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

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

[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 = 0.58, 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 = {4515, 3323, 2264, 2190, 2279, 2391} \[ \frac {a f \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d^2 \sqrt {a^2-b^2}}-\frac {a f \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d^2 \sqrt {a^2-b^2}}+\frac {i a (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d \sqrt {a^2-b^2}}-\frac {i a (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d \sqrt {a^2-b^2}}+\frac {e x}{b}+\frac {f x^2}{2 b} \]

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 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 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 (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) \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 \sqrt {a^2-b^2} d^2}+\frac {(a f) \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 \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 = 1.72, size = 299, normalized size = 1.12 \[ \frac {x (2 e+f x)}{2 b}-\frac {i a \left (-i d \left (2 e \sqrt {b^2-a^2} \tan ^{-1}\left (\frac {i a+b e^{i (c+d x)}}{\sqrt {a^2-b^2}}\right )+f x \sqrt {a^2-b^2} \left (\log \left (1-\frac {b e^{i (c+d x)}}{\sqrt {b^2-a^2}-i a}\right )-\log \left (1+\frac {b e^{i (c+d x)}}{\sqrt {b^2-a^2}+i a}\right )\right )\right )-f \sqrt {a^2-b^2} \text {Li}_2\left (\frac {b e^{i (c+d x)}}{\sqrt {b^2-a^2}-i a}\right )+f \sqrt {a^2-b^2} \text {Li}_2\left (-\frac {b e^{i (c+d x)}}{i a+\sqrt {b^2-a^2}}\right )\right )}{b d^2 \sqrt {-\left (a^2-b^2\right )^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [B]  time = 0.72, size = 1065, normalized size = 3.99 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*(a^2 - b^2)*d^2*f*x^2 + 4*(a^2 - b^2)*d^2*e*x - 2*I*a*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*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*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*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*(a*b*d*e - a*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*b*d*e - a*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*b*d*e - a*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*b*d*e -
a*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*b*d*f*x + a*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*b*d*f*x + a*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*b*d*f*x + a*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*b*d*f*x + a*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))/((a^2*b - b^3)*d^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*sin(d*x+c)/(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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maple [B]  time = 0.20, size = 548, normalized size = 2.05 \[ \frac {f \,x^{2}}{2 b}+\frac {e x}{b}-\frac {2 i a e \arctan \left (\frac {2 i b \,{\mathrm e}^{i \left (d x +c \right )}-2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{d b \sqrt {-a^{2}+b^{2}}}-\frac {a f \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}-\sqrt {-a^{2}+b^{2}}}{i a -\sqrt {-a^{2}+b^{2}}}\right ) x}{d b \sqrt {-a^{2}+b^{2}}}-\frac {a f \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}-\sqrt {-a^{2}+b^{2}}}{i a -\sqrt {-a^{2}+b^{2}}}\right ) c}{d^{2} b \sqrt {-a^{2}+b^{2}}}+\frac {a f \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right ) x}{d b \sqrt {-a^{2}+b^{2}}}+\frac {a f \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right ) c}{d^{2} b \sqrt {-a^{2}+b^{2}}}+\frac {i a f \dilog \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}-\sqrt {-a^{2}+b^{2}}}{i a -\sqrt {-a^{2}+b^{2}}}\right )}{d^{2} b \sqrt {-a^{2}+b^{2}}}-\frac {i a f \dilog \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right )}{d^{2} b \sqrt {-a^{2}+b^{2}}}+\frac {2 i a f c \arctan \left (\frac {2 i b \,{\mathrm e}^{i \left (d x +c \right )}-2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{d^{2} b \sqrt {-a^{2}+b^{2}}} \]

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)

[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 \[ \text {Exception raised: ValueError} \]

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 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 \[ \int \frac {\sin \left (c+d\,x\right )\,\left (e+f\,x\right )}{a+b\,\sin \left (c+d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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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