3.1.100 \(\int \frac {a+a \sin (e+f x)}{(c+d x)^3} \, dx\) [100]

Optimal. Leaf size=123 \[ -\frac {a}{2 d (c+d x)^2}-\frac {a f \cos (e+f x)}{2 d^2 (c+d x)}-\frac {a f^2 \text {Ci}\left (\frac {c f}{d}+f x\right ) \sin \left (e-\frac {c f}{d}\right )}{2 d^3}-\frac {a \sin (e+f x)}{2 d (c+d x)^2}-\frac {a f^2 \cos \left (e-\frac {c f}{d}\right ) \text {Si}\left (\frac {c f}{d}+f x\right )}{2 d^3} \]

[Out]

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

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Rubi [A]
time = 0.14, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {3398, 3378, 3384, 3380, 3383} \begin {gather*} -\frac {a f^2 \text {CosIntegral}\left (\frac {c f}{d}+f x\right ) \sin \left (e-\frac {c f}{d}\right )}{2 d^3}-\frac {a f^2 \cos \left (e-\frac {c f}{d}\right ) \text {Si}\left (x f+\frac {c f}{d}\right )}{2 d^3}-\frac {a f \cos (e+f x)}{2 d^2 (c+d x)}-\frac {a \sin (e+f x)}{2 d (c+d x)^2}-\frac {a}{2 d (c+d x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])/(c + d*x)^3,x]

[Out]

-1/2*a/(d*(c + d*x)^2) - (a*f*Cos[e + f*x])/(2*d^2*(c + d*x)) - (a*f^2*CosIntegral[(c*f)/d + f*x]*Sin[e - (c*f
)/d])/(2*d^3) - (a*Sin[e + f*x])/(2*d*(c + d*x)^2) - (a*f^2*Cos[e - (c*f)/d]*SinIntegral[(c*f)/d + f*x])/(2*d^
3)

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 3398

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, (a + b*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[n, 0] && (EqQ[n, 1] ||
IGtQ[m, 0] || NeQ[a^2 - b^2, 0])

Rubi steps

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

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Mathematica [A]
time = 0.37, size = 104, normalized size = 0.85 \begin {gather*} -\frac {a \left (f^2 (c+d x)^2 \text {Ci}\left (f \left (\frac {c}{d}+x\right )\right ) \sin \left (e-\frac {c f}{d}\right )+d (f (c+d x) \cos (e+f x)+d (1+\sin (e+f x)))+f^2 (c+d x)^2 \cos \left (e-\frac {c f}{d}\right ) \text {Si}\left (f \left (\frac {c}{d}+x\right )\right )\right )}{2 d^3 (c+d x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(a*(f^2*(c + d*x)^2*CosIntegral[f*(c/d + x)]*Sin[e - (c*f)/d] + d*(f*(c + d*x)*Cos[e + f*x] + d*(1 + Sin[
e + f*x])) + f^2*(c + d*x)^2*Cos[e - (c*f)/d]*SinIntegral[f*(c/d + x)]))/(d^3*(c + d*x)^2)

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Maple [A]
time = 0.10, size = 177, normalized size = 1.44

