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

Optimal. Leaf size=311 \[ -\frac {a^2 f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^2 d^2 \sqrt {a^2-b^2}}+\frac {a^2 f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^2 d^2 \sqrt {a^2-b^2}}-\frac {i a^2 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^2 d \sqrt {a^2-b^2}}+\frac {i a^2 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b^2 d \sqrt {a^2-b^2}}-\frac {a e x}{b^2}-\frac {a f x^2}{2 b^2}+\frac {f \sin (c+d x)}{b d^2}-\frac {(e+f x) \cos (c+d x)}{b d} \]

[Out]

-a*e*x/b^2-1/2*a*f*x^2/b^2-(f*x+e)*cos(d*x+c)/b/d+f*sin(d*x+c)/b/d^2-I*a^2*(f*x+e)*ln(1-I*b*exp(I*(d*x+c))/(a-
(a^2-b^2)^(1/2)))/b^2/d/(a^2-b^2)^(1/2)+I*a^2*(f*x+e)*ln(1-I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b^2/d/(a^2-
b^2)^(1/2)-a^2*f*polylog(2,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b^2/d^2/(a^2-b^2)^(1/2)+a^2*f*polylog(2,I*b
*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b^2/d^2/(a^2-b^2)^(1/2)

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Rubi [A]  time = 0.55, antiderivative size = 311, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {4515, 3296, 2637, 3323, 2264, 2190, 2279, 2391} \[ -\frac {a^2 f \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^2 d^2 \sqrt {a^2-b^2}}+\frac {a^2 f \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b^2 d^2 \sqrt {a^2-b^2}}-\frac {i a^2 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^2 d \sqrt {a^2-b^2}}+\frac {i a^2 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b^2 d \sqrt {a^2-b^2}}-\frac {a e x}{b^2}-\frac {a f x^2}{2 b^2}+\frac {f \sin (c+d x)}{b d^2}-\frac {(e+f x) \cos (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*e*x)/b^2) - (a*f*x^2)/(2*b^2) - ((e + f*x)*Cos[c + d*x])/(b*d) - (I*a^2*(e + f*x)*Log[1 - (I*b*E^(I*(c +
d*x)))/(a - Sqrt[a^2 - b^2])])/(b^2*Sqrt[a^2 - b^2]*d) + (I*a^2*(e + f*x)*Log[1 - (I*b*E^(I*(c + d*x)))/(a + S
qrt[a^2 - b^2])])/(b^2*Sqrt[a^2 - b^2]*d) - (a^2*f*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b
^2*Sqrt[a^2 - b^2]*d^2) + (a^2*f*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b^2*Sqrt[a^2 - b^2]
*d^2) + (f*Sin[c + d*x])/(b*d^2)

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2264

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q =
Rt[b^2 - 4*a*c, 2]}, Dist[(2*c)/q, Int[((f + g*x)^m*F^u)/(b - q + 2*c*F^u), x], x] - Dist[(2*c)/q, Int[((f +
g*x)^m*F^u)/(b + q + 2*c*F^u), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[
b^2 - 4*a*c, 0] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2637

