\(\int \frac {(e+f x)^2 \sec (c+d x)}{a+b \sin (c+d x)} \, dx\) [307]

Optimal result

Integrand size = 26, antiderivative size = 667 \[ \int \frac {(e+f x)^2 \sec (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {2 i a (e+f x)^2 \arctan \left (e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d}-\frac {b (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) d}-\frac {b (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) d}+\frac {b (e+f x)^2 \log \left (1+e^{2 i (c+d x)}\right )}{\left (a^2-b^2\right ) d}+\frac {2 i a f (e+f x) \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}-\frac {2 i a f (e+f x) \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}+\frac {2 i b f (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) d^2}+\frac {2 i b f (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) d^2}-\frac {i b f (e+f x) \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}-\frac {2 a f^2 \operatorname {PolyLog}\left (3,-i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^3}+\frac {2 a f^2 \operatorname {PolyLog}\left (3,i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^3}-\frac {2 b f^2 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) d^3}-\frac {2 b f^2 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) d^3}+\frac {b f^2 \operatorname {PolyLog}\left (3,-e^{2 i (c+d x)}\right )}{2 \left (a^2-b^2\right ) d^3} \] Output:

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

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(1461\) vs. \(2(667)=1334\).

Time = 6.91 (sec) , antiderivative size = 1461, normalized size of antiderivative = 2.19 \[ \int \frac {(e+f x)^2 \sec (c+d x)}{a+b \sin (c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

((2*(((2*I)*b*(e + f*x)^3)/f - (3*(a - b)*(1 + E^((2*I)*c))*(e + f*x)^2*Lo 
g[1 - I/E^(I*(c + d*x))])/d + (3*(a + b)*(1 + E^((2*I)*c))*(e + f*x)^2*Log 
[1 + I/E^(I*(c + d*x))])/d + (6*(a + b)*(1 + E^((2*I)*c))*f*(I*d*(e + f*x) 
*PolyLog[2, (-I)/E^(I*(c + d*x))] + f*PolyLog[3, (-I)/E^(I*(c + d*x))]))/d 
^3 - ((6*I)*(a - b)*(1 + E^((2*I)*c))*f*(d*(e + f*x)*PolyLog[2, I/E^(I*(c 
+ d*x))] - I*f*PolyLog[3, I/E^(I*(c + d*x))]))/d^3))/((a^2 - b^2)*(1 + E^( 
(2*I)*c))) + (2*b*((-6*I)*d^3*e^2*E^((2*I)*c)*x - (6*I)*d^3*e*E^((2*I)*c)* 
f*x^2 - (2*I)*d^3*E^((2*I)*c)*f^2*x^3 - 3*d^2*e^2*Log[b - (2*I)*a*E^(I*(c 
+ d*x)) - b*E^((2*I)*(c + d*x))] + 3*d^2*e^2*E^((2*I)*c)*Log[b - (2*I)*a*E 
^(I*(c + d*x)) - b*E^((2*I)*(c + d*x))] - 6*d^2*e*f*x*Log[1 + (b*E^(I*(2*c 
 + d*x)))/(I*a*E^(I*c) - Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + 6*d^2*e*E^((2* 
I)*c)*f*x*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) - Sqrt[(-a^2 + b^2)*E 
^((2*I)*c)])] - 3*d^2*f^2*x^2*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) - 
 Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + 3*d^2*E^((2*I)*c)*f^2*x^2*Log[1 + (b*E 
^(I*(2*c + d*x)))/(I*a*E^(I*c) - Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - 6*d^2* 
e*f*x*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2 
*I)*c)])] + 6*d^2*e*E^((2*I)*c)*f*x*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^( 
I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - 3*d^2*f^2*x^2*Log[1 + (b*E^(I*(2 
*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + 3*d^2*E^((2* 
I)*c)*f^2*x^2*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 +...
 

