3.306 \(\int \frac{\tan ^4(c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=102 \[ \frac{a^4 B \log (a+b \tan (c+d x))}{b^3 d \left (a^2+b^2\right )}+\frac{b B \log (\cos (c+d x))}{d \left (a^2+b^2\right )}+\frac{a B x}{a^2+b^2}-\frac{a B \tan (c+d x)}{b^2 d}+\frac{B \tan ^2(c+d x)}{2 b d} \]

[Out]

(a*B*x)/(a^2 + b^2) + (b*B*Log[Cos[c + d*x]])/((a^2 + b^2)*d) + (a^4*B*Log[a + b*Tan[c + d*x]])/(b^3*(a^2 + b^
2)*d) - (a*B*Tan[c + d*x])/(b^2*d) + (B*Tan[c + d*x]^2)/(2*b*d)

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Rubi [A]  time = 0.290616, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.206, Rules used = {21, 3566, 3647, 3627, 3617, 31, 3475} \[ \frac{a^4 B \log (a+b \tan (c+d x))}{b^3 d \left (a^2+b^2\right )}+\frac{b B \log (\cos (c+d x))}{d \left (a^2+b^2\right )}+\frac{a B x}{a^2+b^2}-\frac{a B \tan (c+d x)}{b^2 d}+\frac{B \tan ^2(c+d x)}{2 b d} \]

Antiderivative was successfully verified.

[In]

Int[(Tan[c + d*x]^4*(a*B + b*B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^2,x]

[Out]

(a*B*x)/(a^2 + b^2) + (b*B*Log[Cos[c + d*x]])/((a^2 + b^2)*d) + (a^4*B*Log[a + b*Tan[c + d*x]])/(b^3*(a^2 + b^
2)*d) - (a*B*Tan[c + d*x])/(b^2*d) + (B*Tan[c + d*x]^2)/(2*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 3647

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*
tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(C*(a + b*Tan[e + f*x])^m*(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 - 1)*(c +
d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f
*x] - (C*m*(b*c - a*d) - b*B*d*(m + n + 1))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}
, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !Intege
rQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3627

Int[((A_) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(a*(A -
C)*x)/(a^2 + b^2), x] + (Dist[(a^2*C + A*b^2)/(a^2 + b^2), Int[(1 + Tan[e + f*x]^2)/(a + b*Tan[e + f*x]), x],
x] - Dist[(b*(A - C))/(a^2 + b^2), Int[Tan[e + f*x], x], x]) /; FreeQ[{a, b, e, f, A, C}, x] && NeQ[a^2*C + A
*b^2, 0] && NeQ[a^2 + b^2, 0] && NeQ[A, C]

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 ^4(c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx &=B \int \frac{\tan ^4(c+d x)}{a+b \tan (c+d x)} \, dx\\ &=\frac{B \tan ^2(c+d x)}{2 b d}+\frac{B \int \frac{\tan (c+d x) \left (-2 a-2 b \tan (c+d x)-2 a \tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx}{2 b}\\ &=-\frac{a B \tan (c+d x)}{b^2 d}+\frac{B \tan ^2(c+d x)}{2 b d}+\frac{B \int \frac{2 a^2+2 \left (a^2-b^2\right ) \tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx}{2 b^2}\\ &=\frac{a B x}{a^2+b^2}-\frac{a B \tan (c+d x)}{b^2 d}+\frac{B \tan ^2(c+d x)}{2 b d}+\frac{\left (a^4 B\right ) \int \frac{1+\tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx}{b^2 \left (a^2+b^2\right )}-\frac{(b B) \int \tan (c+d x) \, dx}{a^2+b^2}\\ &=\frac{a B x}{a^2+b^2}+\frac{b B \log (\cos (c+d x))}{\left (a^2+b^2\right ) d}-\frac{a B \tan (c+d x)}{b^2 d}+\frac{B \tan ^2(c+d x)}{2 b d}+\frac{\left (a^4 B\right ) \operatorname{Subst}\left (\int \frac{1}{a+x} \, dx,x,b \tan (c+d x)\right )}{b^3 \left (a^2+b^2\right ) d}\\ &=\frac{a B x}{a^2+b^2}+\frac{b B \log (\cos (c+d x))}{\left (a^2+b^2\right ) d}+\frac{a^4 B \log (a+b \tan (c+d x))}{b^3 \left (a^2+b^2\right ) d}-\frac{a B \tan (c+d x)}{b^2 d}+\frac{B \tan ^2(c+d x)}{2 b d}\\ \end{align*}

Mathematica [C]  time = 0.412395, size = 108, normalized size = 1.06 \[ \frac{B \left (\frac{2 a^4 \log (a+b \tan (c+d x))}{b^3 \left (a^2+b^2\right )}-\frac{2 a \tan (c+d x)}{b^2}+\frac{\log (-\tan (c+d x)+i)}{-b+i a}-\frac{\log (\tan (c+d x)+i)}{b+i a}+\frac{\tan ^2(c+d x)}{b}\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Tan[c + d*x]^4*(a*B + b*B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^2,x]

[Out]

(B*(Log[I - Tan[c + d*x]]/(I*a - b) - Log[I + Tan[c + d*x]]/(I*a + b) + (2*a^4*Log[a + b*Tan[c + d*x]])/(b^3*(
a^2 + b^2)) - (2*a*Tan[c + d*x])/b^2 + Tan[c + d*x]^2/b))/(2*d)

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Maple [A]  time = 0.034, size = 115, normalized size = 1.1 \begin{align*}{\frac{B \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,bd}}-{\frac{aB\tan \left ( dx+c \right ) }{{b}^{2}d}}-{\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}^{4}B\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{{b}^{3} \left ({a}^{2}+{b}^{2} \right ) d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^4*(a*B+b*B*tan(d*x+c))/(a+b*tan(d*x+c))^2,x)

[Out]

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

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Maxima [A]  time = 1.75869, size = 140, normalized size = 1.37 \begin{align*} \frac{\frac{2 \, B a^{4} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{2} b^{3} + b^{5}} + \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}} + \frac{B b \tan \left (d x + c\right )^{2} - 2 \, B a \tan \left (d x + c\right )}{b^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4*(a*B+b*B*tan(d*x+c))/(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.90752, size = 328, normalized size = 3.22 \begin{align*} \frac{2 \, B a b^{3} d x + B a^{4} \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^{2} + B b^{4}\right )} \tan \left (d x + c\right )^{2} -{\left (B a^{4} - B b^{4}\right )} \log \left (\frac{1}{\tan \left (d x + c\right )^{2} + 1}\right ) - 2 \,{\left (B a^{3} b + B a b^{3}\right )} \tan \left (d x + c\right )}{2 \,{\left (a^{2} b^{3} + b^{5}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**4*(a*B+b*B*tan(d*x+c))/(a+b*tan(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 2.39502, size = 142, normalized size = 1.39 \begin{align*} \frac{\frac{2 \, B a^{4} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{2} b^{3} + b^{5}} + \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}} + \frac{B b \tan \left (d x + c\right )^{2} - 2 \, B a \tan \left (d x + c\right )}{b^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4*(a*B+b*B*tan(d*x+c))/(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

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