3.2.65 \(\int \frac {c+d x}{a+b \sin (e+f x)} \, dx\) [165]

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

[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)

________________________________________________________________________________________

Rubi [A]
time = 0.29, 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 = {3404, 2296, 2221, 2317, 2438} \begin {gather*} -\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}} \end {gather*}

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 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]

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 \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 (e+f x)}\right )}{\sqrt {a^2-b^2} f^2}-\frac {d \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 (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*}

________________________________________________________________________________________

Mathematica [A]
time = 0.03, size = 182, normalized size = 0.78 \begin {gather*} \frac {-i f (c+d x) \left (\log \left (1+\frac {i b e^{i (e+f x)}}{-a+\sqrt {a^2-b^2}}\right )-\log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )\right )-d \text {Li}_2\left (-\frac {i b e^{i (e+f x)}}{-a+\sqrt {a^2-b^2}}\right )+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 {gather*}

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)

________________________________________________________________________________________

Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 491 vs. \(2 (204 ) = 408\).
time = 0.08, size = 492, normalized size = 2.10

method	result	size
risch	\(\frac {2 i c \arctan \left (\frac {2 i b \,{\mathrm e}^{i \left (f x +e \right )}-2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{f \sqrt {-a^{2}+b^{2}}}+\frac {d \ln \left (\frac {-i a -b \,{\mathrm e}^{i \left (f x +e \right )}+\sqrt {-a^{2}+b^{2}}}{-i a +\sqrt {-a^{2}+b^{2}}}\right ) x}{f \sqrt {-a^{2}+b^{2}}}+\frac {d \ln \left (\frac {-i a -b \,{\mathrm e}^{i \left (f x +e \right )}+\sqrt {-a^{2}+b^{2}}}{-i a +\sqrt {-a^{2}+b^{2}}}\right ) e}{f^{2} \sqrt {-a^{2}+b^{2}}}-\frac {d \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (f x +e \right )}+\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right ) x}{f \sqrt {-a^{2}+b^{2}}}-\frac {d \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (f x +e \right )}+\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right ) e}{f^{2} \sqrt {-a^{2}+b^{2}}}+\frac {i d \dilog \left (\frac {i a +b \,{\mathrm e}^{i \left (f x +e \right )}+\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right )}{f^{2} \sqrt {-a^{2}+b^{2}}}-\frac {i d \dilog \left (\frac {-i a -b \,{\mathrm e}^{i \left (f x +e \right )}+\sqrt {-a^{2}+b^{2}}}{-i a +\sqrt {-a^{2}+b^{2}}}\right )}{f^{2} \sqrt {-a^{2}+b^{2}}}-\frac {2 i d e \arctan \left (\frac {2 i b \,{\mathrm e}^{i \left (f x +e \right )}-2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{f^{2} \sqrt {-a^{2}+b^{2}}}\)	\(492\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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))

________________________________________________________________________________________

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((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 de

________________________________________________________________________________________

Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1045 vs. \(2 (204) = 408\).
time = 0.55, size = 1045, normalized size = 4.47 \begin {gather*} \frac {i \, b d \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} {\rm Li}_2\left (\frac {i \, a \cos \left (f x + e\right ) - a \sin \left (f x + e\right ) + {\left (b \cos \left (f x + e\right ) + i \, b \sin \left (f x + e\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b} + 1\right ) - i \, b d \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} {\rm Li}_2\left (\frac {i \, a \cos \left (f x + e\right ) - a \sin \left (f x + e\right ) - {\left (b \cos \left (f x + e\right ) + i \, b \sin \left (f x + e\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b} + 1\right ) - i \, b d \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} {\rm Li}_2\left (\frac {-i \, a \cos \left (f x + e\right ) - a \sin \left (f x + e\right ) + {\left (b \cos \left (f x + e\right ) - i \, b \sin \left (f x + e\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b} + 1\right ) + i \, b d \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} {\rm Li}_2\left (\frac {-i \, a \cos \left (f x + e\right ) - a \sin \left (f x + e\right ) - {\left (b \cos \left (f x + e\right ) - i \, b \sin \left (f x + e\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b} + 1\right ) + {\left (b c f - b d e\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} \log \left (2 \, b \cos \left (f x + e\right ) + 2 i \, b \sin \left (f x + e\right ) + 2 \, b \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} + 2 i \, a\right ) + {\left (b c f - b d e\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} \log \left (2 \, b \cos \left (f x + e\right ) - 2 i \, b \sin \left (f x + e\right ) + 2 \, b \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - 2 i \, a\right ) - {\left (b c f - b d e\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} \log \left (-2 \, b \cos \left (f x + e\right ) + 2 i \, b \sin \left (f x + e\right ) + 2 \, b \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} + 2 i \, a\right ) - {\left (b c f - b d e\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} \log \left (-2 \, b \cos \left (f x + e\right ) - 2 i \, b \sin \left (f x + e\right ) + 2 \, b \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - 2 i \, a\right ) - {\left (b d f x + b d e\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} \log \left (-\frac {i \, a \cos \left (f x + e\right ) - a \sin \left (f x + e\right ) + {\left (b \cos \left (f x + e\right ) + i \, b \sin \left (f x + e\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b}\right ) + {\left (b d f x + b d e\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} \log \left (-\frac {i \, a \cos \left (f x + e\right ) - a \sin \left (f x + e\right ) - {\left (b \cos \left (f x + e\right ) + i \, b \sin \left (f x + e\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b}\right ) - {\left (b d f x + b d e\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} \log \left (-\frac {-i \, a \cos \left (f x + e\right ) - a \sin \left (f x + e\right ) + {\left (b \cos \left (f x + e\right ) - i \, b \sin \left (f x + e\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b}\right ) + {\left (b d f x + b d e\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} \log \left (-\frac {-i \, a \cos \left (f x + e\right ) - a \sin \left (f x + e\right ) - {\left (b \cos \left (f x + e\right ) - i \, b \sin \left (f x + e\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b}\right )}{2 \, {\left (a^{2} - b^{2}\right )} f^{2}} \end {gather*}

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]

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

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

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)

________________________________________________________________________________________