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

Rule 3296

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 3323

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

Rule 4515

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

Rubi steps

\begin {align*} \int \frac {(e+f x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac {\int (e+f x) \sin (c+d x) \, dx}{b}-\frac {a \int \frac {(e+f x) \sin (c+d x)}{a+b \sin (c+d x)} \, dx}{b}\\ &=-\frac {(e+f x) \cos (c+d x)}{b d}-\frac {a \int (e+f x) \, dx}{b^2}+\frac {a^2 \int \frac {e+f x}{a+b \sin (c+d x)} \, dx}{b^2}+\frac {f \int \cos (c+d x) \, dx}{b d}\\ &=-\frac {a e x}{b^2}-\frac {a f x^2}{2 b^2}-\frac {(e+f x) \cos (c+d x)}{b d}+\frac {f \sin (c+d x)}{b d^2}+\frac {\left (2 a^2\right ) \int \frac {e^{i (c+d x)} (e+f x)}{i b+2 a e^{i (c+d x)}-i b e^{2 i (c+d x)}} \, dx}{b^2}\\ &=-\frac {a e x}{b^2}-\frac {a f x^2}{2 b^2}-\frac {(e+f x) \cos (c+d x)}{b d}+\frac {f \sin (c+d x)}{b d^2}-\frac {\left (2 i a^2\right ) \int \frac {e^{i (c+d x)} (e+f x)}{2 a-2 \sqrt {a^2-b^2}-2 i b e^{i (c+d x)}} \, dx}{b \sqrt {a^2-b^2}}+\frac {\left (2 i a^2\right ) \int \frac {e^{i (c+d x)} (e+f x)}{2 a+2 \sqrt {a^2-b^2}-2 i b e^{i (c+d x)}} \, dx}{b \sqrt {a^2-b^2}}\\ &=-\frac {a e x}{b^2}-\frac {a f x^2}{2 b^2}-\frac {(e+f x) \cos (c+d x)}{b d}-\frac {i a^2 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^2 \sqrt {a^2-b^2} d}+\frac {i a^2 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^2 \sqrt {a^2-b^2} d}+\frac {f \sin (c+d x)}{b d^2}+\frac {\left (i a^2 f\right ) \int \log \left (1-\frac {2 i b e^{i (c+d x)}}{2 a-2 \sqrt {a^2-b^2}}\right ) \, dx}{b^2 \sqrt {a^2-b^2} d}-\frac {\left (i a^2 f\right ) \int \log \left (1-\frac {2 i b e^{i (c+d x)}}{2 a+2 \sqrt {a^2-b^2}}\right ) \, dx}{b^2 \sqrt {a^2-b^2} d}\\ &=-\frac {a e x}{b^2}-\frac {a f x^2}{2 b^2}-\frac {(e+f x) \cos (c+d x)}{b d}-\frac {i a^2 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^2 \sqrt {a^2-b^2} d}+\frac {i a^2 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^2 \sqrt {a^2-b^2} d}+\frac {f \sin (c+d x)}{b d^2}+\frac {\left (a^2 f\right ) \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {2 i b x}{2 a-2 \sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{b^2 \sqrt {a^2-b^2} d^2}-\frac {\left (a^2 f\right ) \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {2 i b x}{2 a+2 \sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{b^2 \sqrt {a^2-b^2} d^2}\\ &=-\frac {a e x}{b^2}-\frac {a f x^2}{2 b^2}-\frac {(e+f x) \cos (c+d x)}{b d}-\frac {i a^2 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^2 \sqrt {a^2-b^2} d}+\frac {i a^2 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^2 \sqrt {a^2-b^2} d}-\frac {a^2 f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^2 \sqrt {a^2-b^2} d^2}+\frac {a^2 f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^2 \sqrt {a^2-b^2} d^2}+\frac {f \sin (c+d x)}{b d^2}\\ \end {align*}

