3.113 \(\int \frac {(c+d x)^2}{(a+a \sin (e+f x))^2} \, dx\)

Optimal. Leaf size=243 \[ \frac {4 d (c+d x) \log \left (1-i e^{i (e+f x)}\right )}{3 a^2 f^2}-\frac {d (c+d x) \csc ^2\left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{3 a^2 f^2}-\frac {(c+d x)^2 \cot \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{3 a^2 f}-\frac {(c+d x)^2 \cot \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right ) \csc ^2\left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{6 a^2 f}-\frac {i (c+d x)^2}{3 a^2 f}-\frac {4 i d^2 \text {Li}_2\left (i e^{i (e+f x)}\right )}{3 a^2 f^3}-\frac {2 d^2 \cot \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{3 a^2 f^3} \]

[Out]

-1/3*I*(d*x+c)^2/a^2/f-2/3*d^2*cot(1/2*e+1/4*Pi+1/2*f*x)/a^2/f^3-1/3*(d*x+c)^2*cot(1/2*e+1/4*Pi+1/2*f*x)/a^2/f
-1/3*d*(d*x+c)*csc(1/2*e+1/4*Pi+1/2*f*x)^2/a^2/f^2-1/6*(d*x+c)^2*cot(1/2*e+1/4*Pi+1/2*f*x)*csc(1/2*e+1/4*Pi+1/
2*f*x)^2/a^2/f+4/3*d*(d*x+c)*ln(1-I*exp(I*(f*x+e)))/a^2/f^2-4/3*I*d^2*polylog(2,I*exp(I*(f*x+e)))/a^2/f^3

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Rubi [A]  time = 0.29, antiderivative size = 243, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {3318, 4186, 3767, 8, 4184, 3717, 2190, 2279, 2391} \[ -\frac {4 i d^2 \text {PolyLog}\left (2,i e^{i (e+f x)}\right )}{3 a^2 f^3}+\frac {4 d (c+d x) \log \left (1-i e^{i (e+f x)}\right )}{3 a^2 f^2}-\frac {d (c+d x) \csc ^2\left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{3 a^2 f^2}-\frac {(c+d x)^2 \cot \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{3 a^2 f}-\frac {(c+d x)^2 \cot \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right ) \csc ^2\left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{6 a^2 f}-\frac {i (c+d x)^2}{3 a^2 f}-\frac {2 d^2 \cot \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{3 a^2 f^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/3)*(c + d*x)^2)/(a^2*f) - (2*d^2*Cot[e/2 + Pi/4 + (f*x)/2])/(3*a^2*f^3) - ((c + d*x)^2*Cot[e/2 + Pi/4 + (
f*x)/2])/(3*a^2*f) - (d*(c + d*x)*Csc[e/2 + Pi/4 + (f*x)/2]^2)/(3*a^2*f^2) - ((c + d*x)^2*Cot[e/2 + Pi/4 + (f*
x)/2]*Csc[e/2 + Pi/4 + (f*x)/2]^2)/(6*a^2*f) + (4*d*(c + d*x)*Log[1 - I*E^(I*(e + f*x))])/(3*a^2*f^2) - (((4*I
)/3)*d^2*PolyLog[2, I*E^(I*(e + f*x))])/(a^2*f^3)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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 3318

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*(e + (Pi*a)/(2*b)))/2 + (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 3717

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 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 4184

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]

