3.2.60 \(\int \frac {(a+b \sin (e+f x))^2}{c+d x} \, dx\) [160]

Optimal. Leaf size=156 \[ -\frac {b^2 \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Ci}\left (\frac {2 c f}{d}+2 f x\right )}{2 d}+\frac {a^2 \log (c+d x)}{d}+\frac {b^2 \log (c+d x)}{2 d}+\frac {2 a b \text {Ci}\left (\frac {c f}{d}+f x\right ) \sin \left (e-\frac {c f}{d}\right )}{d}+\frac {2 a b \cos \left (e-\frac {c f}{d}\right ) \text {Si}\left (\frac {c f}{d}+f x\right )}{d}+\frac {b^2 \sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 c f}{d}+2 f x\right )}{2 d} \]

[Out]

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

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Rubi [A]
time = 0.21, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3398, 3384, 3380, 3383, 3393} \begin {gather*} \frac {a^2 \log (c+d x)}{d}+\frac {2 a b \text {CosIntegral}\left (\frac {c f}{d}+f x\right ) \sin \left (e-\frac {c f}{d}\right )}{d}+\frac {2 a b \cos \left (e-\frac {c f}{d}\right ) \text {Si}\left (x f+\frac {c f}{d}\right )}{d}-\frac {b^2 \text {CosIntegral}\left (\frac {2 c f}{d}+2 f x\right ) \cos \left (2 e-\frac {2 c f}{d}\right )}{2 d}+\frac {b^2 \sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (2 x f+\frac {2 c f}{d}\right )}{2 d}+\frac {b^2 \log (c+d x)}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[e + f*x])^2/(c + d*x),x]

[Out]

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

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 3393

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

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

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Mathematica [A]
time = 0.16, size = 134, normalized size = 0.86 \begin {gather*} \frac {-b^2 \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Ci}\left (\frac {2 f (c+d x)}{d}\right )+2 a^2 \log (c+d x)+b^2 \log (c+d x)+4 a b \text {Ci}\left (f \left (\frac {c}{d}+x\right )\right ) \sin \left (e-\frac {c f}{d}\right )+4 a b \cos \left (e-\frac {c f}{d}\right ) \text {Si}\left (f \left (\frac {c}{d}+x\right )\right )+b^2 \sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 f (c+d x)}{d}\right )}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-(b^2*Cos[2*e - (2*c*f)/d]*CosIntegral[(2*f*(c + d*x))/d]) + 2*a^2*Log[c + d*x] + b^2*Log[c + d*x] + 4*a*b*Co
sIntegral[f*(c/d + x)]*Sin[e - (c*f)/d] + 4*a*b*Cos[e - (c*f)/d]*SinIntegral[f*(c/d + x)] + b^2*Sin[2*e - (2*c
*f)/d]*SinIntegral[(2*f*(c + d*x))/d])/(2*d)

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Maple [A]
time = 0.06, size = 221, normalized size = 1.42

