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

Optimal. Leaf size=115 \[ \frac{a (A b-a B)}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac{\left (a^2 A+2 a b B-A b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^2}+\frac{x \left (a^2 (-B)+2 a A b+b^2 B\right )}{\left (a^2+b^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.159216, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {3591, 3531, 3530} \[ \frac{a (A b-a B)}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac{\left (a^2 A+2 a b B-A b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^2}+\frac{x \left (a^2 (-B)+2 a A b+b^2 B\right )}{\left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3591

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

Rule 3531

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

Rule 3530

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c*Log[Re
moveContent[a*Cos[e + f*x] + b*Sin[e + f*x], x]])/(b*f), x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d,
0] && NeQ[a^2 + b^2, 0] && EqQ[a*c + b*d, 0]

Rubi steps

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

Mathematica [C]  time = 1.96514, size = 140, normalized size = 1.22 \[ \frac{\frac{2 \left (\left (a^2 (-A)-2 a b B+A b^2\right ) \log (a+b \tan (c+d x))-\frac{a \left (a^2+b^2\right ) (a B-A b)}{b (a+b \tan (c+d x))}\right )}{\left (a^2+b^2\right )^2}+\frac{(A+i B) \log (-\tan (c+d x)+i)}{(a+i b)^2}+\frac{(A-i B) \log (\tan (c+d x)+i)}{(a-i b)^2}}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.04, size = 305, normalized size = 2.7 \begin{align*}{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ){a}^{2}A}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}-{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) A{b}^{2}}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) Bab}{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+2\,{\frac{A\arctan \left ( \tan \left ( dx+c \right ) \right ) ab}{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}-{\frac{B\arctan \left ( \tan \left ( dx+c \right ) \right ){a}^{2}}{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+{\frac{B\arctan \left ( \tan \left ( dx+c \right ) \right ){b}^{2}}{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+{\frac{Aa}{d \left ({a}^{2}+{b}^{2} \right ) \left ( a+b\tan \left ( dx+c \right ) \right ) }}-{\frac{{a}^{2}B}{d \left ({a}^{2}+{b}^{2} \right ) b \left ( a+b\tan \left ( dx+c \right ) \right ) }}-{\frac{{a}^{2}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) A}{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+{\frac{\ln \left ( a+b\tan \left ( dx+c \right ) \right ) A{b}^{2}}{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}-2\,{\frac{\ln \left ( a+b\tan \left ( dx+c \right ) \right ) Bab}{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)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^2,x)

[Out]

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

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Maxima [A]  time = 1.47768, size = 250, normalized size = 2.17 \begin{align*} -\frac{\frac{2 \,{\left (B a^{2} - 2 \, A a b - B b^{2}\right )}{\left (d x + c\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{2 \,{\left (A a^{2} + 2 \, B a b - A b^{2}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac{{\left (A a^{2} + 2 \, B a b - A 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^{2} - A a b\right )}}{a^{3} b + a b^{3} +{\left (a^{2} b^{2} + b^{4}\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)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.74224, size = 490, normalized size = 4.26 \begin{align*} -\frac{2 \, B a^{2} b - 2 \, A a b^{2} + 2 \,{\left (B a^{3} - 2 \, A a^{2} b - B a b^{2}\right )} d x +{\left (A a^{3} + 2 \, B a^{2} b - A a b^{2} +{\left (A a^{2} b + 2 \, B a b^{2} - A b^{3}\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 ) - 2 \,{\left (B a^{3} - A a^{2} b -{\left (B a^{2} b - 2 \, A a b^{2} - B b^{3}\right )} d x\right )} \tan \left (d x + c\right )}{2 \,{\left ({\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} d \tan \left (d x + c\right ) +{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: AttributeError

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Giac [B]  time = 1.236, size = 325, normalized size = 2.83 \begin{align*} -\frac{\frac{2 \,{\left (B a^{2} - 2 \, A a b - B b^{2}\right )}{\left (d x + c\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac{{\left (A a^{2} + 2 \, B a b - A 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 (A a^{2} b + 2 \, B a b^{2} - A b^{3}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{4} b + 2 \, a^{2} b^{3} + b^{5}} - \frac{2 \,{\left (A a^{2} b^{2} \tan \left (d x + c\right ) + 2 \, B a b^{3} \tan \left (d x + c\right ) - A b^{4} \tan \left (d x + c\right ) - B a^{4} + 2 \, A a^{3} b + B 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)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

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