3.2.61 \(\int \frac {(a+b \sin (e+f x))^2}{(c+d x)^2} \, dx\) [161]

Optimal. Leaf size=183 \[ -\frac {a^2}{d (c+d x)}+\frac {2 a b f \cos \left (e-\frac {c f}{d}\right ) \text {Ci}\left (\frac {c f}{d}+f x\right )}{d^2}+\frac {b^2 f \text {Ci}\left (\frac {2 c f}{d}+2 f x\right ) \sin \left (2 e-\frac {2 c f}{d}\right )}{d^2}-\frac {2 a b \sin (e+f x)}{d (c+d x)}-\frac {b^2 \sin ^2(e+f x)}{d (c+d x)}-\frac {2 a b f \sin \left (e-\frac {c f}{d}\right ) \text {Si}\left (\frac {c f}{d}+f x\right )}{d^2}+\frac {b^2 f \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 c f}{d}+2 f x\right )}{d^2} \]

[Out]

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

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Rubi [A]
time = 0.23, antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {3398, 3378, 3384, 3380, 3383, 3394, 12} \begin {gather*} -\frac {a^2}{d (c+d x)}+\frac {2 a b f \text {CosIntegral}\left (\frac {c f}{d}+f x\right ) \cos \left (e-\frac {c f}{d}\right )}{d^2}-\frac {2 a b f \sin \left (e-\frac {c f}{d}\right ) \text {Si}\left (x f+\frac {c f}{d}\right )}{d^2}-\frac {2 a b \sin (e+f x)}{d (c+d x)}+\frac {b^2 f \text {CosIntegral}\left (\frac {2 c f}{d}+2 f x\right ) \sin \left (2 e-\frac {2 c f}{d}\right )}{d^2}+\frac {b^2 f \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (2 x f+\frac {2 c f}{d}\right )}{d^2}-\frac {b^2 \sin ^2(e+f x)}{d (c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 3394

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Simp[(c + d*x)^(m + 1)*(Sin[e + f*x]^
n/(d*(m + 1))), x] - Dist[f*(n/(d*(m + 1))), Int[ExpandTrigReduce[(c + d*x)^(m + 1), Cos[e + f*x]*Sin[e + f*x]
^(n - 1), x], x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && GeQ[m, -2] && 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)^2} \, dx &=\int \left (\frac {a^2}{(c+d x)^2}+\frac {2 a b \sin (e+f x)}{(c+d x)^2}+\frac {b^2 \sin ^2(e+f x)}{(c+d x)^2}\right ) \, dx\\ &=-\frac {a^2}{d (c+d x)}+(2 a b) \int \frac {\sin (e+f x)}{(c+d x)^2} \, dx+b^2 \int \frac {\sin ^2(e+f x)}{(c+d x)^2} \, dx\\ &=-\frac {a^2}{d (c+d x)}-\frac {2 a b \sin (e+f x)}{d (c+d x)}-\frac {b^2 \sin ^2(e+f x)}{d (c+d x)}+\frac {(2 a b f) \int \frac {\cos (e+f x)}{c+d x} \, dx}{d}+\frac {\left (2 b^2 f\right ) \int \frac {\sin (2 e+2 f x)}{2 (c+d x)} \, dx}{d}\\ &=-\frac {a^2}{d (c+d x)}-\frac {2 a b \sin (e+f x)}{d (c+d x)}-\frac {b^2 \sin ^2(e+f x)}{d (c+d x)}+\frac {\left (b^2 f\right ) \int \frac {\sin (2 e+2 f x)}{c+d x} \, dx}{d}+\frac {\left (2 a b f \cos \left (e-\frac {c f}{d}\right )\right ) \int \frac {\cos \left (\frac {c f}{d}+f x\right )}{c+d x} \, dx}{d}-\frac {\left (2 a b f \sin \left (e-\frac {c f}{d}\right )\right ) \int \frac {\sin \left (\frac {c f}{d}+f x\right )}{c+d x} \, dx}{d}\\ &=-\frac {a^2}{d (c+d x)}+\frac {2 a b f \cos \left (e-\frac {c f}{d}\right ) \text {Ci}\left (\frac {c f}{d}+f x\right )}{d^2}-\frac {2 a b \sin (e+f x)}{d (c+d x)}-\frac {b^2 \sin ^2(e+f x)}{d (c+d x)}-\frac {2 a b f \sin \left (e-\frac {c f}{d}\right ) \text {Si}\left (\frac {c f}{d}+f x\right )}{d^2}+\frac {\left (b^2 f \cos \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}{d}+\frac {\left (b^2 f \sin \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}{d}\\ &=-\frac {a^2}{d (c+d x)}+\frac {2 a b f \cos \left (e-\frac {c f}{d}\right ) \text {Ci}\left (\frac {c f}{d}+f x\right )}{d^2}+\frac {b^2 f \text {Ci}\left (\frac {2 c f}{d}+2 f x\right ) \sin \left (2 e-\frac {2 c f}{d}\right )}{d^2}-\frac {2 a b \sin (e+f x)}{d (c+d x)}-\frac {b^2 \sin ^2(e+f x)}{d (c+d x)}-\frac {2 a b f \sin \left (e-\frac {c f}{d}\right ) \text {Si}\left (\frac {c f}{d}+f x\right )}{d^2}+\frac {b^2 f \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 c f}{d}+2 f x\right )}{d^2}\\ \end {align*}