Rule 4186

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

Rubi steps

\begin {align*} \int \frac {(c+d x)^2}{(a+a \sin (e+f x))^2} \, dx &=\frac {\int (c+d x)^2 \csc ^4\left (\frac {1}{2} \left (e+\frac {\pi }{2}\right )+\frac {f x}{2}\right ) \, dx}{4 a^2}\\ &=-\frac {d (c+d x) \csc ^2\left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{3 a^2 f^2}-\frac {(c+d x)^2 \cot \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right ) \csc ^2\left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{6 a^2 f}+\frac {\int (c+d x)^2 \csc ^2\left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right ) \, dx}{6 a^2}+\frac {d^2 \int \csc ^2\left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right ) \, dx}{3 a^2 f^2}\\ &=-\frac {(c+d x)^2 \cot \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{3 a^2 f}-\frac {d (c+d x) \csc ^2\left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{3 a^2 f^2}-\frac {(c+d x)^2 \cot \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right ) \csc ^2\left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{6 a^2 f}-\frac {\left (2 d^2\right ) \operatorname {Subst}\left (\int 1 \, dx,x,\cot \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )\right )}{3 a^2 f^3}+\frac {(2 d) \int (c+d x) \cot \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right ) \, dx}{3 a^2 f}\\ &=-\frac {i (c+d x)^2}{3 a^2 f}-\frac {2 d^2 \cot \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{3 a^2 f^3}-\frac {(c+d x)^2 \cot \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{3 a^2 f}-\frac {d (c+d x) \csc ^2\left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{3 a^2 f^2}-\frac {(c+d x)^2 \cot \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right ) \csc ^2\left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{6 a^2 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}{3 a^2 f}\\ &=-\frac {i (c+d x)^2}{3 a^2 f}-\frac {2 d^2 \cot \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{3 a^2 f^3}-\frac {(c+d x)^2 \cot \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{3 a^2 f}-\frac {d (c+d x) \csc ^2\left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{3 a^2 f^2}-\frac {(c+d x)^2 \cot \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right ) \csc ^2\left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{6 a^2 f}+\frac {4 d (c+d x) \log \left (1-i e^{i (e+f x)}\right )}{3 a^2 f^2}-\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}{3 a^2 f^2}\\ &=-\frac {i (c+d x)^2}{3 a^2 f}-\frac {2 d^2 \cot \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{3 a^2 f^3}-\frac {(c+d x)^2 \cot \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{3 a^2 f}-\frac {d (c+d x) \csc ^2\left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{3 a^2 f^2}-\frac {(c+d x)^2 \cot \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right ) \csc ^2\left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{6 a^2 f}+\frac {4 d (c+d x) \log \left (1-i e^{i (e+f x)}\right )}{3 a^2 f^2}+\frac {\left (4 i d^2\right ) \operatorname {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{2 i \left (\frac {e}{2}+\frac {f x}{2}\right )}\right )}{3 a^2 f^3}\\ &=-\frac {i (c+d x)^2}{3 a^2 f}-\frac {2 d^2 \cot \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{3 a^2 f^3}-\frac {(c+d x)^2 \cot \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{3 a^2 f}-\frac {d (c+d x) \csc ^2\left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{3 a^2 f^2}-\frac {(c+d x)^2 \cot \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right ) \csc ^2\left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{6 a^2 f}+\frac {4 d (c+d x) \log \left (1-i e^{i (e+f x)}\right )}{3 a^2 f^2}-\frac {4 i d^2 \text {Li}_2\left (i e^{i (e+f x)}\right )}{3 a^2 f^3}\\ \end {align*}

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Mathematica [A]  time = 2.45, size = 175, normalized size = 0.72 \[ \frac {2 \left (c^2 f^2+2 c d f^2 x+d^2 \left (f^2 x^2+2\right )\right ) \tan \left (\frac {1}{4} (2 e+2 f x-\pi )\right )-2 i f (c+d x) \left (f (c+d x)+4 i d \log \left (1-i e^{i (e+f x)}\right )\right )+f (c+d x) \sec ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right ) \left (f (c+d x) \tan \left (\frac {1}{4} (2 e+2 f x-\pi )\right )-2 d\right )-8 i d^2 \text {Li}_2\left (i e^{i (e+f x)}\right )}{6 a^2 f^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*I)*f*(c + d*x)*(f*(c + d*x) + (4*I)*d*Log[1 - I*E^(I*(e + f*x))]) - (8*I)*d^2*PolyLog[2, I*E^(I*(e + f*x)
)] + 2*(c^2*f^2 + 2*c*d*f^2*x + d^2*(2 + f^2*x^2))*Tan[(2*e - Pi + 2*f*x)/4] + f*(c + d*x)*Sec[(2*e - Pi + 2*f
*x)/4]^2*(-2*d + f*(c + d*x)*Tan[(2*e - Pi + 2*f*x)/4]))/(6*a^2*f^3)

