3.3.22 \(\int \frac {(e+f x) \sin (c+d x)}{a+b \sin (c+d x)} \, dx\) [222]

Optimal. Leaf size=267 \[ \frac {e x}{b}+\frac {f x^2}{2 b}+\frac {i a (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d}-\frac {i a (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d}+\frac {a f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d^2}-\frac {a f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d^2} \]

[Out]

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

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Rubi [A]
time = 0.32, antiderivative size = 267, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4611, 3404, 2296, 2221, 2317, 2438} \begin {gather*} \frac {a f \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d^2 \sqrt {a^2-b^2}}-\frac {a f \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d^2 \sqrt {a^2-b^2}}+\frac {i a (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d \sqrt {a^2-b^2}}-\frac {i a (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d \sqrt {a^2-b^2}}+\frac {e x}{b}+\frac {f x^2}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2296

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 2317

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 2438

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 3404

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 4611

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

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Mathematica [A]
time = 2.85, size = 343, normalized size = 1.28 \begin {gather*} \frac {x (2 e+f x)}{2 b}+\frac {a \left (-\frac {2 d e \tan ^{-1}\left (\frac {i a+b e^{i (c+d x)}}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-\frac {e^{i c} f \left (d x \left (\log \left (1+\frac {b e^{i (2 c+d x)}}{i a e^{i c}-\sqrt {\left (-a^2+b^2\right ) e^{2 i c}}}\right )-\log \left (1+\frac {b e^{i (2 c+d x)}}{i a e^{i c}+\sqrt {\left (-a^2+b^2\right ) e^{2 i c}}}\right )\right )-i \text {Li}_2\left (\frac {i b e^{i (2 c+d x)}}{a e^{i c}+i \sqrt {\left (-a^2+b^2\right ) e^{2 i c}}}\right )+i \text {Li}_2\left (-\frac {b e^{i (2 c+d x)}}{i a e^{i c}+\sqrt {\left (-a^2+b^2\right ) e^{2 i c}}}\right )\right )}{\sqrt {\left (-a^2+b^2\right ) e^{2 i c}}}\right )}{b d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 547 vs. \(2 (235 ) = 470\).
time = 0.11, size = 548, normalized size = 2.05

method	result	size
risch	\(\frac {f \,x^{2}}{2 b}+\frac {e x}{b}-\frac {2 i a e \arctan \left (\frac {2 i b \,{\mathrm e}^{i \left (d x +c \right )}-2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{d b \sqrt {-a^{2}+b^{2}}}-\frac {a f \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}-\sqrt {-a^{2}+b^{2}}}{i a -\sqrt {-a^{2}+b^{2}}}\right ) x}{d b \sqrt {-a^{2}+b^{2}}}-\frac {a f \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}-\sqrt {-a^{2}+b^{2}}}{i a -\sqrt {-a^{2}+b^{2}}}\right ) c}{d^{2} b \sqrt {-a^{2}+b^{2}}}+\frac {a f \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right ) x}{d b \sqrt {-a^{2}+b^{2}}}+\frac {a f \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right ) c}{d^{2} b \sqrt {-a^{2}+b^{2}}}-\frac {i a f \dilog \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right )}{d^{2} b \sqrt {-a^{2}+b^{2}}}+\frac {i a f \dilog \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}-\sqrt {-a^{2}+b^{2}}}{i a -\sqrt {-a^{2}+b^{2}}}\right )}{d^{2} b \sqrt {-a^{2}+b^{2}}}+\frac {2 i a f c \arctan \left (\frac {2 i b \,{\mathrm e}^{i \left (d x +c \right )}-2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{d^{2} b \sqrt {-a^{2}+b^{2}}}\)	\(548\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*f*x^2/b+e*x/b-2*I/d/b*a*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))-1/d/b*a
*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-1/d^2/b*a*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+1/d/b*a*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+1/d^2/b*a*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/d^2/b*a*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/d^2/b*a*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/d^2/b*a*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 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*sin(d*x+c)/(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 de

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1057 vs. \(2 (230) = 460\).
time = 0.54, size = 1057, normalized size = 3.96 \begin {gather*} \frac {{\left (a^{2} - b^{2}\right )} d^{2} f x^{2} + 2 \, {\left (a^{2} - b^{2}\right )} d^{2} x e - i \, a b f \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} {\rm Li}_2\left (\frac {i \, a \cos \left (d x + c\right ) - a \sin \left (d x + c\right ) + {\left (b \cos \left (d x + c\right ) + i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b} + 1\right ) + i \, a b f \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} {\rm Li}_2\left (\frac {i \, a \cos \left (d x + c\right ) - a \sin \left (d x + c\right ) - {\left (b \cos \left (d x + c\right ) + i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b} + 1\right ) + i \, a b f \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} {\rm Li}_2\left (\frac {-i \, a \cos \left (d x + c\right ) - a \sin \left (d x + c\right ) + {\left (b \cos \left (d x + c\right ) - i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b} + 1\right ) - i \, a b f \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} {\rm Li}_2\left (\frac {-i \, a \cos \left (d x + c\right ) - a \sin \left (d x + c\right ) - {\left (b \cos \left (d x + c\right ) - i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b} + 1\right ) + {\left (a b c f - a b d e\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} \log \left (2 \, b \cos \left (d x + c\right ) + 2 i \, b \sin \left (d x + c\right ) + 2 \, b \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} + 2 i \, a\right ) + {\left (a b c f - a b d e\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} \log \left (2 \, b \cos \left (d x + c\right ) - 2 i \, b \sin \left (d x + c\right ) + 2 \, b \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - 2 i \, a\right ) - {\left (a b c f - a b d e\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} \log \left (-2 \, b \cos \left (d x + c\right ) + 2 i \, b \sin \left (d x + c\right ) + 2 \, b \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} + 2 i \, a\right ) - {\left (a b c f - a b d e\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} \log \left (-2 \, b \cos \left (d x + c\right ) - 2 i \, b \sin \left (d x + c\right ) + 2 \, b \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - 2 i \, a\right ) + {\left (a b d f x + a b c f\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} \log \left (-\frac {i \, a \cos \left (d x + c\right ) - a \sin \left (d x + c\right ) + {\left (b \cos \left (d x + c\right ) + i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b}\right ) - {\left (a b d f x + a b c f\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} \log \left (-\frac {i \, a \cos \left (d x + c\right ) - a \sin \left (d x + c\right ) - {\left (b \cos \left (d x + c\right ) + i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b}\right ) + {\left (a b d f x + a b c f\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} \log \left (-\frac {-i \, a \cos \left (d x + c\right ) - a \sin \left (d x + c\right ) + {\left (b \cos \left (d x + c\right ) - i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b}\right ) - {\left (a b d f x + a b c f\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} \log \left (-\frac {-i \, a \cos \left (d x + c\right ) - a \sin \left (d x + c\right ) - {\left (b \cos \left (d x + c\right ) - i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b}\right )}{2 \, {\left (a^{2} b - b^{3}\right )} d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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