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Mathematica [A]
time = 0.37, size = 232, normalized size = 1.27 \begin {gather*} \frac {-2 a^2 d-b^2 d+b^2 d \cos (2 (e+f x))+4 a b f (c+d x) \cos \left (e-\frac {c f}{d}\right ) \text {Ci}\left (f \left (\frac {c}{d}+x\right )\right )+2 b^2 f (c+d x) \text {Ci}\left (\frac {2 f (c+d x)}{d}\right ) \sin \left (2 e-\frac {2 c f}{d}\right )-4 a b d \sin (e+f x)-4 a b c f \sin \left (e-\frac {c f}{d}\right ) \text {Si}\left (f \left (\frac {c}{d}+x\right )\right )-4 a b d f x \sin \left (e-\frac {c f}{d}\right ) \text {Si}\left (f \left (\frac {c}{d}+x\right )\right )+2 b^2 c f \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 f (c+d x)}{d}\right )+2 b^2 d f x \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 f (c+d x)}{d}\right )}{2 d^2 (c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.10, size = 301, normalized size = 1.64

method	result	size
derivativedivides	\(\frac {-\frac {a^{2} f^{2}}{\left (c f -d e +d \left (f x +e \right )\right ) d}+2 f^{2} a b \left (-\frac {\sin \left (f x +e \right )}{\left (c f -d e +d \left (f x +e \right )\right ) d}+\frac {\frac {\sinIntegral \left (f x +e +\frac {c f -d e}{d}\right ) \sin \left (\frac {c f -d e}{d}\right )}{d}+\frac {\cosineIntegral \left (f x +e +\frac {c f -d e}{d}\right ) \cos \left (\frac {c f -d e}{d}\right )}{d}}{d}\right )-\frac {f^{2} b^{2}}{2 \left (c f -d e +d \left (f x +e \right )\right ) d}-\frac {f^{2} b^{2} \left (-\frac {2 \cos \left (2 f x +2 e \right )}{\left (c f -d e +d \left (f x +e \right )\right ) d}-\frac {2 \left (\frac {2 \sinIntegral \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \cos \left (\frac {2 c f -2 d e}{d}\right )}{d}-\frac {2 \cosineIntegral \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \sin \left (\frac {2 c f -2 d e}{d}\right )}{d}\right )}{d}\right )}{4}}{f}\)	\(301\)
default	\(\frac {-\frac {a^{2} f^{2}}{\left (c f -d e +d \left (f x +e \right )\right ) d}+2 f^{2} a b \left (-\frac {\sin \left (f x +e \right )}{\left (c f -d e +d \left (f x +e \right )\right ) d}+\frac {\frac {\sinIntegral \left (f x +e +\frac {c f -d e}{d}\right ) \sin \left (\frac {c f -d e}{d}\right )}{d}+\frac {\cosineIntegral \left (f x +e +\frac {c f -d e}{d}\right ) \cos \left (\frac {c f -d e}{d}\right )}{d}}{d}\right )-\frac {f^{2} b^{2}}{2 \left (c f -d e +d \left (f x +e \right )\right ) d}-\frac {f^{2} b^{2} \left (-\frac {2 \cos \left (2 f x +2 e \right )}{\left (c f -d e +d \left (f x +e \right )\right ) d}-\frac {2 \left (\frac {2 \sinIntegral \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \cos \left (\frac {2 c f -2 d e}{d}\right )}{d}-\frac {2 \cosineIntegral \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \sin \left (\frac {2 c f -2 d e}{d}\right )}{d}\right )}{d}\right )}{4}}{f}\)	\(301\)