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Mathematica [B]  time = 7.39, size = 709, normalized size = 2.28 \[ \frac {\frac {2 a^2 d (e+f x) \left (\frac {2 (d e-c f) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-\frac {i f \left (\text {Li}_2\left (\frac {a \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a+i \left (b+\sqrt {b^2-a^2}\right )}\right )+\log \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right ) \log \left (\frac {\sqrt {b^2-a^2}+a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {b^2-a^2}-i a+b}\right )\right )}{\sqrt {b^2-a^2}}+\frac {i f \left (\text {Li}_2\left (\frac {a \left (i \tan \left (\frac {1}{2} (c+d x)\right )+1\right )}{a-i \left (b+\sqrt {b^2-a^2}\right )}\right )+\log \left (1+i \tan \left (\frac {1}{2} (c+d x)\right )\right ) \log \left (\frac {\sqrt {b^2-a^2}+a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {b^2-a^2}+i a+b}\right )\right )}{\sqrt {b^2-a^2}}+\frac {i f \left (\text {Li}_2\left (\frac {a \left (\tan \left (\frac {1}{2} (c+d x)\right )+i\right )}{i a-b+\sqrt {b^2-a^2}}\right )+\log \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right ) \log \left (\frac {\sqrt {b^2-a^2}-a \tan \left (\frac {1}{2} (c+d x)\right )-b}{\sqrt {b^2-a^2}+i a-b}\right )\right )}{\sqrt {b^2-a^2}}-\frac {i f \left (\text {Li}_2\left (\frac {i \tan \left (\frac {1}{2} (c+d x)\right ) a+a}{a+i \left (\sqrt {b^2-a^2}-b\right )}\right )+\log \left (1+i \tan \left (\frac {1}{2} (c+d x)\right )\right ) \log \left (\frac {-\sqrt {b^2-a^2}+a \tan \left (\frac {1}{2} (c+d x)\right )+b}{-\sqrt {b^2-a^2}+i a+b}\right )\right )}{\sqrt {b^2-a^2}}\right )}{i f \log \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right )-i f \log \left (1+i \tan \left (\frac {1}{2} (c+d x)\right )\right )-c f+d e}+a (c+d x) (c f-d (2 e+f x))-2 b d (e+f x) \cos (c+d x)+2 b f \sin (c+d x)}{2 b^2 d^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a*(c + d*x)*(c*f - d*(2*e + f*x)) - 2*b*d*(e + f*x)*Cos[c + d*x] + (2*a^2*d*(e + f*x)*((2*(d*e - c*f)*ArcTan[
(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] - (I*f*(Log[1 - I*Tan[(c + d*x)/2]]*Log[(b + Sqrt[-
a^2 + b^2] + a*Tan[(c + d*x)/2])/((-I)*a + b + Sqrt[-a^2 + b^2])] + PolyLog[2, (a*(1 - I*Tan[(c + d*x)/2]))/(a
+ I*(b + Sqrt[-a^2 + b^2]))]))/Sqrt[-a^2 + b^2] + (I*f*(Log[1 + I*Tan[(c + d*x)/2]]*Log[(b + Sqrt[-a^2 + b^2]
+ a*Tan[(c + d*x)/2])/(I*a + b + Sqrt[-a^2 + b^2])] + PolyLog[2, (a*(1 + I*Tan[(c + d*x)/2]))/(a - I*(b + Sqr
t[-a^2 + b^2]))]))/Sqrt[-a^2 + b^2] + (I*f*(Log[1 - I*Tan[(c + d*x)/2]]*Log[(-b + Sqrt[-a^2 + b^2] - a*Tan[(c
+ d*x)/2])/(I*a - b + Sqrt[-a^2 + b^2])] + PolyLog[2, (a*(I + Tan[(c + d*x)/2]))/(I*a - b + Sqrt[-a^2 + b^2])]
))/Sqrt[-a^2 + b^2] - (I*f*(Log[1 + I*Tan[(c + d*x)/2]]*Log[(b - Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2])/(I*a +
b - Sqrt[-a^2 + b^2])] + PolyLog[2, (a + I*a*Tan[(c + d*x)/2])/(a + I*(-b + Sqrt[-a^2 + b^2]))]))/Sqrt[-a^2 +
b^2]))/(d*e - c*f + I*f*Log[1 - I*Tan[(c + d*x)/2]] - I*f*Log[1 + I*Tan[(c + d*x)/2]]) + 2*b*f*Sin[c + d*x])/
(2*b^2*d^2)

