3.262 \(\int x \sin (a+b (c+d x)^n) \, dx\)

Optimal. Leaf size=243 \[ \frac {i e^{i a} (c+d x)^2 \left (-i b (c+d x)^n\right )^{-2/n} \Gamma \left (\frac {2}{n},-i b (c+d x)^n\right )}{2 d^2 n}-\frac {i e^{i a} c (c+d x) \left (-i b (c+d x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},-i b (c+d x)^n\right )}{2 d^2 n}+\frac {i e^{-i a} c (c+d x) \left (i b (c+d x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},i b (c+d x)^n\right )}{2 d^2 n}-\frac {i e^{-i a} (c+d x)^2 \left (i b (c+d x)^n\right )^{-2/n} \Gamma \left (\frac {2}{n},i b (c+d x)^n\right )}{2 d^2 n} \]

[Out]

-1/2*I*c*exp(I*a)*(d*x+c)*GAMMA(1/n,-I*b*(d*x+c)^n)/d^2/n/((-I*b*(d*x+c)^n)^(1/n))+1/2*I*c*(d*x+c)*GAMMA(1/n,I
*b*(d*x+c)^n)/d^2/exp(I*a)/n/((I*b*(d*x+c)^n)^(1/n))+1/2*I*exp(I*a)*(d*x+c)^2*GAMMA(2/n,-I*b*(d*x+c)^n)/d^2/n/
((-I*b*(d*x+c)^n)^(2/n))-1/2*I*(d*x+c)^2*GAMMA(2/n,I*b*(d*x+c)^n)/d^2/exp(I*a)/n/((I*b*(d*x+c)^n)^(2/n))

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Rubi [A]  time = 0.13, antiderivative size = 243, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3433, 3365, 2208, 3423, 2218} \[ \frac {i e^{i a} (c+d x)^2 \left (-i b (c+d x)^n\right )^{-2/n} \text {Gamma}\left (\frac {2}{n},-i b (c+d x)^n\right )}{2 d^2 n}-\frac {i e^{i a} c (c+d x) \left (-i b (c+d x)^n\right )^{-1/n} \text {Gamma}\left (\frac {1}{n},-i b (c+d x)^n\right )}{2 d^2 n}+\frac {i e^{-i a} c (c+d x) \left (i b (c+d x)^n\right )^{-1/n} \text {Gamma}\left (\frac {1}{n},i b (c+d x)^n\right )}{2 d^2 n}-\frac {i e^{-i a} (c+d x)^2 \left (i b (c+d x)^n\right )^{-2/n} \text {Gamma}\left (\frac {2}{n},i b (c+d x)^n\right )}{2 d^2 n} \]

Antiderivative was successfully verified.

[In]

Int[x*Sin[a + b*(c + d*x)^n],x]

[Out]

((-I/2)*c*E^(I*a)*(c + d*x)*Gamma[n^(-1), (-I)*b*(c + d*x)^n])/(d^2*n*((-I)*b*(c + d*x)^n)^n^(-1)) + ((I/2)*c*
(c + d*x)*Gamma[n^(-1), I*b*(c + d*x)^n])/(d^2*E^(I*a)*n*(I*b*(c + d*x)^n)^n^(-1)) + ((I/2)*E^(I*a)*(c + d*x)^
2*Gamma[2/n, (-I)*b*(c + d*x)^n])/(d^2*n*((-I)*b*(c + d*x)^n)^(2/n)) - ((I/2)*(c + d*x)^2*Gamma[2/n, I*b*(c +
d*x)^n])/(d^2*E^(I*a)*n*(I*b*(c + d*x)^n)^(2/n))

Rule 2208

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_)), x_Symbol] :> -Simp[(F^a*(c + d*x)*Gamma[1/n, -(b*(c + d*x)
^n*Log[F])])/(d*n*(-(b*(c + d*x)^n*Log[F]))^(1/n)), x] /; FreeQ[{F, a, b, c, d, n}, x] &&  !IntegerQ[2/n]

Rule 2218

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

Rule 3365

Int[Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)], x_Symbol] :> Dist[I/2, Int[E^(-(c*I) - d*I*(e + f*x)^n), x],
x] - Dist[I/2, Int[E^(c*I + d*I*(e + f*x)^n), x], x] /; FreeQ[{c, d, e, f, n}, x]

Rule 3423

Int[((e_.)*(x_))^(m_.)*Sin[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> Dist[I/2, Int[(e*x)^m*E^(-(c*I) - d*I*x^n),
x], x] - Dist[I/2, Int[(e*x)^m*E^(c*I + d*I*x^n), x], x] /; FreeQ[{c, d, e, m, n}, x]

Rule 3433

Int[((g_.) + (h_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)])^(p_.), x_Symbol] :
> Module[{k = If[FractionQ[n], Denominator[n], 1]}, Dist[k/f^(m + 1), Subst[Int[ExpandIntegrand[(a + b*Sin[c +
d*x^(k*n)])^p, x^(k - 1)*(f*g - e*h + h*x^k)^m, x], x], x, (e + f*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, f,
g, h}, x] && IGtQ[p, 0] && IGtQ[m, 0]

