\(\int \tan ^2(c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx\) [232]

Optimal result

Integrand size = 29, antiderivative size = 87 \[ \int \tan ^2(c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=-((a A-b B) x)+\frac {(A b+a B) \log (\cos (c+d x))}{d}+\frac {(a A-b B) \tan (c+d x)}{d}+\frac {(A b+a B) \tan ^2(c+d x)}{2 d}+\frac {b B \tan ^3(c+d x)}{3 d} \]

[Out]

-(A*a-B*b)*x+(A*b+B*a)*ln(cos(d*x+c))/d+(A*a-B*b)*tan(d*x+c)/d+1/2*(A*b+B*a)*tan(d*x+c)^2/d+1/3*b*B*tan(d*x+c)
^3/d

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {3673, 3609, 3606, 3556} \[ \int \tan ^2(c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\frac {(a B+A b) \tan ^2(c+d x)}{2 d}+\frac {(a A-b B) \tan (c+d x)}{d}+\frac {(a B+A b) \log (\cos (c+d x))}{d}-x (a A-b B)+\frac {b B \tan ^3(c+d x)}{3 d} \]

[In]

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

[Out]

-((a*A - b*B)*x) + ((A*b + a*B)*Log[Cos[c + d*x]])/d + ((a*A - b*B)*Tan[c + d*x])/d + ((A*b + a*B)*Tan[c + d*x
]^2)/(2*d) + (b*B*Tan[c + d*x]^3)/(3*d)

Rule 3556

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3606

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(a*c - b
*d)*x, x] + (Dist[b*c + a*d, Int[Tan[e + f*x], x], x] + Simp[b*d*(Tan[e + f*x]/f), x]) /; FreeQ[{a, b, c, d, e
, f}, x] && NeQ[b*c - a*d, 0] && NeQ[b*c + a*d, 0]

Rule 3609

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

Rule 3673

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[B*d*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e
+ f*x])^m*Simp[A*c - B*d + (B*c + A*d)*Tan[e + f*x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b
*c - a*d, 0] &&  !LeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {b B \tan ^3(c+d x)}{3 d}+\int \tan ^2(c+d x) (a A-b B+(A b+a B) \tan (c+d x)) \, dx \\ & = \frac {(A b+a B) \tan ^2(c+d x)}{2 d}+\frac {b B \tan ^3(c+d x)}{3 d}+\int \tan (c+d x) (-A b-a B+(a A-b B) \tan (c+d x)) \, dx \\ & = -((a A-b B) x)+\frac {(a A-b B) \tan (c+d x)}{d}+\frac {(A b+a B) \tan ^2(c+d x)}{2 d}+\frac {b B \tan ^3(c+d x)}{3 d}+(-A b-a B) \int \tan (c+d x) \, dx \\ & = -((a A-b B) x)+\frac {(A b+a B) \log (\cos (c+d x))}{d}+\frac {(a A-b B) \tan (c+d x)}{d}+\frac {(A b+a B) \tan ^2(c+d x)}{2 d}+\frac {b B \tan ^3(c+d x)}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.62 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.99 \[ \int \tan ^2(c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\frac {(-6 a A+6 b B) \arctan (\tan (c+d x))+6 (A b+a B) \log (\cos (c+d x))+6 (a A-b B) \tan (c+d x)+3 (A b+a B) \tan ^2(c+d x)+2 b B \tan ^3(c+d x)}{6 d} \]

[In]

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

[Out]

((-6*a*A + 6*b*B)*ArcTan[Tan[c + d*x]] + 6*(A*b + a*B)*Log[Cos[c + d*x]] + 6*(a*A - b*B)*Tan[c + d*x] + 3*(A*b
+ a*B)*Tan[c + d*x]^2 + 2*b*B*Tan[c + d*x]^3)/(6*d)

