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

Optimal. Leaf size=212 \[ -\frac {i f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d^2}-\frac {i f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b d^2}+\frac {(e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}+\frac {(e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {i (e+f x)^2}{2 b f} \]

[Out]

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

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Rubi [A]  time = 0.28, antiderivative size = 212, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4519, 2190, 2279, 2391} \[ -\frac {i f \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d^2}-\frac {i f \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d^2}+\frac {(e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}+\frac {(e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {i (e+f x)^2}{2 b f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/2)*(e + f*x)^2)/(b*f) + ((e + f*x)*Log[1 - (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*d) + ((e + f*
x)*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*d) - (I*f*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a - Sq
rt[a^2 - b^2])])/(b*d^2) - (I*f*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(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 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 4519

Int[(Cos[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sin[(c_.) + (d_.)*(x_)]), x_Symbol] :>
-Simp[(I*(e + f*x)^(m + 1))/(b*f*(m + 1)), x] + (Int[((e + f*x)^m*E^(I*(c + d*x)))/(a - Rt[a^2 - b^2, 2] - I*b
*E^(I*(c + d*x))), x] + Int[((e + f*x)^m*E^(I*(c + d*x)))/(a + Rt[a^2 - b^2, 2] - I*b*E^(I*(c + d*x))), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && PosQ[a^2 - b^2]

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 197, normalized size = 0.93 \[ -\frac {i \left (d (e+f x) \left (2 i f \log \left (1+\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}-a}\right )+2 i f \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )+d e+d f x\right )+2 f^2 \text {Li}_2\left (-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}-a}\right )+2 f^2 \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )\right )}{2 b d^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.63, size = 781, normalized size = 3.68 \[ \frac {i \, f {\rm Li}_2\left (-\frac {2 i \, a \cos \left (d x + c\right ) + 2 \, a \sin \left (d x + c\right ) + 2 \, {\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}}} + 2 \, b}{2 \, b} + 1\right ) + i \, f {\rm Li}_2\left (-\frac {2 i \, a \cos \left (d x + c\right ) + 2 \, a \sin \left (d x + c\right ) - 2 \, {\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}}} + 2 \, b}{2 \, b} + 1\right ) - i \, f {\rm Li}_2\left (-\frac {-2 i \, a \cos \left (d x + c\right ) + 2 \, a \sin \left (d x + c\right ) + 2 \, {\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}}} + 2 \, b}{2 \, b} + 1\right ) - i \, f {\rm Li}_2\left (-\frac {-2 i \, a \cos \left (d x + c\right ) + 2 \, a \sin \left (d x + c\right ) - 2 \, {\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}}} + 2 \, b}{2 \, b} + 1\right ) + {\left (d e - c f\right )} \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 (d e - c f\right )} \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 (d e - c f\right )} \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 (d e - c f\right )} \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 (d f x + c f\right )} \log \left (\frac {2 i \, a \cos \left (d x + c\right ) + 2 \, a \sin \left (d x + c\right ) + 2 \, {\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}}} + 2 \, b}{2 \, b}\right ) + {\left (d f x + c f\right )} \log \left (\frac {2 i \, a \cos \left (d x + c\right ) + 2 \, a \sin \left (d x + c\right ) - 2 \, {\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}}} + 2 \, b}{2 \, b}\right ) + {\left (d f x + c f\right )} \log \left (\frac {-2 i \, a \cos \left (d x + c\right ) + 2 \, a \sin \left (d x + c\right ) + 2 \, {\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}}} + 2 \, b}{2 \, b}\right ) + {\left (d f x + c f\right )} \log \left (\frac {-2 i \, a \cos \left (d x + c\right ) + 2 \, a \sin \left (d x + c\right ) - 2 \, {\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}}} + 2 \, b}{2 \, b}\right )}{2 \, b d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(I*f*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) + I*f*dilog(-1/2*(2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) - 2*(b*cos(d*x + c) - I*b*si
n(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1) - I*f*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) - I*f*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) + (d*e - c*f)
*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) + (d*e - c*f)*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) + (d*e - c*f)*log(-2*b*cos(d*x + c) + 2*I*b*s
in(d*x + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) + 2*I*a) + (d*e - c*f)*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) + (d*f*x + c*f)*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) + (d*f*x + c*f)*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) + (d*f*x + c*f)*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) + (d*f*x + c*f)*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))*sq
rt(-(a^2 - b^2)/b^2) + 2*b)/b))/(b*d^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.24, size = 1006, normalized size = 4.75 \[ -\frac {i f \,c^{2}}{d^{2} b}+\frac {i e x}{b}+\frac {i 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 ) a^{2}}{d^{2} b \left (-a^{2}+b^{2}\right )}-\frac {i b 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} \left (-a^{2}+b^{2}\right )}+\frac {b 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 \left (-a^{2}+b^{2}\right )}+\frac {b 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} \left (-a^{2}+b^{2}\right )}-\frac {i f \,x^{2}}{2 b}+\frac {b 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 \left (-a^{2}+b^{2}\right )}+\frac {b 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} \left (-a^{2}+b^{2}\right )}-\frac {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 ) a^{2} x}{d b \left (-a^{2}+b^{2}\right )}-\frac {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 ) a^{2} c}{d^{2} b \left (-a^{2}+b^{2}\right )}+\frac {e \ln \left (i b \,{\mathrm e}^{2 i \left (d x +c \right )}-2 a \,{\mathrm e}^{i \left (d x +c \right )}-i b \right )}{d b}-\frac {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 ) a^{2} x}{d b \left (-a^{2}+b^{2}\right )}-\frac {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 ) a^{2} c}{d^{2} b \left (-a^{2}+b^{2}\right )}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right ) e}{d b}-\frac {i b 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} \left (-a^{2}+b^{2}\right )}+\frac {i 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 ) a^{2}}{d^{2} b \left (-a^{2}+b^{2}\right )}-\frac {2 i f c x}{d b}-\frac {f c \ln \left (i b \,{\mathrm e}^{2 i \left (d x +c \right )}-2 a \,{\mathrm e}^{i \left (d x +c \right )}-i b \right )}{d^{2} b}+\frac {2 f c \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{d^{2} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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