\(\int \tan (c+d x) (a+b \tan (c+d x)) (B \tan (c+d x)+C \tan ^2(c+d x)) \, dx\) [1]

Optimal result

Integrand size = 36, antiderivative size = 87 \[ \int \tan (c+d x) (a+b \tan (c+d x)) \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=-((a B-b C) x)+\frac {(b B+a C) \log (\cos (c+d x))}{d}+\frac {(a B-b C) \tan (c+d x)}{d}+\frac {(b B+a C) \tan ^2(c+d x)}{2 d}+\frac {b C \tan ^3(c+d x)}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {3713, 3673, 3609, 3606, 3556} \[ \int \tan (c+d x) (a+b \tan (c+d x)) \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\frac {(a C+b B) \tan ^2(c+d x)}{2 d}+\frac {(a B-b C) \tan (c+d x)}{d}+\frac {(a C+b B) \log (\cos (c+d x))}{d}-x (a B-b C)+\frac {b C \tan ^3(c+d x)}{3 d} \]

[In]

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

[Out]

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

Rule 3713

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[1/b^2, Int[(a + b*Tan[e + f*x])
^(m + 1)*(c + d*Tan[e + f*x])^n*(b*B - a*C + b*C*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

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

Mathematica [A] (verified)

Time = 0.65 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.99 \[ \int \tan (c+d x) (a+b \tan (c+d x)) \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\frac {(-6 a B+6 b C) \arctan (\tan (c+d x))+6 (b B+a C) \log (\cos (c+d x))+6 (a B-b C) \tan (c+d x)+3 (b B+a C) \tan ^2(c+d x)+2 b C \tan ^3(c+d x)}{6 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.01

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.98 \[ \int \tan (c+d x) (a+b \tan (c+d x)) \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\frac {2 \, C b \tan \left (d x + c\right )^{3} - 6 \, {\left (B a - C b\right )} d x + 3 \, {\left (C a + B b\right )} \tan \left (d x + c\right )^{2} + 3 \, {\left (C a + B b\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) + 6 \, {\left (B a - C b\right )} \tan \left (d x + c\right )}{6 \, d} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.60 \[ \int \tan (c+d x) (a+b \tan (c+d x)) \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\begin {cases} - B a x + \frac {B a \tan {\left (c + d x \right )}}{d} - \frac {B b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {B b \tan ^{2}{\left (c + d x \right )}}{2 d} - \frac {C a \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {C a \tan ^{2}{\left (c + d x \right )}}{2 d} + C b x + \frac {C b \tan ^{3}{\left (c + d x \right )}}{3 d} - \frac {C b \tan {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \tan {\left (c \right )}\right ) \left (B \tan {\left (c \right )} + C \tan ^{2}{\left (c \right )}\right ) \tan {\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.99 \[ \int \tan (c+d x) (a+b \tan (c+d x)) \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\frac {2 \, C b \tan \left (d x + c\right )^{3} + 3 \, {\left (C a + B b\right )} \tan \left (d x + c\right )^{2} - 6 \, {\left (B a - C b\right )} {\left (d x + c\right )} - 3 \, {\left (C a + B b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 6 \, {\left (B a - C b\right )} \tan \left (d x + c\right )}{6 \, d} \]

[In]

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

[Out]

1/6*(2*C*b*tan(d*x + c)^3 + 3*(C*a + B*b)*tan(d*x + c)^2 - 6*(B*a - C*b)*(d*x + c) - 3*(C*a + B*b)*log(tan(d*x
+ c)^2 + 1) + 6*(B*a - C*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 = 0.91 (sec) , antiderivative size = 937, normalized size of antiderivative = 10.77 \[ \int \tan (c+d x) (a+b \tan (c+d x)) \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\text {Too large to display} \]

[In]

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

[Out]

-1/6*(6*B*a*d*x*tan(d*x)^3*tan(c)^3 - 6*C*b*d*x*tan(d*x)^3*tan(c)^3 - 3*C*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*B*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*B*a*d*x*tan(d*x)^2*tan(c)^2 + 18*C*b*d*x*tan(d*x)^2*tan(c)^2 - 3*C*a*tan(d*x)^3*tan(c)^3 - 3*B*b*tan(d*x)
^3*tan(c)^3 + 9*C*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*B*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*B*a*tan(d*x)^3*tan(c)^2 - 6*C*b*tan(d*x)^3*tan(c)^2
+ 6*B*a*tan(d*x)^2*tan(c)^3 - 6*C*b*tan(d*x)^2*tan(c)^3 + 18*B*a*d*x*tan(d*x)*tan(c) - 18*C*b*d*x*tan(d*x)*tan
(c) - 3*C*a*tan(d*x)^3*tan(c) - 3*B*b*tan(d*x)^3*tan(c) + 3*C*a*tan(d*x)^2*tan(c)^2 + 3*B*b*tan(d*x)^2*tan(c)^
2 - 3*C*a*tan(d*x)*tan(c)^3 - 3*B*b*tan(d*x)*tan(c)^3 + 2*C*b*tan(d*x)^3 - 9*C*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*B*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*B*a*tan(d*x)^2*tan(c) + 18*C*b*tan(d*x)^2*tan(c) - 12*B*a*tan(d*x)*tan(c)^2 + 18*C*b*tan(d*x)*tan(c)^2 + 2*C
*b*tan(c)^3 - 6*B*a*d*x + 6*C*b*d*x + 3*C*a*tan(d*x)^2 + 3*B*b*tan(d*x)^2 - 3*C*a*tan(d*x)*tan(c) - 3*B*b*tan(
d*x)*tan(c) + 3*C*a*tan(c)^2 + 3*B*b*tan(c)^2 + 3*C*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*B*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*B*a*tan(d*x) - 6*C*b*tan(d*x) + 6*B*a*tan(c) - 6*C*b*tan(c
) + 3*C*a + 3*B*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 = 8.20 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.97 \[ \int \tan (c+d x) (a+b \tan (c+d x)) \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (B\,a-C\,b\right )-\ln \left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )\,\left (\frac {B\,b}{2}+\frac {C\,a}{2}\right )+{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {B\,b}{2}+\frac {C\,a}{2}\right )-d\,x\,\left (B\,a-C\,b\right )+\frac {C\,b\,{\mathrm {tan}\left (c+d\,x\right )}^3}{3}}{d} \]

[In]

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

[Out]

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