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fricas [B]  time = 0.92, size = 876, normalized size = 3.60 \[ \frac {d^{2} f^{2} x^{2} + c^{2} f^{2} + 2 \, c d f + {\left (d^{2} f^{2} x^{2} + 2 \, c d f^{2} x + c^{2} f^{2} + 2 \, d^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left (c d f^{2} + d^{2} f\right )} x + 2 \, {\left (d^{2} f^{2} x^{2} + c^{2} f^{2} + c d f + d^{2} + {\left (2 \, c d f^{2} + d^{2} f\right )} x\right )} \cos \left (f x + e\right ) - {\left (2 i \, d^{2} \cos \left (f x + e\right )^{2} - 2 i \, d^{2} \cos \left (f x + e\right ) - 4 i \, d^{2} + {\left (-2 i \, d^{2} \cos \left (f x + e\right ) - 4 i \, d^{2}\right )} \sin \left (f x + e\right )\right )} {\rm Li}_2\left (i \, \cos \left (f x + e\right ) - \sin \left (f x + e\right )\right ) - {\left (-2 i \, d^{2} \cos \left (f x + e\right )^{2} + 2 i \, d^{2} \cos \left (f x + e\right ) + 4 i \, d^{2} + {\left (2 i \, d^{2} \cos \left (f x + e\right ) + 4 i \, d^{2}\right )} \sin \left (f x + e\right )\right )} {\rm Li}_2\left (-i \, \cos \left (f x + e\right ) - \sin \left (f x + e\right )\right ) + 2 \, {\left (2 \, d^{2} e - 2 \, c d f - {\left (d^{2} e - c d f\right )} \cos \left (f x + e\right )^{2} + {\left (d^{2} e - c d f\right )} \cos \left (f x + e\right ) + {\left (2 \, d^{2} e - 2 \, c d f + {\left (d^{2} e - c d f\right )} \cos \left (f x + e\right )\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 (2 \, d^{2} f x + 2 \, d^{2} e - {\left (d^{2} f x + d^{2} e\right )} \cos \left (f x + e\right )^{2} + {\left (d^{2} f x + d^{2} e\right )} \cos \left (f x + e\right ) + {\left (2 \, d^{2} f x + 2 \, d^{2} e + {\left (d^{2} f x + d^{2} e\right )} \cos \left (f x + e\right )\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 (2 \, d^{2} f x + 2 \, d^{2} e - {\left (d^{2} f x + d^{2} e\right )} \cos \left (f x + e\right )^{2} + {\left (d^{2} f x + d^{2} e\right )} \cos \left (f x + e\right ) + {\left (2 \, d^{2} f x + 2 \, d^{2} e + {\left (d^{2} f x + d^{2} e\right )} \cos \left (f x + e\right )\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 (2 \, d^{2} e - 2 \, c d f - {\left (d^{2} e - c d f\right )} \cos \left (f x + e\right )^{2} + {\left (d^{2} e - c d f\right )} \cos \left (f