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

Optimal. Leaf size=157 \[ -\frac{a^2 (A b-a B)}{b^2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac{a \left (2 A b^3-a B \left (a^2+3 b^2\right )\right ) \log (a+b \tan (c+d x))}{b^2 d \left (a^2+b^2\right )^2}-\frac{\left (a^2 (-B)+2 a A b+b^2 B\right ) \log (\cos (c+d x))}{d \left (a^2+b^2\right )^2}-\frac{x \left (a^2 A+2 a b B-A b^2\right )}{\left (a^2+b^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.273318, antiderivative size = 157, 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 = {3604, 3626, 3617, 31, 3475} \[ -\frac{a^2 (A b-a B)}{b^2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac{a \left (2 A b^3-a B \left (a^2+3 b^2\right )\right ) \log (a+b \tan (c+d x))}{b^2 d \left (a^2+b^2\right )^2}-\frac{\left (a^2 (-B)+2 a A b+b^2 B\right ) \log (\cos (c+d x))}{d \left (a^2+b^2\right )^2}-\frac{x \left (a^2 A+2 a b B-A b^2\right )}{\left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3604

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

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))^2} \, dx &=-\frac{a^2 (A b-a B)}{b^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac{\int \frac{-a (A b-a B)+b (A b-a B) \tan (c+d x)+\left (a^2+b^2\right ) B \tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx}{b \left (a^2+b^2\right )}\\ &=-\frac{\left (a^2 A-A b^2+2 a b B\right ) x}{\left (a^2+b^2\right )^2}-\frac{a^2 (A b-a B)}{b^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac{\left (2 a A b-a^2 B+b^2 B\right ) \int \tan (c+d x) \, dx}{\left (a^2+b^2\right )^2}-\frac{\left (a \left (2 A b^3-a \left (a^2+3 b^2\right ) B\right )\right ) \int \frac{1+\tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx}{b \left (a^2+b^2\right )^2}\\ &=-\frac{\left (a^2 A-A b^2+2 a b B\right ) x}{\left (a^2+b^2\right )^2}-\frac{\left (2 a A b-a^2 B+b^2 B\right ) \log (\cos (c+d x))}{\left (a^2+b^2\right )^2 d}-\frac{a^2 (A b-a B)}{b^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))}-\frac{\left (a \left (2 A b^3-a \left (a^2+3 b^2\right ) B\right )\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 )^2 d}\\ &=-\frac{\left (a^2 A-A b^2+2 a b B\right ) x}{\left (a^2+b^2\right )^2}-\frac{\left (2 a A b-a^2 B+b^2 B\right ) \log (\cos (c+d x))}{\left (a^2+b^2\right )^2 d}-\frac{a \left (2 A b^3-a \left (a^2+3 b^2\right ) B\right ) \log (a+b \tan (c+d x))}{b^2 \left (a^2+b^2\right )^2 d}-\frac{a^2 (A b-a B)}{b^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))}\\ \end{align*}

Mathematica [C]  time = 1.99352, size = 323, normalized size = 2.06 \[ \frac{-2 i a \left (a B \left (a^2+3 b^2\right )-2 A b^3\right ) \tan ^{-1}(\tan (c+d x)) (a+b \tan (c+d x))+a \left (a \left (a B \left (a^2+3 b^2\right )-2 A b^3\right ) \log \left ((a \cos (c+d x)+b \sin (c+d x))^2\right )-2 B \left (a^2+b^2\right )^2 \log (\cos (c+d x))+2 (a+i b)^2 (c+d x) \left (-A b^2+a B (2 b+i a)\right )\right )+b \tan (c+d x) \left (2 (a+i b) \left (a^2 b (A+B (c+d x+i))+i a^3 B (c+d x+i)-a b^2 (A (c+d x+i)-2 i B (c+d x))-i A b^3 (c+d x)\right )+a \left (a B \left (a^2+3 b^2\right )-2 A b^3\right ) \log \left ((a \cos (c+d x)+b \sin (c+d x))^2\right )-2 B \left (a^2+b^2\right )^2 \log (\cos (c+d x))\right )}{2 b^2 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.042, size = 313, normalized size = 2. \begin{align*}{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) Aab}{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}-{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ){a}^{2}B}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ){b}^{2}B}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}-{\frac{A\arctan \left ( \tan \left ( dx+c \right ) \right ){a}^{2}}{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+{\frac{A\arctan \left ( \tan \left ( dx+c \right ) \right ){b}^{2}}{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}-2\,{\frac{B\arctan \left ( \tan \left ( dx+c \right ) \right ) ab}{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}-{\frac{{a}^{2}A}{bd \left ({a}^{2}+{b}^{2} \right ) \left ( a+b\tan \left ( dx+c \right ) \right ) }}+{\frac{B{a}^{3}}{{b}^{2}d \left ({a}^{2}+{b}^{2} \right ) \left ( a+b\tan \left ( dx+c \right ) \right ) }}-2\,{\frac{ab\ln \left ( a+b\tan \left ( dx+c \right ) \right ) A}{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+{\frac{{a}^{4}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) B}{d \left ({a}^{2}+{b}^{2} \right ) ^{2}{b}^{2}}}+3\,{\frac{{a}^{2}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) B}{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}} \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))^2,x)

[Out]

