3.1.85 \(\int \frac {(a+b x^3) \sin (c+d x)}{x^3} \, dx\) [85]

Optimal. Leaf size=70 \[ -\frac {b \cos (c+d x)}{d}-\frac {a d \cos (c+d x)}{2 x}-\frac {1}{2} a d^2 \text {Ci}(d x) \sin (c)-\frac {a \sin (c+d x)}{2 x^2}-\frac {1}{2} a d^2 \cos (c) \text {Si}(d x) \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {3420, 2718, 3378, 3384, 3380, 3383} \begin {gather*} -\frac {1}{2} a d^2 \sin (c) \text {CosIntegral}(d x)-\frac {1}{2} a d^2 \cos (c) \text {Si}(d x)-\frac {a \sin (c+d x)}{2 x^2}-\frac {a d \cos (c+d x)}{2 x}-\frac {b \cos (c+d x)}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x^3)*Sin[c + d*x])/x^3,x]

[Out]

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

Rule 2718

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

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^3} \, dx &=\int \left (b \sin (c+d x)+\frac {a \sin (c+d x)}{x^3}\right ) \, dx\\ &=a \int \frac {\sin (c+d x)}{x^3} \, dx+b \int \sin (c+d x) \, dx\\ &=-\frac {b \cos (c+d x)}{d}-\frac {a \sin (c+d x)}{2 x^2}+\frac {1}{2} (a d) \int \frac {\cos (c+d x)}{x^2} \, dx\\ &=-\frac {b \cos (c+d x)}{d}-\frac {a d \cos (c+d x)}{2 x}-\frac {a \sin (c+d x)}{2 x^2}-\frac {1}{2} \left (a d^2\right ) \int \frac {\sin (c+d x)}{x} \, dx\\ &=-\frac {b \cos (c+d x)}{d}-\frac {a d \cos (c+d x)}{2 x}-\frac {a \sin (c+d x)}{2 x^2}-\frac {1}{2} \left (a d^2 \cos (c)\right ) \int \frac {\sin (d x)}{x} \, dx-\frac {1}{2} \left (a d^2 \sin (c)\right ) \int \frac {\cos (d x)}{x} \, dx\\ &=-\frac {b \cos (c+d x)}{d}-\frac {a d \cos (c+d x)}{2 x}-\frac {1}{2} a d^2 \text {Ci}(d x) \sin (c)-\frac {a \sin (c+d x)}{2 x^2}-\frac {1}{2} a d^2 \cos (c) \text {Si}(d x)\\ \end {align*}

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Mathematica [A]
time = 0.11, size = 66, normalized size = 0.94 \begin {gather*} \frac {1}{2} \left (-\frac {2 b \cos (c+d x)}{d}-\frac {a d \cos (c+d x)}{x}-a d^2 \text {Ci}(d x) \sin (c)-\frac {a \sin (c+d x)}{x^2}-a d^2 \cos (c) \text {Si}(d x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.09, size = 65, normalized size = 0.93

