3.307 \(\int \frac{\tan ^3(c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=83 \[ -\frac{a^3 B \log (a+b \tan (c+d x))}{b^2 d \left (a^2+b^2\right )}+\frac{a B \log (\cos (c+d x))}{d \left (a^2+b^2\right )}-\frac{b B x}{a^2+b^2}+\frac{B \tan (c+d x)}{b d} \]

[Out]

-((b*B*x)/(a^2 + b^2)) + (a*B*Log[Cos[c + d*x]])/((a^2 + b^2)*d) - (a^3*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 = 0.174478, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {21, 3566, 3626, 3617, 31, 3475} \[ -\frac{a^3 B \log (a+b \tan (c+d x))}{b^2 d \left (a^2+b^2\right )}+\frac{a B \log (\cos (c+d x))}{d \left (a^2+b^2\right )}-\frac{b B x}{a^2+b^2}+\frac{B \tan (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

Int[(Tan[c + d*x]^3*(a*B + b*B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^2,x]

[Out]

-((b*B*x)/(a^2 + b^2)) + (a*B*Log[Cos[c + d*x]])/((a^2 + b^2)*d) - (a^3*B*Log[a + b*Tan[c + d*x]])/(b^2*(a^2 +
b^2)*d) + (B*Tan[c + d*x])/(b*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 3566

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(m + n - 1)), x] + Dist[1/(d*(m + n -
1)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(
1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
&& IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || IntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0]
&& NeQ[a, 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 ^3(c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx &=B \int \frac{\tan ^3(c+d x)}{a+b \tan (c+d x)} \, dx\\ &=\frac{B \tan (c+d x)}{b d}+\frac{B \int \frac{-a-b \tan (c+d x)-a \tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx}{b}\\ &=-\frac{b B x}{a^2+b^2}+\frac{B \tan (c+d x)}{b d}-\frac{(a B) \int \tan (c+d x) \, dx}{a^2+b^2}-\frac{\left (a^3 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{b B x}{a^2+b^2}+\frac{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^3 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{b B x}{a^2+b^2}+\frac{a B \log (\cos (c+d x))}{\left (a^2+b^2\right ) d}-\frac{a^3 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 = 0.381524, size = 92, normalized size = 1.11 \[ -\frac{B \left (\frac{2 a^3 \log (a+b \tan (c+d x))}{b^2 \left (a^2+b^2\right )}+\frac{\log (-\tan (c+d x)+i)}{a+i b}+\frac{\log (\tan (c+d x)+i)}{a-i b}-\frac{2 \tan (c+d x)}{b}\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Tan[c + d*x]^3*(a*B + b*B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^2,x]

[Out]

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

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Maple [A]  time = 0.033, size = 98, normalized size = 1.2 \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{B\arctan \left ( \tan \left ( dx+c \right ) \right ) b}{d \left ({a}^{2}+{b}^{2} \right ) }}-{\frac{{a}^{3}B\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)^3*(a*B+b*B*tan(d*x+c))/(a+b*tan(d*x+c))^2,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/d/(a^2+b^2)*B*arctan(tan(d*x+c))*b-a^3*B*ln(a+b*tan(
d*x+c))/b^2/(a^2+b^2)/d

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Maxima [A]  time = 1.72996, size = 120, normalized size = 1.45 \begin{align*} -\frac{\frac{2 \, B a^{3} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{2} b^{2} + b^{4}} + \frac{2 \,{\left (d x + c\right )} B b}{a^{2} + b^{2}} + \frac{B a \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)^3*(a*B+b*B*tan(d*x+c))/(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/2*(2*B*a^3*log(b*tan(d*x + c) + a)/(a^2*b^2 + b^4) + 2*(d*x + c)*B*b/(a^2 + b^2) + B*a*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 = 1.87438, size = 277, normalized size = 3.34 \begin{align*} -\frac{2 \, B b^{3} d x + B a^{3} \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} + B a b^{2}\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)^3*(a*B+b*B*tan(d*x+c))/(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/2*(2*B*b^3*d*x + B*a^3*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 +
B*a*b^2)*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 [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)**3*(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 = 1.78158, size = 122, normalized size = 1.47 \begin{align*} -\frac{\frac{2 \, B a^{3} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{2} b^{2} + b^{4}} + \frac{2 \,{\left (d x + c\right )} B b}{a^{2} + b^{2}} + \frac{B a \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)^3*(a*B+b*B*tan(d*x+c))/(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

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