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

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Maxima [A]  time = 1.5503, size = 266, normalized size = 1.69 \begin{align*} -\frac{\frac{2 \,{\left (A a^{2} + 2 \, B a b - A b^{2}\right )}{\left (d x + c\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac{2 \,{\left (B a^{4} + 3 \, B a^{2} b^{2} - 2 \, A a b^{3}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6}} + \frac{{\left (B a^{2} - 2 \, A a b - B b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac{2 \,{\left (B a^{3} - A a^{2} b\right )}}{a^{3} b^{2} + a b^{4} +{\left (a^{2} b^{3} + b^{5}\right )} \tan \left (d x + c\right )}}{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))^2,x, algorithm="maxima")

[Out]

-1/2*(2*(A*a^2 + 2*B*a*b - A*b^2)*(d*x + c)/(a^4 + 2*a^2*b^2 + b^4) - 2*(B*a^4 + 3*B*a^2*b^2 - 2*A*a*b^3)*log(
b*tan(d*x + c) + a)/(a^4*b^2 + 2*a^2*b^4 + b^6) + (B*a^2 - 2*A*a*b - B*b^2)*log(tan(d*x + c)^2 + 1)/(a^4 + 2*a
^2*b^2 + b^4) - 2*(B*a^3 - A*a^2*b)/(a^3*b^2 + a*b^4 + (a^2*b^3 + b^5)*tan(d*x + c)))/d

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Fricas [B]  time = 2.09051, size = 682, normalized size = 4.34 \begin{align*} \frac{2 \, B a^{3} b^{2} - 2 \, A a^{2} b^{3} - 2 \,{\left (A a^{3} b^{2} + 2 \, B a^{2} b^{3} - A a b^{4}\right )} d x +{\left (B a^{5} + 3 \, B a^{3} b^{2} - 2 \, A a^{2} b^{3} +{\left (B a^{4} b + 3 \, B a^{2} b^{3} - 2 \, A a b^{4}\right )} \tan \left (d x + c\right )\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^{5} + 2 \, B a^{3} b^{2} + B a b^{4} +{\left (B a^{4} b + 2 \, B a^{2} b^{3} + B b^{5}\right )} \tan \left (d x + c\right )\right )} \log \left (\frac{1}{\tan \left (d x + c\right )^{2} + 1}\right ) - 2 \,{\left (B a^{4} b - A a^{3} b^{2} +{\left (A a^{2} b^{3} + 2 \, B a b^{4} - A b^{5}\right )} d x\right )} \tan \left (d x + c\right )}{2 \,{\left ({\left (a^{4} b^{3} + 2 \, a^{2} b^{5} + b^{7}\right )} d \tan \left (d x + c\right ) +{\left (a^{5} b^{2} + 2 \, a^{3} b^{4} + a b^{6}\right )} d\right )}} \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))^2,x, algorithm="fricas")

[Out]

1/2*(2*B*a^3*b^2 - 2*A*a^2*b^3 - 2*(A*a^3*b^2 + 2*B*a^2*b^3 - A*a*b^4)*d*x + (B*a^5 + 3*B*a^3*b^2 - 2*A*a^2*b^
3 + (B*a^4*b + 3*B*a^2*b^3 - 2*A*a*b^4)*tan(d*x + c))*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^5 + 2*B*a^3*b^2 + B*a*b^4 + (B*a^4*b + 2*B*a^2*b^3 + B*b^5)*tan(d*x + c))*log(1/(tan(
d*x + c)^2 + 1)) - 2*(B*a^4*b - A*a^3*b^2 + (A*a^2*b^3 + 2*B*a*b^4 - A*b^5)*d*x)*tan(d*x + c))/((a^4*b^3 + 2*a
^2*b^5 + b^7)*d*tan(d*x + c) + (a^5*b^2 + 2*a^3*b^4 + a*b^6)*d)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \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))**2,x)

[Out]

Exception raised: AttributeError

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Giac [A]  time = 1.44776, size = 329, normalized size = 2.1 \begin{align*} -\frac{\frac{2 \,{\left (A a^{2} + 2 \, B a b - A b^{2}\right )}{\left (d x + c\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{{\left (B a^{2} - 2 \, A a b - B b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac{2 \,{\left (B a^{4} + 3 \, B a^{2} b^{2} - 2 \, A a b^{3}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6}} + \frac{2 \,{\left (B a^{4} \tan \left (d x + c\right ) + 3 \, B a^{2} b^{2} \tan \left (d x + c\right ) - 2 \, A a b^{3} \tan \left (d x + c\right ) + A a^{4} + 2 \, B a^{3} b - A a^{2} b^{2}\right )}}{{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )}{\left (b \tan \left (d x + c\right ) + a\right )}}}{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))^2,x, algorithm="giac")

[Out]

-1/2*(2*(A*a^2 + 2*B*a*b - A*b^2)*(d*x + c)/(a^4 + 2*a^2*b^2 + b^4) + (B*a^2 - 2*A*a*b - B*b^2)*log(tan(d*x +
c)^2 + 1)/(a^4 + 2*a^2*b^2 + b^4) - 2*(B*a^4 + 3*B*a^2*b^2 - 2*A*a*b^3)*log(abs(b*tan(d*x + c) + a))/(a^4*b^2
+ 2*a^2*b^4 + b^6) + 2*(B*a^4*tan(d*x + c) + 3*B*a^2*b^2*tan(d*x + c) - 2*A*a*b^3*tan(d*x + c) + A*a^4 + 2*B*a
^3*b - A*a^2*b^2)/((a^4*b + 2*a^2*b^3 + b^5)*(b*tan(d*x + c) + a)))/d