\(\int \frac {(e+f x) \sec ^3(c+d x)}{a+a \sin (c+d x)} \, dx\) [283]

Optimal result

Integrand size = 26, antiderivative size = 241 \[ \int \frac {(e+f x) \sec ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {3 i (e+f x) \arctan \left (e^{i (c+d x)}\right )}{4 a d}+\frac {3 i f \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{8 a d^2}-\frac {3 i f \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{8 a d^2}-\frac {3 f \sec (c+d x)}{8 a d^2}-\frac {f \sec ^3(c+d x)}{12 a d^2}-\frac {(e+f x) \sec ^4(c+d x)}{4 a d}+\frac {f \tan (c+d x)}{4 a d^2}+\frac {3 (e+f x) \sec (c+d x) \tan (c+d x)}{8 a d}+\frac {(e+f x) \sec ^3(c+d x) \tan (c+d x)}{4 a d}+\frac {f \tan ^3(c+d x)}{12 a d^2} \] Output:

-3/4*I*(f*x+e)*arctan(exp(I*(d*x+c)))/a/d+3/8*I*f*polylog(2,-I*exp(I*(d*x+ 
c)))/a/d^2-3/8*I*f*polylog(2,I*exp(I*(d*x+c)))/a/d^2-3/8*f*sec(d*x+c)/a/d^ 
2-1/12*f*sec(d*x+c)^3/a/d^2-1/4*(f*x+e)*sec(d*x+c)^4/a/d+1/4*f*tan(d*x+c)/ 
a/d^2+3/8*(f*x+e)*sec(d*x+c)*tan(d*x+c)/a/d+1/4*(f*x+e)*sec(d*x+c)^3*tan(d 
*x+c)/a/d+1/12*f*tan(d*x+c)^3/a/d^2
 

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(1171\) vs. \(2(241)=482\).

Time = 13.46 (sec) , antiderivative size = 1171, normalized size of antiderivative = 4.86 \[ \int \frac {(e+f x) \sec ^3(c+d x)}{a+a \sin (c+d x)} \, dx =\text {Too large to display} \] Input:

Integrate[((e + f*x)*Sec[c + d*x]^3)/(a + a*Sin[c + d*x]),x]
 

Output:

(-6*d*e - f + 6*c*f - 6*f*(c + d*x))/(24*d^2*(a + a*Sin[c + d*x])) + (-(d* 
e) + c*f - f*(c + d*x))/(8*d^2*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2*(a 
+ a*Sin[c + d*x])) + (f*Sin[(c + d*x)/2])/(12*d^2*(Cos[(c + d*x)/2] + Sin[ 
(c + d*x)/2])*(a + a*Sin[c + d*x])) + (7*f*Sin[(c + d*x)/2]*(Cos[(c + d*x) 
/2] + Sin[(c + d*x)/2]))/(12*d^2*(a + a*Sin[c + d*x])) + (3*(c + d*x)*(2*d 
*e - 2*c*f + f*(c + d*x))*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2)/(16*d^2 
*(a + a*Sin[c + d*x])) + (3*e*((-c - d*x)/2 - Log[Cos[(c + d*x)/2] - Sin[( 
c + d*x)/2]])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2)/(8*d*(a + a*Sin[c + 
 d*x])) - (3*c*f*((-c - d*x)/2 - Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]) 
*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2)/(8*d^2*(a + a*Sin[c + d*x])) - ( 
3*e*((c + d*x)/2 - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])*(Cos[(c + d*x 
)/2] + Sin[(c + d*x)/2])^2)/(8*d*(a + a*Sin[c + d*x])) + (3*c*f*((c + d*x) 
/2 - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])*(Cos[(c + d*x)/2] + Sin[(c 
+ d*x)/2])^2)/(8*d^2*(a + a*Sin[c + d*x])) - (3*f*((c + d*x)^2/(4*E^((I/4) 
*Pi)) - (((-3*I)/4)*Pi*(c + d*x) - Pi*Log[1 + E^((-I)*(c + d*x))] - 2*(-1/ 
4*Pi + (c + d*x)/2)*Log[1 - E^((2*I)*(-1/4*Pi + (c + d*x)/2))] + Pi*Log[Co 
s[(c + d*x)/2]] - (Pi*Log[-Sin[Pi/4 + (-c - d*x)/2]])/2 + I*PolyLog[2, E^( 
(2*I)*(-1/4*Pi + (c + d*x)/2))])/Sqrt[2])*(Cos[(c + d*x)/2] + Sin[(c + d*x 
)/2])^2)/(4*Sqrt[2]*d^2*(a + a*Sin[c + d*x])) - (3*f*((E^((I/4)*Pi)*(c + d 
*x)^2)/4 + ((-1/4*I)*Pi*(c + d*x) - Pi*Log[1 + E^((-I)*(c + d*x))] - 2*...
 

