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

Optimal. Leaf size=234 \[ -\frac {i (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{f \sqrt {a^2-b^2}}+\frac {i (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}+a}\right )}{f \sqrt {a^2-b^2}}-\frac {d \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{f^2 \sqrt {a^2-b^2}}+\frac {d \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{f^2 \sqrt {a^2-b^2}} \]

[Out]

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

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Rubi [A]  time = 0.45, antiderivative size = 234, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {3323, 2264, 2190, 2279, 2391} \[ -\frac {d \text {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{f^2 \sqrt {a^2-b^2}}+\frac {d \text {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}+a}\right )}{f^2 \sqrt {a^2-b^2}}-\frac {i (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{f \sqrt {a^2-b^2}}+\frac {i (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}+a}\right )}{f \sqrt {a^2-b^2}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)/(a + b*Sin[e + f*x]),x]

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 182, normalized size = 0.78 \[ \frac {-i f (c+d x) \left (\log \left (1+\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}-a}\right )-\log \left (1-\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}+a}\right )\right )-d \text {Li}_2\left (-\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}-a}\right )+d \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{f^2 \sqrt {a^2-b^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.64, size = 1009, normalized size = 4.31 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.13, size = 501, normalized size = 2.14 \[ \frac {2 i c \arctan \left (\frac {2 i b \,{\mathrm e}^{i \left (f x +e \right )}-2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{f \sqrt {-a^{2}+b^{2}}}+\frac {d \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (f x +e \right )}-\sqrt {-a^{2}+b^{2}}}{i a -\sqrt {-a^{2}+b^{2}}}\right ) x}{f \sqrt {-a^{2}+b^{2}}}+\frac {d \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (f x +e \right )}-\sqrt {-a^{2}+b^{2}}}{i a -\sqrt {-a^{2}+b^{2}}}\right ) e}{f^{2} \sqrt {-a^{2}+b^{2}}}-\frac {d \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (f x +e \right )}+\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right ) x}{f \sqrt {-a^{2}+b^{2}}}-\frac {d \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (f x +e \right )}+\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right ) e}{f^{2} \sqrt {-a^{2}+b^{2}}}-\frac {i d \dilog \left (\frac {i a +b \,{\mathrm e}^{i \left (f x +e \right )}-\sqrt {-a^{2}+b^{2}}}{i a -\sqrt {-a^{2}+b^{2}}}\right )}{f^{2} \sqrt {-a^{2}+b^{2}}}+\frac {i d \dilog \left (\frac {i a +b \,{\mathrm e}^{i \left (f x +e \right )}+\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right )}{f^{2} \sqrt {-a^{2}+b^{2}}}-\frac {2 i d e \arctan \left (\frac {2 i b \,{\mathrm e}^{i \left (f x +e \right )}-2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{f^{2} \sqrt {-a^{2}+b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*I/f*c/(-a^2+b^2)^(1/2)*arctan(1/2*(2*I*b*exp(I*(f*x+e))-2*a)/(-a^2+b^2)^(1/2))+1/f*d/(-a^2+b^2)^(1/2)*ln((I*
a+b*exp(I*(f*x+e))-(-a^2+b^2)^(1/2))/(I*a-(-a^2+b^2)^(1/2)))*x+1/f^2*d/(-a^2+b^2)^(1/2)*ln((I*a+b*exp(I*(f*x+e
))-(-a^2+b^2)^(1/2))/(I*a-(-a^2+b^2)^(1/2)))*e-1/f*d/(-a^2+b^2)^(1/2)*ln((I*a+b*exp(I*(f*x+e))+(-a^2+b^2)^(1/2
))/(I*a+(-a^2+b^2)^(1/2)))*x-1/f^2*d/(-a^2+b^2)^(1/2)*ln((I*a+b*exp(I*(f*x+e))+(-a^2+b^2)^(1/2))/(I*a+(-a^2+b^
2)^(1/2)))*e-I/f^2*d/(-a^2+b^2)^(1/2)*dilog((I*a+b*exp(I*(f*x+e))-(-a^2+b^2)^(1/2))/(I*a-(-a^2+b^2)^(1/2)))+I/
f^2*d/(-a^2+b^2)^(1/2)*dilog((I*a+b*exp(I*(f*x+e))+(-a^2+b^2)^(1/2))/(I*a+(-a^2+b^2)^(1/2)))-2*I/f^2*d*e/(-a^2
+b^2)^(1/2)*arctan(1/2*(2*I*b*exp(I*(f*x+e))-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((d*x+c)/(a+b*sin(f*x+e)),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 \[ \int \frac {c+d\,x}{a+b\,\sin \left (e+f\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)/(a + b*sin(e + f*x)),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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