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.01

method	result	size
norman	\(\left (-a A +B b \right ) x +\frac {\left (a A -B b \right ) \tan \left (d x +c \right )}{d}+\frac {\left (A b +B a \right ) \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}+\frac {b B \left (\tan ^{3}\left (d x +c \right )\right )}{3 d}-\frac {\left (A b +B a \right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d}\)	\(88\)
parts	\(\frac {\left (A b +B a \right ) \left (\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}\right )}{d}+\frac {B b \left (\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-\tan \left (d x +c \right )+\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}+\frac {a A \left (\tan \left (d x +c \right )-\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\)	\(91\)
derivativedivides	\(\frac {\frac {b B \left (\tan ^{3}\left (d x +c \right )\right )}{3}+\frac {A b \left (\tan ^{2}\left (d x +c \right )\right )}{2}+\frac {B \left (\tan ^{2}\left (d x +c \right )\right ) a}{2}+A \tan \left (d x +c \right ) a -b B \tan \left (d x +c \right )+\frac {\left (-A b -B a \right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+\left (-a A +B b \right ) \arctan \left (\tan \left (d x +c \right )\right )}{d}\)	\(99\)
default	\(\frac {\frac {b B \left (\tan ^{3}\left (d x +c \right )\right )}{3}+\frac {A b \left (\tan ^{2}\left (d x +c \right )\right )}{2}+\frac {B \left (\tan ^{2}\left (d x +c \right )\right ) a}{2}+A \tan \left (d x +c \right ) a -b B \tan \left (d x +c \right )+\frac {\left (-A b -B a \right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+\left (-a A +B b \right ) \arctan \left (\tan \left (d x +c \right )\right )}{d}\)	\(99\)
parallelrisch	\(-\frac {-2 b B \left (\tan ^{3}\left (d x +c \right )\right )+6 A x a d -3 A b \left (\tan ^{2}\left (d x +c \right )\right )-6 B b d x -3 B \left (\tan ^{2}\left (d x +c \right )\right ) a +3 A \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) b -6 A \tan \left (d x +c \right ) a +3 B \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a +6 b B \tan \left (d x +c \right )}{6 d}\)	\(105\)
risch	\(-i A b x -i B a x -A a x +B b x -\frac {2 i A b c}{d}-\frac {2 i a B c}{d}+\frac {2 i \left (-3 i A b \,{\mathrm e}^{4 i \left (d x +c \right )}-3 i B a \,{\mathrm e}^{4 i \left (d x +c \right )}+3 A a \,{\mathrm e}^{4 i \left (d x +c \right )}-6 B b \,{\mathrm e}^{4 i \left (d x +c \right )}-3 i A b \,{\mathrm e}^{2 i \left (d x +c \right )}-3 i B a \,{\mathrm e}^{2 i \left (d x +c \right )}+6 A a \,{\mathrm e}^{2 i \left (d x +c \right )}-6 B b \,{\mathrm e}^{2 i \left (d x +c \right )}+3 a A -4 B b \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) A b}{d}+\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) B}{d}\)	\(213\)

[In]

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

[Out]

(-A*a+B*b)*x+(A*a-B*b)*tan(d*x+c)/d+1/2*(A*b+B*a)*tan(d*x+c)^2/d+1/3*b*B*tan(d*x+c)^3/d-1/2*(A*b+B*a)/d*ln(1+t
an(d*x+c)^2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.98 \[ \int \tan ^2(c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\frac {2 \, B b \tan \left (d x + c\right )^{3} - 6 \, {\left (A a - B b\right )} d x + 3 \, {\left (B a + A b\right )} \tan \left (d x + c\right )^{2} + 3 \, {\left (B a + A b\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) + 6 \, {\left (A a - B b\right )} \tan \left (d x + c\right )}{6 \, d} \]

[In]

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

[Out]

1/6*(2*B*b*tan(d*x + c)^3 - 6*(A*a - B*b)*d*x + 3*(B*a + A*b)*tan(d*x + c)^2 + 3*(B*a + A*b)*log(1/(tan(d*x +
c)^2 + 1)) + 6*(A*a - B*b)*tan(d*x + c))/d

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.56 \[ \int \tan ^2(c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\begin {cases} - A a x + \frac {A a \tan {\left (c + d x \right )}}{d} - \frac {A b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {A b \tan ^{2}{\left (c + d x \right )}}{2 d} - \frac {B a \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {B a \tan ^{2}{\left (c + d x \right )}}{2 d} + B b x + \frac {B b \tan ^{3}{\left (c + d x \right )}}{3 d} - \frac {B b \tan {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (A + B \tan {\left (c \right )}\right ) \left (a + b \tan {\left (c \right )}\right ) \tan ^{2}{\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(tan(d*x+c)**2*(a+b*tan(d*x+c))*(A+B*tan(d*x+c)),x)

[Out]

Piecewise((-A*a*x + A*a*tan(c + d*x)/d - A*b*log(tan(c + d*x)**2 + 1)/(2*d) + A*b*tan(c + d*x)**2/(2*d) - B*a*
log(tan(c + d*x)**2 + 1)/(2*d) + B*a*tan(c + d*x)**2/(2*d) + B*b*x + B*b*tan(c + d*x)**3/(3*d) - B*b*tan(c + d
*x)/d, Ne(d, 0)), (x*(A + B*tan(c))*(a + b*tan(c))*tan(c)**2, True))

