\(\int \frac {(c+d x)^2}{a-a \sin (e+f x)} \, dx\) [118]

Optimal result

Integrand size = 21, antiderivative size = 112 \[ \int \frac {(c+d x)^2}{a-a \sin (e+f x)} \, dx=-\frac {i (c+d x)^2}{a f}+\frac {4 d (c+d x) \log \left (1+i e^{i (e+f x)}\right )}{a f^2}-\frac {4 i d^2 \operatorname {PolyLog}\left (2,-i e^{i (e+f x)}\right )}{a f^3}+\frac {(c+d x)^2 \tan \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{a f} \]

[Out]

-I*(d*x+c)^2/a/f+4*d*(d*x+c)*ln(1+I*exp(I*(f*x+e)))/a/f^2-4*I*d^2*polylog(2,-I*exp(I*(f*x+e)))/a/f^3+(d*x+c)^2
*tan(1/2*e+1/4*Pi+1/2*f*x)/a/f

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3399, 4269, 3798, 2221, 2317, 2438} \[ \int \frac {(c+d x)^2}{a-a \sin (e+f x)} \, dx=\frac {4 d (c+d x) \log \left (1+i e^{i (e+f x)}\right )}{a f^2}+\frac {(c+d x)^2 \tan \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{a f}-\frac {i (c+d x)^2}{a f}-\frac {4 i d^2 \operatorname {PolyLog}\left (2,-i e^{i (e+f x)}\right )}{a f^3} \]

[In]

Int[(c + d*x)^2/(a - a*Sin[e + f*x]),x]

[Out]

((-I)*(c + d*x)^2)/(a*f) + (4*d*(c + d*x)*Log[1 + I*E^(I*(e + f*x))])/(a*f^2) - ((4*I)*d^2*PolyLog[2, (-I)*E^(
I*(e + f*x))])/(a*f^3) + ((c + d*x)^2*Tan[e/2 + Pi/4 + (f*x)/2])/(a*f)

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 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 3399

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(2*a)^n, Int[(c
+ d*x)^m*Sin[(1/2)*(e + Pi*(a/(2*b))) + f*(x/2)]^(2*n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[a^2
- b^2, 0] && IntegerQ[n] && (GtQ[n, 0] || IGtQ[m, 0])

Rule 3798

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Simp[I*((c + d*x)^(m + 1)/(d*(
m + 1))), x] - Dist[2*I, Int[(c + d*x)^m*E^(2*I*k*Pi)*(E^(2*I*(e + f*x))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x))))
, x], x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 4269

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(c + d*x)^m)*(Cot[e + f*x]/f), x
] + Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\int (c+d x)^2 \csc ^2\left (\frac {1}{2} \left (e-\frac {\pi }{2}\right )+\frac {f x}{2}\right ) \, dx}{2 a} \\ & = \frac {(c+d x)^2 \tan \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{a f}+\frac {(2 d) \int (c+d x) \cot \left (\frac {e}{2}-\frac {\pi }{4}+\frac {f x}{2}\right ) \, dx}{a f} \\ & = -\frac {i (c+d x)^2}{a f}+\frac {(c+d x)^2 \tan \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{a f}-\frac {(4 d) \int \frac {e^{2 i \left (\frac {e}{2}+\frac {f x}{2}\right )} (c+d x)}{1+i e^{2 i \left (\frac {e}{2}+\frac {f x}{2}\right )}} \, dx}{a f} \\ & = -\frac {i (c+d x)^2}{a f}+\frac {4 d (c+d x) \log \left (1+i e^{i (e+f x)}\right )}{a f^2}+\frac {(c+d x)^2 \tan \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{a f}-\frac {\left (4 d^2\right ) \int \log \left (1+i e^{2 i \left (\frac {e}{2}+\frac {f x}{2}\right )}\right ) \, dx}{a f^2} \\ & = -\frac {i (c+d x)^2}{a f}+\frac {4 d (c+d x) \log \left (1+i e^{i (e+f x)}\right )}{a f^2}+\frac {(c+d x)^2 \tan \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{a f}+\frac {\left (4 i d^2\right ) \text {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{2 i \left (\frac {e}{2}+\frac {f x}{2}\right )}\right )}{a f^3} \\ & = -\frac {i (c+d x)^2}{a f}+\frac {4 d (c+d x) \log \left (1+i e^{i (e+f x)}\right )}{a f^2}-\frac {4 i d^2 \operatorname {PolyLog}\left (2,-i e^{i (e+f x)}\right )}{a f^3}+\frac {(c+d x)^2 \tan \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{a f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.53 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.82 \[ \int \frac {(c+d x)^2}{a-a \sin (e+f x)} \, dx=\frac {-4 i d^2 \operatorname {PolyLog}\left (2,-i e^{i (e+f x)}\right )+f (c+d x) \left (-i f (c+d x)+4 d \log \left (1+i e^{i (e+f x)}\right )+f (c+d x) \tan \left (\frac {1}{4} (2 e+\pi +2 f x)\right )\right )}{a f^3} \]

[In]

Integrate[(c + d*x)^2/(a - a*Sin[e + f*x]),x]

[Out]

((-4*I)*d^2*PolyLog[2, (-I)*E^(I*(e + f*x))] + f*(c + d*x)*((-I)*f*(c + d*x) + 4*d*Log[1 + I*E^(I*(e + f*x))]
+ f*(c + d*x)*Tan[(2*e + Pi + 2*f*x)/4]))/(a*f^3)

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 253 vs. \(2 (98 ) = 196\).

