-\frac {(1+m)^2}{n^2}} n}{1+m}}}{x}-e^{\frac {a (1+m)}{\sqrt {-\frac {(1+m)^2}{n^2}} n}} x^{-1+\frac {2 (1+m)}{n}}\right ) \, dx,x,c x^n\right )}{2 \sqrt {-\frac {(1+m)^2}{n^2}} n^2}\\ &=-\frac {e^{\frac {a (1+m)}{\sqrt {-\frac {(1+m)^2}{n^2}} n}} x^{1+m} \left (c x^n\right )^{\frac {1+m}{n}}}{4 \sqrt {-\frac {(1+m)^2}{n^2}} n}+\frac {e^{\frac {a \sqrt {-\frac {(1+m)^2}{n^2}} n}{1+m}} (1+m) x^{1+m} \left (c x^n\right )^{-\frac {1+m}{n}} \log (x)}{2 \sqrt {-\frac {(1+m)^2}{n^2}} n}\\ \end {align*}

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Mathematica [F]  time = 0.36, size = 0, normalized size = 0.00 \[ \int x^m \sin \left (a+\sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [C]  time = 0.76, size = 62, normalized size = 0.47 \[ \frac {{\left (i \, x^{2} x^{2 \, m} + {\left (-2 i \, m - 2 i\right )} e^{\left (\frac {2 \, {\left (i \, a n - {\left (m + 1\right )} \log \relax (c)\right )}}{n}\right )} \log \relax (x)\right )} e^{\left (-\frac {i \, a n - {\left (m + 1\right )} \log \relax (c)}{n}\right )}}{4 \, {\left (m + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(I*x^2*x^(2*m) + (-2*I*m - 2*I)*e^(2*(I*a*n - (m + 1)*log(c))/n)*log(x))*e^(-(I*a*n - (m + 1)*log(c))/n)/(
m + 1)

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giac [C]  time = 2.08, size = 272, normalized size = 2.05 \[ \frac {-i \, m n^{2} x x^{m} e^{\left (i \, a - \frac {n {\left | m n + n \right |} \log \relax (x) + {\left | m n + n \right |} \log \relax (c)}{n^{2}}\right )} + i \, m n^{2} x x^{m} e^{\left (-i \, a + \frac {n {\left | m n + n \right |} \log \relax (x) + {\left | m n + n \right |} \log \relax (c)}{n^{2}}\right )} - i \, n^{2} x x^{m} e^{\left (i \, a - \frac {n {\left | m n + n \right |} \log \relax (x) + {\left | m n + n \right |} \log \relax (c)}{n^{2}}\right )} - i \, n x x^{m} {\left | m n + n \right |} e^{\left (i \, a - \frac {n {\left | m n + n \right |} \log \relax (x) + {\left | m n + n \right |} \log \relax (c)}{n^{2}}\right )} + i \, n^{2} x x^{m} e^{\left (-i \, a + \frac {n {\left | m n + n \right |} \log \relax (x) + {\left | m n + n \right |} \log \relax (c)}{n^{2}}\right )} - i \, n x x^{m} {\left | m n + n \right |} e^{\left (-i \, a + \frac {n {\left | m n + n \right |} \log \relax (x) + {\left | m n + n \right |} \log \relax (c)}{n^{2}}\right )}}{2 \, {\left (m^{2} n^{2} + 2 \, m n^{2} - {\left (m n + n\right )}^{2} + n^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(-I*m*n^2*x*x^m*e^(I*a - (n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c))/n^2) + I*m*n^2*x*x^m*e^(-I*a + (n*a
bs(m*n + n)*log(x) + abs(m*n + n)*log(c))/n^2) - I*n^2*x*x^m*e^(I*a - (n*abs(m*n + n)*log(x) + abs(m*n + n)*lo
g(c))/n^2) - I*n*x*x^m*abs(m*n + n)*e^(I*a - (n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c))/n^2) + I*n^2*x*x^m*
e^(-I*a + (n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c))/n^2) - I*n*x*x^m*abs(m*n + n)*e^(-I*a + (n*abs(m*n + n
)*log(x) + abs(m*n + n)*log(c))/n^2))/(m^2*n^2 + 2*m*n^2 - (m*n + n)^2 + n^2)

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int x^{m} \sin \left (a +\ln \left (c \,x^{n}\right ) \sqrt {-\frac {\left (1+m \right )^{2}}{n^{2}}}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.39, size = 82, normalized size = 0.62 \[ \frac {c^{\frac {2 \, m}{n} + \frac {2}{n}} x e^{\left (m \log \relax (x) + \frac {m \log \left (x^{n}\right )}{n} + \frac {\log \left (x^{n}\right )}{n}\right )} \sin \relax (a) + 2 \, {\left (m \sin \relax (a) + \sin \relax (a)\right )} \log \relax (x)}{4 \, {\left (c^{\frac {m}{n} + \frac {1}{n}} m + c^{\frac {m}{n} + \frac {1}{n}}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(c^(2*m/n + 2/n)*x*e^(m*log(x) + m*log(x^n)/n + log(x^n)/n)*sin(a) + 2*(m*sin(a) + sin(a))*log(x))/(c^(m/n
+ 1/n)*m + c^(m/n + 1/n))

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mupad [B]  time = 3.94, size = 135, normalized size = 1.02 \[ \frac {x\,x^m\,{\mathrm {e}}^{-a\,1{}\mathrm {i}}\,\frac {1}{{\left (c\,x^n\right )}^{\sqrt {-\frac {2\,m}{n^2}-\frac {1}{n^2}-\frac {m^2}{n^2}}\,1{}\mathrm {i}}}\,1{}\mathrm {i}}{2\,m+2-n\,\sqrt {-\frac {{\left (m+1\right )}^2}{n^2}}\,2{}\mathrm {i}}-\frac {x\,x^m\,{\mathrm {e}}^{a\,1{}\mathrm {i}}\,{\left (c\,x^n\right )}^{\sqrt {-\frac {2\,m}{n^2}-\frac {1}{n^2}-\frac {m^2}{n^2}}\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,m+2+n\,\sqrt {-\frac {{\left (m+1\right )}^2}{n^2}}\,2{}\mathrm {i}} \]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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