\(\int (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx\) [242]

Optimal result

Integrand size = 23, antiderivative size = 87 \[ \int (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\left (a^2 A-A b^2-2 a b B\right ) x-\frac {\left (2 a A b+a^2 B-b^2 B\right ) \log (\cos (c+d x))}{d}+\frac {b (A b+a B) \tan (c+d x)}{d}+\frac {B (a+b \tan (c+d x))^2}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3609, 3606, 3556} \[ \int (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=-\frac {\left (a^2 B+2 a A b-b^2 B\right ) \log (\cos (c+d x))}{d}+x \left (a^2 A-2 a b B-A b^2\right )+\frac {b (a B+A b) \tan (c+d x)}{d}+\frac {B (a+b \tan (c+d x))^2}{2 d} \]

[In]

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

[Out]

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

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.48 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.10 \[ \int (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {(a+i b)^2 (-i A+B) \log (i-\tan (c+d x))+(a-i b)^2 (i A+B) \log (i+\tan (c+d x))+2 b (A b+2 a B) \tan (c+d x)+b^2 B \tan ^2(c+d x)}{2 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.03

method	result	size
norman	\(\left (A \,a^{2}-A \,b^{2}-2 B a b \right ) x +\frac {b \left (A b +2 B a \right ) \tan \left (d x +c \right )}{d}+\frac {B \,b^{2} \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}+\frac {\left (2 A a b +B \,a^{2}-B \,b^{2}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d}\)	\(90\)
derivativedivides	\(\frac {\frac {B \,b^{2} \left (\tan ^{2}\left (d x +c \right )\right )}{2}+A \,b^{2} \tan \left (d x +c \right )+2 B a b \tan \left (d x +c \right )+\frac {\left (2 A a b +B \,a^{2}-B \,b^{2}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+\left (A \,a^{2}-A \,b^{2}-2 B a b \right ) \arctan \left (\tan \left (d x +c \right )\right )}{d}\)	\(97\)
default	\(\frac {\frac {B \,b^{2} \left (\tan ^{2}\left (d x +c \right )\right )}{2}+A \,b^{2} \tan \left (d x +c \right )+2 B a b \tan \left (d x +c \right )+\frac {\left (2 A a b +B \,a^{2}-B \,b^{2}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+\left (A \,a^{2}-A \,b^{2}-2 B a b \right ) \arctan \left (\tan \left (d x +c \right )\right )}{d}\)	\(97\)
parts	\(A \,a^{2} x +\frac {\left (A \,b^{2}+2 B a b \right ) \left (\tan \left (d x +c \right )-\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}+\frac {\left (2 A a b +B \,a^{2}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d}+\frac {B \,b^{2} \left (\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}\right )}{d}\)	\(98\)
parallelrisch	\(\frac {2 A x \,a^{2} d -2 A \,b^{2} d x -4 B a b d x +B \,b^{2} \left (\tan ^{2}\left (d x +c \right )\right )+2 A \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a b +2 A \,b^{2} \tan \left (d x +c \right )+B \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a^{2}-B \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) b^{2}+4 B a b \tan \left (d x +c \right )}{2 d}\)	\(115\)
risch	\(\frac {4 i A a b c}{d}-\frac {2 i B \,b^{2} c}{d}+\frac {2 i a^{2} B c}{d}+A \,a^{2} x -A \,b^{2} x -2 B a b x +\frac {2 i b \left (A b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 B a \,{\mathrm e}^{2 i \left (d x +c \right )}-i B b \,{\mathrm e}^{2 i \left (d x +c \right )}+A b +2 B a \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-i B \,b^{2} x +2 i A a b x +i B \,a^{2} x -\frac {2 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) A a b}{d}-\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) B}{d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) B \,b^{2}}{d}\)	\(204\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.05 \[ \int (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {B b^{2} \tan \left (d x + c\right )^{2} + 2 \, {\left (A a^{2} - 2 \, B a b - A b^{2}\right )} d x - {\left (B a^{2} + 2 \, A a b - B b^{2}\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) + 2 \, {\left (2 \, B a b + A b^{2}\right )} \tan \left (d x + c\right )}{2 \, d} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.64 \[ \int (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\begin {cases} A a^{2} x + \frac {A a b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} - A b^{2} x + \frac {A b^{2} \tan {\left (c + d x \right )}}{d} + \frac {B a^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - 2 B a b x + \frac {2 B a b \tan {\left (c + d x \right )}}{d} - \frac {B b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {B b^{2} \tan ^{2}{\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (A + B \tan {\left (c \right )}\right ) \left (a + b \tan {\left (c \right )}\right )^{2} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.05 \[ \int (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {B b^{2} \tan \left (d x + c\right )^{2} + 2 \, {\left (A a^{2} - 2 \, B a b - A b^{2}\right )} {\left (d x + c\right )} + {\left (B a^{2} + 2 \, A a b - B b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \, {\left (2 \, B a b + A b^{2}\right )} \tan \left (d x + c\right )}{2 \, d} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 811 vs. \(2 (85) = 170\).

Time = 0.70 (sec) , antiderivative size = 811, normalized size of antiderivative = 9.32 \[ \int (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/2*(2*A*a^2*d*x*tan(d*x)^2*tan(c)^2 - 4*B*a*b*d*x*tan(d*x)^2*tan(c)^2 - 2*A*b^2*d*x*tan(d*x)^2*tan(c)^2 - B*a
^2*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 - 2*A*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 + B*b^2*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 - 4*A*a^2*d*x*tan(d*x)*tan(c) + 8*B*a*b*d*x*tan(d*
x)*tan(c) + 4*A*b^2*d*x*tan(d*x)*tan(c) + B*b^2*tan(d*x)^2*tan(c)^2 + 2*B*a^2*log(4*(tan(d*x)^2*tan(c)^2 - 2*t
an(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) + 4*A*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) - 2*
B*b^2*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))*t
an(d*x)*tan(c) - 4*B*a*b*tan(d*x)^2*tan(c) - 2*A*b^2*tan(d*x)^2*tan(c) - 4*B*a*b*tan(d*x)*tan(c)^2 - 2*A*b^2*t
an(d*x)*tan(c)^2 + 2*A*a^2*d*x - 4*B*a*b*d*x - 2*A*b^2*d*x + B*b^2*tan(d*x)^2 + B*b^2*tan(c)^2 - B*a^2*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)) - 2*A*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)) + B*b^2*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)) + 4*B*a*b*
tan(d*x) + 2*A*b^2*tan(d*x) + 4*B*a*b*tan(c) + 2*A*b^2*tan(c) + B*b^2)/(d*tan(d*x)^2*tan(c)^2 - 2*d*tan(d*x)*t
an(c) + d)

Mupad [B] (verification not implemented)

Time = 8.35 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.05 \[ \int (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {\ln \left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )\,\left (\frac {B\,a^2}{2}+A\,a\,b-\frac {B\,b^2}{2}\right )}{d}-x\,\left (-A\,a^2+2\,B\,a\,b+A\,b^2\right )+\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (A\,b^2+2\,B\,a\,b\right )}{d}+\frac {B\,b^2\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2\,d} \]

[In]

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

[Out]

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