method	result	size
derivativedivides	\(d^{2} \left (a \left (-\frac {\sin \left (d x +c \right )}{2 d^{2} x^{2}}-\frac {\cos \left (d x +c \right )}{2 d x}-\frac {\sinIntegral \left (d x \right ) \cos \left (c \right )}{2}-\frac {\cosineIntegral \left (d x \right ) \sin \left (c \right )}{2}\right )-\frac {b \cos \left (d x +c \right )}{d^{3}}\right )\)	\(65\)
default	\(d^{2} \left (a \left (-\frac {\sin \left (d x +c \right )}{2 d^{2} x^{2}}-\frac {\cos \left (d x +c \right )}{2 d x}-\frac {\sinIntegral \left (d x \right ) \cos \left (c \right )}{2}-\frac {\cosineIntegral \left (d x \right ) \sin \left (c \right )}{2}\right )-\frac {b \cos \left (d x +c \right )}{d^{3}}\right )\)	\(65\)
risch	\(\frac {i \cos \left (c \right ) \expIntegral \left (1, i d x \right ) a \,d^{2}}{4}-\frac {i \cos \left (c \right ) \expIntegral \left (1, -i d x \right ) a \,d^{2}}{4}+\frac {\sin \left (c \right ) \expIntegral \left (1, i d x \right ) a \,d^{2}}{4}+\frac {\sin \left (c \right ) \expIntegral \left (1, -i d x \right ) a \,d^{2}}{4}-\frac {i \left (-2 i a \,d^{6} x^{3}-4 i b \,d^{4} x^{4}\right ) \cos \left (d x +c \right )}{4 d^{5} x^{4}}-\frac {a \sin \left (d x +c \right )}{2 x^{2}}\)	\(112\)
meijerg	\(\frac {b \sin \left (c \right ) \sin \left (d x \right )}{d}+\frac {b \sqrt {\pi }\, \cos \left (c \right ) \left (\frac {1}{\sqrt {\pi }}-\frac {\cos \left (d x \right )}{\sqrt {\pi }}\right )}{d}+\frac {a \sqrt {\pi }\, \sin \left (c \right ) d^{2} \left (-\frac {4}{\sqrt {\pi }\, x^{2} d^{2}}-\frac {2 \left (2 \gamma -3+2 \ln \left (x \right )+\ln \left (d^{2}\right )\right )}{\sqrt {\pi }}+\frac {-6 d^{2} x^{2}+4}{\sqrt {\pi }\, x^{2} d^{2}}+\frac {4 \gamma }{\sqrt {\pi }}+\frac {4 \ln \left (2\right )}{\sqrt {\pi }}+\frac {4 \ln \left (\frac {d x}{2}\right )}{\sqrt {\pi }}-\frac {4 \cos \left (d x \right )}{\sqrt {\pi }\, d^{2} x^{2}}+\frac {4 \sin \left (d x \right )}{\sqrt {\pi }\, d x}-\frac {4 \cosineIntegral \left (d x \right )}{\sqrt {\pi }}\right )}{8}+\frac {a \sqrt {\pi }\, \cos \left (c \right ) d^{2} \left (-\frac {4 \cos \left (d x \right )}{d x \sqrt {\pi }}-\frac {4 \sin \left (d x \right )}{d^{2} x^{2} \sqrt {\pi }}-\frac {4 \sinIntegral \left (d x \right )}{\sqrt {\pi }}\right )}{8}\)	\(211\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C] Result contains complex when optimal does not.
time = 0.54, size = 1146, normalized size = 16.37 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(((I*exp_integral_e(3, I*d*x) - I*exp_integral_e(3, -I*d*x))*cos(c)^3 + (I*exp_integral_e(3, I*d*x) - I*ex
p_integral_e(3, -I*d*x))*cos(c)*sin(c)^2 + (exp_integral_e(3, I*d*x) + exp_integral_e(3, -I*d*x))*sin(c)^3 + (