method	result	size
derivativedivides	\(\frac {a \,f^{3} \left (-\frac {\sin \left (f x +e \right )}{2 \left (c f -d e +d \left (f x +e \right )\right )^{2} d}+\frac {-\frac {\cos \left (f x +e \right )}{\left (c f -d e +d \left (f x +e \right )\right ) d}-\frac {\frac {\sinIntegral \left (f x +e +\frac {c f -d e}{d}\right ) \cos \left (\frac {c f -d e}{d}\right )}{d}-\frac {\cosineIntegral \left (f x +e +\frac {c f -d e}{d}\right ) \sin \left (\frac {c f -d e}{d}\right )}{d}}{d}}{2 d}\right )-\frac {a \,f^{3}}{2 \left (c f -d e +d \left (f x +e \right )\right )^{2} d}}{f}\)	\(177\)
default	\(\frac {a \,f^{3} \left (-\frac {\sin \left (f x +e \right )}{2 \left (c f -d e +d \left (f x +e \right )\right )^{2} d}+\frac {-\frac {\cos \left (f x +e \right )}{\left (c f -d e +d \left (f x +e \right )\right ) d}-\frac {\frac {\sinIntegral \left (f x +e +\frac {c f -d e}{d}\right ) \cos \left (\frac {c f -d e}{d}\right )}{d}-\frac {\cosineIntegral \left (f x +e +\frac {c f -d e}{d}\right ) \sin \left (\frac {c f -d e}{d}\right )}{d}}{d}}{2 d}\right )-\frac {a \,f^{3}}{2 \left (c f -d e +d \left (f x +e \right )\right )^{2} d}}{f}\)	\(177\)
risch	\(-\frac {a}{2 d \left (d x +c \right )^{2}}+\frac {i f^{2} a \,{\mathrm e}^{\frac {i \left (c f -d e \right )}{d}} \expIntegral \left (1, i f x +i e +\frac {i \left (c f -d e \right )}{d}\right )}{4 d^{3}}-\frac {i f^{2} a \,{\mathrm e}^{-\frac {i \left (c f -d e \right )}{d}} \expIntegral \left (1, -i f x -i e -\frac {i c f -i d e}{d}\right )}{4 d^{3}}+\frac {i a \left (-2 i d^{3} f^{3} x^{3}-6 i c \,d^{2} f^{3} x^{2}-6 i c^{2} d \,f^{3} x -2 i c^{3} f^{3}\right ) \cos \left (f x +e \right )}{4 d^{2} \left (d x +c \right )^{2} \left (-d^{2} x^{2} f^{2}-2 c d \,f^{2} x -c^{2} f^{2}\right )}-\frac {a \left (-2 d^{2} x^{2} f^{2}-4 c d \,f^{2} x -2 c^{2} f^{2}\right ) \sin \left (f x +e \right )}{4 d \left (d x +c \right )^{2} \left (-d^{2} x^{2} f^{2}-2 c d \,f^{2} x -c^{2} f^{2}\right )}\)	\(292\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C] Result contains complex when optimal does not.
time = 0.41, size = 279, normalized size = 2.27 \begin {gather*} -\frac {\frac {a f^{3}}{{\left (f x + e\right )}^{2} d^{3} + c^{2} d f^{2} - 2 \, c d^{2} f e + d^{3} e^{2} + 2 \, {\left (c d^{2} f - d^{3} e\right )} {\left (f x + e\right )}} - \frac {{\left (f^{3} {\left (-i \, E_{3}\left (\frac {i \, {\left (f x + e\right )} d + i \, c f - i \, d e}{d}\right ) + i \, E_{3}\left (-\frac {i \, {\left (f x + e\right )} d + i \, c f - i \, d e}{d}\right )\right )} \cos \left (\frac {c f - d e}{d}\right ) + f^{3} {\left (E_{3}\left (\frac {i \, {\left (f x + e\right )} d + i \, c f - i \, d e}{d}\right ) + E_{3}\left (-\frac {i \, {\left (f x + e\right )} d + i \, c f - i \, d e}{d}\right )\right )} \sin \left (\frac {c f - d e}{d}\right )\right )} a}{{\left (f x + e\right )}^{2} d^{3} + c^{2} d f^{2} - 2 \, c d^{2} f e + d^{3} e^{2} + 2 \, {\left (c d^{2} f - d^{3} e\right )} {\left (f x + e\right )}}}{2 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(a*f^3/((f*x + e)^2*d^3 + c^2*d*f^2 - 2*c*d^2*f*e + d^3*e^2 + 2*(c*d^2*f - d^3*e)*(f*x + e)) - (f^3*(-I*e
xp_integral_e(3, (I*(f*x + e)*d + I*c*f - I*d*e)/d) + I*exp_integral_e(3, -(I*(f*x + e)*d + I*c*f - I*d*e)/d))
*cos((c*f - d*e)/d) + f^3*(exp_integral_e(3, (I*(f*x + e)*d + I*c*f - I*d*e)/d) + exp_integral_e(3, -(I*(f*x +
e)*d + I*c*f - I*d*e)/d))*sin((c*f - d*e)/d))*a/((f*x + e)^2*d^3 + c^2*d*f^2 - 2*c*d^2*f*e + d^3*e^2 + 2*(c*d
^2*f - d^3*e)*(f*x + e)))/f