x + e\right ) + {\left (2 \, d^{2} e - 2 \, c d f + {\left (d^{2} e - c d f\right )} \cos \left (f x + e\right )\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} + c^{2} f^{2} - 2 \, c d f + 2 \, {\left (c d f^{2} - d^{2} f\right )} x - {\left (d^{2} f^{2} x^{2} + 2 \, c d f^{2} x + c^{2} f^{2} + 2 \, d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{3 \, {\left (a^{2} f^{3} \cos \left (f x + e\right )^{2} - a^{2} f^{3} \cos \left (f x + e\right ) - 2 \, a^{2} f^{3} - {\left (a^{2} f^{3} \cos \left (f x + e\right ) + 2 \, a^{2} f^{3}\right )} \sin \left (f x + e\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(d^2*f^2*x^2 + c^2*f^2 + 2*c*d*f + (d^2*f^2*x^2 + 2*c*d*f^2*x + c^2*f^2 + 2*d^2)*cos(f*x + e)^2 + 2*(c*d*f
^2 + d^2*f)*x + 2*(d^2*f^2*x^2 + c^2*f^2 + c*d*f + d^2 + (2*c*d*f^2 + d^2*f)*x)*cos(f*x + e) - (2*I*d^2*cos(f*
x + e)^2 - 2*I*d^2*cos(f*x + e) - 4*I*d^2 + (-2*I*d^2*cos(f*x + e) - 4*I*d^2)*sin(f*x + e))*dilog(I*cos(f*x +
e) - sin(f*x + e)) - (-2*I*d^2*cos(f*x + e)^2 + 2*I*d^2*cos(f*x + e) + 4*I*d^2 + (2*I*d^2*cos(f*x + e) + 4*I*d
^2)*sin(f*x + e))*dilog(-I*cos(f*x + e) - sin(f*x + e)) + 2*(2*d^2*e - 2*c*d*f - (d^2*e - c*d*f)*cos(f*x + e)^
2 + (d^2*e - c*d*f)*cos(f*x + e) + (2*d^2*e - 2*c*d*f + (d^2*e - c*d*f)*cos(f*x + e))*sin(f*x + e))*log(cos(f*
x + e) + I*sin(f*x + e) + I) - 2*(2*d^2*f*x + 2*d^2*e - (d^2*f*x + d^2*e)*cos(f*x + e)^2 + (d^2*f*x + d^2*e)*c
os(f*x + e) + (2*d^2*f*x + 2*d^2*e + (d^2*f*x + d^2*e)*cos(f*x + e))*sin(f*x + e))*log(I*cos(f*x + e) + sin(f*
x + e) + 1) - 2*(2*d^2*f*x + 2*d^2*e - (d^2*f*x + d^2*e)*cos(f*x + e)^2 + (d^2*f*x + d^2*e)*cos(f*x + e) + (2*
d^2*f*x + 2*d^2*e + (d^2*f*x + d^2*e)*cos(f*x + e))*sin(f*x + e))*log(-I*cos(f*x + e) + sin(f*x + e) + 1) + 2*
(2*d^2*e - 2*c*d*f - (d^2*e - c*d*f)*cos(f*x + e)^2 + (d^2*e - c*d*f)*cos(f*x + e) + (2*d^2*e - 2*c*d*f + (d^2
*e - c*d*f)*cos(f*x + e))*sin(f*x + e))*log(-cos(f*x + e) + I*sin(f*x + e) + I) - (d^2*f^2*x^2 + c^2*f^2 - 2*c
*d*f + 2*(c*d*f^2 - d^2*f)*x - (d^2*f^2*x^2 + 2*c*d*f^2*x + c^2*f^2 + 2*d^2)*cos(f*x + e))*sin(f*x + e))/(a^2*