Time = 0.21 (sec) , antiderivative size = 254, normalized size of antiderivative = 2.27

method	result	size
risch	\(\frac {2 d^{2} x^{2}+4 c d x +2 c^{2}}{f a \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}-\frac {4 \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right ) c d}{a \,f^{2}}+\frac {4 \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right ) c d}{a \,f^{2}}-\frac {2 i d^{2} x^{2}}{a f}-\frac {4 i d^{2} e x}{a \,f^{2}}-\frac {2 i d^{2} e^{2}}{a \,f^{3}}+\frac {4 d^{2} \ln \left (1+i {\mathrm e}^{i \left (f x +e \right )}\right ) x}{a \,f^{2}}+\frac {4 d^{2} \ln \left (1+i {\mathrm e}^{i \left (f x +e \right )}\right ) e}{a \,f^{3}}-\frac {4 i d^{2} \operatorname {Li}_{2}\left (-i {\mathrm e}^{i \left (f x +e \right )}\right )}{a \,f^{3}}+\frac {4 e \,d^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{a \,f^{3}}-\frac {4 e \,d^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}{a \,f^{3}}\)	\(254\)

[In]

int((d*x+c)^2/(a-a*sin(f*x+e)),x,method=_RETURNVERBOSE)

[Out]

2*(d^2*x^2+2*c*d*x+c^2)/f/a/(exp(I*(f*x+e))-I)-4/a/f^2*ln(exp(I*(f*x+e)))*c*d+4/a/f^2*ln(exp(I*(f*x+e))-I)*c*d
-2*I/a/f*d^2*x^2-4*I/a/f^2*d^2*e*x-2*I/a/f^3*d^2*e^2+4/a/f^2*d^2*ln(1+I*exp(I*(f*x+e)))*x+4/a/f^3*d^2*ln(1+I*e
xp(I*(f*x+e)))*e-4*I*d^2*polylog(2,-I*exp(I*(f*x+e)))/a/f^3+4/a/f^3*e*d^2*ln(exp(I*(f*x+e)))-4/a/f^3*e*d^2*ln(
exp(I*(f*x+e))-I)

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 496 vs. \(2 (93) = 186\).

Time = 0.32 (sec) , antiderivative size = 496, normalized size of antiderivative = 4.43 \[ \int \frac {(c+d x)^2}{a-a \sin (e+f x)} \, dx=\frac {d^{2} f^{2} x^{2} + 2 \, c d f^{2} x + c^{2} f^{2} + {\left (d^{2} f^{2} x^{2} + 2 \, c d f^{2} x + c^{2} f^{2}\right )} \cos \left (f x + e\right ) + 2 \, {\left (i \, d^{2} \cos \left (f x + e\right ) - i \, d^{2} \sin \left (f x + e\right ) + i \, d^{2}\right )} {\rm Li}_2\left (i \, \cos \left (f x + e\right ) + \sin \left (f x + e\right )\right ) + 2 \, {\left (-i \, d^{2} \cos \left (f x + e\right ) + i \, d^{2} \sin \left (f x + e\right ) - i \, d^{2}\right )} {\rm Li}_2\left (-i \, \cos \left (f x + e\right ) + \sin \left (f x + e\right )\right ) - 2 \, {\left (d^{2} e - c d f + {\left (d^{2} e - c d f\right )} \cos \left (f x + e\right ) - {\left (d^{2} e - c d f\right )} \sin \left (f x + e\right )\right )} \log \left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right ) + i\right ) + 2 \, {\left (d^{2} f x + d^{2} e + {\left (d^{2} f x + d^{2} e\right )} \cos \left (f x + e\right ) - {\left (d^{2} f x + d^{2} e\right )} \sin \left (f x + e\right )\right )} \log \left (i \, \cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right ) + 2 \, {\left (d^{2} f x + d^{2} e + {\left (d^{2} f x + d^{2} e\right )} \cos \left (f x + e\right ) - {\left (d^{2} f x + d^{2} e\right )} \sin \left (f x + e\right )\right )} \log \left (-i \, \cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right ) - 2 \, {\left (d^{2} e - c d f + {\left (d^{2} e - c d f\right )} \cos \left (f x + e\right ) - {\left (d^{2} e - c d f\right )} \sin \left (f x + e\right )\right )} \log \left (-\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right ) + i\right ) + {\left (d^{2} f^{2} x^{2} + 2 \, c d f^{2} x + c^{2} f^{2}\right )} \sin \left (f x + e\right )}{a f^{3} \cos \left (f x + e\right ) - a f^{3} \sin \left (f x + e\right ) + a f^{3}} \]

