\(\int (c+d x)^3 \sec ^2(a+b x) \tan (a+b x) \, dx\) [292]

Optimal result

Integrand size = 22, antiderivative size = 115 \[ \int (c+d x)^3 \sec ^2(a+b x) \tan (a+b x) \, dx=\frac {3 i d (c+d x)^2}{2 b^2}-\frac {3 d^2 (c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{b^3}+\frac {3 i d^3 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^4}+\frac {(c+d x)^3 \sec ^2(a+b x)}{2 b}-\frac {3 d (c+d x)^2 \tan (a+b x)}{2 b^2} \] Output:

3/2*I*d*(d*x+c)^2/b^2-3*d^2*(d*x+c)*ln(1+exp(2*I*(b*x+a)))/b^3+3/2*I*d^3*p 
olylog(2,-exp(2*I*(b*x+a)))/b^4+1/2*(d*x+c)^3*sec(b*x+a)^2/b-3/2*d*(d*x+c) 
^2*tan(b*x+a)/b^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. \(286\) vs. \(2(115)=230\).

Time = 6.27 (sec) , antiderivative size = 286, normalized size of antiderivative = 2.49 \[ \int (c+d x)^3 \sec ^2(a+b x) \tan (a+b x) \, dx=\frac {(c+d x)^3 \sec ^2(a+b x)}{2 b}-\frac {3 c d^2 \sec (a) (\cos (a) \log (\cos (a) \cos (b x)-\sin (a) \sin (b x))+b x \sin (a))}{b^3 \left (\cos ^2(a)+\sin ^2(a)\right )}-\frac {3 d^3 \csc (a) \left (b^2 e^{-i \arctan (\cot (a))} x^2-\frac {\cot (a) \left (i b x (-\pi -2 \arctan (\cot (a)))-\pi \log \left (1+e^{-2 i b x}\right )-2 (b x-\arctan (\cot (a))) \log \left (1-e^{2 i (b x-\arctan (\cot (a)))}\right )+\pi \log (\cos (b x))-2 \arctan (\cot (a)) \log (\sin (b x-\arctan (\cot (a))))+i \operatorname {PolyLog}\left (2,e^{2 i (b x-\arctan (\cot (a)))}\right )\right )}{\sqrt {1+\cot ^2(a)}}\right ) \sec (a)}{2 b^4 \sqrt {\csc ^2(a) \left (\cos ^2(a)+\sin ^2(a)\right )}}-\frac {3 \sec (a) \sec (a+b x) \left (c^2 d \sin (b x)+2 c d^2 x \sin (b x)+d^3 x^2 \sin (b x)\right )}{2 b^2} \] Input:

Integrate[(c + d*x)^3*Sec[a + b*x]^2*Tan[a + b*x],x]
 

Output:

((c + d*x)^3*Sec[a + b*x]^2)/(2*b) - (3*c*d^2*Sec[a]*(Cos[a]*Log[Cos[a]*Co 
s[b*x] - Sin[a]*Sin[b*x]] + b*x*Sin[a]))/(b^3*(Cos[a]^2 + Sin[a]^2)) - (3* 
d^3*Csc[a]*((b^2*x^2)/E^(I*ArcTan[Cot[a]]) - (Cot[a]*(I*b*x*(-Pi - 2*ArcTa 
n[Cot[a]]) - Pi*Log[1 + E^((-2*I)*b*x)] - 2*(b*x - ArcTan[Cot[a]])*Log[1 - 
 E^((2*I)*(b*x - ArcTan[Cot[a]]))] + Pi*Log[Cos[b*x]] - 2*ArcTan[Cot[a]]*L 
og[Sin[b*x - ArcTan[Cot[a]]]] + I*PolyLog[2, E^((2*I)*(b*x - ArcTan[Cot[a] 
]))]))/Sqrt[1 + Cot[a]^2])*Sec[a])/(2*b^4*Sqrt[Csc[a]^2*(Cos[a]^2 + Sin[a] 
^2)]) - (3*Sec[a]*Sec[a + b*x]*(c^2*d*Sin[b*x] + 2*c*d^2*x*Sin[b*x] + d^3* 
x^2*Sin[b*x]))/(2*b^2)
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.11, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {4909, 3042, 4672, 25, 3042, 4202, 2620, 2715, 2838}

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.