Rubi [A] (verified)

Time = 1.14 (sec) , antiderivative size = 225, normalized size of antiderivative = 0.93, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5042, 3042, 4673, 3042, 4673, 3042, 4669, 2715, 2838, 4909, 3042, 4254, 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.

\(\displaystyle \int \frac {(e+f x) \sec ^3(c+d x)}{a \sin (c+d x)+a} \, dx\)

\(\Big \downarrow \) 5042

\(\displaystyle \frac {\int (e+f x) \sec ^5(c+d x)dx}{a}-\frac {\int (e+f x) \sec ^4(c+d x) \tan (c+d x)dx}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int (e+f x) \csc \left (c+d x+\frac {\pi }{2}\right )^5dx}{a}-\frac {\int (e+f x) \sec ^4(c+d x) \tan (c+d x)dx}{a}\)

\(\Big \downarrow \) 4673

\(\displaystyle \frac {\frac {3}{4} \int (e+f x) \sec ^3(c+d x)dx-\frac {f \sec ^3(c+d x)}{12 d^2}+\frac {(e+f x) \tan (c+d x) \sec ^3(c+d x)}{4 d}}{a}-\frac {\int (e+f x) \sec ^4(c+d x) \tan (c+d x)dx}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3}{4} \int (e+f x) \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {f \sec ^3(c+d x)}{12 d^2}+\frac {(e+f x) \tan (c+d x) \sec ^3(c+d x)}{4 d}}{a}-\frac {\int (e+f x) \sec ^4(c+d x) \tan (c+d x)dx}{a}\)

\(\Big \downarrow \) 4673

\(\displaystyle \frac {\frac {3}{4} \left (\frac {1}{2} \int (e+f x) \sec (c+d x)dx-\frac {f \sec (c+d x)}{2 d^2}+\frac {(e+f x) \tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {f \sec ^3(c+d x)}{12 d^2}+\frac {(e+f x) \tan (c+d x) \sec ^3(c+d x)}{4 d}}{a}-\frac {\int (e+f x) \sec ^4(c+d x) \tan (c+d x)dx}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3}{4} \left (\frac {1}{2} \int (e+f x) \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {f \sec (c+d x)}{2 d^2}+\frac {(e+f x) \tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {f \sec ^3(c+d x)}{12 d^2}+\frac {(e+f x) \tan (c+d x) \sec ^3(c+d x)}{4 d}}{a}-\frac {\int (e+f x) \sec ^4(c+d x) \tan (c+d x)dx}{a}\)

\(\Big \downarrow \) 4669

\(\displaystyle -\frac {\int (e+f x) \sec ^4(c+d x) \tan (c+d x)dx}{a}+\frac {\frac {3}{4} \left (\frac {1}{2} \left (-\frac {f \int \log \left (1-i e^{i (c+d x)}\right )dx}{d}+\frac {f \int \log \left (1+i e^{i (c+d x)}\right )dx}{d}-\frac {2 i (e+f x) \arctan \left (e^{i (c+d x)}\right )}{d}\right )-\frac {f \sec (c+d x)}{2 d^2}+\frac {(e+f x) \tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {f \sec ^3(c+d x)}{12 d^2}+\frac {(e+f x) \tan (c+d x) \sec ^3(c+d x)}{4 d}}{a}\)

