3.1.84 \(\int \frac {(a+b x^3) \sin (c+d x)}{x^2} \, dx\) [84]

Optimal. Leaf size=56 \[ -\frac {b x \cos (c+d x)}{d}+a d \cos (c) \text {Ci}(d x)+\frac {b \sin (c+d x)}{d^2}-\frac {a \sin (c+d x)}{x}-a d \sin (c) \text {Si}(d x) \]

[Out]

a*d*Ci(d*x)*cos(c)-b*x*cos(d*x+c)/d-a*d*Si(d*x)*sin(c)+b*sin(d*x+c)/d^2-a*sin(d*x+c)/x

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Rubi [A]
time = 0.08, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {3420, 3378, 3384, 3380, 3383, 3377, 2717} \begin {gather*} a d \cos (c) \text {CosIntegral}(d x)-a d \sin (c) \text {Si}(d x)-\frac {a \sin (c+d x)}{x}+\frac {b \sin (c+d x)}{d^2}-\frac {b x \cos (c+d x)}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x^3)*Sin[c + d*x])/x^2,x]

[Out]

-((b*x*Cos[c + d*x])/d) + a*d*Cos[c]*CosIntegral[d*x] + (b*Sin[c + d*x])/d^2 - (a*Sin[c + d*x])/x - a*d*Sin[c]
*SinIntegral[d*x]

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3377

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(-(c + d*x)^m)*(Cos[e + f*x]/f), x]
+ Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3378

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(c + d*x)^(m + 1)*(Sin[e + f*x]/(d*(m
+ 1))), x] - Dist[f/(d*(m + 1)), Int[(c + d*x)^(m + 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && LtQ[
m, -1]

Rule 3380

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3383

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rule 3384

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[c*(f/d) + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[c*(f/d) + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 3420

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*Sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Int[ExpandIntegran
d[Sin[c + d*x], (e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^3\right ) \sin (c+d x)}{x^2} \, dx &=\int \left (\frac {a \sin (c+d x)}{x^2}+b x \sin (c+d x)\right ) \, dx\\ &=a \int \frac {\sin (c+d x)}{x^2} \, dx+b \int x \sin (c+d x) \, dx\\ &=-\frac {b x \cos (c+d x)}{d}-\frac {a \sin (c+d x)}{x}+\frac {b \int \cos (c+d x) \, dx}{d}+(a d) \int \frac {\cos (c+d x)}{x} \, dx\\ &=-\frac {b x \cos (c+d x)}{d}+\frac {b \sin (c+d x)}{d^2}-\frac {a \sin (c+d x)}{x}+(a d \cos (c)) \int \frac {\cos (d x)}{x} \, dx-(a d \sin (c)) \int \frac {\sin (d x)}{x} \, dx\\ &=-\frac {b x \cos (c+d x)}{d}+a d \cos (c) \text {Ci}(d x)+\frac {b \sin (c+d x)}{d^2}-\frac {a \sin (c+d x)}{x}-a d \sin (c) \text {Si}(d x)\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 56, normalized size = 1.00 \begin {gather*} -\frac {b x \cos (c+d x)}{d}+a d \cos (c) \text {Ci}(d x)+\frac {b \sin (c+d x)}{d^2}-\frac {a \sin (c+d x)}{x}-a d \sin (c) \text {Si}(d x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^3)*Sin[c + d*x])/x^2,x]

[Out]

-((b*x*Cos[c + d*x])/d) + a*d*Cos[c]*CosIntegral[d*x] + (b*Sin[c + d*x])/d^2 - (a*Sin[c + d*x])/x - a*d*Sin[c]
*SinIntegral[d*x]

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Maple [A]
time = 0.08, size = 79, normalized size = 1.41