risch	\(-\frac {f a b \,{\mathrm e}^{\frac {i \left (c f -d e \right )}{d}} \expIntegral \left (1, i f x +i e +\frac {i \left (c f -d e \right )}{d}\right )}{d^{2}}-\frac {a^{2}}{d \left (d x +c \right )}-\frac {b^{2}}{2 d \left (d x +c \right )}-\frac {i b^{2} f \,{\mathrm e}^{\frac {2 i \left (c f -d e \right )}{d}} \expIntegral \left (1, 2 i f x +2 i e +\frac {2 i \left (c f -d e \right )}{d}\right )}{2 d^{2}}+\frac {i f \,b^{2} {\mathrm e}^{-\frac {2 i \left (c f -d e \right )}{d}} \expIntegral \left (1, -2 i f x -2 i e -\frac {2 \left (i c f -i d e \right )}{d}\right )}{2 d^{2}}-\frac {a b f \,{\mathrm e}^{-\frac {i \left (c f -d e \right )}{d}} \expIntegral \left (1, -i f x -i e -\frac {i c f -i d e}{d}\right )}{d^{2}}-\frac {a b \left (-2 d x f -2 c f \right ) \sin \left (f x +e \right )}{d \left (d x +c \right ) \left (-d x f -c f \right )}+\frac {b^{2} \left (-2 d x f -2 c f \right ) \cos \left (2 f x +2 e \right )}{4 d \left (d x +c \right ) \left (-d x f -c f \right )}\)	\(324\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C] Result contains complex when optimal does not.
time = 0.50, size = 395, normalized size = 2.16 \begin {gather*} -\frac {\frac {4 \, a^{2} f^{2}}{{\left (f x + e\right )} d^{2} + c d f - d^{2} e} - \frac {4 \, {\left (f^{2} {\left (-i \, E_{2}\left (\frac {i \, {\left (f x + e\right )} d + i \, c f - i \, d e}{d}\right ) + i \, E_{2}\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^{2} {\left (E_{2}\left (\frac {i \, {\left (f x + e\right )} d + i \, c f - i \, d e}{d}\right ) + E_{2}\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}{{\left (f x + e\right )} d^{2} + c d f - d^{2} e} - \frac {{\left (f^{2} {\left (E_{2}\left (\frac {2 \, {\left (-i \, {\left (f x + e\right )} d - i \, c f + i \, d e\right )}}{d}\right ) + E_{2}\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^{2} {\left (i \, E_{2}\left (\frac {2 \, {\left (-i \, {\left (f x + e\right )} d - i \, c f + i \, d e\right )}}{d}\right ) - i \, E_{2}\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^{2}\right )} b^{2}}{{\left (f x + e\right )} d^{2} + c d f - d^{2} e}}{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)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.37, size = 286, normalized size = 1.56 \begin {gather*} \frac {2 \, b^{2} d \cos \left (f x + e\right )^{2} - 4 \, a b d \sin \left (f x + e\right ) + 2 \, {\left (b^{2} d f x + b^{2} c f\right )} \cos \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 ) - 4 \, {\left (a b d f x + a b c f\right )} \sin \left (-\frac {c f - d e}{d}\right ) \operatorname {Si}\left (\frac {d f x + c f}{d}\right ) - 2 \, {\left (a^{2} + b^{2}\right )} d + 2 \, {\left ({\left (a b d f x + a b c f\right )} \operatorname {Ci}\left (\frac {d f x + c f}{d}\right ) + {\left (a b d f x + a b c f\right )} \operatorname {Ci}\left (-\frac {d f x + c f}{d}\right )\right )} \cos \left (-\frac {c f - d e}{d}\right ) + {\left ({\left (b^{2} d f x + b^{2} c f\right )} \operatorname {Ci}\left (\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) + {\left (b^{2} d f x + b^{2} c f\right )} \operatorname {Ci}\left (-\frac {2 \, {\left (d f x + c f\right )}}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (c f - d e\right )}}{d}\right )}{2 \, {\left (d^{3} x + c d^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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}}{\left (c + d x\right )^{2}}\, 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)**2,x)