method	result	size
derivativedivides	\(\frac {\frac {a^{2} f \ln \left (c f -d e +d \left (f x +e \right )\right )}{d}+2 f a b \left (\frac {\sinIntegral \left (f x +e +\frac {c f -d e}{d}\right ) \cos \left (\frac {c f -d e}{d}\right )}{d}-\frac {\cosineIntegral \left (f x +e +\frac {c f -d e}{d}\right ) \sin \left (\frac {c f -d e}{d}\right )}{d}\right )+\frac {f \,b^{2} \ln \left (c f -d e +d \left (f x +e \right )\right )}{2 d}-\frac {f \,b^{2} \left (\frac {2 \sinIntegral \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \sin \left (\frac {2 c f -2 d e}{d}\right )}{d}+\frac {2 \cosineIntegral \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \cos \left (\frac {2 c f -2 d e}{d}\right )}{d}\right )}{4}}{f}\)	\(221\)
default	\(\frac {\frac {a^{2} f \ln \left (c f -d e +d \left (f x +e \right )\right )}{d}+2 f a b \left (\frac {\sinIntegral \left (f x +e +\frac {c f -d e}{d}\right ) \cos \left (\frac {c f -d e}{d}\right )}{d}-\frac {\cosineIntegral \left (f x +e +\frac {c f -d e}{d}\right ) \sin \left (\frac {c f -d e}{d}\right )}{d}\right )+\frac {f \,b^{2} \ln \left (c f -d e +d \left (f x +e \right )\right )}{2 d}-\frac {f \,b^{2} \left (\frac {2 \sinIntegral \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \sin \left (\frac {2 c f -2 d e}{d}\right )}{d}+\frac {2 \cosineIntegral \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \cos \left (\frac {2 c f -2 d e}{d}\right )}{d}\right )}{4}}{f}\)	\(221\)
risch	\(-\frac {i a b \,{\mathrm e}^{\frac {i \left (c f -d e \right )}{d}} \expIntegral \left (1, i f x +i e +\frac {i \left (c f -d e \right )}{d}\right )}{d}+\frac {a^{2} \ln \left (d x +c \right )}{d}+\frac {b^{2} \ln \left (d x +c \right )}{2 d}+\frac {b^{2} {\mathrm e}^{\frac {2 i \left (c f -d e \right )}{d}} \expIntegral \left (1, 2 i f x +2 i e +\frac {2 i \left (c f -d e \right )}{d}\right )}{4 d}+\frac {b^{2} {\mathrm e}^{-\frac {2 i \left (c f -d e \right )}{d}} \expIntegral \left (1, -2 i f x -2 i e -\frac {2 \left (i c f -i d e \right )}{d}\right )}{4 d}+\frac {i a b \,{\mathrm e}^{-\frac {i \left (c f -d e \right )}{d}} \expIntegral \left (1, -i f x -i e -\frac {i c f -i d e}{d}\right )}{d}\)	\(229\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(a^2*f*ln(c*f-d*e+d*(f*x+e))/d+2*f*a*b*(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)+1/2*f*b^2*ln(c*f-d*e+d*(f*x+e))/d-1/4*f*b^2*(2*Si(2*f*x+2*e+2*(c*f-d*e)/d)*sin(2*(c*f-d*e)/d)
/d+2*Ci(2*f*x+2*e+2*(c*f-d*e)/d)*cos(2*(c*f-d*e)/d)/d))

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Maxima [C] Result contains complex when optimal does not.
time = 0.40, size = 358, normalized size = 2.29 \begin {gather*} \frac {\frac {4 \, a^{2} f \log \left (c + \frac {{\left (f x + e\right )} d}{f} - \frac {d e}{f}\right )}{d} + \frac {4 \, {\left (f {\left (-i \, E_{1}\left (\frac {i \, {\left (f x + e\right )} d + i \, c f - i \, d e}{d}\right ) + i \, E_{1}\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 {\left (E_{1}\left (\frac {i \, {\left (f x + e\right )} d + i \, c f - i \, d e}{d}\right ) + E_{1}\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 b}{d} + \frac {{\left (f {\left (E_{1}\left (\frac {2 \, {\left (-i \, {\left (f x + e\right )} d - i \, c f + i \, d e\right )}}{d}\right ) + E_{1}\left (-\frac {2 \, {\left (-i \, {\left (f x + e\right )} d - i \, c f + i \, d e\right )}}{d}\right )\right )} \cos \left (\frac {2 \, {\left (c f - d e\right )}}{d}\right ) + f {\left (-i \, E_{1}\left (\frac {2 \, {\left (-i \, {\left (f x + e\right )} d - i \, c f + i \, d e\right )}}{d}\right ) + i \, E_{1}\left (-\frac {2 \, {\left (-i \, {\left (f x + e\right )} d - i \, c f + i \, d e\right )}}{d}\right )\right )} \sin \left (\frac {2 \, {\left (c f - d e\right )}}{d}\right ) + 2 \, f \log \left ({\left (f x + e\right )} d + c f - d e\right )\right )} b^{2}}{d}}{4 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*a^2*f*log(c + (f*x + e)*d/f - d*e/f)/d + 4*(f*(-I*exp_integral_e(1, (I*(f*x + e)*d + I*c*f - I*d*e)/d)
+ I*exp_integral_e(1, -(I*(f*x + e)*d + I*c*f - I*d*e)/d))*cos((c*f - d*e)/d) + f*(exp_integral_e(1, (I*(f*x +
e)*d + I*c*f - I*d*e)/d) + exp_integral_e(1, -(I*(f*x + e)*d + I*c*f - I*d*e)/d))*sin((c*f - d*e)/d))*a*b/d +
(f*(exp_integral_e(1, 2*(-I*(f*x + e)*d - I*c*f + I*d*e)/d) + exp_integral_e(1, -2*(-I*(f*x + e)*d - I*c*f +
I*d*e)/d))*cos(2*(c*f - d*e)/d) + f*(-I*exp_integral_e(1, 2*(-I*(f*x + e)*d - I*c*f + I*d*e)/d) + I*exp_integr
al_e(1, -2*(-I*(f*x + e)*d - I*c*f + I*d*e)/d))*sin(2*(c*f - d*e)/d) + 2*f*log((f*x + e)*d + c*f - d*e))*b^2/d
)/f