method	result	size
derivativedivides	\(d \left (a \left (-\frac {\sin \left (d x +c \right )}{d x}-\sinIntegral \left (d x \right ) \sin \left (c \right )+\cosineIntegral \left (d x \right ) \cos \left (c \right )\right )+\frac {3 b c \cos \left (d x +c \right )}{d^{3}}+\frac {\left (2 c +1\right ) b \left (\sin \left (d x +c \right )-\left (d x +c \right ) \cos \left (d x +c \right )\right )}{d^{3}}\right )\)	\(79\)
default	\(d \left (a \left (-\frac {\sin \left (d x +c \right )}{d x}-\sinIntegral \left (d x \right ) \sin \left (c \right )+\cosineIntegral \left (d x \right ) \cos \left (c \right )\right )+\frac {3 b c \cos \left (d x +c \right )}{d^{3}}+\frac {\left (2 c +1\right ) b \left (\sin \left (d x +c \right )-\left (d x +c \right ) \cos \left (d x +c \right )\right )}{d^{3}}\right )\)	\(79\)
risch	\(-\frac {d \cos \left (c \right ) a \expIntegral \left (1, -i d x \right )}{2}-\frac {d \cos \left (c \right ) a \expIntegral \left (1, i d x \right )}{2}-\frac {i d \sin \left (c \right ) a \expIntegral \left (1, -i d x \right )}{2}+\frac {i d \sin \left (c \right ) a \expIntegral \left (1, i d x \right )}{2}+\frac {b \left (2 d^{4} x^{4}+6 c \,d^{3} x^{3}\right ) \cos \left (d x +c \right )}{2 d^{4} x^{2} \left (-d x -3 c \right )}-\frac {\left (-2 a \,d^{5} x^{2}-6 a c \,d^{4} x +2 d^{3} x^{3} b +6 c \,d^{2} x^{2} b \right ) \sin \left (d x +c \right )}{2 d^{4} x^{2} \left (-d x -3 c \right )}\)	\(160\)
meijerg	\(\frac {2 b \sqrt {\pi }\, \sin \left (c \right ) \left (-\frac {1}{2 \sqrt {\pi }}+\frac {\cos \left (d x \right )}{2 \sqrt {\pi }}+\frac {d x \sin \left (d x \right )}{2 \sqrt {\pi }}\right )}{d^{2}}+\frac {2 b \sqrt {\pi }\, \cos \left (c \right ) \left (-\frac {d x \cos \left (d x \right )}{2 \sqrt {\pi }}+\frac {\sin \left (d x \right )}{2 \sqrt {\pi }}\right )}{d^{2}}+\frac {a \sqrt {\pi }\, \sin \left (c \right ) d^{2} \left (-\frac {4 d^{2} \cos \left (x \sqrt {d^{2}}\right )}{x \left (d^{2}\right )^{\frac {3}{2}} \sqrt {\pi }}-\frac {4 \sinIntegral \left (x \sqrt {d^{2}}\right )}{\sqrt {\pi }}\right )}{4 \sqrt {d^{2}}}+\frac {a \sqrt {\pi }\, \cos \left (c \right ) d \left (\frac {4 \gamma -4+4 \ln \left (x \right )+4 \ln \left (d \right )}{\sqrt {\pi }}+\frac {4}{\sqrt {\pi }}-\frac {4 \gamma }{\sqrt {\pi }}-\frac {4 \ln \left (2\right )}{\sqrt {\pi }}-\frac {4 \ln \left (\frac {d x}{2}\right )}{\sqrt {\pi }}-\frac {4 \sin \left (d x \right )}{\sqrt {\pi }\, d x}+\frac {4 \cosineIntegral \left (d x \right )}{\sqrt {\pi }}\right )}{4}\)	\(205\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)*sin(d*x+c)/x^2,x,method=_RETURNVERBOSE)

[Out]

d*(a*(-sin(d*x+c)/d/x-Si(d*x)*sin(c)+Ci(d*x)*cos(c))+3/d^3*b*c*cos(d*x+c)+(2*c+1)/d^3*b*(sin(d*x+c)-(d*x+c)*co
s(d*x+c)))

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Maxima [C] Result contains complex when optimal does not.
time = 0.69, size = 69, normalized size = 1.23 \begin {gather*} \frac {{\left (a {\left (\Gamma \left (-1, i \, d x\right ) + \Gamma \left (-1, -i \, d x\right )\right )} \cos \left (c\right ) + a {\left (-i \, \Gamma \left (-1, i \, d x\right ) + i \, \Gamma \left (-1, -i \, d x\right )\right )} \sin \left (c\right )\right )} d^{3} - 2 \, b d x \cos \left (d x + c\right ) + 2 \, b \sin \left (d x + c\right )}{2 \, d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)*sin(d*x+c)/x^2,x, algorithm="maxima")

[Out]

1/2*((a*(gamma(-1, I*d*x) + gamma(-1, -I*d*x))*cos(c) + a*(-I*gamma(-1, I*d*x) + I*gamma(-1, -I*d*x))*sin(c))*
d^3 - 2*b*d*x*cos(d*x + c) + 2*b*sin(d*x + c))/d^2

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Fricas [A]
time = 0.37, size = 79, normalized size = 1.41 \begin {gather*} -\frac {2 \, a d^{3} x \sin \left (c\right ) \operatorname {Si}\left (d x\right ) + 2 \, b d x^{2} \cos \left (d x + c\right ) - {\left (a d^{3} x \operatorname {Ci}\left (d x\right ) + a d^{3} x \operatorname {Ci}\left (-d x\right )\right )} \cos \left (c\right ) + 2 \, {\left (a d^{2} - b x\right )} \sin \left (d x + c\right )}{2 \, d^{2} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)*sin(d*x+c)/x^2,x, algorithm="fricas")

[Out]

-1/2*(2*a*d^3*x*sin(c)*sin_integral(d*x) + 2*b*d*x^2*cos(d*x + c) - (a*d^3*x*cos_integral(d*x) + a*d^3*x*cos_i
ntegral(-d*x))*cos(c) + 2*(a*d^2 - b*x)*sin(d*x + c))/(d^2*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)*sin(d*x+c)/x**2,x)

[Out]