\(\Big \downarrow \) 2715

\(\displaystyle -\frac {\int (e+f x) \sec ^4(c+d x) \tan (c+d x)dx}{a}+\frac {\frac {3}{4} \left (\frac {1}{2} \left (\frac {i f \int e^{-i (c+d x)} \log \left (1-i e^{i (c+d x)}\right )de^{i (c+d x)}}{d^2}-\frac {i f \int e^{-i (c+d x)} \log \left (1+i e^{i (c+d x)}\right )de^{i (c+d x)}}{d^2}-\frac {2 i (e+f x) \arctan \left (e^{i (c+d x)}\right )}{d}\right )-\frac {f \sec (c+d x)}{2 d^2}+\frac {(e+f x) \tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {f \sec ^3(c+d x)}{12 d^2}+\frac {(e+f x) \tan (c+d x) \sec ^3(c+d x)}{4 d}}{a}\)

\(\Big \downarrow \) 2838

\(\displaystyle -\frac {\int (e+f x) \sec ^4(c+d x) \tan (c+d x)dx}{a}+\frac {\frac {3}{4} \left (\frac {1}{2} \left (-\frac {2 i (e+f x) \arctan \left (e^{i (c+d x)}\right )}{d}+\frac {i f \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d^2}-\frac {i f \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d^2}\right )-\frac {f \sec (c+d x)}{2 d^2}+\frac {(e+f x) \tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {f \sec ^3(c+d x)}{12 d^2}+\frac {(e+f x) \tan (c+d x) \sec ^3(c+d x)}{4 d}}{a}\)

\(\Big \downarrow \) 4909

\(\displaystyle -\frac {\frac {(e+f x) \sec ^4(c+d x)}{4 d}-\frac {f \int \sec ^4(c+d x)dx}{4 d}}{a}+\frac {\frac {3}{4} \left (\frac {1}{2} \left (-\frac {2 i (e+f x) \arctan \left (e^{i (c+d x)}\right )}{d}+\frac {i f \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d^2}-\frac {i f \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d^2}\right )-\frac {f \sec (c+d x)}{2 d^2}+\frac {(e+f x) \tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {f \sec ^3(c+d x)}{12 d^2}+\frac {(e+f x) \tan (c+d x) \sec ^3(c+d x)}{4 d}}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {(e+f x) \sec ^4(c+d x)}{4 d}-\frac {f \int \csc \left (c+d x+\frac {\pi }{2}\right )^4dx}{4 d}}{a}+\frac {\frac {3}{4} \left (\frac {1}{2} \left (-\frac {2 i (e+f x) \arctan \left (e^{i (c+d x)}\right )}{d}+\frac {i f \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d^2}-\frac {i f \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d^2}\right )-\frac {f \sec (c+d x)}{2 d^2}+\frac {(e+f x) \tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {f \sec ^3(c+d x)}{12 d^2}+\frac {(e+f x) \tan (c+d x) \sec ^3(c+d x)}{4 d}}{a}\)

\(\Big \downarrow \) 4254

\(\displaystyle -\frac {\frac {f \int \left (\tan ^2(c+d x)+1\right )d(-\tan (c+d x))}{4 d^2}+\frac {(e+f x) \sec ^4(c+d x)}{4 d}}{a}+\frac {\frac {3}{4} \left (\frac {1}{2} \left (-\frac {2 i (e+f x) \arctan \left (e^{i (c+d x)}\right )}{d}+\frac {i f \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d^2}-\frac {i f \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d^2}\right )-\frac {f \sec (c+d x)}{2 d^2}+\frac {(e+f x) \tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {f \sec ^3(c+d x)}{12 d^2}+\frac {(e+f x) \tan (c+d x) \sec ^3(c+d x)}{4 d}}{a}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\frac {f \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{4 d^2}+\frac {(e+f x) \sec ^4(c+d x)}{4 d}}{a}+\frac {\frac {3}{4} \left (\frac {1}{2} \left (-\frac {2 i (e+f x) \arctan \left (e^{i (c+d x)}\right )}{d}+\frac {i f \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d^2}-\frac {i f \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d^2}\right )-\frac {f \sec (c+d x)}{2 d^2}+\frac {(e+f x) \tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {f \sec ^3(c+d x)}{12 d^2}+\frac {(e+f x) \tan (c+d x) \sec ^3(c+d x)}{4 d}}{a}\)

Input:

Int[((e + f*x)*Sec[c + d*x]^3)/(a + a*Sin[c + d*x]),x]
 

Output:

(-1/12*(f*Sec[c + d*x]^3)/d^2 + ((e + f*x)*Sec[c + d*x]^3*Tan[c + d*x])/(4 
*d) + (3*((((-2*I)*(e + f*x)*ArcTan[E^(I*(c + d*x))])/d + (I*f*PolyLog[2, 
(-I)*E^(I*(c + d*x))])/d^2 - (I*f*PolyLog[2, I*E^(I*(c + d*x))])/d^2)/2 - 
(f*Sec[c + d*x])/(2*d^2) + ((e + f*x)*Sec[c + d*x]*Tan[c + d*x])/(2*d)))/4 
)/a - (((e + f*x)*Sec[c + d*x]^4)/(4*d) + (f*(-Tan[c + d*x] - Tan[c + d*x] 
^3/3))/(4*d^2))/a
 

Defintions of rubi rules used

rule 2009

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

rule 2715

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[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 2838

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 3042

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

rule 4254

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

rule 4669

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

rule 4673

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

rule 4909

Int[((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b_.)*(x_)]^(n_.)*Tan[(a_.) + (b 
_.)*(x_)]^(p_.), x_Symbol] :> Simp[(c + d*x)^m*(Sec[a + b*x]^n/(b*n)), x] - 
 Simp[d*(m/(b*n))   Int[(c + d*x)^(m - 1)*Sec[a + b*x]^n, x], x] /; FreeQ[{ 
a, b, c, d, n}, x] && EqQ[p, 1] && GtQ[m, 0]
 

rule 5042

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

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 433 vs. \(2 (210 ) = 420\).

Time = 2.60 (sec) , antiderivative size = 434, normalized size of antiderivative = 1.80

method	result	size
risch	\(-\frac {i \left (-18 i d f x \,{\mathrm e}^{2 i \left (d x +c \right )}+9 d f x \,{\mathrm e}^{5 i \left (d x +c \right )}-8 i f \,{\mathrm e}^{3 i \left (d x +c \right )}+18 i d f x \,{\mathrm e}^{4 i \left (d x +c \right )}+9 d e \,{\mathrm e}^{5 i \left (d x +c \right )}-18 i d e \,{\mathrm e}^{2 i \left (d x +c \right )}+6 d f x \,{\mathrm e}^{3 i \left (d x +c \right )}+i {\mathrm e}^{i \left (d x +c \right )} f -9 i f \,{\mathrm e}^{5 i \left (d x +c \right )}+6 d e \,{\mathrm e}^{3 i \left (d x +c \right )}+18 f \,{\mathrm e}^{4 i \left (d x +c \right )}+9 d f x \,{\mathrm e}^{i \left (d x +c \right )}+18 i d e \,{\mathrm e}^{4 i \left (d x +c \right )}+9 d e \,{\mathrm e}^{i \left (d x +c \right )}+22 f \,{\mathrm e}^{2 i \left (d x +c \right )}+4 f \right )}{12 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{4} d^{2} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{2} a}-\frac {3 i e \arctan \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{4 d a}+\frac {3 f \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) x}{8 d a}+\frac {3 f \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) c}{8 d^{2} a}-\frac {3 i f \operatorname {polylog}\left (2, i {\mathrm e}^{i \left (d x +c \right )}\right )}{8 a \,d^{2}}-\frac {3 f \ln \left (1+i {\mathrm e}^{i \left (d x +c \right )}\right ) x}{8 d a}-\frac {3 f \ln \left (1+i {\mathrm e}^{i \left (d x +c \right )}\right ) c}{8 d^{2} a}+\frac {3 i f \operatorname {polylog}\left (2, -i {\mathrm e}^{i \left (d x +c \right )}\right )}{8 a \,d^{2}}+\frac {3 i f c \arctan \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{4 d^{2} a}\)	\(434\)

