3.2.69 \(\int \frac {(c+d x)^2}{(a+b \sin (e+f x))^2} \, dx\) [169]

Optimal. Leaf size=671 \[ \frac {i (c+d x)^2}{\left (a^2-b^2\right ) f}-\frac {2 d (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) f^2}-\frac {i a (c+d x)^2 \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f}-\frac {2 d (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) f^2}+\frac {i a (c+d x)^2 \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f}+\frac {2 i d^2 \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) f^3}-\frac {2 a d (c+d x) \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f^2}+\frac {2 i d^2 \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) f^3}+\frac {2 a d (c+d x) \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f^2}-\frac {2 i a d^2 \text {Li}_3\left (\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f^3}+\frac {2 i a d^2 \text {Li}_3\left (\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f^3}+\frac {b (c+d x)^2 \cos (e+f x)}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))} \]

[Out]

I*(d*x+c)^2/(a^2-b^2)/f-2*d*(d*x+c)*ln(1-I*b*exp(I*(f*x+e))/(a-(a^2-b^2)^(1/2)))/(a^2-b^2)/f^2-I*a*(d*x+c)^2*l
n(1-I*b*exp(I*(f*x+e))/(a-(a^2-b^2)^(1/2)))/(a^2-b^2)^(3/2)/f-2*d*(d*x+c)*ln(1-I*b*exp(I*(f*x+e))/(a+(a^2-b^2)
^(1/2)))/(a^2-b^2)/f^2+I*a*(d*x+c)^2*ln(1-I*b*exp(I*(f*x+e))/(a+(a^2-b^2)^(1/2)))/(a^2-b^2)^(3/2)/f+2*I*d^2*po
lylog(2,I*b*exp(I*(f*x+e))/(a-(a^2-b^2)^(1/2)))/(a^2-b^2)/f^3-2*a*d*(d*x+c)*polylog(2,I*b*exp(I*(f*x+e))/(a-(a
^2-b^2)^(1/2)))/(a^2-b^2)^(3/2)/f^2+2*I*d^2*polylog(2,I*b*exp(I*(f*x+e))/(a+(a^2-b^2)^(1/2)))/(a^2-b^2)/f^3+2*
a*d*(d*x+c)*polylog(2,I*b*exp(I*(f*x+e))/(a+(a^2-b^2)^(1/2)))/(a^2-b^2)^(3/2)/f^2-2*I*a*d^2*polylog(3,I*b*exp(
I*(f*x+e))/(a-(a^2-b^2)^(1/2)))/(a^2-b^2)^(3/2)/f^3+2*I*a*d^2*polylog(3,I*b*exp(I*(f*x+e))/(a+(a^2-b^2)^(1/2))
)/(a^2-b^2)^(3/2)/f^3+b*(d*x+c)^2*cos(f*x+e)/(a^2-b^2)/f/(a+b*sin(f*x+e))

