3.268 \(\int \frac{\tan ^2(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx\)

Optimal. Leaf size=101 \[ \frac{a^2 (A b-a B) \log (a+b \tan (c+d x))}{b^2 d \left (a^2+b^2\right )}-\frac{(A b-a B) \log (\cos (c+d x))}{d \left (a^2+b^2\right )}-\frac{x (a A+b B)}{a^2+b^2}+\frac{B \tan (c+d x)}{b d} \]

[Out]

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

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Rubi [A]  time = 0.197371, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {3606, 3626, 3617, 31, 3475} \[ \frac{a^2 (A b-a B) \log (a+b \tan (c+d x))}{b^2 d \left (a^2+b^2\right )}-\frac{(A b-a B) \log (\cos (c+d x))}{d \left (a^2+b^2\right )}-\frac{x (a A+b B)}{a^2+b^2}+\frac{B \tan (c+d x)}{b d} \]

Antiderivative was successfully verified.

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

Rule 3606

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

Rule 3626

Int[((A_) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2)/((a_.) + (b_.)*tan[(e_.) + (f_.)*
(x_)]), x_Symbol] :> Simp[((a*A + b*B - a*C)*x)/(a^2 + b^2), x] + (Dist[(A*b^2 - a*b*B + a^2*C)/(a^2 + b^2), I
nt[(1 + Tan[e + f*x]^2)/(a + b*Tan[e + f*x]), x], x] - Dist[(A*b - a*B - b*C)/(a^2 + b^2), Int[Tan[e + f*x], x
], x]) /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0] && NeQ[a^2 + b^2, 0] && NeQ[A*b - a
*B - b*C, 0]

Rule 3617

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[
A/(b*f), Subst[Int[(a + x)^m, x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, A, C, m}, x] && EqQ[A, C]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 3475

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

Rubi steps

\begin{align*} \int \frac{\tan ^2(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx &=\frac{B \tan (c+d x)}{b d}+\frac{\int \frac{-a B-b B \tan (c+d x)+(A b-a B) \tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx}{b}\\ &=-\frac{(a A+b B) x}{a^2+b^2}+\frac{B \tan (c+d x)}{b d}+\frac{(A b-a B) \int \tan (c+d x) \, dx}{a^2+b^2}+\frac{\left (a^2 (A b-a B)\right ) \int \frac{1+\tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx}{b \left (a^2+b^2\right )}\\ &=-\frac{(a A+b B) x}{a^2+b^2}-\frac{(A b-a B) \log (\cos (c+d x))}{\left (a^2+b^2\right ) d}+\frac{B \tan (c+d x)}{b d}+\frac{\left (a^2 (A b-a B)\right ) \operatorname{Subst}\left (\int \frac{1}{a+x} \, dx,x,b \tan (c+d x)\right )}{b^2 \left (a^2+b^2\right ) d}\\ &=-\frac{(a A+b B) x}{a^2+b^2}-\frac{(A b-a B) \log (\cos (c+d x))}{\left (a^2+b^2\right ) d}+\frac{a^2 (A b-a B) \log (a+b \tan (c+d x))}{b^2 \left (a^2+b^2\right ) d}+\frac{B \tan (c+d x)}{b d}\\ \end{align*}