Input:

int((f*x+e)*sec(d*x+c)^3/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

-1/12*I*(-18*I*d*f*x*exp(2*I*(d*x+c))+9*d*f*x*exp(5*I*(d*x+c))-8*I*f*exp(3 
*I*(d*x+c))+18*I*d*f*x*exp(4*I*(d*x+c))+9*d*e*exp(5*I*(d*x+c))-18*I*d*e*ex 
p(2*I*(d*x+c))+6*d*f*x*exp(3*I*(d*x+c))+I*exp(I*(d*x+c))*f-9*I*f*exp(5*I*( 
d*x+c))+6*d*e*exp(3*I*(d*x+c))+18*f*exp(4*I*(d*x+c))+9*d*f*x*exp(I*(d*x+c) 
)+18*I*d*e*exp(4*I*(d*x+c))+9*d*e*exp(I*(d*x+c))+22*f*exp(2*I*(d*x+c))+4*f 
)/(exp(I*(d*x+c))+I)^4/d^2/(exp(I*(d*x+c))-I)^2/a-3/4*I/d/a*e*arctan(exp(I 
*(d*x+c)))+3/8/d/a*f*ln(1-I*exp(I*(d*x+c)))*x+3/8/d^2/a*f*ln(1-I*exp(I*(d* 
x+c)))*c-3/8*I*f*polylog(2,I*exp(I*(d*x+c)))/a/d^2-3/8/d/a*f*ln(1+I*exp(I* 
(d*x+c)))*x-3/8/d^2/a*f*ln(1+I*exp(I*(d*x+c)))*c+3/8*I*f*polylog(2,-I*exp( 
I*(d*x+c)))/a/d^2+3/4*I/d^2/a*f*c*arctan(exp(I*(d*x+c)))
 

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. 792 vs. \(2 (206) = 412\).

Time = 0.18 (sec) , antiderivative size = 792, normalized size of antiderivative = 3.29 \[ \int \frac {(e+f x) \sec ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Too large to display} \] Input:

integrate((f*x+e)*sec(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="fricas")
 

Output:

-1/48*(8*f*cos(d*x + c)^3 - 6*d*f*x + 18*(d*f*x + d*e)*cos(d*x + c)^2 - 6* 
d*e + 14*f*cos(d*x + c) + 9*(I*f*cos(d*x + c)^2*sin(d*x + c) + I*f*cos(d*x 
 + c)^2)*dilog(I*cos(d*x + c) + sin(d*x + c)) + 9*(I*f*cos(d*x + c)^2*sin( 
d*x + c) + I*f*cos(d*x + c)^2)*dilog(I*cos(d*x + c) - sin(d*x + c)) + 9*(- 
I*f*cos(d*x + c)^2*sin(d*x + c) - I*f*cos(d*x + c)^2)*dilog(-I*cos(d*x + c 
) + sin(d*x + c)) + 9*(-I*f*cos(d*x + c)^2*sin(d*x + c) - I*f*cos(d*x + c) 
^2)*dilog(-I*cos(d*x + c) - sin(d*x + c)) - 9*((d*e - c*f)*cos(d*x + c)^2* 
sin(d*x + c) + (d*e - c*f)*cos(d*x + c)^2)*log(cos(d*x + c) + I*sin(d*x + 
c) + I) + 9*((d*e - c*f)*cos(d*x + c)^2*sin(d*x + c) + (d*e - c*f)*cos(d*x 
 + c)^2)*log(cos(d*x + c) - I*sin(d*x + c) + I) - 9*((d*f*x + c*f)*cos(d*x 
 + c)^2*sin(d*x + c) + (d*f*x + c*f)*cos(d*x + c)^2)*log(I*cos(d*x + c) + 
sin(d*x + c) + 1) + 9*((d*f*x + c*f)*cos(d*x + c)^2*sin(d*x + c) + (d*f*x 
+ c*f)*cos(d*x + c)^2)*log(I*cos(d*x + c) - sin(d*x + c) + 1) - 9*((d*f*x 
+ c*f)*cos(d*x + c)^2*sin(d*x + c) + (d*f*x + c*f)*cos(d*x + c)^2)*log(-I* 
cos(d*x + c) + sin(d*x + c) + 1) + 9*((d*f*x + c*f)*cos(d*x + c)^2*sin(d*x 
 + c) + (d*f*x + c*f)*cos(d*x + c)^2)*log(-I*cos(d*x + c) - sin(d*x + c) + 
 1) - 9*((d*e - c*f)*cos(d*x + c)^2*sin(d*x + c) + (d*e - c*f)*cos(d*x + c 
)^2)*log(-cos(d*x + c) + I*sin(d*x + c) + I) + 9*((d*e - c*f)*cos(d*x + c) 
^2*sin(d*x + c) + (d*e - c*f)*cos(d*x + c)^2)*log(-cos(d*x + c) - I*sin(d* 
x + c) + I) - 2*(9*d*f*x + 9*d*e - 5*f*cos(d*x + c))*sin(d*x + c))/(a*d...
 

