\(\int \frac {(e+f x)^2 \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx\) [311]

Optimal result

Integrand size = 28, antiderivative size = 659 \[ \int \frac {(e+f x)^2 \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {i a (e+f x)^2}{\left (a^2-b^2\right ) d}-\frac {4 i b f (e+f x) \arctan \left (e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}+\frac {i b^2 (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}-\frac {i b^2 (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}+\frac {2 a f (e+f x) \log \left (1+e^{2 i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}+\frac {2 i b f^2 \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^3}-\frac {2 i b f^2 \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^3}+\frac {2 b^2 f (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^2}-\frac {2 b^2 f (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^2}-\frac {i a f^2 \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right )}{\left (a^2-b^2\right ) d^3}+\frac {2 i b^2 f^2 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^3}-\frac {2 i b^2 f^2 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^3}-\frac {b (e+f x)^2 \sec (c+d x)}{\left (a^2-b^2\right ) d}+\frac {a (e+f x)^2 \tan (c+d x)}{\left (a^2-b^2\right ) d} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 8.91 (sec) , antiderivative size = 1122, normalized size of antiderivative = 1.70 \[ \int \frac {(e+f x)^2 \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

(I*b^2*(-2*Sqrt[a^2 - b^2]*d*f*(e + f*x)*PolyLog[2, (b*E^(I*(c + d*x)))/(( 
-I)*a + Sqrt[-a^2 + b^2])] + 2*Sqrt[a^2 - b^2]*d*f*(e + f*x)*PolyLog[2, -( 
(b*E^(I*(c + d*x)))/(I*a + Sqrt[-a^2 + b^2]))] - I*(d^2*(2*Sqrt[-a^2 + b^2 
]*e^2*ArcTan[(I*a + b*E^(I*(c + d*x)))/Sqrt[a^2 - b^2]] + Sqrt[a^2 - b^2]* 
f*x*(2*e + f*x)*(Log[1 - (b*E^(I*(c + d*x)))/((-I)*a + Sqrt[-a^2 + b^2])] 
- Log[1 + (b*E^(I*(c + d*x)))/(I*a + Sqrt[-a^2 + b^2])])) + 2*Sqrt[a^2 - b 
^2]*f^2*PolyLog[3, (b*E^(I*(c + d*x)))/((-I)*a + Sqrt[-a^2 + b^2])] - 2*Sq 
rt[a^2 - b^2]*f^2*PolyLog[3, -((b*E^(I*(c + d*x)))/(I*a + Sqrt[-a^2 + b^2] 
))])))/(Sqrt[-(a^2 - b^2)^2]*(-a^2 + b^2)*d^3) + (b*(e + f*x)^2*Sec[c])/(( 
-a^2 + b^2)*d) + (2*a*e*f*Sec[c]*(Cos[c]*Log[Cos[c]*Cos[d*x] - Sin[c]*Sin[ 
d*x]] + d*x*Sin[c]))/((a^2 - b^2)*d^2*(Cos[c]^2 + Sin[c]^2)) + ((4*I)*b*e* 
f*ArcTan[((-I)*Sin[c] - I*Cos[c]*Tan[(d*x)/2])/Sqrt[Cos[c]^2 + Sin[c]^2]]) 
/((a^2 - b^2)*d^2*Sqrt[Cos[c]^2 + Sin[c]^2]) + (a*f^2*Csc[c]*((d^2*x^2)/E^ 
(I*ArcTan[Cot[c]]) - (Cot[c]*(I*d*x*(-Pi - 2*ArcTan[Cot[c]]) - Pi*Log[1 + 
E^((-2*I)*d*x)] - 2*(d*x - ArcTan[Cot[c]])*Log[1 - E^((2*I)*(d*x - ArcTan[ 
Cot[c]]))] + Pi*Log[Cos[d*x]] - 2*ArcTan[Cot[c]]*Log[Sin[d*x - ArcTan[Cot[ 
c]]]] + I*PolyLog[2, E^((2*I)*(d*x - ArcTan[Cot[c]]))]))/Sqrt[1 + Cot[c]^2 
])*Sec[c])/((a^2 - b^2)*d^3*Sqrt[Csc[c]^2*(Cos[c]^2 + Sin[c]^2)]) + (2*b*f 
^2*(-((Csc[c]*((d*x - ArcTan[Cot[c]])*(Log[1 - E^(I*(d*x - ArcTan[Cot[c]]) 
)] - Log[1 + E^(I*(d*x - ArcTan[Cot[c]]))]) + I*(PolyLog[2, -E^(I*(d*x ...
 