Maxima [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.99 \[ \int \tan ^2(c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\frac {2 \, B b \tan \left (d x + c\right )^{3} + 3 \, {\left (B a + A b\right )} \tan \left (d x + c\right )^{2} - 6 \, {\left (A a - B b\right )} {\left (d x + c\right )} - 3 \, {\left (B a + A b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 6 \, {\left (A a - B b\right )} \tan \left (d x + c\right )}{6 \, d} \]

[In]

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

[Out]

1/6*(2*B*b*tan(d*x + c)^3 + 3*(B*a + A*b)*tan(d*x + c)^2 - 6*(A*a - B*b)*(d*x + c) - 3*(B*a + A*b)*log(tan(d*x
+ c)^2 + 1) + 6*(A*a - B*b)*tan(d*x + c))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 937 vs. \(2 (83) = 166\).

Time = 0.85 (sec) , antiderivative size = 937, normalized size of antiderivative = 10.77 \[ \int \tan ^2(c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]

[In]

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

[Out]

-1/6*(6*A*a*d*x*tan(d*x)^3*tan(c)^3 - 6*B*b*d*x*tan(d*x)^3*tan(c)^3 - 3*B*a*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan
(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)^3*tan(c)^3 - 3*A*b*log(4*(tan(d*
x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)^3*tan(c)^3
- 18*A*a*d*x*tan(d*x)^2*tan(c)^2 + 18*B*b*d*x*tan(d*x)^2*tan(c)^2 - 3*B*a*tan(d*x)^3*tan(c)^3 - 3*A*b*tan(d*x)
^3*tan(c)^3 + 9*B*a*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + ta
n(c)^2 + 1))*tan(d*x)^2*tan(c)^2 + 9*A*b*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c
)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)^2*tan(c)^2 + 6*A*a*tan(d*x)^3*tan(c)^2 - 6*B*b*tan(d*x)^3*tan(c)^2
+ 6*A*a*tan(d*x)^2*tan(c)^3 - 6*B*b*tan(d*x)^2*tan(c)^3 + 18*A*a*d*x*tan(d*x)*tan(c) - 18*B*b*d*x*tan(d*x)*tan
(c) - 3*B*a*tan(d*x)^3*tan(c) - 3*A*b*tan(d*x)^3*tan(c) + 3*B*a*tan(d*x)^2*tan(c)^2 + 3*A*b*tan(d*x)^2*tan(c)^
2 - 3*B*a*tan(d*x)*tan(c)^3 - 3*A*b*tan(d*x)*tan(c)^3 + 2*B*b*tan(d*x)^3 - 9*B*a*log(4*(tan(d*x)^2*tan(c)^2 -
2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)*tan(c) - 9*A*b*log(4*(tan(d
*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)*tan(c) - 1
2*A*a*tan(d*x)^2*tan(c) + 18*B*b*tan(d*x)^2*tan(c) - 12*A*a*tan(d*x)*tan(c)^2 + 18*B*b*tan(d*x)*tan(c)^2 + 2*B
*b*tan(c)^3 - 6*A*a*d*x + 6*B*b*d*x + 3*B*a*tan(d*x)^2 + 3*A*b*tan(d*x)^2 - 3*B*a*tan(d*x)*tan(c) - 3*A*b*tan(
d*x)*tan(c) + 3*B*a*tan(c)^2 + 3*A*b*tan(c)^2 + 3*B*a*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan
(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1)) + 3*A*b*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(ta
n(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1)) + 6*A*a*tan(d*x) - 6*B*b*tan(d*x) + 6*A*a*tan(c) - 6*B*b*tan(c
) + 3*B*a + 3*A*b)/(d*tan(d*x)^3*tan(c)^3 - 3*d*tan(d*x)^2*tan(c)^2 + 3*d*tan(d*x)*tan(c) - d)

Mupad [B] (verification not implemented)

Time = 7.28 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.97 \[ \int \tan ^2(c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (A\,a-B\,b\right )-\ln \left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )\,\left (\frac {A\,b}{2}+\frac {B\,a}{2}\right )+{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {A\,b}{2}+\frac {B\,a}{2}\right )-d\,x\,\left (A\,a-B\,b\right )+\frac {B\,b\,{\mathrm {tan}\left (c+d\,x\right )}^3}{3}}{d} \]

[In]

int(tan(c + d*x)^2*(A + B*tan(c + d*x))*(a + b*tan(c + d*x)),x)

[Out]

(tan(c + d*x)*(A*a - B*b) - log(tan(c + d*x)^2 + 1)*((A*b)/2 + (B*a)/2) + tan(c + d*x)^2*((A*b)/2 + (B*a)/2) -
d*x*(A*a - B*b) + (B*b*tan(c + d*x)^3)/3)/d