________________________________________________________________________________________

Rubi [A]
time = 0.76, antiderivative size = 671, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 10, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3405, 3404, 2296, 2221, 2611, 2320, 6724, 4615, 2317, 2438} \begin {gather*} -\frac {2 a d (c+d x) \text {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{f^2 \left (a^2-b^2\right )^{3/2}}+\frac {2 a d (c+d x) \text {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}+a}\right )}{f^2 \left (a^2-b^2\right )^{3/2}}+\frac {2 i d^2 \text {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{f^3 \left (a^2-b^2\right )}+\frac {2 i d^2 \text {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}+a}\right )}{f^3 \left (a^2-b^2\right )}-\frac {2 i a d^2 \text {PolyLog}\left (3,\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{f^3 \left (a^2-b^2\right )^{3/2}}+\frac {2 i a d^2 \text {PolyLog}\left (3,\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}+a}\right )}{f^3 \left (a^2-b^2\right )^{3/2}}-\frac {2 d (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{f^2 \left (a^2-b^2\right )}-\frac {2 d (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}+a}\right )}{f^2 \left (a^2-b^2\right )}-\frac {i a (c+d x)^2 \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{f \left (a^2-b^2\right )^{3/2}}+\frac {i a (c+d x)^2 \log \left (1-\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}+a}\right )}{f \left (a^2-b^2\right )^{3/2}}+\frac {b (c+d x)^2 \cos (e+f x)}{f \left (a^2-b^2\right ) (a+b \sin (e+f x))}+\frac {i (c+d x)^2}{f \left (a^2-b^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(I*(c + d*x)^2)/((a^2 - b^2)*f) - (2*d*(c + d*x)*Log[1 - (I*b*E^(I*(e + f*x)))/(a - Sqrt[a^2 - b^2])])/((a^2 -
b^2)*f^2) - (I*a*(c + d*x)^2*Log[1 - (I*b*E^(I*(e + f*x)))/(a - Sqrt[a^2 - b^2])])/((a^2 - b^2)^(3/2)*f) - (2
*d*(c + d*x)*Log[1 - (I*b*E^(I*(e + f*x)))/(a + Sqrt[a^2 - b^2])])/((a^2 - b^2)*f^2) + (I*a*(c + d*x)^2*Log[1
- (I*b*E^(I*(e + f*x)))/(a + Sqrt[a^2 - b^2])])/((a^2 - b^2)^(3/2)*f) + ((2*I)*d^2*PolyLog[2, (I*b*E^(I*(e + f
*x)))/(a - Sqrt[a^2 - b^2])])/((a^2 - b^2)*f^3) - (2*a*d*(c + d*x)*PolyLog[2, (I*b*E^(I*(e + f*x)))/(a - Sqrt[
a^2 - b^2])])/((a^2 - b^2)^(3/2)*f^2) + ((2*I)*d^2*PolyLog[2, (I*b*E^(I*(e + f*x)))/(a + Sqrt[a^2 - b^2])])/((
a^2 - b^2)*f^3) + (2*a*d*(c + d*x)*PolyLog[2, (I*b*E^(I*(e + f*x)))/(a + Sqrt[a^2 - b^2])])/((a^2 - b^2)^(3/2)
*f^2) - ((2*I)*a*d^2*PolyLog[3, (I*b*E^(I*(e + f*x)))/(a - Sqrt[a^2 - b^2])])/((a^2 - b^2)^(3/2)*f^3) + ((2*I)
*a*d^2*PolyLog[3, (I*b*E^(I*(e + f*x)))/(a + Sqrt[a^2 - b^2])])/((a^2 - b^2)^(3/2)*f^3) + (b*(c + d*x)^2*Cos[e
+ f*x])/((a^2 - b^2)*f*(a + b*Sin[e + f*x]))

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 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

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 2611

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(
f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Dist[g*(m/(b*c*n*Log[F])), Int[(f + g*
x)^(m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

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]

Rule 3405

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

Rule 4615

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]