Rubi [A] (verified)

Time = 2.49 (sec) , antiderivative size = 561, normalized size of antiderivative = 0.85, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.393, Rules used = {5044, 3042, 3804, 2694, 27, 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.

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

\(\Big \downarrow \) 5044

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3804

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

\(\Big \downarrow \) 2694

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {i a f^2 \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right )}{d^3}+\frac {2 a f (e+f x) \log \left (1+e^{2 i (c+d x)}\right )}{d^2}+\frac {a (e+f x)^2 \tan (c+d x)}{d}-\frac {i a (e+f x)^2}{d}-\frac {4 i b f (e+f x) \arctan \left (e^{i (c+d x)}\right )}{d^2}+\frac {2 i b f^2 \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d^3}-\frac {2 i b f^2 \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d^3}-\frac {b (e+f x)^2 \sec (c+d x)}{d}}{a^2-b^2}-\frac {2 b^2 \left (\frac {i b \left (\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}-\frac {f \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d^2}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}-\frac {f \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d^2}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}\right )}{a^2-b^2}\)

Input:

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

Output:

(-2*b^2*(((-1/2*I)*b*(((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)))/Sqrt[a^2 - b^2] + ((I/2)*b*(((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)))/Sqrt[a^2 - 
 b^2]))/(a^2 - b^2) + (((-I)*a*(e + f*x)^2)/d - ((4*I)*b*f*(e + f*x)*ArcTa 
n[E^(I*(c + d*x))])/d^2 + (2*a*f*(e + f*x)*Log[1 + E^((2*I)*(c + d*x))])/d 
^2 + ((2*I)*b*f^2*PolyLog[2, (-I)*E^(I*(c + d*x))])/d^3 - ((2*I)*b*f^2*Pol 
yLog[2, I*E^(I*(c + d*x))])/d^3 - (I*a*f^2*PolyLog[2, -E^((2*I)*(c + d*x)) 
])/d^3 - (b*(e + f*x)^2*Sec[c + d*x])/d + (a*(e + f*x)^2*Tan[c + d*x])/d)/ 
(a^2 - b^2)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 2694

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]}, Simp[2*(c/q)   Int 
[(f + g*x)^m*(F^u/(b - q + 2*c*F^u)), x], x] - Simp[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 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 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3804

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Sy 
mbol] :> Simp[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 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]

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

Input:

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

Output:

int((f*x+e)^2*sec(d*x+c)^2/(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. 2659 vs. \(2 (574) = 1148\).

Time = 0.36 (sec) , antiderivative size = 2659, normalized size of antiderivative = 4.03 \[ \int \frac {(e+f x)^2 \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/2*(2*b^3*f^2*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*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^3*f^2*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*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^3*f^2*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c 
)*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^3*f^2*sqrt(-(a^2 - b^2)/b^2 
)*cos(d*x + c)*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^2*b - b^3)*d^ 
2*f^2*x^2 + 4*(a^2*b - b^3)*d^2*e*f*x + 2*(a^2*b - b^3)*d^2*e^2 - 2*I*(a^3 
 - a^2*b - a*b^2 + b^3)*f^2*cos(d*x + c)*dilog(I*cos(d*x + c) + sin(d*x + 
c)) + 2*I*(a^3 + a^2*b - a*b^2 - b^3)*f^2*cos(d*x + c)*dilog(I*cos(d*x + c 
) - sin(d*x + c)) + 2*I*(a^3 - a^2*b - a*b^2 + b^3)*f^2*cos(d*x + c)*dilog 
(-I*cos(d*x + c) + sin(d*x + c)) - 2*I*(a^3 + a^2*b - a*b^2 - b^3)*f^2*cos 
(d*x + c)*dilog(-I*cos(d*x + c) - sin(d*x + c)) + 2*(I*b^3*d*f^2*x + I*b^3 
*d*e*f)*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*dilog((I*a*cos(d*x + c) - a*si 
n(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^3*d*f^2*x - I*b^3*d*e*f)*sqrt(-(a^2 - b^2)/b^2)*cos(d* 
x + c)*dilog((I*a*cos(d*x + c) - a*sin(d*x + c) - (b*cos(d*x + c) + I*b*si 
n(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) + 2*(-I*b^3*d*f^2*x - I*...
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate((f*x+e)^2*sec(d*x+c)^2/(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]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