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Fricas [A]
time = 0.42, size = 231, normalized size = 1.88 \begin {gather*} -\frac {2 \, a d^{2} \sin \left (f x + e\right ) + 2 \, a d^{2} + 2 \, {\left (a d^{2} f^{2} x^{2} + 2 \, a c d f^{2} x + a c^{2} f^{2}\right )} \cos \left (-\frac {c f - d e}{d}\right ) \operatorname {Si}\left (\frac {d f x + c f}{d}\right ) + 2 \, {\left (a d^{2} f x + a c d f\right )} \cos \left (f x + e\right ) + {\left ({\left (a d^{2} f^{2} x^{2} + 2 \, a c d f^{2} x + a c^{2} f^{2}\right )} \operatorname {Ci}\left (\frac {d f x + c f}{d}\right ) + {\left (a d^{2} f^{2} x^{2} + 2 \, a c d f^{2} x + a c^{2} f^{2}\right )} \operatorname {Ci}\left (-\frac {d f x + c f}{d}\right )\right )} \sin \left (-\frac {c f - d e}{d}\right )}{4 \, {\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*a*d^2*sin(f*x + e) + 2*a*d^2 + 2*(a*d^2*f^2*x^2 + 2*a*c*d*f^2*x + a*c^2*f^2)*cos(-(c*f - d*e)/d)*sin_i
ntegral((d*f*x + c*f)/d) + 2*(a*d^2*f*x + a*c*d*f)*cos(f*x + e) + ((a*d^2*f^2*x^2 + 2*a*c*d*f^2*x + a*c^2*f^2)
*cos_integral((d*f*x + c*f)/d) + (a*d^2*f^2*x^2 + 2*a*c*d*f^2*x + a*c^2*f^2)*cos_integral(-(d*f*x + c*f)/d))*s
in(-(c*f - d*e)/d))/(d^5*x^2 + 2*c*d^4*x + c^2*d^3)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a \left (\int \frac {\sin {\left (e + f x \right )}}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx + \int \frac {1}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*(Integral(sin(e + f*x)/(c**3 + 3*c**2*d*x + 3*c*d**2*x**2 + d**3*x**3), x) + Integral(1/(c**3 + 3*c**2*d*x +
3*c*d**2*x**2 + d**3*x**3), x))