f^3*cos(f*x + e)^2 - a^2*f^3*cos(f*x + e) - 2*a^2*f^3 - (a^2*f^3*cos(f*x + e) + 2*a^2*f^3)*sin(f*x + e))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d x + c\right )}^{2}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.95, size = 421, normalized size = 1.73 \[ -\frac {2 i \left (i d^{2} f^{2} x^{2}+3 d^{2} f^{2} x^{2} {\mathrm e}^{i \left (f x +e \right )}+2 i c d \,f^{2} x +2 i f \,d^{2} x \,{\mathrm e}^{i \left (f x +e \right )}+6 c d \,f^{2} x \,{\mathrm e}^{i \left (f x +e \right )}+2 f \,d^{2} x \,{\mathrm e}^{2 i \left (f x +e \right )}+i c^{2} f^{2}+2 i f c d \,{\mathrm e}^{i \left (f x +e \right )}-2 i d^{2} {\mathrm e}^{2 i \left (f x +e \right )}+3 c^{2} f^{2} {\mathrm e}^{i \left (f x +e \right )}+2 f c d \,{\mathrm e}^{2 i \left (f x +e \right )}+2 i d^{2}+4 d^{2} {\mathrm e}^{i \left (f x +e \right )}\right )}{3 \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )^{3} f^{3} a^{2}}+\frac {4 \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right ) c d}{3 a^{2} f^{2}}-\frac {4 \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right ) c d}{3 a^{2} f^{2}}-\frac {2 i d^{2} x^{2}}{3 a^{2} f}-\frac {4 i d^{2} e x}{3 a^{2} f^{2}}-\frac {2 i d^{2} e^{2}}{3 a^{2} f^{3}}+\frac {4 d^{2} \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right ) x}{3 a^{2} f^{2}}+\frac {4 d^{2} \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right ) e}{3 a^{2} f^{3}}-\frac {4 i d^{2} \polylog \left (2, i {\mathrm e}^{i \left (f x +e \right )}\right )}{3 a^{2} f^{3}}-\frac {4 d^{2} e \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}{3 a^{2} f^{3}}+\frac {4 d^{2} e \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{3 a^{2} f^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*I*(I*d^2*f^2*x^2+3*d^2*f^2*x^2*exp(I*(f*x+e))+2*I*c*d*f^2*x+2*I*f*d^2*x*exp(I*(f*x+e))+6*c*d*f^2*x*exp(I*
(f*x+e))+2*f*d^2*x*exp(2*I*(f*x+e))+I*c^2*f^2+2*I*f*c*d*exp(I*(f*x+e))-2*I*d^2*exp(2*I*(f*x+e))+3*c^2*f^2*exp(
I*(f*x+e))+2*f*c*d*exp(2*I*(f*x+e))+2*I*d^2+4*d^2*exp(I*(f*x+e)))/(exp(I*(f*x+e))+I)^3/f^3/a^2+4/3/a^2/f^2*ln(
exp(I*(f*x+e))+I)*c*d-4/3/a^2/f^2*ln(exp(I*(f*x+e)))*c*d-2/3*I/a^2/f*d^2*x^2-4/3*I/a^2/f^2*d^2*e*x-2/3*I/a^2/f
^3*d^2*e^2+4/3/a^2/f^2*d^2*ln(1-I*exp(I*(f*x+e)))*x+4/3/a^2/f^3*d^2*ln(1-I*exp(I*(f*x+e)))*e-4/3*I*d^2*polylog
(2,I*exp(I*(f*x+e)))/a^2/f^3-4/3/a^2/f^3*d^2*e*ln(exp(I*(f*x+e))+I)+4/3/a^2/f^3*d^2*e*ln(exp(I*(f*x+e)))