Integral((a + b*x**3)*sin(c + d*x)/x**2, x)

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Giac [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 4.68, size = 489, normalized size = 8.73 \begin {gather*} -\frac {a d^{3} x \Re \left ( \operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} + a d^{3} x \Re \left ( \operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} + 2 \, a d^{3} x \Im \left ( \operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right ) - 2 \, a d^{3} x \Im \left ( \operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right ) + 4 \, a d^{3} x \operatorname {Si}\left (d x\right ) \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right ) - a d^{3} x \Re \left ( \operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, d x\right )^{2} - a d^{3} x \Re \left ( \operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, d x\right )^{2} + a d^{3} x \Re \left ( \operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, c\right )^{2} + a d^{3} x \Re \left ( \operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, c\right )^{2} + 2 \, b d x^{2} \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} + 2 \, a d^{3} x \Im \left ( \operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, c\right ) - 2 \, a d^{3} x \Im \left ( \operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, c\right ) + 4 \, a d^{3} x \operatorname {Si}\left (d x\right ) \tan \left (\frac {1}{2} \, c\right ) - a d^{3} x \Re \left ( \operatorname {Ci}\left (d x\right ) \right ) - a d^{3} x \Re \left ( \operatorname {Ci}\left (-d x\right ) \right ) - 2 \, b d x^{2} \tan \left (\frac {1}{2} \, d x\right )^{2} - 8 \, b d x^{2} \tan \left (\frac {1}{2} \, d x\right ) \tan \left (\frac {1}{2} \, c\right ) - 4 \, a d^{2} \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right ) - 2 \, b d x^{2} \tan \left (\frac {1}{2} \, c\right )^{2} - 4 \, a d^{2} \tan \left (\frac {1}{2} \, d x\right ) \tan \left (\frac {1}{2} \, c\right )^{2} + 4 \, b x \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right ) + 4 \, b x \tan \left (\frac {1}{2} \, d x\right ) \tan \left (\frac {1}{2} \, c\right )^{2} + 2 \, b d x^{2} + 4 \, a d^{2} \tan \left (\frac {1}{2} \, d x\right ) + 4 \, a d^{2} \tan \left (\frac {1}{2} \, c\right ) - 4 \, b x \tan \left (\frac {1}{2} \, d x\right ) - 4 \, b x \tan \left (\frac {1}{2} \, c\right )}{2 \, {\left (d^{2} x \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} + d^{2} x \tan \left (\frac {1}{2} \, d x\right )^{2} + d^{2} x \tan \left (\frac {1}{2} \, c\right )^{2} + d^{2} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)*sin(d*x+c)/x^2,x, algorithm="giac")

[Out]

-1/2*(a*d^3*x*real_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 + a*d^3*x*real_part(cos_integral(-d*x))
*tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*a*d^3*x*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c) - 2*a*d^3*x*im
ag_part(cos_integral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c) + 4*a*d^3*x*sin_integral(d*x)*tan(1/2*d*x)^2*tan(1/2*c)
- a*d^3*x*real_part(cos_integral(d*x))*tan(1/2*d*x)^2 - a*d^3*x*real_part(cos_integral(-d*x))*tan(1/2*d*x)^2 +
a*d^3*x*real_part(cos_integral(d*x))*tan(1/2*c)^2 + a*d^3*x*real_part(cos_integral(-d*x))*tan(1/2*c)^2 + 2*b*
d*x^2*tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*a*d^3*x*imag_part(cos_integral(d*x))*tan(1/2*c) - 2*a*d^3*x*imag_part(co
s_integral(-d*x))*tan(1/2*c) + 4*a*d^3*x*sin_integral(d*x)*tan(1/2*c) - a*d^3*x*real_part(cos_integral(d*x)) -
a*d^3*x*real_part(cos_integral(-d*x)) - 2*b*d*x^2*tan(1/2*d*x)^2 - 8*b*d*x^2*tan(1/2*d*x)*tan(1/2*c) - 4*a*d^
2*tan(1/2*d*x)^2*tan(1/2*c) - 2*b*d*x^2*tan(1/2*c)^2 - 4*a*d^2*tan(1/2*d*x)*tan(1/2*c)^2 + 4*b*x*tan(1/2*d*x)^
2*tan(1/2*c) + 4*b*x*tan(1/2*d*x)*tan(1/2*c)^2 + 2*b*d*x^2 + 4*a*d^2*tan(1/2*d*x) + 4*a*d^2*tan(1/2*c) - 4*b*x
*tan(1/2*d*x) - 4*b*x*tan(1/2*c))/(d^2*x*tan(1/2*d*x)^2*tan(1/2*c)^2 + d^2*x*tan(1/2*d*x)^2 + d^2*x*tan(1/2*c)
^2 + d^2*x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\sin \left (c+d\,x\right )\,\left (b\,x^3+a\right )}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(c + d*x)*(a + b*x^3))/x^2,x)

[Out]

int((sin(c + d*x)*(a + b*x^3))/x^2, x)

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