Rule 6724

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin {align*} \int \frac {(c+d x)^2}{(a+b \sin (e+f x))^2} \, dx &=\frac {b (c+d x)^2 \cos (e+f x)}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))}+\frac {a \int \frac {(c+d x)^2}{a+b \sin (e+f x)} \, dx}{a^2-b^2}-\frac {(2 b d) \int \frac {(c+d x) \cos (e+f x)}{a+b \sin (e+f x)} \, dx}{\left (a^2-b^2\right ) f}\\ &=\frac {i (c+d x)^2}{\left (a^2-b^2\right ) f}+\frac {b (c+d x)^2 \cos (e+f x)}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))}+\frac {(2 a) \int \frac {e^{i (e+f x)} (c+d x)^2}{i b+2 a e^{i (e+f x)}-i b e^{2 i (e+f x)}} \, dx}{a^2-b^2}-\frac {(2 b d) \int \frac {e^{i (e+f x)} (c+d x)}{a-\sqrt {a^2-b^2}-i b e^{i (e+f x)}} \, dx}{\left (a^2-b^2\right ) f}-\frac {(2 b d) \int \frac {e^{i (e+f x)} (c+d x)}{a+\sqrt {a^2-b^2}-i b e^{i (e+f x)}} \, dx}{\left (a^2-b^2\right ) f}\\ &=\frac {i (c+d x)^2}{\left (a^2-b^2\right ) f}-\frac {2 d (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) f^2}-\frac {2 d (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) f^2}+\frac {b (c+d x)^2 \cos (e+f x)}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))}-\frac {(2 i a b) \int \frac {e^{i (e+f x)} (c+d x)^2}{2 a-2 \sqrt {a^2-b^2}-2 i b e^{i (e+f x)}} \, dx}{\left (a^2-b^2\right )^{3/2}}+\frac {(2 i a b) \int \frac {e^{i (e+f x)} (c+d x)^2}{2 a+2 \sqrt {a^2-b^2}-2 i b e^{i (e+f x)}} \, dx}{\left (a^2-b^2\right )^{3/2}}+\frac {\left (2 d^2\right ) \int \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right ) \, dx}{\left (a^2-b^2\right ) f^2}+\frac {\left (2 d^2\right ) \int \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right ) \, dx}{\left (a^2-b^2\right ) f^2}\\ &=\frac {i (c+d x)^2}{\left (a^2-b^2\right ) f}-\frac {2 d (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) f^2}-\frac {i a (c+d x)^2 \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f}-\frac {2 d (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) f^2}+\frac {i a (c+d x)^2 \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f}+\frac {b (c+d x)^2 \cos (e+f x)}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))}-\frac {\left (2 i d^2\right ) \text {Subst}\left (\int \frac {\log \left (1-\frac {i b x}{a-\sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (e+f x)}\right )}{\left (a^2-b^2\right ) f^3}-\frac {\left (2 i d^2\right ) \text {Subst}\left (\int \frac {\log \left (1-\frac {i b x}{a+\sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (e+f x)}\right )}{\left (a^2-b^2\right ) f^3}+\frac {(2 i a d) \int (c+d x) \log \left (1-\frac {2 i b e^{i (e+f x)}}{2 a-2 \sqrt {a^2-b^2}}\right ) \, dx}{\left (a^2-b^2\right )^{3/2} f}-\frac {(2 i a d) \int (c+d x) \log \left (1-\frac {2 i b e^{i (e+f x)}}{2 a+2 \sqrt {a^2-b^2}}\right ) \, dx}{\left (a^2-b^2\right )^{3/2} f}\\ &=\frac {i (c+d x)^2}{\left (a^2-b^2\right ) f}-\frac {2 d (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) f^2}-\frac {i a (c+d x)^2 \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f}-\frac {2 d (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) f^2}+\frac {i a (c+d x)^2 \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f}+\frac {2 i d^2 \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) f^3}-\frac {2 a d (c+d x) \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f^2}+\frac {2 i d^2 \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) f^3}+\frac {2 a d (c+d x) \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f^2}+\frac {b (c+d x)^2 \cos (e+f x)}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))}+\frac {\left (2 a d^2\right ) \int \text {Li}_2\left (\frac {2 i b e^{i (e+f x)}}{2 a-2 \sqrt {a^2-b^2}}\right ) \, dx}{\left (a^2-b^2\right )^{3/2} f^2}-\frac {\left (2 a d^2\right ) \int \text {Li}_2\left (\frac {2 i b e^{i (e+f x)}}{2 a+2 \sqrt {a^2-b^2}}\right ) \, dx}{\left (a^2-b^2\right )^{3/2} f^2}\\ &=\frac {i (c+d x)^2}{\left (a^2-b^2\right ) f}-\frac {2 d (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) f^2}-\frac {i a (c+d x)^2 \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f}-\frac {2 d (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) f^2}+\frac {i a (c+d x)^2 \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f}+\frac {2 i d^2 \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) f^3}-\frac {2 a d (c+d x) \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f^2}+\frac {2 i d^2 \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) f^3}+\frac {2 a d (c+d x) \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f^2}+\frac {b (c+d x)^2 \cos (e+f x)}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))}-\frac {\left (2 i a d^2\right ) \text {Subst}\left (\int \frac {\text {Li}_2\left (\frac {i b x}{a-\sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (e+f x)}\right )}{\left (a^2-b^2\right )^{3/2} f^3}+\frac {\left (2 i a d^2\right ) \text {Subst}\left (\int \frac {\text {Li}_2\left (\frac {i b x}{a+\sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (e+f x)}\right )}{\left (a^2-b^2\right )^{3/2} f^3}\\ &=\frac {i (c+d x)^2}{\left (a^2-b^2\right ) f}-\frac {2 d (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) f^2}-\frac {i a (c+d x)^2 \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f}-\frac {2 d (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) f^2}+\frac {i a (c+d x)^2 \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f}+\frac {2 i d^2 \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) f^3}-\frac {2 a d (c+d x) \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f^2}+\frac {2 i d^2 \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) f^3}+\frac {2 a d (c+d x) \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f^2}-\frac {2 i a d^2 \text {Li}_3\left (\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f^3}+\frac {2 i a d^2 \text {Li}_3\left (\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f^3}+\frac {b (c+d x)^2 \cos (e+f x)}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 1.11, size = 530, normalized size = 0.79 \begin {gather*} \frac {i f^2 (c+d x)^2-2 d f (c+d x) \log \left (1+\frac {i b e^{i (e+f x)}}{-a+\sqrt {a^2-b^2}}\right )-2 d f (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )+2 i d^2 \text {Li}_2\left (-\frac {i b e^{i (e+f x)}}{-a+\sqrt {a^2-b^2}}\right )+2 i d^2 \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )-\frac {i a \left (f^2 (c+d x)^2 \log \left (1+\frac {i b e^{i (e+f x)}}{-a+\sqrt {a^2-b^2}}\right )-f^2 (c+d x)^2 \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )-2 i d f (c+d x) \text {Li}_2\left (-\frac {i b e^{i (e+f x)}}{-a+\sqrt {a^2-b^2}}\right )+2 i d f (c+d x) \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )+2 d^2 \text {Li}_3\left (\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )-2 d^2 \text {Li}_3\left (\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )\right )}{\sqrt {a^2-b^2}}+\frac {b f^2 (c+d x)^2 \cos (e+f x)}{a+b \sin (e+f x)}}{\left (a^2-b^2\right ) f^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(I*f^2*(c + d*x)^2 - 2*d*f*(c + d*x)*Log[1 + (I*b*E^(I*(e + f*x)))/(-a + Sqrt[a^2 - b^2])] - 2*d*f*(c + d*x)*L
og[1 - (I*b*E^(I*(e + f*x)))/(a + Sqrt[a^2 - b^2])] + (2*I)*d^2*PolyLog[2, ((-I)*b*E^(I*(e + f*x)))/(-a + Sqrt
[a^2 - b^2])] + (2*I)*d^2*PolyLog[2, (I*b*E^(I*(e + f*x)))/(a + Sqrt[a^2 - b^2])] - (I*a*(f^2*(c + d*x)^2*Log[
1 + (I*b*E^(I*(e + f*x)))/(-a + Sqrt[a^2 - b^2])] - f^2*(c + d*x)^2*Log[1 - (I*b*E^(I*(e + f*x)))/(a + Sqrt[a^
2 - b^2])] - (2*I)*d*f*(c + d*x)*PolyLog[2, ((-I)*b*E^(I*(e + f*x)))/(-a + Sqrt[a^2 - b^2])] + (2*I)*d*f*(c +
d*x)*PolyLog[2, (I*b*E^(I*(e + f*x)))/(a + Sqrt[a^2 - b^2])] + 2*d^2*PolyLog[3, (I*b*E^(I*(e + f*x)))/(a - Sqr
t[a^2 - b^2])] - 2*d^2*PolyLog[3, (I*b*E^(I*(e + f*x)))/(a + Sqrt[a^2 - b^2])]))/Sqrt[a^2 - b^2] + (b*f^2*(c +
d*x)^2*Cos[e + f*x])/(a + b*Sin[e + f*x]))/((a^2 - b^2)*f^3)