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Fricas [A]
time = 0.37, size = 194, normalized size = 1.24 \begin {gather*} \frac {2 \, b^{2} \sin \left (-\frac {2 \, {\left (c f - d e\right )}}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) + 8 \, a b \cos \left (-\frac {c f - d e}{d}\right ) \operatorname {Si}\left (\frac {d f x + c f}{d}\right ) - {\left (b^{2} \operatorname {Ci}\left (\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) + b^{2} \operatorname {Ci}\left (-\frac {2 \, {\left (d f x + c f\right )}}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (c f - d e\right )}}{d}\right ) + 2 \, {\left (2 \, a^{2} + b^{2}\right )} \log \left (d x + c\right ) + 4 \, {\left (a b \operatorname {Ci}\left (\frac {d f x + c f}{d}\right ) + a b \operatorname {Ci}\left (-\frac {d f x + c f}{d}\right )\right )} \sin \left (-\frac {c f - d e}{d}\right )}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*b^2*sin(-2*(c*f - d*e)/d)*sin_integral(2*(d*f*x + c*f)/d) + 8*a*b*cos(-(c*f - d*e)/d)*sin_integral((d*f
*x + c*f)/d) - (b^2*cos_integral(2*(d*f*x + c*f)/d) + b^2*cos_integral(-2*(d*f*x + c*f)/d))*cos(-2*(c*f - d*e)
/d) + 2*(2*a^2 + b^2)*log(d*x + c) + 4*(a*b*cos_integral((d*f*x + c*f)/d) + a*b*cos_integral(-(d*f*x + c*f)/d)
)*sin(-(c*f - d*e)/d))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 3.90, size = 7397, normalized size = 47.42 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*a*b*imag_part(cos_integral(f*x + c*f/d))*tan(c*f/d)^2*tan(1/2*c*f/d)^2*tan(1/2*e)^2*tan(e)^2 - 4*a*b*im
ag_part(cos_integral(-f*x - c*f/d))*tan(c*f/d)^2*tan(1/2*c*f/d)^2*tan(1/2*e)^2*tan(e)^2 + 4*a^2*log(abs(d*x +
c))*tan(c*f/d)^2*tan(1/2*c*f/d)^2*tan(1/2*e)^2*tan(e)^2 + 2*b^2*log(abs(d*x + c))*tan(c*f/d)^2*tan(1/2*c*f/d)^
2*tan(1/2*e)^2*tan(e)^2 - b^2*real_part(cos_integral(2*f*x + 2*c*f/d))*tan(c*f/d)^2*tan(1/2*c*f/d)^2*tan(1/2*e
)^2*tan(e)^2 - b^2*real_part(cos_integral(-2*f*x - 2*c*f/d))*tan(c*f/d)^2*tan(1/2*c*f/d)^2*tan(1/2*e)^2*tan(e)
^2 + 8*a*b*sin_integral((d*f*x + c*f)/d)*tan(c*f/d)^2*tan(1/2*c*f/d)^2*tan(1/2*e)^2*tan(e)^2 - 2*b^2*imag_part
(cos_integral(2*f*x + 2*c*f/d))*tan(c*f/d)^2*tan(1/2*c*f/d)^2*tan(1/2*e)^2*tan(e) + 2*b^2*imag_part(cos_integr