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Giac [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 1.98, size = 6157, normalized size = 50.06 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(a*d^2*f^2*x^2*imag_part(cos_integral(f*x + c*f/d))*tan(1/2*f*x)^2*tan(1/2*c*f/d)^2*tan(1/2*e)^2 - a*d^2*
f^2*x^2*imag_part(cos_integral(-f*x - c*f/d))*tan(1/2*f*x)^2*tan(1/2*c*f/d)^2*tan(1/2*e)^2 + 2*a*d^2*f^2*x^2*s
in_integral((d*f*x + c*f)/d)*tan(1/2*f*x)^2*tan(1/2*c*f/d)^2*tan(1/2*e)^2 - 2*a*d^2*f^2*x^2*real_part(cos_inte
gral(f*x + c*f/d))*tan(1/2*f*x)^2*tan(1/2*c*f/d)^2*tan(1/2*e) - 2*a*d^2*f^2*x^2*real_part(cos_integral(-f*x -
c*f/d))*tan(1/2*f*x)^2*tan(1/2*c*f/d)^2*tan(1/2*e) + 2*a*d^2*f^2*x^2*real_part(cos_integral(f*x + c*f/d))*tan(
1/2*f*x)^2*tan(1/2*c*f/d)*tan(1/2*e)^2 + 2*a*d^2*f^2*x^2*real_part(cos_integral(-f*x - c*f/d))*tan(1/2*f*x)^2*
tan(1/2*c*f/d)*tan(1/2*e)^2 + 2*a*c*d*f^2*x*imag_part(cos_integral(f*x + c*f/d))*tan(1/2*f*x)^2*tan(1/2*c*f/d)
^2*tan(1/2*e)^2 - 2*a*c*d*f^2*x*imag_part(cos_integral(-f*x - c*f/d))*tan(1/2*f*x)^2*tan(1/2*c*f/d)^2*tan(1/2*
e)^2 + 4*a*c*d*f^2*x*sin_integral((d*f*x + c*f)/d)*tan(1/2*f*x)^2*tan(1/2*c*f/d)^2*tan(1/2*e)^2 - a*d^2*f^2*x^
2*imag_part(cos_integral(f*x + c*f/d))*tan(1/2*f*x)^2*tan(1/2*c*f/d)^2 + a*d^2*f^2*x^2*imag_part(cos_integral(
-f*x - c*f/d))*tan(1/2*f*x)^2*tan(1/2*c*f/d)^2 - 2*a*d^2*f^2*x^2*sin_integral((d*f*x + c*f)/d)*tan(1/2*f*x)^2*
tan(1/2*c*f/d)^2 + 4*a*d^2*f^2*x^2*imag_part(cos_integral(f*x + c*f/d))*tan(1/2*f*x)^2*tan(1/2*c*f/d)*tan(1/2*
e) - 4*a*d^2*f^2*x^2*imag_part(cos_integral(-f*x - c*f/d))*tan(1/2*f*x)^2*tan(1/2*c*f/d)*tan(1/2*e) + 8*a*d^2*
f^2*x^2*sin_integral((d*f*x + c*f)/d)*tan(1/2*f*x)^2*tan(1/2*c*f/d)*tan(1/2*e) - 4*a*c*d*f^2*x*real_part(cos_i
ntegral(f*x + c*f/d))*tan(1/2*f*x)^2*tan(1/2*c*f/d)^2*tan(1/2*e) - 4*a*c*d*f^2*x*real_part(cos_integral(-f*x -
c*f/d))*tan(1/2*f*x)^2*tan(1/2*c*f/d)^2*tan(1/2*e) - a*d^2*f^2*x^2*imag_part(cos_integral(f*x + c*f/d))*tan(1
/2*f*x)^2*tan(1/2*e)^2 + a*d^2*f^2*x^2*imag_part(cos_integral(-f*x - c*f/d))*tan(1/2*f*x)^2*tan(1/2*e)^2 - 2*a
*d^2*f^2*x^2*sin_integral((d*f*x + c*f)/d)*tan(1/2*f*x)^2*tan(1/2*e)^2 + 4*a*c*d*f^2*x*real_part(cos_integral(
f*x + c*f/d))*tan(1/2*f*x)^2*tan(1/2*c*f/d)*tan(1/2*e)^2 + 4*a*c*d*f^2*x*real_part(cos_integral(-f*x - c*f/d))
*tan(1/2*f*x)^2*tan(1/2*c*f/d)*tan(1/2*e)^2 + a*d^2*f^2*x^2*imag_part(cos_integral(f*x + c*f/d))*tan(1/2*c*f/d
)^2*tan(1/2*e)^2 - a*d^2*f^2*x^2*imag_part(cos_integral(-f*x - c*f/d))*tan(1/2*c*f/d)^2*tan(1/2*e)^2 + 2*a*d^2
*f^2*x^2*sin_integral((d*f*x + c*f)/d)*tan(1/2*c*f/d)^2*tan(1/2*e)^2 + a*c^2*f^2*imag_part(cos_integral(f*x +