I*exp_integral_e(3, I*d*x) - I*exp_integral_e(3, -I*d*x))*cos(c) + ((exp_integral_e(3, I*d*x) + exp_integral_e
(3, -I*d*x))*cos(c)^2 + exp_integral_e(3, I*d*x) + exp_integral_e(3, -I*d*x))*sin(c))*b*c^3/((d*x + c)^2*(cos(
c)^2 + sin(c)^2)*d^3 - 2*(c*cos(c)^2 + c*sin(c)^2)*(d*x + c)*d^3 + (c^2*cos(c)^2 + c^2*sin(c)^2)*d^3) - ((I*ex
p_integral_e(3, I*d*x) - I*exp_integral_e(3, -I*d*x))*cos(c)^3 + (I*exp_integral_e(3, I*d*x) - I*exp_integral_
e(3, -I*d*x))*cos(c)*sin(c)^2 + (exp_integral_e(3, I*d*x) + exp_integral_e(3, -I*d*x))*sin(c)^3 + (I*exp_integ
ral_e(3, I*d*x) - I*exp_integral_e(3, -I*d*x))*cos(c) + ((exp_integral_e(3, I*d*x) + exp_integral_e(3, -I*d*x)
)*cos(c)^2 + exp_integral_e(3, I*d*x) + exp_integral_e(3, -I*d*x))*sin(c))*a/(c^2*cos(c)^2 + c^2*sin(c)^2 + (d
*x + c)^2*(cos(c)^2 + sin(c)^2) - 2*(c*cos(c)^2 + c*sin(c)^2)*(d*x + c)) - (2*((b*cos(c)^2 + b*sin(c)^2)*(d*x
+ c)^3 - 3*(b*c*cos(c)^2 + b*c*sin(c)^2)*(d*x + c)^2 + 3*(b*c^2*cos(c)^2 + b*c^2*sin(c)^2)*(d*x + c))*cos(d*x
+ c)^3 - 3*(b*c^3*(exp_integral_e(4, I*d*x) + exp_integral_e(4, -I*d*x))*cos(c)^3 + b*c^3*(exp_integral_e(4, I
*d*x) + exp_integral_e(4, -I*d*x))*cos(c)*sin(c)^2 + b*c^3*(-I*exp_integral_e(4, I*d*x) + I*exp_integral_e(4,
-I*d*x))*sin(c)^3 + b*c^3*(exp_integral_e(4, I*d*x) + exp_integral_e(4, -I*d*x))*cos(c) + (b*c^3*(-I*exp_integ
ral_e(4, I*d*x) + I*exp_integral_e(4, -I*d*x))*cos(c)^2 + b*c^3*(-I*exp_integral_e(4, I*d*x) + I*exp_integral_
e(4, -I*d*x)))*sin(c))*cos(d*x + c)^2 - (3*b*c^3*(exp_integral_e(4, I*d*x) + exp_integral_e(4, -I*d*x))*cos(c)
^3 + 3*b*c^3*(exp_integral_e(4, I*d*x) + exp_integral_e(4, -I*d*x))*cos(c)*sin(c)^2 + 3*b*c^3*(-I*exp_integral
_e(4, I*d*x) + I*exp_integral_e(4, -I*d*x))*sin(c)^3 + 3*b*c^3*(exp_integral_e(4, I*d*x) + exp_integral_e(4, -
I*d*x))*cos(c) - 2*((b*cos(c)^2 + b*sin(c)^2)*(d*x + c)^3 - 3*(b*c*cos(c)^2 + b*c*sin(c)^2)*(d*x + c)^2 + 3*(b
*c^2*cos(c)^2 + b*c^2*sin(c)^2)*(d*x + c))*cos(d*x + c) + 3*(b*c^3*(-I*exp_integral_e(4, I*d*x) + I*exp_integr
al_e(4, -I*d*x))*cos(c)^2 + b*c^3*(-I*exp_integral_e(4, I*d*x) + I*exp_integral_e(4, -I*d*x)))*sin(c))*sin(d*x
+ c)^2 + 2*((b*cos(c)^2 + b*sin(c)^2)*(d*x + c)^3 - 3*(b*c*cos(c)^2 + b*c*sin(c)^2)*(d*x + c)^2 + 3*(b*c^2*co
s(c)^2 + b*c^2*sin(c)^2)*(d*x + c))*cos(d*x + c))/(((d*x + c)^3*(cos(c)^2 + sin(c)^2)*d^3 - 3*(c*cos(c)^2 + c*
sin(c)^2)*(d*x + c)^2*d^3 + 3*(c^2*cos(c)^2 + c^2*sin(c)^2)*(d*x + c)*d^3 - (c^3*cos(c)^2 + c^3*sin(c)^2)*d^3)
*cos(d*x + c)^2 + ((d*x + c)^3*(cos(c)^2 + sin(c)^2)*d^3 - 3*(c*cos(c)^2 + c*sin(c)^2)*(d*x + c)^2*d^3 + 3*(c^
2*cos(c)^2 + c^2*sin(c)^2)*(d*x + c)*d^3 - (c^3*cos(c)^2 + c^3*sin(c)^2)*d^3)*sin(d*x + c)^2))*d^2