al(-2*f*x - 2*c*f/d))*tan(c*f/d)^2*tan(1/2*c*f/d)^2*tan(1/2*e)^2*tan(e) - 4*b^2*sin_integral(2*(d*f*x + c*f)/d
)*tan(c*f/d)^2*tan(1/2*c*f/d)^2*tan(1/2*e)^2*tan(e) - 8*a*b*real_part(cos_integral(f*x + c*f/d))*tan(c*f/d)^2*
tan(1/2*c*f/d)^2*tan(1/2*e)*tan(e)^2 - 8*a*b*real_part(cos_integral(-f*x - c*f/d))*tan(c*f/d)^2*tan(1/2*c*f/d)
^2*tan(1/2*e)*tan(e)^2 + 8*a*b*real_part(cos_integral(f*x + c*f/d))*tan(c*f/d)^2*tan(1/2*c*f/d)*tan(1/2*e)^2*t
an(e)^2 + 8*a*b*real_part(cos_integral(-f*x - c*f/d))*tan(c*f/d)^2*tan(1/2*c*f/d)*tan(1/2*e)^2*tan(e)^2 + 2*b^
2*imag_part(cos_integral(2*f*x + 2*c*f/d))*tan(c*f/d)*tan(1/2*c*f/d)^2*tan(1/2*e)^2*tan(e)^2 - 2*b^2*imag_part
(cos_integral(-2*f*x - 2*c*f/d))*tan(c*f/d)*tan(1/2*c*f/d)^2*tan(1/2*e)^2*tan(e)^2 + 4*b^2*sin_integral(2*(d*f
*x + c*f)/d)*tan(c*f/d)*tan(1/2*c*f/d)^2*tan(1/2*e)^2*tan(e)^2 + 4*a*b*imag_part(cos_integral(f*x + c*f/d))*ta
n(c*f/d)^2*tan(1/2*c*f/d)^2*tan(1/2*e)^2 - 4*a*b*imag_part(cos_integral(-f*x - c*f/d))*tan(c*f/d)^2*tan(1/2*c*
f/d)^2*tan(1/2*e)^2 + 4*a^2*log(abs(d*x + c))*tan(c*f/d)^2*tan(1/2*c*f/d)^2*tan(1/2*e)^2 + 2*b^2*log(abs(d*x +
c))*tan(c*f/d)^2*tan(1/2*c*f/d)^2*tan(1/2*e)^2 + b^2*real_part(cos_integral(2*f*x + 2*c*f/d))*tan(c*f/d)^2*ta
n(1/2*c*f/d)^2*tan(1/2*e)^2 + b^2*real_part(cos_integral(-2*f*x - 2*c*f/d))*tan(c*f/d)^2*tan(1/2*c*f/d)^2*tan(
1/2*e)^2 + 8*a*b*sin_integral((d*f*x + c*f)/d)*tan(c*f/d)^2*tan(1/2*c*f/d)^2*tan(1/2*e)^2 - 4*b^2*real_part(co
s_integral(2*f*x + 2*c*f/d))*tan(c*f/d)*tan(1/2*c*f/d)^2*tan(1/2*e)^2*tan(e) - 4*b^2*real_part(cos_integral(-2
*f*x - 2*c*f/d))*tan(c*f/d)*tan(1/2*c*f/d)^2*tan(1/2*e)^2*tan(e) - 4*a*b*imag_part(cos_integral(f*x + c*f/d))*
tan(c*f/d)^2*tan(1/2*c*f/d)^2*tan(e)^2 + 4*a*b*imag_part(cos_integral(-f*x - c*f/d))*tan(c*f/d)^2*tan(1/2*c*f/
d)^2*tan(e)^2 + 4*a^2*log(abs(d*x + c))*tan(c*f/d)^2*tan(1/2*c*f/d)^2*tan(e)^2 + 2*b^2*log(abs(d*x + c))*tan(c
*f/d)^2*tan(1/2*c*f/d)^2*tan(e)^2 - b^2*real_part(cos_integral(2*f*x + 2*c*f/d))*tan(c*f/d)^2*tan(1/2*c*f/d)^2
*tan(e)^2 - b^2*real_part(cos_integral(-2*f*x - 2*c*f/d))*tan(c*f/d)^2*tan(1/2*c*f/d)^2*tan(e)^2 - 8*a*b*sin_i
ntegral((d*f*x + c*f)/d)*tan(c*f/d)^2*tan(1/2*c*f/d)^2*tan(e)^2 + 16*a*b*imag_part(cos_integral(f*x + c*f/d))*