________________________________________________________________________________________

Maple [F]
time = 0.71, size = 0, normalized size = 0.00 \[\int \frac {\left (d x +c \right )^{2}}{\left (a +b \sin \left (f x +e \right )\right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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)^2/(a+b*sin(f*x+e))^2,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. 3197 vs. \(2 (593) = 1186\).
time = 0.60, size = 3197, normalized size = 4.76 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(a*b^2*d^2*sin(f*x + e) + a^2*b*d^2)*sqrt(-(a^2 - b^2)/b^2)*polylog(3, -(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) - 2*(a*b^2*d^2*sin(f*x + e) + a^2*b*d^2)*
sqrt(-(a^2 - b^2)/b^2)*polylog(3, -(I*a*cos(f*x + e) + a*sin(f*x + e) - (b*cos(f*x + e) - I*b*sin(f*x + e))*sq
rt(-(a^2 - b^2)/b^2))/b) + 2*(a*b^2*d^2*sin(f*x + e) + a^2*b*d^2)*sqrt(-(a^2 - b^2)/b^2)*polylog(3, -(-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) - 2*(a*b^2*d^2*sin
(f*x + e) + a^2*b*d^2)*sqrt(-(a^2 - b^2)/b^2)*polylog(3, -(-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) + 2*((a^2*b - b^3)*d^2*f^2*x^2 + 2*(a^2*b - b^3)*c*d*f^2*x +
(a^2*b - b^3)*c^2*f^2)*cos(f*x + e) - 2*(-I*(a^2*b - b^3)*d^2*sin(f*x + e) - I*(a^3 - a*b^2)*d^2 + (-I*a^2*b*d
^2*f*x - I*a^2*b*c*d*f + (-I*a*b^2*d^2*f*x - I*a*b^2*c*d*f)*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2))*dilog((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 + 1) - 2*(-I*
(a^2*b - b^3)*d^2*sin(f*x + e) - I*(a^3 - a*b^2)*d^2 + (I*a^2*b*d^2*f*x + I*a^2*b*c*d*f + (I*a*b^2*d^2*f*x + I
*a*b^2*c*d*f)*sin(f*x + e))*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) - 2*(I*(a^2*b - b^3)*d^2*sin(f*x + e) + I*(a^3 - a*b^2
)*d^2 + (I*a^2*b*d^2*f*x + I*a^2*b*c*d*f + (I*a*b^2*d^2*f*x + I*a*b^2*c*d*f)*sin(f*x + e))*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) - 2*(I*(a^2*b - b^3)*d^2*sin(f*x + e) + I*(a^3 - a*b^2)*d^2 + (-I*a^2*b*d^2*f*x - I*a^2*b*c*d*f + (-
I*a*b^2*d^2*f*x - I*a*b^2*c*d*f)*sin(f*x + e))*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) - (2*(a^3 - a*b^2)*c*d*f - 2*(a^3
- a*b^2)*d^2*e + 2*((a^2*b - b^3)*c*d*f - (a^2*b - b^3)*d^2*e)*sin(f*x + e) - (a^2*b*c^2*f^2 - 2*a^2*b*c*d*f*e
+ a^2*b*d^2*e^2 + (a*b^2*c^2*f^2 - 2*a*b^2*c*d*f*e + a*b^2*d^2*e^2)*sin(f*x + 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) - (2*(a^3 - a*b^2)*c*d*f - 2*(a^3
- a*b^2)*d^2*e + 2*((a^2*b - b^3)*c*d*f - (a^2*b - b^3)*d^2*e)*sin(f*x + e) - (a^2*b*c^2*f^2 - 2*a^2*b*c*d*f*