\(\displaystyle \int (c+d x)^3 \tan (a+b x) \sec ^2(a+b x) \, dx\)

\(\Big \downarrow \) 4909

\(\displaystyle \frac {(c+d x)^3 \sec ^2(a+b x)}{2 b}-\frac {3 d \int (c+d x)^2 \sec ^2(a+b x)dx}{2 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(c+d x)^3 \sec ^2(a+b x)}{2 b}-\frac {3 d \int (c+d x)^2 \csc \left (a+b x+\frac {\pi }{2}\right )^2dx}{2 b}\)

\(\Big \downarrow \) 4672

\(\displaystyle \frac {(c+d x)^3 \sec ^2(a+b x)}{2 b}-\frac {3 d \left (\frac {2 d \int -((c+d x) \tan (a+b x))dx}{b}+\frac {(c+d x)^2 \tan (a+b x)}{b}\right )}{2 b}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {(c+d x)^3 \sec ^2(a+b x)}{2 b}-\frac {3 d \left (\frac {(c+d x)^2 \tan (a+b x)}{b}-\frac {2 d \int (c+d x) \tan (a+b x)dx}{b}\right )}{2 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(c+d x)^3 \sec ^2(a+b x)}{2 b}-\frac {3 d \left (\frac {(c+d x)^2 \tan (a+b x)}{b}-\frac {2 d \int (c+d x) \tan (a+b x)dx}{b}\right )}{2 b}\)

\(\Big \downarrow \) 4202

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

Input:

Int[(c + d*x)^3*Sec[a + b*x]^2*Tan[a + b*x],x]
 

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 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 4202

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[I 
*((c + d*x)^(m + 1)/(d*(m + 1))), x] - Simp[2*I   Int[(c + d*x)^m*(E^(2*I*( 
e + f*x))/(1 + E^(2*I*(e + f*x)))), x], x] /; FreeQ[{c, d, e, f}, x] && IGt 
Q[m, 0]
 

rule 4672

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp 
[(-(c + d*x)^m)*(Cot[e + f*x]/f), x] + Simp[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 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]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 300 vs. \(2 (101 ) = 202\).

Time = 2.54 (sec) , antiderivative size = 301, normalized size of antiderivative = 2.62

method	result	size
risch	\(\frac {2 b \,d^{3} x^{3} {\mathrm e}^{2 i \left (b x +a \right )}-3 i d^{3} x^{2} {\mathrm e}^{2 i \left (b x +a \right )}+6 b c \,d^{2} x^{2} {\mathrm e}^{2 i \left (b x +a \right )}-6 i c \,d^{2} x \,{\mathrm e}^{2 i \left (b x +a \right )}+6 b \,c^{2} d x \,{\mathrm e}^{2 i \left (b x +a \right )}-3 i c^{2} d \,{\mathrm e}^{2 i \left (b x +a \right )}-3 i d^{3} x^{2}+2 b \,c^{3} {\mathrm e}^{2 i \left (b x +a \right )}-6 i c \,d^{2} x -3 i c^{2} d}{b^{2} \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{2}}-\frac {3 d^{2} c \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )}{b^{3}}+\frac {6 d^{2} c \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {3 i d^{3} x^{2}}{b^{2}}+\frac {6 i d^{3} a x}{b^{3}}+\frac {3 i d^{3} a^{2}}{b^{4}}-\frac {3 d^{3} \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right ) x}{b^{3}}+\frac {3 i d^{3} \operatorname {polylog}\left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{2 b^{4}}-\frac {6 d^{3} a \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}\)	\(301\)

Input:

int((d*x+c)^3*sec(b*x+a)^2*tan(b*x+a),x,method=_RETURNVERBOSE)
 

Output:

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

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. 540 vs. \(2 (98) = 196\).