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fricas [B]  time = 0.67, size = 1167, normalized size = 3.75 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*(a^3 - a*b^2)*d^2*f*x^2 - 2*I*a^2*b*f*sqrt(-(a^2 - b^2)/b^2)*dilog(-1/2*(2*I*a*cos(d*x + c) + 2*a*sin(
d*x + c) + 2*(b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1) + 2*I*a^2*b*f*sqrt(-(a^2
- b^2)/b^2)*dilog(-1/2*(2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) - 2*(b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(
a^2 - b^2)/b^2) + 2*b)/b + 1) + 2*I*a^2*b*f*sqrt(-(a^2 - b^2)/b^2)*dilog(-1/2*(-2*I*a*cos(d*x + c) + 2*a*sin(d
*x + c) + 2*(b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1) - 2*I*a^2*b*f*sqrt(-(a^2
- b^2)/b^2)*dilog(-1/2*(-2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) - 2*(b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(
a^2 - b^2)/b^2) + 2*b)/b + 1) + 4*(a^3 - a*b^2)*d^2*e*x - 4*(a^2*b - b^3)*f*sin(d*x + c) - 2*(a^2*b*d*e - a^2*
b*c*f)*sqrt(-(a^2 - b^2)/b^2)*log(2*b*cos(d*x + c) + 2*I*b*sin(d*x + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) + 2*I*a)
- 2*(a^2*b*d*e - a^2*b*c*f)*sqrt(-(a^2 - b^2)/b^2)*log(2*b*cos(d*x + c) - 2*I*b*sin(d*x + c) + 2*b*sqrt(-(a^2
- b^2)/b^2) - 2*I*a) + 2*(a^2*b*d*e - a^2*b*c*f)*sqrt(-(a^2 - b^2)/b^2)*log(-2*b*cos(d*x + c) + 2*I*b*sin(d*x
+ c) + 2*b*sqrt(-(a^2 - b^2)/b^2) + 2*I*a) + 2*(a^2*b*d*e - a^2*b*c*f)*sqrt(-(a^2 - b^2)/b^2)*log(-2*b*cos(d*x
+ c) - 2*I*b*sin(d*x + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) - 2*I*a) - 2*(a^2*b*d*f*x + a^2*b*c*f)*sqrt(-(a^2 - b^
2)/b^2)*log(1/2*(2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) + 2*(b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^
2)/b^2) + 2*b)/b) + 2*(a^2*b*d*f*x + a^2*b*c*f)*sqrt(-(a^2 - b^2)/b^2)*log(1/2*(2*I*a*cos(d*x + c) + 2*a*sin(d
*x + c) - 2*(b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) - 2*(a^2*b*d*f*x + a^2*b*c*f)
*sqrt(-(a^2 - b^2)/b^2)*log(1/2*(-2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) + 2*(b*cos(d*x + c) + I*b*sin(d*x + c)
)*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) + 2*(a^2*b*d*f*x + a^2*b*c*f)*sqrt(-(a^2 - b^2)/b^2)*log(1/2*(-2*I*a*cos(d*
x + c) + 2*a*sin(d*x + c) - 2*(b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) + 4*((a^2*b
- b^3)*d*f*x + (a^2*b - b^3)*d*e)*cos(d*x + c))/((a^2*b^2 - b^4)*d^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.52, size = 625, normalized size = 2.01 \[ -\frac {a f \,x^{2}}{2 b^{2}}-\frac {a e x}{b^{2}}-\frac {\left (d f x +d e +i f \right ) {\mathrm e}^{i \left (d x +c \right )}}{2 d^{2} b}-\frac {\left (d f x +d e -i f \right ) {\mathrm e}^{-i \left (d x +c \right )}}{2 d^{2} b}+\frac {2 i a^{2} e \arctan \left (\frac {2 i b \,{\mathrm e}^{i \left (d x +c \right )}-2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{b^{2} d \sqrt {-a^{2}+b^{2}}}+\frac {a^{2} f \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}-\sqrt {-a^{2}+b^{2}}}{i a -\sqrt {-a^{2}+b^{2}}}\right ) x}{b^{2} d \sqrt {-a^{2}+b^{2}}}+\frac {a^{2} f \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}-\sqrt {-a^{2}+b^{2}}}{i a -\sqrt {-a^{2}+b^{2}}}\right ) c}{b^{2} d^{2} \sqrt {-a^{2}+b^{2}}}-\frac {a^{2} f \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right ) x}{b^{2} d \sqrt {-a^{2}+b^{2}}}-\frac {a^{2} f \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right ) c}{b^{2} d^{2} \sqrt {-a^{2}+b^{2}}}-\frac {i a^{2} f \dilog \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}-\sqrt {-a^{2}+b^{2}}}{i a -\sqrt {-a^{2}+b^{2}}}\right )}{b^{2} d^{2} \sqrt {-a^{2}+b^{2}}}+\frac {i a^{2} f \dilog \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right )}{b^{2} d^{2} \sqrt {-a^{2}+b^{2}}}-\frac {2 i a^{2} f c \arctan \left (\frac {2 i b \,{\mathrm e}^{i \left (d x +c \right )}-2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{b^{2} d^{2} \sqrt {-a^{2}+b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a*f*x^2/b^2-a*e*x/b^2-1/2*(d*f*x+I*f+d*e)/d^2/b*exp(I*(d*x+c))-1/2*(d*f*x-I*f+d*e)/d^2/b*exp(-I*(d*x+c))+
2*I*a^2/b^2/d*e/(-a^2+b^2)^(1/2)*arctan(1/2*(2*I*b*exp(I*(d*x+c))-2*a)/(-a^2+b^2)^(1/2))+a^2/b^2/d*f/(-a^2+b^2
)^(1/2)*ln((I*a+b*exp(I*(d*x+c))-(-a^2+b^2)^(1/2))/(I*a-(-a^2+b^2)^(1/2)))*x+a^2/b^2/d^2*f/(-a^2+b^2)^(1/2)*ln
((I*a+b*exp(I*(d*x+c))-(-a^2+b^2)^(1/2))/(I*a-(-a^2+b^2)^(1/2)))*c-a^2/b^2/d*f/(-a^2+b^2)^(1/2)*ln((I*a+b*exp(
I*(d*x+c))+(-a^2+b^2)^(1/2))/(I*a+(-a^2+b^2)^(1/2)))*x-a^2/b^2/d^2*f/(-a^2+b^2)^(1/2)*ln((I*a+b*exp(I*(d*x+c))
+(-a^2+b^2)^(1/2))/(I*a+(-a^2+b^2)^(1/2)))*c-I*a^2/b^2/d^2*f/(-a^2+b^2)^(1/2)*dilog((I*a+b*exp(I*(d*x+c))-(-a^
2+b^2)^(1/2))/(I*a-(-a^2+b^2)^(1/2)))+I*a^2/b^2/d^2*f/(-a^2+b^2)^(1/2)*dilog((I*a+b*exp(I*(d*x+c))+(-a^2+b^2)^
(1/2))/(I*a+(-a^2+b^2)^(1/2)))-2*I*a^2/b^2/d^2*f*c/(-a^2+b^2)^(1/2)*arctan(1/2*(2*I*b*exp(I*(d*x+c))-2*a)/(-a^
2+b^2)^(1/2))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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