\text{Hanged}
 

Reduce [F]

\[ \int \frac {(e+f x)^2 \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {-2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \cos \left (d x +c \right ) b^{2} e^{2}+\cos \left (d x +c \right ) \left (\int \frac {\sec \left (d x +c \right )^{2} x^{2}}{\sin \left (d x +c \right ) b +a}d x \right ) a^{4} d \,f^{2}-2 \cos \left (d x +c \right ) \left (\int \frac {\sec \left (d x +c \right )^{2} x^{2}}{\sin \left (d x +c \right ) b +a}d x \right ) a^{2} b^{2} d \,f^{2}+\cos \left (d x +c \right ) \left (\int \frac {\sec \left (d x +c \right )^{2} x^{2}}{\sin \left (d x +c \right ) b +a}d x \right ) b^{4} d \,f^{2}+2 \cos \left (d x +c \right ) \left (\int \frac {\sec \left (d x +c \right )^{2} x}{\sin \left (d x +c \right ) b +a}d x \right ) a^{4} d e f -4 \cos \left (d x +c \right ) \left (\int \frac {\sec \left (d x +c \right )^{2} x}{\sin \left (d x +c \right ) b +a}d x \right ) a^{2} b^{2} d e f +2 \cos \left (d x +c \right ) \left (\int \frac {\sec \left (d x +c \right )^{2} x}{\sin \left (d x +c \right ) b +a}d x \right ) b^{4} d e f +\cos \left (d x +c \right ) a^{2} b \,e^{2}-\cos \left (d x +c \right ) b^{3} e^{2}+\sin \left (d x +c \right ) a^{3} e^{2}-\sin \left (d x +c \right ) a \,b^{2} e^{2}-a^{2} b \,e^{2}+b^{3} e^{2}}{\cos \left (d x +c \right ) d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )} \] Input:

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

Output:

( - 2*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*c 
os(c + d*x)*b**2*e**2 + cos(c + d*x)*int((sec(c + d*x)**2*x**2)/(sin(c + d 
*x)*b + a),x)*a**4*d*f**2 - 2*cos(c + d*x)*int((sec(c + d*x)**2*x**2)/(sin 
(c + d*x)*b + a),x)*a**2*b**2*d*f**2 + cos(c + d*x)*int((sec(c + d*x)**2*x 
**2)/(sin(c + d*x)*b + a),x)*b**4*d*f**2 + 2*cos(c + d*x)*int((sec(c + d*x 
)**2*x)/(sin(c + d*x)*b + a),x)*a**4*d*e*f - 4*cos(c + d*x)*int((sec(c + d 
*x)**2*x)/(sin(c + d*x)*b + a),x)*a**2*b**2*d*e*f + 2*cos(c + d*x)*int((se 
c(c + d*x)**2*x)/(sin(c + d*x)*b + a),x)*b**4*d*e*f + cos(c + d*x)*a**2*b* 
e**2 - cos(c + d*x)*b**3*e**2 + sin(c + d*x)*a**3*e**2 - sin(c + d*x)*a*b* 
*2*e**2 - a**2*b*e**2 + b**3*e**2)/(cos(c + d*x)*d*(a**4 - 2*a**2*b**2 + b 
**4))