c*f/d))*tan(1/2*f*x)^2*tan(1/2*c*f/d)^2*tan(1/2*e)^2 - a*c^2*f^2*imag_part(cos_integral(-f*x - c*f/d))*tan(1/2
*f*x)^2*tan(1/2*c*f/d)^2*tan(1/2*e)^2 + 2*a*c^2*f^2*sin_integral((d*f*x + c*f)/d)*tan(1/2*f*x)^2*tan(1/2*c*f/d
)^2*tan(1/2*e)^2 - 2*a*d^2*f^2*x^2*real_part(cos_integral(f*x + c*f/d))*tan(1/2*f*x)^2*tan(1/2*c*f/d) - 2*a*d^
2*f^2*x^2*real_part(cos_integral(-f*x - c*f/d))*tan(1/2*f*x)^2*tan(1/2*c*f/d) - 2*a*c*d*f^2*x*imag_part(cos_in
tegral(f*x + c*f/d))*tan(1/2*f*x)^2*tan(1/2*c*f/d)^2 + 2*a*c*d*f^2*x*imag_part(cos_integral(-f*x - c*f/d))*tan
(1/2*f*x)^2*tan(1/2*c*f/d)^2 - 4*a*c*d*f^2*x*sin_integral((d*f*x + c*f)/d)*tan(1/2*f*x)^2*tan(1/2*c*f/d)^2 + 2
*a*d^2*f^2*x^2*real_part(cos_integral(f*x + c*f/d))*tan(1/2*f*x)^2*tan(1/2*e) + 2*a*d^2*f^2*x^2*real_part(cos_
integral(-f*x - c*f/d))*tan(1/2*f*x)^2*tan(1/2*e) + 8*a*c*d*f^2*x*imag_part(cos_integral(f*x + c*f/d))*tan(1/2
*f*x)^2*tan(1/2*c*f/d)*tan(1/2*e) - 8*a*c*d*f^2*x*imag_part(cos_integral(-f*x - c*f/d))*tan(1/2*f*x)^2*tan(1/2
*c*f/d)*tan(1/2*e) + 16*a*c*d*f^2*x*sin_integral((d*f*x + c*f)/d)*tan(1/2*f*x)^2*tan(1/2*c*f/d)*tan(1/2*e) - 2
*a*d^2*f^2*x^2*real_part(cos_integral(f*x + c*f/d))*tan(1/2*c*f/d)^2*tan(1/2*e) - 2*a*d^2*f^2*x^2*real_part(co
s_integral(-f*x - c*f/d))*tan(1/2*c*f/d)^2*tan(1/2*e) - 2*a*c^2*f^2*real_part(cos_integral(f*x + c*f/d))*tan(1
/2*f*x)^2*tan(1/2*c*f/d)^2*tan(1/2*e) - 2*a*c^2*f^2*real_part(cos_integral(-f*x - c*f/d))*tan(1/2*f*x)^2*tan(1
/2*c*f/d)^2*tan(1/2*e) - 2*a*c*d*f^2*x*imag_part(cos_integral(f*x + c*f/d))*tan(1/2*f*x)^2*tan(1/2*e)^2 + 2*a*
c*d*f^2*x*imag_part(cos_integral(-f*x - c*f/d))*tan(1/2*f*x)^2*tan(1/2*e)^2 - 4*a*c*d*f^2*x*sin_integral((d*f*
x + c*f)/d)*tan(1/2*f*x)^2*tan(1/2*e)^2 + 2*a*d^2*f^2*x^2*real_part(cos_integral(f*x + c*f/d))*tan(1/2*c*f/d)*
tan(1/2*e)^2 + 2*a*d^2*f^2*x^2*real_part(cos_integral(-f*x - c*f/d))*tan(1/2*c*f/d)*tan(1/2*e)^2 + 2*a*c^2*f^2
*real_part(cos_integral(f*x + c*f/d))*tan(1/2*f*x)^2*tan(1/2*c*f/d)*tan(1/2*e)^2 + 2*a*c^2*f^2*real_part(cos_i
ntegral(-f*x - c*f/d))*tan(1/2*f*x)^2*tan(1/2*c*f/d)*tan(1/2*e)^2 + 2*a*c*d*f^2*x*imag_part(cos_integral(f*x +
c*f/d))*tan(1/2*c*f/d)^2*tan(1/2*e)^2 - 2*a*c*d*f^2*x*imag_part(cos_integral(-f*x - c*f/d))*tan(1/2*c*f/d)^2*
tan(1/2*e)^2 + 4*a*c*d*f^2*x*sin_integral((d*f*x + c*f)/d)*tan(1/2*c*f/d)^2*tan(1/2*e)^2 + 2*a*d^2*f*x*tan(1/2
*f*x)^2*tan(1/2*c*f/d)^2*tan(1/2*e)^2 + a*d^2*f^2*x^2*imag_part(cos_integral(f*x + c*f/d))*tan(1/2*f*x)^2 - a*
d^2*f^2*x^2*imag_part(cos_integral(-f*x - c*f/d))*tan(1/2*f*x)^2 + 2*a*d^2*f^2*x^2*sin_integral((d*f*x + c*f)/
d)*tan(1/2*f*x)^2 - 4*a*c*d*f^2*x*real_part(cos...

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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