Time = 0.11 (sec) , antiderivative size = 540, normalized size of antiderivative = 4.70 \[ \int (c+d x)^3 \sec ^2(a+b x) \tan (a+b x) \, dx=\frac {b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3} - 3 i \, d^{3} \cos \left (b x + a\right )^{2} {\rm Li}_2\left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) + 3 i \, d^{3} \cos \left (b x + a\right )^{2} {\rm Li}_2\left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) + 3 i \, d^{3} \cos \left (b x + a\right )^{2} {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) - 3 i \, d^{3} \cos \left (b x + a\right )^{2} {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) - 3 \, {\left (b c d^{2} - a d^{3}\right )} \cos \left (b x + a\right )^{2} \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) - 3 \, {\left (b c d^{2} - a d^{3}\right )} \cos \left (b x + a\right )^{2} \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right ) - 3 \, {\left (b d^{3} x + a d^{3}\right )} \cos \left (b x + a\right )^{2} \log \left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) - 3 \, {\left (b d^{3} x + a d^{3}\right )} \cos \left (b x + a\right )^{2} \log \left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) - 3 \, {\left (b d^{3} x + a d^{3}\right )} \cos \left (b x + a\right )^{2} \log \left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) - 3 \, {\left (b d^{3} x + a d^{3}\right )} \cos \left (b x + a\right )^{2} \log \left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) - 3 \, {\left (b c d^{2} - a d^{3}\right )} \cos \left (b x + a\right )^{2} \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) - 3 \, {\left (b c d^{2} - a d^{3}\right )} \cos \left (b x + a\right )^{2} \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right ) - 3 \, {\left (b^{2} d^{3} x^{2} + 2 \, b^{2} c d^{2} x + b^{2} c^{2} d\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right )}{2 \, b^{4} \cos \left (b x + a\right )^{2}} \] Input:

integrate((d*x+c)^3*sec(b*x+a)^2*tan(b*x+a),x, algorithm="fricas")
 

Output:

1/2*(b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + b^3*c^3 - 3*I*d^3*cos 
(b*x + a)^2*dilog(I*cos(b*x + a) + sin(b*x + a)) + 3*I*d^3*cos(b*x + a)^2* 
dilog(I*cos(b*x + a) - sin(b*x + a)) + 3*I*d^3*cos(b*x + a)^2*dilog(-I*cos 
(b*x + a) + sin(b*x + a)) - 3*I*d^3*cos(b*x + a)^2*dilog(-I*cos(b*x + a) - 
 sin(b*x + a)) - 3*(b*c*d^2 - a*d^3)*cos(b*x + a)^2*log(cos(b*x + a) + I*s 
in(b*x + a) + I) - 3*(b*c*d^2 - a*d^3)*cos(b*x + a)^2*log(cos(b*x + a) - I 
*sin(b*x + a) + I) - 3*(b*d^3*x + a*d^3)*cos(b*x + a)^2*log(I*cos(b*x + a) 
 + sin(b*x + a) + 1) - 3*(b*d^3*x + a*d^3)*cos(b*x + a)^2*log(I*cos(b*x + 
a) - sin(b*x + a) + 1) - 3*(b*d^3*x + a*d^3)*cos(b*x + a)^2*log(-I*cos(b*x 
 + a) + sin(b*x + a) + 1) - 3*(b*d^3*x + a*d^3)*cos(b*x + a)^2*log(-I*cos( 
b*x + a) - sin(b*x + a) + 1) - 3*(b*c*d^2 - a*d^3)*cos(b*x + a)^2*log(-cos 
(b*x + a) + I*sin(b*x + a) + I) - 3*(b*c*d^2 - a*d^3)*cos(b*x + a)^2*log(- 
cos(b*x + a) - I*sin(b*x + a) + I) - 3*(b^2*d^3*x^2 + 2*b^2*c*d^2*x + b^2* 
c^2*d)*cos(b*x + a)*sin(b*x + a))/(b^4*cos(b*x + a)^2)
 

