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

Optimal. Leaf size=81 \[ \frac{a^2 B \log (a \cos (c+d x)+b \sin (c+d x))}{b d \left (a^2+b^2\right )}+\frac{a^3 B x}{b^2 \left (a^2+b^2\right )}-\frac{a B x}{b^2}-\frac{B \log (\cos (c+d x))}{b d} \]

[Out]

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

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Rubi [A]  time = 0.115725, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.147, Rules used = {21, 3541, 3475, 3484, 3530} \[ \frac{a^2 B \log (a \cos (c+d x)+b \sin (c+d x))}{b d \left (a^2+b^2\right )}+\frac{a^3 B x}{b^2 \left (a^2+b^2\right )}-\frac{a B x}{b^2}-\frac{B \log (\cos (c+d x))}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 3541

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

Rule 3475

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

Rule 3484

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[(a*x)/(a^2 + b^2), x] + Dist[b/(a^2 + b^2),
Int[(b - a*Tan[c + d*x])/(a + b*Tan[c + d*x]), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 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 ^2(c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx &=B \int \frac{\tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx\\ &=-\frac{a B x}{b^2}+\frac{\left (a^2 B\right ) \int \frac{1}{a+b \tan (c+d x)} \, dx}{b^2}+\frac{B \int \tan (c+d x) \, dx}{b}\\ &=-\frac{a B x}{b^2}+\frac{a^3 B x}{b^2 \left (a^2+b^2\right )}-\frac{B \log (\cos (c+d x))}{b d}+\frac{\left (a^2 B\right ) \int \frac{b-a \tan (c+d x)}{a+b \tan (c+d x)} \, dx}{b \left (a^2+b^2\right )}\\ &=-\frac{a B x}{b^2}+\frac{a^3 B x}{b^2 \left (a^2+b^2\right )}-\frac{B \log (\cos (c+d x))}{b d}+\frac{a^2 B \log (a \cos (c+d x)+b \sin (c+d x))}{b \left (a^2+b^2\right ) d}\\ \end{align*}

Mathematica [C]  time = 0.082385, size = 79, normalized size = 0.98 \[ \frac{B \left (2 a^2 \log (a+b \tan (c+d x))+b (b+i a) \log (-\tan (c+d x)+i)+b (b-i a) \log (\tan (c+d x)+i)\right )}{2 b d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.032, size = 83, normalized size = 1. \begin{align*}{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) Bb}{2\,d \left ({a}^{2}+{b}^{2} \right ) }}-{\frac{B\arctan \left ( \tan \left ( dx+c \right ) \right ) a}{d \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{{a}^{2}B\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{d \left ({a}^{2}+{b}^{2} \right ) b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.7828, size = 101, normalized size = 1.25 \begin{align*} \frac{\frac{2 \, B a^{2} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{2} b + b^{3}} - \frac{2 \,{\left (d x + c\right )} B a}{a^{2} + b^{2}} + \frac{B b \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.76961, size = 224, normalized size = 2.77 \begin{align*} -\frac{2 \, B a b d x - B a^{2} \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^{2} + B b^{2}\right )} \log \left (\frac{1}{\tan \left (d x + c\right )^{2} + 1}\right )}{2 \,{\left (a^{2} b + b^{3}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: AttributeError

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Giac [A]  time = 1.47351, size = 103, normalized size = 1.27 \begin{align*} \frac{\frac{2 \, B a^{2} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{2} b + b^{3}} - \frac{2 \,{\left (d x + c\right )} B a}{a^{2} + b^{2}} + \frac{B b \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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