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Fricas [A]
time = 0.37, size = 84, normalized size = 1.20 \begin {gather*} -\frac {2 \, a d^{3} x^{2} \cos \left (c\right ) \operatorname {Si}\left (d x\right ) + 2 \, a d \sin \left (d x + c\right ) + 2 \, {\left (a d^{2} x + 2 \, b x^{2}\right )} \cos \left (d x + c\right ) + {\left (a d^{3} x^{2} \operatorname {Ci}\left (d x\right ) + a d^{3} x^{2} \operatorname {Ci}\left (-d x\right )\right )} \sin \left (c\right )}{4 \, d x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*a*d^3*x^2*cos(c)*sin_integral(d*x) + 2*a*d*sin(d*x + c) + 2*(a*d^2*x + 2*b*x^2)*cos(d*x + c) + (a*d^3*
x^2*cos_integral(d*x) + a*d^3*x^2*cos_integral(-d*x))*sin(c))/(d*x^2)

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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^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 5.13, size = 564, normalized size = 8.06 \begin {gather*} \frac {a d^{3} x^{2} \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^{2} \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} + 2 \, a d^{3} x^{2} \operatorname {Si}\left (d x\right ) \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} - 2 \, a d^{3} x^{2} \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^{2} \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 ) - a d^{3} x^{2} \Im \left ( \operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, d x\right )^{2} + a d^{3} x^{2} \Im \left ( \operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, d x\right )^{2} - 2 \, a d^{3} x^{2} \operatorname {Si}\left (d x\right ) \tan \left (\frac {1}{2} \, d x\right )^{2} + a d^{3} x^{2} \Im \left ( \operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, c\right )^{2} - a d^{3} x^{2} \Im \left ( \operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, c\right )^{2} + 2 \, a d^{3} x^{2} \operatorname {Si}\left (d x\right ) \tan \left (\frac {1}{2} \, c\right )^{2} - 2 \, a d^{3} x^{2} \Re \left ( \operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, c\right ) - 2 \, a d^{3} x^{2} \Re \left ( \operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, c\right ) - 2 \, a d^{2} x \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} - a d^{3} x^{2} \Im \left ( \operatorname {Ci}\left (d x\right ) \right ) + a d^{3} x^{2} \Im \left ( \operatorname {Ci}\left (-d x\right ) \right ) - 2 \, a d^{3} x^{2} \operatorname {Si}\left (d x\right ) - 4 \, b x^{2} \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} + 2 \, a d^{2} x \tan \left (\frac {1}{2} \, d x\right )^{2} + 8 \, a d^{2} x \tan \left (\frac {1}{2} \, d x\right ) \tan \left (\frac {1}{2} \, c\right ) + 2 \, a d^{2} x \tan \left (\frac {1}{2} \, c\right )^{2} + 4 \, b x^{2} \tan \left (\frac {1}{2} \, d x\right )^{2} + 16 \, b x^{2} \tan \left (\frac {1}{2} \, d x\right ) \tan \left (\frac {1}{2} \, c\right ) + 4 \, a d \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right ) + 4 \, b x^{2} \tan \left (\frac {1}{2} \, c\right )^{2} + 4 \, a d \tan \left (\frac {1}{2} \, d x\right ) \tan \left (\frac {1}{2} \, c\right )^{2} - 2 \, a d^{2} x - 4 \, b x^{2} - 4 \, a d \tan \left (\frac {1}{2} \, d x\right ) - 4 \, a d \tan \left (\frac {1}{2} \, c\right )}{4 \, {\left (d x^{2} \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} + d x^{2} \tan \left (\frac {1}{2} \, d x\right )^{2} + d x^{2} \tan \left (\frac {1}{2} \, c\right )^{2} + d x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(a*d^3*x^2*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 - a*d^3*x^2*imag_part(cos_integral(-d*
x))*tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*a*d^3*x^2*sin_integral(d*x)*tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*a*d^3*x^2*real
_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c) - 2*a*d^3*x^2*real_part(cos_integral(-d*x))*tan(1/2*d*x)^2*
tan(1/2*c) - a*d^3*x^2*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2 + a*d^3*x^2*imag_part(cos_integral(-d*x))*t
an(1/2*d*x)^2 - 2*a*d^3*x^2*sin_integral(d*x)*tan(1/2*d*x)^2 + a*d^3*x^2*imag_part(cos_integral(d*x))*tan(1/2*
c)^2 - a*d^3*x^2*imag_part(cos_integral(-d*x))*tan(1/2*c)^2 + 2*a*d^3*x^2*sin_integral(d*x)*tan(1/2*c)^2 - 2*a
*d^3*x^2*real_part(cos_integral(d*x))*tan(1/2*c) - 2*a*d^3*x^2*real_part(cos_integral(-d*x))*tan(1/2*c) - 2*a*
d^2*x*tan(1/2*d*x)^2*tan(1/2*c)^2 - a*d^3*x^2*imag_part(cos_integral(d*x)) + a*d^3*x^2*imag_part(cos_integral(
-d*x)) - 2*a*d^3*x^2*sin_integral(d*x) - 4*b*x^2*tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*a*d^2*x*tan(1/2*d*x)^2 + 8*a*
d^2*x*tan(1/2*d*x)*tan(1/2*c) + 2*a*d^2*x*tan(1/2*c)^2 + 4*b*x^2*tan(1/2*d*x)^2 + 16*b*x^2*tan(1/2*d*x)*tan(1/
2*c) + 4*a*d*tan(1/2*d*x)^2*tan(1/2*c) + 4*b*x^2*tan(1/2*c)^2 + 4*a*d*tan(1/2*d*x)*tan(1/2*c)^2 - 2*a*d^2*x -
4*b*x^2 - 4*a*d*tan(1/2*d*x) - 4*a*d*tan(1/2*c))/(d*x^2*tan(1/2*d*x)^2*tan(1/2*c)^2 + d*x^2*tan(1/2*d*x)^2 + d
*x^2*tan(1/2*c)^2 + d*x^2)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sin \left (c+d\,x\right )\,\left (b\,x^3+a\right )}{x^3} \,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^3,x)

[Out]

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

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