Sympy [F]

\[ \int (c+d x)^3 \sec ^2(a+b x) \tan (a+b x) \, dx=\int \left (c + d x\right )^{3} \tan {\left (a + b x \right )} \sec ^{2}{\left (a + b x \right )}\, dx \] Input:

integrate((d*x+c)**3*sec(b*x+a)**2*tan(b*x+a),x)
 

Output:

Integral((c + d*x)**3*tan(a + b*x)*sec(a + b*x)**2, x)
 

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 668 vs. \(2 (98) = 196\).

Time = 0.28 (sec) , antiderivative size = 668, normalized size of antiderivative = 5.81 \[ \int (c+d x)^3 \sec ^2(a+b x) \tan (a+b x) \, dx =\text {Too large to display} \] Input:

integrate((d*x+c)^3*sec(b*x+a)^2*tan(b*x+a),x, algorithm="maxima")
 

Output:

-(6*b^2*c^2*d + 6*(b*d^3*x + b*c*d^2 + (b*d^3*x + b*c*d^2)*cos(4*b*x + 4*a 
) + 2*(b*d^3*x + b*c*d^2)*cos(2*b*x + 2*a) - (-I*b*d^3*x - I*b*c*d^2)*sin( 
4*b*x + 4*a) - 2*(-I*b*d^3*x - I*b*c*d^2)*sin(2*b*x + 2*a))*arctan2(sin(2* 
b*x + 2*a), cos(2*b*x + 2*a) + 1) - 6*(b^2*d^3*x^2 + 2*b^2*c*d^2*x)*cos(4* 
b*x + 4*a) - 2*(-2*I*b^3*d^3*x^3 - 2*I*b^3*c^3 - 3*b^2*c^2*d + 3*(-2*I*b^3 
*c*d^2 + b^2*d^3)*x^2 + 6*(-I*b^3*c^2*d + b^2*c*d^2)*x)*cos(2*b*x + 2*a) - 
 3*(d^3*cos(4*b*x + 4*a) + 2*d^3*cos(2*b*x + 2*a) + I*d^3*sin(4*b*x + 4*a) 
 + 2*I*d^3*sin(2*b*x + 2*a) + d^3)*dilog(-e^(2*I*b*x + 2*I*a)) - 3*(I*b*d^ 
3*x + I*b*c*d^2 + (I*b*d^3*x + I*b*c*d^2)*cos(4*b*x + 4*a) + 2*(I*b*d^3*x 
+ I*b*c*d^2)*cos(2*b*x + 2*a) - (b*d^3*x + b*c*d^2)*sin(4*b*x + 4*a) - 2*( 
b*d^3*x + b*c*d^2)*sin(2*b*x + 2*a))*log(cos(2*b*x + 2*a)^2 + sin(2*b*x + 
2*a)^2 + 2*cos(2*b*x + 2*a) + 1) - 6*(I*b^2*d^3*x^2 + 2*I*b^2*c*d^2*x)*sin 
(4*b*x + 4*a) - 2*(2*b^3*d^3*x^3 + 2*b^3*c^3 - 3*I*b^2*c^2*d + 3*(2*b^3*c* 
d^2 + I*b^2*d^3)*x^2 + 6*(b^3*c^2*d + I*b^2*c*d^2)*x)*sin(2*b*x + 2*a))/(- 
2*I*b^4*cos(4*b*x + 4*a) - 4*I*b^4*cos(2*b*x + 2*a) + 2*b^4*sin(4*b*x + 4* 
a) + 4*b^4*sin(2*b*x + 2*a) - 2*I*b^4)
 

Giac [F]

\[ \int (c+d x)^3 \sec ^2(a+b x) \tan (a+b x) \, dx=\int { {\left (d x + c\right )}^{3} \sec \left (b x + a\right )^{2} \tan \left (b x + a\right ) \,d x } \] Input:

integrate((d*x+c)^3*sec(b*x+a)^2*tan(b*x+a),x, algorithm="giac")
 

