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

Optimal. Leaf size=87 \[ -\frac {\left (a^2 B+2 a A b-b^2 B\right ) \log (\cos (c+d x))}{d}+x \left (a^2 A-2 a b B-A b^2\right )+\frac {b (a B+A b) \tan (c+d x)}{d}+\frac {B (a+b \tan (c+d x))^2}{2 d} \]

[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

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Rubi [A]  time = 0.08, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3528, 3525, 3475} \[ -\frac {\left (a^2 B+2 a A b-b^2 B\right ) \log (\cos (c+d x))}{d}+x \left (a^2 A-2 a b B-A b^2\right )+\frac {b (a B+A b) \tan (c+d x)}{d}+\frac {B (a+b \tan (c+d x))^2}{2 d} \]

Antiderivative was successfully verified.

[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 3475

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

Rule 3525

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 3528

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*} \int (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx &=\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*}

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Mathematica [C]  time = 0.43, size = 96, normalized size = 1.10 \[ \frac {2 b (2 a B+A b) \tan (c+d x)+(a-i b)^2 (B+i A) \log (\tan (c+d x)+i)+(a+i b)^2 (B-i A) \log (-\tan (c+d x)+i)+b^2 B \tan ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 1.14, size = 91, normalized size = 1.05 \[ \frac {B b^{2} \tan \left (d x + c\right )^{2} + 2 \, {\left (A a^{2} - 2 \, B a b - A b^{2}\right )} d x - {\left (B a^{2} + 2 \, A a b - B b^{2}\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) + 2 \, {\left (2 \, B a b + A b^{2}\right )} \tan \left (d x + c\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [B]  time = 0.95, size = 901, normalized size = 10.36 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1
)/(tan(c)^2 + 1))*tan(d*x)^2*tan(c)^2 - 2*A*a*b*log(4*(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*
tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(c)^2 + 1))*tan(d*x)^2*tan(c)^2 + B*b^2*log(4*(tan(d*x)^4*t
an(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1)/(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)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan
(d*x)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(c)^2 + 1))*tan(d*x)*tan(c) + 4*A*a*b*log(4*(tan(d*x)^4*tan(c)^2 - 2*tan(
d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(c)^2 + 1))*tan(d*x)*tan(c) - 2*
B*b^2*log(4*(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c)
+ 1)/(tan(c)^2 + 1))*tan(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*tan(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*ta
n(c)^2 - B*a^2*log(4*(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x
)*tan(c) + 1)/(tan(c)^2 + 1)) - 2*A*a*b*log(4*(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2
+ tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(c)^2 + 1)) + B*b^2*log(4*(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c
) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1)/(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)*tan(c) + d)

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maple [A]  time = 0.02, size = 151, normalized size = 1.74 \[ \frac {b^{2} B \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}+\frac {A \tan \left (d x +c \right ) b^{2}}{d}+\frac {2 B \tan \left (d x +c \right ) a b}{d}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) A a b}{d}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a^{2} B}{2 d}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) b^{2} B}{2 d}+\frac {A \arctan \left (\tan \left (d x +c \right )\right ) a^{2}}{d}-\frac {A \arctan \left (\tan \left (d x +c \right )\right ) b^{2}}{d}-\frac {2 B \arctan \left (\tan \left (d x +c \right )\right ) a b}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.97, size = 91, normalized size = 1.05 \[ \frac {B b^{2} \tan \left (d x + c\right )^{2} + 2 \, {\left (A a^{2} - 2 \, B a b - A b^{2}\right )} {\left (d x + c\right )} + {\left (B a^{2} + 2 \, A a b - B b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \, {\left (2 \, B a b + A b^{2}\right )} \tan \left (d x + c\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 6.23, size = 91, normalized size = 1.05 \[ \frac {\ln \left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )\,\left (\frac {B\,a^2}{2}+A\,a\,b-\frac {B\,b^2}{2}\right )}{d}-x\,\left (-A\,a^2+2\,B\,a\,b+A\,b^2\right )+\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (A\,b^2+2\,B\,a\,b\right )}{d}+\frac {B\,b^2\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.30, size = 143, normalized size = 1.64 \[ \begin {cases} A a^{2} x + \frac {A a b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} - A b^{2} x + \frac {A b^{2} \tan {\left (c + d x \right )}}{d} + \frac {B a^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - 2 B a b x + \frac {2 B a b \tan {\left (c + d x \right )}}{d} - \frac {B b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {B b^{2} \tan ^{2}{\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (A + B \tan {\relax (c )}\right ) \left (a + b \tan {\relax (c )}\right )^{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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