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

Optimal. Leaf size=111 \[ -\frac{A b-a B}{d \left (a^2+b^2\right ) (a+b \tan (c+d x))}+\frac{\left (a^2 (-B)+2 a A b+b^2 B\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^2}+\frac{x \left (a^2 A+2 a b B-A b^2\right )}{\left (a^2+b^2\right )^2} \]

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

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Rubi [A]  time = 0.136639, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3529, 3531, 3530} \[ -\frac{A b-a B}{d \left (a^2+b^2\right ) (a+b \tan (c+d x))}+\frac{\left (a^2 (-B)+2 a A b+b^2 B\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^2}+\frac{x \left (a^2 A+2 a b B-A b^2\right )}{\left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

Rule 3529

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

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

Mathematica [C]  time = 1.84137, size = 190, normalized size = 1.71 \[ \frac{\frac{B ((-b-i a) \log (-\tan (c+d x)+i)+i (a+i b) \log (\tan (c+d x)+i)+2 b \log (a+b \tan (c+d x)))}{a^2+b^2}-(A b-a B) \left (\frac{2 b \left (\frac{a^2+b^2}{a+b \tan (c+d x)}-2 a \log (a+b \tan (c+d x))\right )}{\left (a^2+b^2\right )^2}+\frac{i \log (-\tan (c+d x)+i)}{(a+i b)^2}-\frac{i \log (\tan (c+d x)+i)}{(a-i b)^2}\right )}{2 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.04, size = 301, normalized size = 2.7 \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{Ab}{d \left ({a}^{2}+{b}^{2} \right ) \left ( a+b\tan \left ( dx+c \right ) \right ) }}+{\frac{aB}{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}^{2}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) B}{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+{\frac{\ln \left ( a+b\tan \left ( dx+c \right ) \right ){b}^{2}B}{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Maxima [A]  time = 1.52135, size = 239, normalized size = 2.15 \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^{2} - 2 \, A a b - B 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 (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 - A b\right )}}{a^{3} + a b^{2} +{\left (a^{2} b + b^{3}\right )} \tan \left (d x + c\right )}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [A]  time = 1.70246, size = 489, normalized size = 4.41 \begin{align*} \frac{2 \, B a b^{2} - 2 \, A b^{3} + 2 \,{\left (A a^{3} + 2 \, B a^{2} b - A a b^{2}\right )} d x -{\left (B a^{3} - 2 \, A a^{2} b - B a b^{2} +{\left (B a^{2} b - 2 \, A a b^{2} - B 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^{2} b - A a b^{2} -{\left (A a^{2} b + 2 \, B a b^{2} - A 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((A+B*tan(d*x+c))/(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/2*(2*B*a*b^2 - 2*A*b^3 + 2*(A*a^3 + 2*B*a^2*b - A*a*b^2)*d*x - (B*a^3 - 2*A*a^2*b - B*a*b^2 + (B*a^2*b - 2*A
*a*b^2 - B*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^2*b - A*a*b^2 - (A*a^2*b + 2*B*a*b^2 - A*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((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 = 1.27155, size = 316, normalized size = 2.85 \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^{2} b - 2 \, A a b^{2} - B 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 (B a^{2} b \tan \left (d x + c\right ) - 2 \, A a b^{2} \tan \left (d x + c\right ) - B b^{3} \tan \left (d x + c\right ) + 2 \, B a^{3} - 3 \, A a^{2} b - A b^{3}\right )}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\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((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^2*b - 2*A*a*b^2 - B*b^3)*log(abs(b*tan(d*x + c) + a))/(a^4*b + 2*a^2
*b^3 + b^5) + 2*(B*a^2*b*tan(d*x + c) - 2*A*a*b^2*tan(d*x + c) - B*b^3*tan(d*x + c) + 2*B*a^3 - 3*A*a^2*b - A*
b^3)/((a^4 + 2*a^2*b^2 + b^4)*(b*tan(d*x + c) + a)))/d