Output:

integrate((d*x + c)^3*sec(b*x + a)^2*tan(b*x + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int (c+d x)^3 \sec ^2(a+b x) \tan (a+b x) \, dx=\int \frac {\mathrm {tan}\left (a+b\,x\right )\,{\left (c+d\,x\right )}^3}{{\cos \left (a+b\,x\right )}^2} \,d x \] Input:

int((tan(a + b*x)*(c + d*x)^3)/cos(a + b*x)^2,x)
 

Output:

int((tan(a + b*x)*(c + d*x)^3)/cos(a + b*x)^2, x)
 

Reduce [F]

\[ \int (c+d x)^3 \sec ^2(a+b x) \tan (a+b x) \, dx=\text {too large to display} \] Input:

int((d*x+c)^3*sec(b*x+a)^2*tan(b*x+a),x)
 

Output:

( - 14*cos(a + b*x)*sin(a + b*x)*tan((a + b*x)/2)**4*b**4*c*d**2*x**3 + 9* 
cos(a + b*x)*sin(a + b*x)*tan((a + b*x)/2)**4*b**2*c**2*d - 12*cos(a + b*x 
)*sin(a + b*x)*tan((a + b*x)/2)**4*b**2*c*d**2*x - 6*cos(a + b*x)*sin(a + 
b*x)*tan((a + b*x)/2)**4*b**2*d**3*x**2 + 6*cos(a + b*x)*sin(a + b*x)*tan( 
(a + b*x)/2)**4*d**3 + 28*cos(a + b*x)*sin(a + b*x)*tan((a + b*x)/2)**2*b* 
*4*c*d**2*x**3 - 18*cos(a + b*x)*sin(a + b*x)*tan((a + b*x)/2)**2*b**2*c** 
2*d + 24*cos(a + b*x)*sin(a + b*x)*tan((a + b*x)/2)**2*b**2*c*d**2*x + 12* 
cos(a + b*x)*sin(a + b*x)*tan((a + b*x)/2)**2*b**2*d**3*x**2 - 12*cos(a + 
b*x)*sin(a + b*x)*tan((a + b*x)/2)**2*d**3 - 14*cos(a + b*x)*sin(a + b*x)* 
b**4*c*d**2*x**3 + 9*cos(a + b*x)*sin(a + b*x)*b**2*c**2*d - 12*cos(a + b* 
x)*sin(a + b*x)*b**2*c*d**2*x - 6*cos(a + b*x)*sin(a + b*x)*b**2*d**3*x**2 
 + 6*cos(a + b*x)*sin(a + b*x)*d**3 - 42*cos(a + b*x)*tan((a + b*x)/2)**4* 
b**3*c*d**2*x**2 - 14*cos(a + b*x)*tan((a + b*x)/2)**4*b**3*d**3*x**3 + 12 
*cos(a + b*x)*tan((a + b*x)/2)**4*b*c*d**2 - 12*cos(a + b*x)*tan((a + b*x) 
/2)**4*b*d**3*x + 84*cos(a + b*x)*tan((a + b*x)/2)**2*b**3*c*d**2*x**2 + 2 
8*cos(a + b*x)*tan((a + b*x)/2)**2*b**3*d**3*x**3 - 24*cos(a + b*x)*tan((a 
 + b*x)/2)**2*b*c*d**2 + 24*cos(a + b*x)*tan((a + b*x)/2)**2*b*d**3*x - 42 
*cos(a + b*x)*b**3*c*d**2*x**2 - 14*cos(a + b*x)*b**3*d**3*x**3 + 12*cos(a 
 + b*x)*b*c*d**2 - 12*cos(a + b*x)*b*d**3*x + 12*int(x**2/(tan((a + b*x)/2 
)**6 - 3*tan((a + b*x)/2)**4 + 3*tan((a + b*x)/2)**2 - 1),x)*sin(a + b*...