Rubi steps

\begin {align*} \int x \sin \left (a+b (c+d x)^n\right ) \, dx &=\frac {\operatorname {Subst}\left (\int \left (-c \sin \left (a+b x^n\right )+x \sin \left (a+b x^n\right )\right ) \, dx,x,c+d x\right )}{d^2}\\ &=\frac {\operatorname {Subst}\left (\int x \sin \left (a+b x^n\right ) \, dx,x,c+d x\right )}{d^2}-\frac {c \operatorname {Subst}\left (\int \sin \left (a+b x^n\right ) \, dx,x,c+d x\right )}{d^2}\\ &=\frac {i \operatorname {Subst}\left (\int e^{-i a-i b x^n} x \, dx,x,c+d x\right )}{2 d^2}-\frac {i \operatorname {Subst}\left (\int e^{i a+i b x^n} x \, dx,x,c+d x\right )}{2 d^2}-\frac {(i c) \operatorname {Subst}\left (\int e^{-i a-i b x^n} \, dx,x,c+d x\right )}{2 d^2}+\frac {(i c) \operatorname {Subst}\left (\int e^{i a+i b x^n} \, dx,x,c+d x\right )}{2 d^2}\\ &=-\frac {i c e^{i a} (c+d x) \left (-i b (c+d x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},-i b (c+d x)^n\right )}{2 d^2 n}+\frac {i c e^{-i a} (c+d x) \left (i b (c+d x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},i b (c+d x)^n\right )}{2 d^2 n}+\frac {i e^{i a} (c+d x)^2 \left (-i b (c+d x)^n\right )^{-2/n} \Gamma \left (\frac {2}{n},-i b (c+d x)^n\right )}{2 d^2 n}-\frac {i e^{-i a} (c+d x)^2 \left (i b (c+d x)^n\right )^{-2/n} \Gamma \left (\frac {2}{n},i b (c+d x)^n\right )}{2 d^2 n}\\ \end {align*}

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Mathematica [A]  time = 0.86, size = 192, normalized size = 0.79 \[ \frac {(c+d x) \left ((\sin (a)-i \cos (a)) \left (-i b (c+d x)^n\right )^{-2/n} \left (c \left (-i b (c+d x)^n\right )^{\frac {1}{n}} \Gamma \left (\frac {1}{n},-i b (c+d x)^n\right )-(c+d x) \Gamma \left (\frac {2}{n},-i b (c+d x)^n\right )\right )+(\sin (a)+i \cos (a)) \left (i b (c+d x)^n\right )^{-2/n} \left (c \left (i b (c+d x)^n\right )^{\frac {1}{n}} \Gamma \left (\frac {1}{n},i b (c+d x)^n\right )-(c+d x) \Gamma \left (\frac {2}{n},i b (c+d x)^n\right )\right )\right )}{2 d^2 n} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Sin[a + b*(c + d*x)^n],x]

[Out]

((c + d*x)*(((c*((-I)*b*(c + d*x)^n)^n^(-1)*Gamma[n^(-1), (-I)*b*(c + d*x)^n] - (c + d*x)*Gamma[2/n, (-I)*b*(c
+ d*x)^n])*((-I)*Cos[a] + Sin[a]))/((-I)*b*(c + d*x)^n)^(2/n) + ((c*(I*b*(c + d*x)^n)^n^(-1)*Gamma[n^(-1), I*
b*(c + d*x)^n] - (c + d*x)*Gamma[2/n, I*b*(c + d*x)^n])*(I*Cos[a] + Sin[a]))/(I*b*(c + d*x)^n)^(2/n)))/(2*d^2*
n)

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fricas [F]  time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x \sin \left ({\left (d x + c\right )}^{n} b + a\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sin(a+b*(d*x+c)^n),x, algorithm="fricas")

[Out]

integral(x*sin((d*x + c)^n*b + a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \sin \left ({\left (d x + c\right )}^{n} b + a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sin(a+b*(d*x+c)^n),x, algorithm="giac")

[Out]

integrate(x*sin((d*x + c)^n*b + a), x)

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maple [F]  time = 0.14, size = 0, normalized size = 0.00 \[ \int x \sin \left (a +b \left (d x +c \right )^{n}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*sin(a+b*(d*x+c)^n),x)

[Out]

int(x*sin(a+b*(d*x+c)^n),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \sin \left ({\left (d x + c\right )}^{n} b + a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sin(a+b*(d*x+c)^n),x, algorithm="maxima")

[Out]

integrate(x*sin((d*x + c)^n*b + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x\,\sin \left (a+b\,{\left (c+d\,x\right )}^n\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*sin(a + b*(c + d*x)^n),x)

[Out]

int(x*sin(a + b*(c + d*x)^n), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \sin {\left (a + b \left (c + d x\right )^{n} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sin(a+b*(d*x+c)**n),x)

[Out]

Integral(x*sin(a + b*(c + d*x)**n), x)

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