[In]

integrate((d*x+c)^2/(a-a*sin(f*x+e)),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {(c+d x)^2}{a-a \sin (e+f x)} \, dx=- \frac {\int \frac {c^{2}}{\sin {\left (e + f x \right )} - 1}\, dx + \int \frac {d^{2} x^{2}}{\sin {\left (e + f x \right )} - 1}\, dx + \int \frac {2 c d x}{\sin {\left (e + f x \right )} - 1}\, dx}{a} \]

[In]

integrate((d*x+c)**2/(a-a*sin(f*x+e)),x)

[Out]

-(Integral(c**2/(sin(e + f*x) - 1), x) + Integral(d**2*x**2/(sin(e + f*x) - 1), x) + Integral(2*c*d*x/(sin(e +
f*x) - 1), x))/a

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 311 vs. \(2 (93) = 186\).

Time = 0.28 (sec) , antiderivative size = 311, normalized size of antiderivative = 2.78 \[ \int \frac {(c+d x)^2}{a-a \sin (e+f x)} \, dx=-\frac {2 \, {\left (i \, c^{2} f^{2} - 2 \, {\left (c d f \cos \left (f x + e\right ) + i \, c d f \sin \left (f x + e\right ) - i \, c d f\right )} \arctan \left (\sin \left (f x + e\right ) - 1, \cos \left (f x + e\right )\right ) - 2 \, {\left (d^{2} f x \cos \left (f x + e\right ) + i \, d^{2} f x \sin \left (f x + e\right ) - i \, d^{2} f x\right )} \arctan \left (\cos \left (f x + e\right ), -\sin \left (f x + e\right ) + 1\right ) + {\left (d^{2} f^{2} x^{2} + 2 \, c d f^{2} x\right )} \cos \left (f x + e\right ) + 2 \, {\left (d^{2} \cos \left (f x + e\right ) + i \, d^{2} \sin \left (f x + e\right ) - i \, d^{2}\right )} {\rm Li}_2\left (-i \, e^{\left (i \, f x + i \, e\right )}\right ) + {\left (d^{2} f x + c d f + {\left (i \, d^{2} f x + i \, c d f\right )} \cos \left (f x + e\right ) - {\left (d^{2} f x + c d f\right )} \sin \left (f x + e\right )\right )} \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} - 2 \, \sin \left (f x + e\right ) + 1\right ) + {\left (i \, d^{2} f^{2} x^{2} + 2 i \, c d f^{2} x\right )} \sin \left (f x + e\right )\right )}}{-i \, a f^{3} \cos \left (f x + e\right ) + a f^{3} \sin \left (f x + e\right ) - a f^{3}} \]

[In]

integrate((d*x+c)^2/(a-a*sin(f*x+e)),x, algorithm="maxima")

[Out]

-2*(I*c^2*f^2 - 2*(c*d*f*cos(f*x + e) + I*c*d*f*sin(f*x + e) - I*c*d*f)*arctan2(sin(f*x + e) - 1, cos(f*x + e)
) - 2*(d^2*f*x*cos(f*x + e) + I*d^2*f*x*sin(f*x + e) - I*d^2*f*x)*arctan2(cos(f*x + e), -sin(f*x + e) + 1) + (
d^2*f^2*x^2 + 2*c*d*f^2*x)*cos(f*x + e) + 2*(d^2*cos(f*x + e) + I*d^2*sin(f*x + e) - I*d^2)*dilog(-I*e^(I*f*x
+ I*e)) + (d^2*f*x + c*d*f + (I*d^2*f*x + I*c*d*f)*cos(f*x + e) - (d^2*f*x + c*d*f)*sin(f*x + e))*log(cos(f*x
+ e)^2 + sin(f*x + e)^2 - 2*sin(f*x + e) + 1) + (I*d^2*f^2*x^2 + 2*I*c*d*f^2*x)*sin(f*x + e))/(-I*a*f^3*cos(f*
x + e) + a*f^3*sin(f*x + e) - a*f^3)

Giac [F]

\[ \int \frac {(c+d x)^2}{a-a \sin (e+f x)} \, dx=\int { -\frac {{\left (d x + c\right )}^{2}}{a \sin \left (f x + e\right ) - a} \,d x } \]

[In]

integrate((d*x+c)^2/(a-a*sin(f*x+e)),x, algorithm="giac")

[Out]

integrate(-(d*x + c)^2/(a*sin(f*x + e) - a), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x)^2}{a-a \sin (e+f x)} \, dx=\int \frac {{\left (c+d\,x\right )}^2}{a-a\,\sin \left (e+f\,x\right )} \,d x \]

[In]

int((c + d*x)^2/(a - a*sin(e + f*x)),x)

[Out]

int((c + d*x)^2/(a - a*sin(e + f*x)), x)