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 1135 vs. \(2 (192) = 384\).
time = 4.29, size = 1135, normalized size = 6.20 \begin {gather*} \frac {{\left (4 \, {\left (d x + c\right )} a b {\left (\frac {c f}{d x + c} - f - \frac {d e}{d x + c}\right )} f^{2} \cos \left (\frac {c f - d e}{d}\right ) \operatorname {Ci}\left (-\frac {{\left (d x + c\right )} {\left (\frac {c f}{d x + c} - f - \frac {d e}{d x + c}\right )} - c f + d e}{d}\right ) - 4 \, a b c f^{3} \cos \left (\frac {c f - d e}{d}\right ) \operatorname {Ci}\left (-\frac {{\left (d x + c\right )} {\left (\frac {c f}{d x + c} - f - \frac {d e}{d x + c}\right )} - c f + d e}{d}\right ) + 4 \, a b d f^{2} \cos \left (\frac {c f - d e}{d}\right ) \operatorname {Ci}\left (-\frac {{\left (d x + c\right )} {\left (\frac {c f}{d x + c} - f - \frac {d e}{d x + c}\right )} - c f + d e}{d}\right ) e - 2 \, {\left (d x + c\right )} b^{2} {\left (\frac {c f}{d x + c} - f - \frac {d e}{d x + c}\right )} f^{2} \operatorname {Ci}\left (-\frac {2 \, {\left ({\left (d x + c\right )} {\left (\frac {c f}{d x + c} - f - \frac {d e}{d x + c}\right )} - c f + d e\right )}}{d}\right ) \sin \left (\frac {2 \, {\left (c f - d e\right )}}{d}\right ) + 2 \, b^{2} c f^{3} \operatorname {Ci}\left (-\frac {2 \, {\left ({\left (d x + c\right )} {\left (\frac {c f}{d x + c} - f - \frac {d e}{d x + c}\right )} - c f + d e\right )}}{d}\right ) \sin \left (\frac {2 \, {\left (c f - d e\right )}}{d}\right ) - 2 \, b^{2} d f^{2} \operatorname {Ci}\left (-\frac {2 \, {\left ({\left (d x + c\right )} {\left (\frac {c f}{d x + c} - f - \frac {d e}{d x + c}\right )} - c f + d e\right )}}{d}\right ) e \sin \left (\frac {2 \, {\left (c f - d e\right )}}{d}\right ) + 4 \, {\left (d x + c\right )} a b {\left (\frac {c f}{d x + c} - f - \frac {d e}{d x + c}\right )} f^{2} \sin \left (\frac {c f - d e}{d}\right ) \operatorname {Si}\left (-\frac {{\left (d x + c\right )} {\left (\frac {c f}{d x + c} - f - \frac {d e}{d x + c}\right )} - c f + d e}{d}\right ) - 4 \, a b c f^{3} \sin \left (\frac {c f - d e}{d}\right ) \operatorname {Si}\left (-\frac {{\left (d x + c\right )} {\left (\frac {c f}{d x + c} - f - \frac {d e}{d x + c}\right )} - c f + d e}{d}\right ) + 4 \, a b d f^{2} e \sin \left (\frac {c f - d e}{d}\right ) \operatorname {Si}\left (-\frac {{\left (d x + c\right )} {\left (\frac {c f}{d x + c} - f - \frac {d e}{d x + c}\right )} - c f + d e}{d}\right ) + 2 \, {\left (d x + c\right )} b^{2} {\left (\frac {c f}{d x + c} - f - \frac {d e}{d x + c}\right )} f^{2} \cos \left (\frac {2 \, {\left (c f - d e\right )}}{d}\right ) \operatorname {Si}\left (-\frac {2 \, {\left ({\left (d x + c\right )} {\left (\frac {c f}{d x + c} - f - \frac {d e}{d x + c}\right )} - c f + d e\right )}}{d}\right ) - 2 \, b^{2} c f^{3} \cos \left (\frac {2 \, {\left (c f - d e\right )}}{d}\right ) \operatorname {Si}\left (-\frac {2 \, {\left ({\left (d x + c\right )} {\left (\frac {c f}{d x + c} - f - \frac {d e}{d x + c}\right )} - c f + d e\right )}}{d}\right ) + 2 \, b^{2} d f^{2} \cos \left (\frac {2 \, {\left (c f - d e\right )}}{d}\right ) e \operatorname {Si}\left (-\frac {2 \, {\left ({\left (d x + c\right )} {\left (\frac {c f}{d x + c} - f - \frac {d e}{d x + c}\right )} - c f + d e\right )}}{d}\right ) - b^{2} d f^{2} \cos \left (\frac {2 \, {\left (d x + c\right )} {\left (\frac {c f}{d x + c} - f - \frac {d e}{d x + c}\right )}}{d}\right ) - 4 \, a b d f^{2} \sin \left (\frac {{\left (d x + c\right )} {\left (\frac {c f}{d x + c} - f - \frac {d e}{d x + c}\right )}}{d}\right ) + 2 \, a^{2} d f^{2} + b^{2} d f^{2}\right )} d^{2}}{2 \, {\left ({\left (d x + c\right )} d^{4} {\left (\frac {c f}{d x + c} - f - \frac {d e}{d x + c}\right )} - c d^{4} f + d^{5} e\right )} 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)^2,x, algorithm="giac")