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maxima [B]  time = 1.41, size = 832, normalized size = 3.42 \[ \frac {-2 i \, c^{2} f^{2} - 4 i \, d^{2} + {\left (4 \, c d f \cos \left (3 \, f x + 3 \, e\right ) + 12 i \, c d f \cos \left (2 \, f x + 2 \, e\right ) - 12 \, c d f \cos \left (f x + e\right ) + 4 i \, c d f \sin \left (3 \, f x + 3 \, e\right ) - 12 \, c d f \sin \left (2 \, f x + 2 \, e\right ) - 12 i \, c d f \sin \left (f x + e\right ) - 4 i \, c d f\right )} \arctan \left (\sin \left (f x + e\right ) + 1, \cos \left (f x + e\right )\right ) - {\left (4 \, d^{2} f x \cos \left (3 \, f x + 3 \, e\right ) + 12 i \, d^{2} f x \cos \left (2 \, f x + 2 \, e\right ) - 12 \, d^{2} f x \cos \left (f x + e\right ) + 4 i \, d^{2} f x \sin \left (3 \, f x + 3 \, e\right ) - 12 \, d^{2} f x \sin \left (2 \, f x + 2 \, e\right ) - 12 i \, d^{2} f x \sin \left (f x + e\right ) - 4 i \, d^{2} f x\right )} \arctan \left (\cos \left (f x + e\right ), \sin \left (f x + e\right ) + 1\right ) - 2 \, {\left (d^{2} f^{2} x^{2} + 2 \, c d f^{2} x\right )} \cos \left (3 \, f x + 3 \, e\right ) + {\left (-6 i \, d^{2} f^{2} x^{2} - 4 \, c d f + 4 i \, d^{2} - 4 \, {\left (3 i \, c d f^{2} + d^{2} f\right )} x\right )} \cos \left (2 \, f x + 2 \, e\right ) - {\left (6 \, c^{2} f^{2} + 4 i \, d^{2} f x + 4 i \, c d f + 8 \, d^{2}\right )} \cos \left (f x + e\right ) - {\left (4 \, d^{2} \cos \left (3 \, f x + 3 \, e\right ) + 12 i \, d^{2} \cos \left (2 \, f x + 2 \, e\right ) - 12 \, d^{2} \cos \left (f x + e\right ) + 4 i \, d^{2} \sin \left (3 \, f x + 3 \, e\right ) - 12 \, d^{2} \sin \left (2 \, f x + 2 \, e\right ) - 12 i \, d^{2} \sin \left (f x + e\right ) - 4 i \, d^{2}\right )} {\rm Li}_2\left (i \, e^{\left (i \, f x + i \, e\right )}\right ) - {\left (2 \, d^{2} f x + 2 \, c d f - {\left (-2 i \, d^{2} f x - 2 i \, c d f\right )} \cos \left (3 \, f x + 3 \, e\right ) - 6 \, {\left (d^{2} f x + c d f\right )} \cos \left (2 \, f x + 2 \, e\right ) - {\left (6 i \, d^{2} f x + 6 i \, c d f\right )} \cos \left (f x + e\right ) - 2 \, {\left (d^{2} f x + c d f\right )} \sin \left (3 \, f x + 3 \, e\right ) - {\left (6 i \, d^{2} f x + 6 i \, c d f\right )} \sin \left (2 \, f x + 2 \, e\right ) + 6 \, {\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 (-2 i \, d^{2} f^{2} x^{2} - 4 i \, c d f^{2} x\right )} \sin \left (3 \, f x + 3 \, e\right ) + {\left (6 \, d^{2} f^{2} x^{2} - 4 i \, c d f - 4 \, d^{2} + {\left (12 \, c d f^{2} - 4 i \, d^{2} f\right )} x\right )} \sin \left (2 \, f x + 2 \, e\right ) + {\left (-6 i \, c^{2} f^{2} + 4 \, d^{2} f x + 4 \, c d f - 8 i \, d^{2}\right )} \sin \left (f x + e\right )}{-3 i \, a^{2} f^{3} \cos \left (3 \, f x + 3 \, e\right ) + 9 \, a^{2} f^{3} \cos \left (2 \, f x + 2 \, e\right ) + 9 i \, a^{2} f^{3} \cos \left (f x + e\right ) + 3 \, a^{2} f^{3} \sin \left (3 \, f x + 3 \, e\right ) + 9 i \, a^{2} f^{3} \sin \left (2 \, f x + 2 \, e\right ) - 9 \, a^{2} f^{3} \sin \left (f x + e\right ) - 3 \, a^{2} f^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-2*I*c^2*f^2 - 4*I*d^2 + (4*c*d*f*cos(3*f*x + 3*e) + 12*I*c*d*f*cos(2*f*x + 2*e) - 12*c*d*f*cos(f*x + e) + 4*
I*c*d*f*sin(3*f*x + 3*e) - 12*c*d*f*sin(2*f*x + 2*e) - 12*I*c*d*f*sin(f*x + e) - 4*I*c*d*f)*arctan2(sin(f*x +
e) + 1, cos(f*x + e)) - (4*d^2*f*x*cos(3*f*x + 3*e) + 12*I*d^2*f*x*cos(2*f*x + 2*e) - 12*d^2*f*x*cos(f*x + e)
+ 4*I*d^2*f*x*sin(3*f*x + 3*e) - 12*d^2*f*x*sin(2*f*x + 2*e) - 12*I*d^2*f*x*sin(f*x + e) - 4*I*d^2*f*x)*arctan
2(cos(f*x + e), sin(f*x + e) + 1) - 2*(d^2*f^2*x^2 + 2*c*d*f^2*x)*cos(3*f*x + 3*e) + (-6*I*d^2*f^2*x^2 - 4*c*d
*f + 4*I*d^2 - 4*(3*I*c*d*f^2 + d^2*f)*x)*cos(2*f*x + 2*e) - (6*c^2*f^2 + 4*I*d^2*f*x + 4*I*c*d*f + 8*d^2)*cos
(f*x + e) - (4*d^2*cos(3*f*x + 3*e) + 12*I*d^2*cos(2*f*x + 2*e) - 12*d^2*cos(f*x + e) + 4*I*d^2*sin(3*f*x + 3*
e) - 12*d^2*sin(2*f*x + 2*e) - 12*I*d^2*sin(f*x + e) - 4*I*d^2)*dilog(I*e^(I*f*x + I*e)) - (2*d^2*f*x + 2*c*d*
f - (-2*I*d^2*f*x - 2*I*c*d*f)*cos(3*f*x + 3*e) - 6*(d^2*f*x + c*d*f)*cos(2*f*x + 2*e) - (6*I*d^2*f*x + 6*I*c*
d*f)*cos(f*x + e) - 2*(d^2*f*x + c*d*f)*sin(3*f*x + 3*e) - (6*I*d^2*f*x + 6*I*c*d*f)*sin(2*f*x + 2*e) + 6*(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) + (-2*I*d^2*f^2*x^2 - 4*
I*c*d*f^2*x)*sin(3*f*x + 3*e) + (6*d^2*f^2*x^2 - 4*I*c*d*f - 4*d^2 + (12*c*d*f^2 - 4*I*d^2*f)*x)*sin(2*f*x + 2
*e) + (-6*I*c^2*f^2 + 4*d^2*f*x + 4*c*d*f - 8*I*d^2)*sin(f*x + e))/(-3*I*a^2*f^3*cos(3*f*x + 3*e) + 9*a^2*f^3*
cos(2*f*x + 2*e) + 9*I*a^2*f^3*cos(f*x + e) + 3*a^2*f^3*sin(3*f*x + 3*e) + 9*I*a^2*f^3*sin(2*f*x + 2*e) - 9*a^
2*f^3*sin(f*x + e) - 3*a^2*f^3)

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mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.00 \[ \text {Hanged} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

\text{Hanged}

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {c^{2}}{\sin ^{2}{\left (e + f x \right )} + 2 \sin {\left (e + f x \right )} + 1}\, dx + \int \frac {d^{2} x^{2}}{\sin ^{2}{\left (e + f x \right )} + 2 \sin {\left (e + f x \right )} + 1}\, dx + \int \frac {2 c d x}{\sin ^{2}{\left (e + f x \right )} + 2 \sin {\left (e + f x \right )} + 1}\, dx}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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