Rubi [A] (verified)

Time = 2.17 (sec) , antiderivative size = 591, normalized size of antiderivative = 0.89, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {5044, 5030, 2620, 3011, 2720, 7143, 7293, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 5044

\(\displaystyle \frac {\int (e+f x)^2 \sec (c+d x) (a-b \sin (c+d x))dx}{a^2-b^2}-\frac {b^2 \int \frac {(e+f x)^2 \cos (c+d x)}{a+b \sin (c+d x)}dx}{a^2-b^2}\)

\(\Big \downarrow \) 5030

\(\displaystyle \frac {\int (e+f x)^2 \sec (c+d x) (a-b \sin (c+d x))dx}{a^2-b^2}-\frac {b^2 \left (\int \frac {e^{i (c+d x)} (e+f x)^2}{a-i b e^{i (c+d x)}-\sqrt {a^2-b^2}}dx+\int \frac {e^{i (c+d x)} (e+f x)^2}{a-i b e^{i (c+d x)}+\sqrt {a^2-b^2}}dx-\frac {i (e+f x)^3}{3 b f}\right )}{a^2-b^2}\)

\(\Big \downarrow \) 2620

\(\displaystyle \frac {\int (e+f x)^2 \sec (c+d x) (a-b \sin (c+d x))dx}{a^2-b^2}-\frac {b^2 \left (-\frac {2 f \int (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )dx}{b d}-\frac {2 f \int (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )dx}{b d}+\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}+\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {i (e+f x)^3}{3 b f}\right )}{a^2-b^2}\)

\(\Big \downarrow \) 3011

\(\displaystyle \frac {\int (e+f x)^2 \sec (c+d x) (a-b \sin (c+d x))dx}{a^2-b^2}-\frac {b^2 \left (-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}-\frac {i f \int \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )dx}{d}\right )}{b d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}-\frac {i f \int \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )dx}{d}\right )}{b d}+\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}+\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {i (e+f x)^3}{3 b f}\right )}{a^2-b^2}\)

\(\Big \downarrow \) 2720

\(\displaystyle \frac {\int (e+f x)^2 \sec (c+d x) (a-b \sin (c+d x))dx}{a^2-b^2}-\frac {b^2 \left (-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}-\frac {f \int e^{-i (c+d x)} \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )de^{i (c+d x)}}{d^2}\right )}{b d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}-\frac {f \int e^{-i (c+d x)} \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )de^{i (c+d x)}}{d^2}\right )}{b d}+\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}+\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {i (e+f x)^3}{3 b f}\right )}{a^2-b^2}\)

\(\Big \downarrow \) 7143

\(\displaystyle \frac {\int (e+f x)^2 \sec (c+d x) (a-b \sin (c+d x))dx}{a^2-b^2}-\frac {b^2 \left (-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}-\frac {f \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d^2}\right )}{b d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}-\frac {f \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d^2}\right )}{b d}+\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}+\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {i (e+f x)^3}{3 b f}\right )}{a^2-b^2}\)

\(\Big \downarrow \) 7293

\(\displaystyle \frac {\int \left (a (e+f x)^2 \sec (c+d x)-b (e+f x)^2 \tan (c+d x)\right )dx}{a^2-b^2}-\frac {b^2 \left (-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}-\frac {f \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d^2}\right )}{b d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}-\frac {f \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d^2}\right )}{b d}+\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}+\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {i (e+f x)^3}{3 b f}\right )}{a^2-b^2}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {2 i a (e+f x)^2 \arctan \left (e^{i (c+d x)}\right )}{d}-\frac {2 a f^2 \operatorname {PolyLog}\left (3,-i e^{i (c+d x)}\right )}{d^3}+\frac {2 a f^2 \operatorname {PolyLog}\left (3,i e^{i (c+d x)}\right )}{d^3}+\frac {2 i a f (e+f x) \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d^2}-\frac {2 i a f (e+f x) \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d^2}+\frac {b f^2 \operatorname {PolyLog}\left (3,-e^{2 i (c+d x)}\right )}{2 d^3}-\frac {i b f (e+f x) \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right )}{d^2}+\frac {b (e+f x)^2 \log \left (1+e^{2 i (c+d x)}\right )}{d}-\frac {i b (e+f x)^3}{3 f}}{a^2-b^2}-\frac {b^2 \left (-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}-\frac {f \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d^2}\right )}{b d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}-\frac {f \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d^2}\right )}{b d}+\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}+\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {i (e+f x)^3}{3 b f}\right )}{a^2-b^2}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2620

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] - Si 
mp[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 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{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 3011

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] + Simp[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 5030

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 5044

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

rule 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> 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]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

Maple [F]

\[\int \frac {\left (f x +e \right )^{2} \sec \left (d x +c \right )}{a +b \sin \left (d x +c \right )}d x\]

Input:

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

Output:

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

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. 2035 vs. \(2 (588) = 1176\).