Mathematica [C]  time = 0.555355, size = 118, normalized size = 1.17 \[ \frac{\frac{2 a^2 (A b-a B) \log (a+b \tan (c+d x))}{b^2 \left (a^2+b^2\right )}+\frac{i (A+i B) \log (-\tan (c+d x)+i)}{a+i b}-\frac{(B+i A) \log (\tan (c+d x)+i)}{a-i b}+\frac{2 B \tan (c+d x)}{b}}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.035, size = 179, normalized size = 1.8 \begin{align*}{\frac{B\tan \left ( dx+c \right ) }{bd}}+{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) Ab}{2\,d \left ({a}^{2}+{b}^{2} \right ) }}-{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) aB}{2\,d \left ({a}^{2}+{b}^{2} \right ) }}-{\frac{A\arctan \left ( \tan \left ( dx+c \right ) \right ) a}{d \left ({a}^{2}+{b}^{2} \right ) }}-{\frac{B\arctan \left ( \tan \left ( dx+c \right ) \right ) b}{d \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{{a}^{2}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) A}{bd \left ({a}^{2}+{b}^{2} \right ) }}-{\frac{B{a}^{3}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{ \left ({a}^{2}+{b}^{2} \right ){b}^{2}d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.49143, size = 147, normalized size = 1.46 \begin{align*} -\frac{\frac{2 \,{\left (A a + B b\right )}{\left (d x + c\right )}}{a^{2} + b^{2}} + \frac{2 \,{\left (B a^{3} - A a^{2} b\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{2} b^{2} + b^{4}} + \frac{{\left (B a - A b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} - \frac{2 \, B \tan \left (d x + c\right )}{b}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.97233, size = 333, normalized size = 3.3 \begin{align*} -\frac{2 \,{\left (A a b^{2} + B b^{3}\right )} d x +{\left (B a^{3} - A a^{2} b\right )} \log \left (\frac{b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) -{\left (B a^{3} - A a^{2} b + B a b^{2} - A b^{3}\right )} \log \left (\frac{1}{\tan \left (d x + c\right )^{2} + 1}\right ) - 2 \,{\left (B a^{2} b + B b^{3}\right )} \tan \left (d x + c\right )}{2 \,{\left (a^{2} b^{2} + b^{4}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 8.95012, size = 1015, normalized size = 10.05 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x*(A + B*tan(c))*tan(c), Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), (-I*A*d*x*tan(c + d*x)/(-2*b*d*tan(c
+ d*x) + 2*I*b*d) - A*d*x/(-2*b*d*tan(c + d*x) + 2*I*b*d) - A*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(-2*b*d*ta
n(c + d*x) + 2*I*b*d) + I*A*log(tan(c + d*x)**2 + 1)/(-2*b*d*tan(c + d*x) + 2*I*b*d) + I*A/(-2*b*d*tan(c + d*x
) + 2*I*b*d) + 3*B*d*x*tan(c + d*x)/(-2*b*d*tan(c + d*x) + 2*I*b*d) - 3*I*B*d*x/(-2*b*d*tan(c + d*x) + 2*I*b*d
) - I*B*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(-2*b*d*tan(c + d*x) + 2*I*b*d) - B*log(tan(c + d*x)**2 + 1)/(-2
*b*d*tan(c + d*x) + 2*I*b*d) - 2*B*tan(c + d*x)**2/(-2*b*d*tan(c + d*x) + 2*I*b*d) - 3*B/(-2*b*d*tan(c + d*x)
+ 2*I*b*d), Eq(a, -I*b)), (-I*A*d*x*tan(c + d*x)/(2*b*d*tan(c + d*x) + 2*I*b*d) + A*d*x/(2*b*d*tan(c + d*x) +
2*I*b*d) + A*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(2*b*d*tan(c + d*x) + 2*I*b*d) + I*A*log(tan(c + d*x)**2 +
1)/(2*b*d*tan(c + d*x) + 2*I*b*d) + I*A/(2*b*d*tan(c + d*x) + 2*I*b*d) - 3*B*d*x*tan(c + d*x)/(2*b*d*tan(c + d
*x) + 2*I*b*d) - 3*I*B*d*x/(2*b*d*tan(c + d*x) + 2*I*b*d) - I*B*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(2*b*d*t
an(c + d*x) + 2*I*b*d) + B*log(tan(c + d*x)**2 + 1)/(2*b*d*tan(c + d*x) + 2*I*b*d) + 2*B*tan(c + d*x)**2/(2*b*
d*tan(c + d*x) + 2*I*b*d) + 3*B/(2*b*d*tan(c + d*x) + 2*I*b*d), Eq(a, I*b)), ((-A*x + A*tan(c + d*x)/d - B*log
(tan(c + d*x)**2 + 1)/(2*d) + B*tan(c + d*x)**2/(2*d))/a, Eq(b, 0)), (x*(A + B*tan(c))*tan(c)**2/(a + b*tan(c)
), Eq(d, 0)), (2*A*a**2*b*log(a/b + tan(c + d*x))/(2*a**2*b**2*d + 2*b**4*d) - 2*A*a*b**2*d*x/(2*a**2*b**2*d +
2*b**4*d) + A*b**3*log(tan(c + d*x)**2 + 1)/(2*a**2*b**2*d + 2*b**4*d) - 2*B*a**3*log(a/b + tan(c + d*x))/(2*
a**2*b**2*d + 2*b**4*d) + 2*B*a**2*b*tan(c + d*x)/(2*a**2*b**2*d + 2*b**4*d) - B*a*b**2*log(tan(c + d*x)**2 +
1)/(2*a**2*b**2*d + 2*b**4*d) - 2*B*b**3*d*x/(2*a**2*b**2*d + 2*b**4*d) + 2*B*b**3*tan(c + d*x)/(2*a**2*b**2*d
+ 2*b**4*d), True))

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Giac [A]  time = 1.44274, size = 149, normalized size = 1.48 \begin{align*} -\frac{\frac{2 \,{\left (A a + B b\right )}{\left (d x + c\right )}}{a^{2} + b^{2}} + \frac{{\left (B a - A b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} + \frac{2 \,{\left (B a^{3} - A a^{2} b\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{2} b^{2} + b^{4}} - \frac{2 \, B \tan \left (d x + c\right )}{b}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*(A*a + B*b)*(d*x + c)/(a^2 + b^2) + (B*a - A*b)*log(tan(d*x + c)^2 + 1)/(a^2 + b^2) + 2*(B*a^3 - A*a^2
*b)*log(abs(b*tan(d*x + c) + a))/(a^2*b^2 + b^4) - 2*B*tan(d*x + c)/b)/d