Sympy [F]

\[ \int \frac {(e+f x) \sec ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {e \sec ^{3}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx + \int \frac {f x \sec ^{3}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \] Input:

integrate((f*x+e)*sec(d*x+c)**3/(a+a*sin(d*x+c)),x)
 

Output:

(Integral(e*sec(c + d*x)**3/(sin(c + d*x) + 1), x) + Integral(f*x*sec(c + 
d*x)**3/(sin(c + d*x) + 1), x))/a
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(e+f x) \sec ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate((f*x+e)*sec(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="maxima")
 

Output:

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un 
defined.
 

Giac [F]

\[ \int \frac {(e+f x) \sec ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\int { \frac {{\left (f x + e\right )} \sec \left (d x + c\right )^{3}}{a \sin \left (d x + c\right ) + a} \,d x } \] Input:

integrate((f*x+e)*sec(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="giac")
 

Output:

integrate((f*x + e)*sec(d*x + c)^3/(a*sin(d*x + c) + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e+f x) \sec ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Hanged} \] Input:

int((e + f*x)/(cos(c + d*x)^3*(a + a*sin(c + d*x))),x)
 

Output:

\text{Hanged}
 

Reduce [F]

\[ \int \frac {(e+f x) \sec ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {8 \left (\int \frac {\sec \left (d x +c \right )^{3} x}{\sin \left (d x +c \right )+1}d x \right ) \sin \left (d x +c \right )^{3} d f +8 \left (\int \frac {\sec \left (d x +c \right )^{3} x}{\sin \left (d x +c \right )+1}d x \right ) \sin \left (d x +c \right )^{2} d f -8 \left (\int \frac {\sec \left (d x +c \right )^{3} x}{\sin \left (d x +c \right )+1}d x \right ) \sin \left (d x +c \right ) d f -8 \left (\int \frac {\sec \left (d x +c \right )^{3} x}{\sin \left (d x +c \right )+1}d x \right ) d f -3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{3} e -3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} e +3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right ) e +3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) e +3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{3} e +3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} e -3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right ) e -3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) e -3 \sin \left (d x +c \right )^{3} e -6 \sin \left (d x +c \right )^{2} e +5 e}{8 a d \left (\sin \left (d x +c \right )^{3}+\sin \left (d x +c \right )^{2}-\sin \left (d x +c \right )-1\right )} \] Input:

int((f*x+e)*sec(d*x+c)^3/(a+a*sin(d*x+c)),x)
 

Output:

(8*int((sec(c + d*x)**3*x)/(sin(c + d*x) + 1),x)*sin(c + d*x)**3*d*f + 8*i 
nt((sec(c + d*x)**3*x)/(sin(c + d*x) + 1),x)*sin(c + d*x)**2*d*f - 8*int(( 
sec(c + d*x)**3*x)/(sin(c + d*x) + 1),x)*sin(c + d*x)*d*f - 8*int((sec(c + 
 d*x)**3*x)/(sin(c + d*x) + 1),x)*d*f - 3*log(tan((c + d*x)/2) - 1)*sin(c 
+ d*x)**3*e - 3*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*e + 3*log(tan((c 
 + d*x)/2) - 1)*sin(c + d*x)*e + 3*log(tan((c + d*x)/2) - 1)*e + 3*log(tan 
((c + d*x)/2) + 1)*sin(c + d*x)**3*e + 3*log(tan((c + d*x)/2) + 1)*sin(c + 
 d*x)**2*e - 3*log(tan((c + d*x)/2) + 1)*sin(c + d*x)*e - 3*log(tan((c + d 
*x)/2) + 1)*e - 3*sin(c + d*x)**3*e - 6*sin(c + d*x)**2*e + 5*e)/(8*a*d*(s 
in(c + d*x)**3 + sin(c + d*x)**2 - sin(c + d*x) - 1))