tan(c*f/d)^2*tan(1/2*c*f/d)*tan(1/2*e)*tan(e)^2 - 16*a*b*imag_part(cos_integral(-f*x - c*f/d))*tan(c*f/d)^2*ta
n(1/2*c*f/d)*tan(1/2*e)*tan(e)^2 + 32*a*b*sin_integral((d*f*x + c*f)/d)*tan(c*f/d)^2*tan(1/2*c*f/d)*tan(1/2*e)
*tan(e)^2 - 4*a*b*imag_part(cos_integral(f*x + c*f/d))*tan(c*f/d)^2*tan(1/2*e)^2*tan(e)^2 + 4*a*b*imag_part(co
s_integral(-f*x - c*f/d))*tan(c*f/d)^2*tan(1/2*e)^2*tan(e)^2 + 4*a^2*log(abs(d*x + c))*tan(c*f/d)^2*tan(1/2*e)
^2*tan(e)^2 + 2*b^2*log(abs(d*x + c))*tan(c*f/d)^2*tan(1/2*e)^2*tan(e)^2 - b^2*real_part(cos_integral(2*f*x +
2*c*f/d))*tan(c*f/d)^2*tan(1/2*e)^2*tan(e)^2 - b^2*real_part(cos_integral(-2*f*x - 2*c*f/d))*tan(c*f/d)^2*tan(
1/2*e)^2*tan(e)^2 - 8*a*b*sin_integral((d*f*x + c*f)/d)*tan(c*f/d)^2*tan(1/2*e)^2*tan(e)^2 + 4*a*b*imag_part(c
os_integral(f*x + c*f/d))*tan(1/2*c*f/d)^2*tan(1/2*e)^2*tan(e)^2 - 4*a*b*imag_part(cos_integral(-f*x - c*f/d))
*tan(1/2*c*f/d)^2*tan(1/2*e)^2*tan(e)^2 + 4*a^2*log(abs(d*x + c))*tan(1/2*c*f/d)^2*tan(1/2*e)^2*tan(e)^2 + 2*b
^2*log(abs(d*x + c))*tan(1/2*c*f/d)^2*tan(1/2*e)^2*tan(e)^2 + b^2*real_part(cos_integral(2*f*x + 2*c*f/d))*tan
(1/2*c*f/d)^2*tan(1/2*e)^2*tan(e)^2 + b^2*real_part(cos_integral(-2*f*x - 2*c*f/d))*tan(1/2*c*f/d)^2*tan(1/2*e
)^2*tan(e)^2 + 8*a*b*sin_integral((d*f*x + c*f)/d)*tan(1/2*c*f/d)^2*tan(1/2*e)^2*tan(e)^2 - 8*a*b*real_part(co
s_integral(f*x + c*f/d))*tan(c*f/d)^2*tan(1/2*c*f/d)^2*tan(1/2*e) - 8*a*b*real_part(cos_integral(-f*x - c*f/d)
)*tan(c*f/d)^2*tan(1/2*c*f/d)^2*tan(1/2*e) + 8*a*b*real_part(cos_integral(f*x + c*f/d))*tan(c*f/d)^2*tan(1/2*c
*f/d)*tan(1/2*e)^2 + 8*a*b*real_part(cos_integral(-f*x - c*f/d))*tan(c*f/d)^2*tan(1/2*c*f/d)*tan(1/2*e)^2 - 2*
b^2*imag_part(cos_integral(2*f*x + 2*c*f/d))*tan(c*f/d)*tan(1/2*c*f/d)^2*tan(1/2*e)^2 + 2*b^2*imag_part(cos_in
tegral(-2*f*x - 2*c*f/d))*tan(c*f/d)*tan(1/2*c*f/d)^2*tan(1/2*e)^2 - 4*b^2*sin_integral(2*(d*f*x + c*f)/d)*tan
(c*f/d)*tan(1/2*c*f/d)^2*tan(1/2*e)^2 - 2*b^2*imag_part(cos_integral(2*f*x + 2*c*f/d))*tan(c*f/d)^2*tan(1/2*c*
f/d)^2*tan(e) + 2*b^2*imag_part(cos_integral(-2...

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,\sin \left (e+f\,x\right )\right )}^2}{c+d\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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