[Out]

1/2*(4*(d*x + c)*a*b*(c*f/(d*x + c) - f - d*e/(d*x + c))*f^2*cos((c*f - d*e)/d)*cos_integral(-((d*x + c)*(c*f/
(d*x + c) - f - d*e/(d*x + c)) - c*f + d*e)/d) - 4*a*b*c*f^3*cos((c*f - d*e)/d)*cos_integral(-((d*x + c)*(c*f/
(d*x + c) - f - d*e/(d*x + c)) - c*f + d*e)/d) + 4*a*b*d*f^2*cos((c*f - d*e)/d)*cos_integral(-((d*x + c)*(c*f/
(d*x + c) - f - d*e/(d*x + c)) - c*f + d*e)/d)*e - 2*(d*x + c)*b^2*(c*f/(d*x + c) - f - d*e/(d*x + c))*f^2*cos
_integral(-2*((d*x + c)*(c*f/(d*x + c) - f - d*e/(d*x + c)) - c*f + d*e)/d)*sin(2*(c*f - d*e)/d) + 2*b^2*c*f^3
*cos_integral(-2*((d*x + c)*(c*f/(d*x + c) - f - d*e/(d*x + c)) - c*f + d*e)/d)*sin(2*(c*f - d*e)/d) - 2*b^2*d
*f^2*cos_integral(-2*((d*x + c)*(c*f/(d*x + c) - f - d*e/(d*x + c)) - c*f + d*e)/d)*e*sin(2*(c*f - d*e)/d) + 4
*(d*x + c)*a*b*(c*f/(d*x + c) - f - d*e/(d*x + c))*f^2*sin((c*f - d*e)/d)*sin_integral(-((d*x + c)*(c*f/(d*x +
c) - f - d*e/(d*x + c)) - c*f + d*e)/d) - 4*a*b*c*f^3*sin((c*f - d*e)/d)*sin_integral(-((d*x + c)*(c*f/(d*x +
c) - f - d*e/(d*x + c)) - c*f + d*e)/d) + 4*a*b*d*f^2*e*sin((c*f - d*e)/d)*sin_integral(-((d*x + c)*(c*f/(d*x
+ c) - f - d*e/(d*x + c)) - c*f + d*e)/d) + 2*(d*x + c)*b^2*(c*f/(d*x + c) - f - d*e/(d*x + c))*f^2*cos(2*(c*
f - d*e)/d)*sin_integral(-2*((d*x + c)*(c*f/(d*x + c) - f - d*e/(d*x + c)) - c*f + d*e)/d) - 2*b^2*c*f^3*cos(2
*(c*f - d*e)/d)*sin_integral(-2*((d*x + c)*(c*f/(d*x + c) - f - d*e/(d*x + c)) - c*f + d*e)/d) + 2*b^2*d*f^2*c
os(2*(c*f - d*e)/d)*e*sin_integral(-2*((d*x + c)*(c*f/(d*x + c) - f - d*e/(d*x + c)) - c*f + d*e)/d) - b^2*d*f
^2*cos(2*(d*x + c)*(c*f/(d*x + c) - f - d*e/(d*x + c))/d) - 4*a*b*d*f^2*sin((d*x + c)*(c*f/(d*x + c) - f - d*e
/(d*x + c))/d) + 2*a^2*d*f^2 + b^2*d*f^2)*d^2/(((d*x + c)*d^4*(c*f/(d*x + c) - f - d*e/(d*x + c)) - c*d^4*f +
d^5*e)*f)

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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}{{\left (c+d\,x\right )}^2} \,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)^2,x)

[Out]

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

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