Time = 0.32 (sec) , antiderivative size = 2035, normalized size of antiderivative = 3.05 \[ \int \frac {(e+f x)^2 \sec (c+d x)}{a+b \sin (c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/2*(2*b*f^2*polylog(3, -(I*a*cos(d*x + c) + a*sin(d*x + c) + (b*cos(d*x 
+ c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b) + 2*b*f^2*polylog(3, - 
(I*a*cos(d*x + c) + a*sin(d*x + c) - (b*cos(d*x + c) - I*b*sin(d*x + c))*s 
qrt(-(a^2 - b^2)/b^2))/b) + 2*b*f^2*polylog(3, -(-I*a*cos(d*x + c) + a*sin 
(d*x + c) + (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b) 
 + 2*b*f^2*polylog(3, -(-I*a*cos(d*x + c) + a*sin(d*x + c) - (b*cos(d*x + 
c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b) + 2*(a - b)*f^2*polylog( 
3, I*cos(d*x + c) + sin(d*x + c)) - 2*(a + b)*f^2*polylog(3, I*cos(d*x + c 
) - sin(d*x + c)) + 2*(a - b)*f^2*polylog(3, -I*cos(d*x + c) + sin(d*x + c 
)) - 2*(a + b)*f^2*polylog(3, -I*cos(d*x + c) - sin(d*x + c)) - 2*(I*b*d*f 
^2*x + I*b*d*e*f)*dilog((I*a*cos(d*x + c) - a*sin(d*x + c) + (b*cos(d*x + 
c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) - 2*(I*b*d*f^2*x 
 + I*b*d*e*f)*dilog((I*a*cos(d*x + c) - a*sin(d*x + c) - (b*cos(d*x + c) + 
 I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) - 2*(-I*b*d*f^2*x - 
I*b*d*e*f)*dilog((-I*a*cos(d*x + c) - a*sin(d*x + c) + (b*cos(d*x + c) - I 
*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) - 2*(-I*b*d*f^2*x - I* 
b*d*e*f)*dilog((-I*a*cos(d*x + c) - a*sin(d*x + c) - (b*cos(d*x + c) - I*b 
*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) - 2*(-I*(a - b)*d*f^2*x 
- I*(a - b)*d*e*f)*dilog(I*cos(d*x + c) + sin(d*x + c)) - 2*(-I*(a + b)*d* 
f^2*x - I*(a + b)*d*e*f)*dilog(I*cos(d*x + c) - sin(d*x + c)) - 2*(I*(a...
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {(e+f x)^2 \sec (c+d x)}{a+b \sin (c+d x)} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f 
or more de
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(e+f x)^2 \sec (c+d x)}{a+b \sin (c+d x)} \, dx=\text {Hanged} \] Input:

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

Output:

\text{Hanged}
                                                                                    
                                                                                    
 

Reduce [F]

\[ \int \frac {(e+f x)^2 \sec (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\left (\int \frac {\sec \left (d x +c \right ) x^{2}}{\sin \left (d x +c \right ) b +a}d x \right ) a^{2} d \,f^{2}-\left (\int \frac {\sec \left (d x +c \right ) x^{2}}{\sin \left (d x +c \right ) b +a}d x \right ) b^{2} d \,f^{2}+2 \left (\int \frac {\sec \left (d x +c \right ) x}{\sin \left (d x +c \right ) b +a}d x \right ) a^{2} d e f -2 \left (\int \frac {\sec \left (d x +c \right ) x}{\sin \left (d x +c \right ) b +a}d x \right ) b^{2} d e f -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a \,e^{2}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b \,e^{2}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a \,e^{2}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b \,e^{2}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) b \,e^{2}}{d \left (a^{2}-b^{2}\right )} \] Input:

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

Output:

(int((sec(c + d*x)*x**2)/(sin(c + d*x)*b + a),x)*a**2*d*f**2 - int((sec(c 
+ d*x)*x**2)/(sin(c + d*x)*b + a),x)*b**2*d*f**2 + 2*int((sec(c + d*x)*x)/ 
(sin(c + d*x)*b + a),x)*a**2*d*e*f - 2*int((sec(c + d*x)*x)/(sin(c + d*x)* 
b + a),x)*b**2*d*e*f - log(tan((c + d*x)/2) - 1)*a*e**2 + log(tan((c + d*x 
)/2) - 1)*b*e**2 + log(tan((c + d*x)/2) + 1)*a*e**2 + log(tan((c + d*x)/2) 
 + 1)*b*e**2 - log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*b*e** 
2)/(d*(a**2 - b**2))