e + a^2*b*d^2*e^2 + (a*b^2*c^2*f^2 - 2*a*b^2*c*d*f*e + a*b^2*d^2*e^2)*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2))*lo
g(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*(a^3 - a*b^2)*c*d*f - 2*(a^
3 - a*b^2)*d^2*e + 2*((a^2*b - b^3)*c*d*f - (a^2*b - b^3)*d^2*e)*sin(f*x + e) + (a^2*b*c^2*f^2 - 2*a^2*b*c*d*f
*e + a^2*b*d^2*e^2 + (a*b^2*c^2*f^2 - 2*a*b^2*c*d*f*e + a*b^2*d^2*e^2)*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2))*l
og(-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*(a^3 - a*b^2)*c*d*f - 2*(
a^3 - a*b^2)*d^2*e + 2*((a^2*b - b^3)*c*d*f - (a^2*b - b^3)*d^2*e)*sin(f*x + e) + (a^2*b*c^2*f^2 - 2*a^2*b*c*d
*f*e + a^2*b*d^2*e^2 + (a*b^2*c^2*f^2 - 2*a*b^2*c*d*f*e + a*b^2*d^2*e^2)*sin(f*x + 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) - (2*(a^3 - a*b^2)*d^2*f*x +
2*(a^3 - a*b^2)*d^2*e + 2*((a^2*b - b^3)*d^2*f*x + (a^2*b - b^3)*d^2*e)*sin(f*x + e) + (a^2*b*d^2*f^2*x^2 + 2
*a^2*b*c*d*f^2*x + 2*a^2*b*c*d*f*e - a^2*b*d^2*e^2 + (a*b^2*d^2*f^2*x^2 + 2*a*b^2*c*d*f^2*x + 2*a*b^2*c*d*f*e
- a*b^2*d^2*e^2)*sin(f*x + 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) - (2*(a^3 - a*b^2)*d^2*f*x + 2*(a^3 - a*b^2)*d^2*e + 2*(
(a^2*b - b^3)*d^2*f*x + (a^2*b - b^3)*d^2*e)*sin(f*x + e) - (a^2*b*d^2*f^2*x^2 + 2*a^2*b*c*d*f^2*x + 2*a^2*b*c
*d*f*e - a^2*b*d^2*e^2 + (a*b^2*d^2*f^2*x^2 + 2*a*b^2*c*d*f^2*x + 2*a*b^2*c*d*f*e - a*b^2*d^2*e^2)*sin(f*x + 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) - (2*(a^3 - a*b^2)*d^2*f*x + 2*(a^3 - a*b^2)*d^2*e + 2*((a^2*b - b^3)*d^2*f*x + (a^2
*b - b^3)*d^2*e)*sin(f*x + e) + (a^2*b*d^2*f^2*x^2 + 2*a^2*b*c*d*f^2*x + 2*a^2*b*c*d*f*e - a^2*b*d^2*e^2 + (a*
b^2*d^2*f^2*x^2 + 2*a*b^2*c*d*f^2*x + 2*a*b^2*c*d*f*e - a*b^2*d^2*e^2)*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2))*l
og(-(-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) -
(2*(a^3 - a*b^2)*d^2*f*x + 2*(a^3 - a*b^2)*d^2*e + 2*((a^2*b - b^3)*d^2*f*x + (a^2*b - b^3)*d^2*e)*sin(f*x +
e) - (a^2*b*d^2*f^2*x^2 + 2*a^2*b*c*d*f^2*x + 2*a^2*b*c*d*f*e - a^2*b*d^2*e^2 + (a*b^2*d^2*f^2*x^2 + 2*a*b^2*c
*d*f^2*x + 2*a*b^2*c*d*f*e - a*b^2*d^2*e^2)*sin(f*x + 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^4*b - 2*a^2*b^3 + b^5)*
f^3*sin(f*x + e) + (a^5 - 2*a^3*b^2 + a*b^4)*f^3)

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Mupad [F(-1)]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \text {Hanged} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

\text{Hanged}

________________________________________________________________________________________