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

Optimal. Leaf size=155 \[ -\frac{a^2 \tan ^2(c+d x)}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}+\frac{\left (2 a^2+b^2\right ) \tan (c+d x)}{b^2 d \left (a^2+b^2\right )}-\frac{2 a^3 \left (a^2+2 b^2\right ) \log (a+b \tan (c+d x))}{b^3 d \left (a^2+b^2\right )^2}+\frac{2 a b \log (\cos (c+d x))}{d \left (a^2+b^2\right )^2}+\frac{x \left (a^2-b^2\right )}{\left (a^2+b^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.297089, antiderivative size = 155, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3565, 3647, 3626, 3617, 31, 3475} \[ -\frac{a^2 \tan ^2(c+d x)}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}+\frac{\left (2 a^2+b^2\right ) \tan (c+d x)}{b^2 d \left (a^2+b^2\right )}-\frac{2 a^3 \left (a^2+2 b^2\right ) \log (a+b \tan (c+d x))}{b^3 d \left (a^2+b^2\right )^2}+\frac{2 a b \log (\cos (c+d x))}{d \left (a^2+b^2\right )^2}+\frac{x \left (a^2-b^2\right )}{\left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3565

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[((b*c - a*d)^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 + d^2)), x] - D
ist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a^2*d*(b*d*(
m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c*(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3
*a*b^2*d)*Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*(n + 1)))*Tan[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && Gt
Q[m, 2] && LtQ[n, -1] && IntegerQ[2*m]

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

Mathematica [C]  time = 2.60801, size = 329, normalized size = 2.12 \[ \frac{2 i a^3 \left (a^2+2 b^2\right ) \tan ^{-1}(\tan (c+d x)) (a+b \tan (c+d x))+b^2 \left (a^2+b^2\right )^2 \tan ^2(c+d x)+a \left ((a+i b)^2 \left (-4 a^2 b-2 i a^3+2 i a b^2+b^3\right ) (c+d x)+2 a \left (a^2+b^2\right )^2 \log (\cos (c+d x))+a^3 \left (-\left (a^2+2 b^2\right )\right ) \log \left ((a \cos (c+d x)+b \sin (c+d x))^2\right )\right )+b \tan (c+d x) \left (2 a \left (a^2+b^2\right )^2 \log (\cos (c+d x))-a^3 \left (a^2+2 b^2\right ) \log \left ((a \cos (c+d x)+b \sin (c+d x))^2\right )-4 i a^3 b^2 c+a^2 b^3 c-4 i a^3 b^2 d x+a^2 b^3 d x+3 a^3 b^2-2 i a^5 c-2 i a^5 d x+2 a^5+a b^4-b^5 c-b^5 d x\right )}{b^3 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*((a + I*b)^2*((-2*I)*a^3 - 4*a^2*b + (2*I)*a*b^2 + b^3)*(c + d*x) + 2*a*(a^2 + b^2)^2*Log[Cos[c + d*x]] - a
^3*(a^2 + 2*b^2)*Log[(a*Cos[c + d*x] + b*Sin[c + d*x])^2]) + b*(2*a^5 + 3*a^3*b^2 + a*b^4 - (2*I)*a^5*c - (4*I
)*a^3*b^2*c + a^2*b^3*c - b^5*c - (2*I)*a^5*d*x - (4*I)*a^3*b^2*d*x + a^2*b^3*d*x - b^5*d*x + 2*a*(a^2 + b^2)^
2*Log[Cos[c + d*x]] - a^3*(a^2 + 2*b^2)*Log[(a*Cos[c + d*x] + b*Sin[c + d*x])^2])*Tan[c + d*x] + b^2*(a^2 + b^
2)^2*Tan[c + d*x]^2 + (2*I)*a^3*(a^2 + 2*b^2)*ArcTan[Tan[c + d*x]]*(a + b*Tan[c + d*x]))/(b^3*(a^2 + b^2)^2*d*
(a + b*Tan[c + d*x]))

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Maple [A]  time = 0.024, size = 183, normalized size = 1.2 \begin{align*}{\frac{\tan \left ( dx+c \right ) }{{b}^{2}d}}+{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){a}^{2}}{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}-{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){b}^{2}}{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}-{\frac{ab\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}-{\frac{{a}^{4}}{d{b}^{3} \left ({a}^{2}+{b}^{2} \right ) \left ( a+b\tan \left ( dx+c \right ) \right ) }}-2\,{\frac{{a}^{5}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{d{b}^{3} \left ({a}^{2}+{b}^{2} \right ) ^{2}}}-4\,{\frac{{a}^{3}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{bd \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)^4/(a+b*tan(d*x+c))^2,x)

[Out]

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

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Maxima [A]  time = 1.5861, size = 221, normalized size = 1.43 \begin{align*} -\frac{\frac{a^{4}}{a^{3} b^{3} + a b^{5} +{\left (a^{2} b^{4} + b^{6}\right )} \tan \left (d x + c\right )} + \frac{a b \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac{{\left (a^{2} - b^{2}\right )}{\left (d x + c\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{2 \,{\left (a^{5} + 2 \, a^{3} b^{2}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{4} b^{3} + 2 \, a^{2} b^{5} + b^{7}} - \frac{\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)^4/(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

-(a^4/(a^3*b^3 + a*b^5 + (a^2*b^4 + b^6)*tan(d*x + c)) + a*b*log(tan(d*x + c)^2 + 1)/(a^4 + 2*a^2*b^2 + b^4) -
(a^2 - b^2)*(d*x + c)/(a^4 + 2*a^2*b^2 + b^4) + 2*(a^5 + 2*a^3*b^2)*log(b*tan(d*x + c) + a)/(a^4*b^3 + 2*a^2*
b^5 + b^7) - tan(d*x + c)/b^2)/d

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Fricas [A]  time = 2.36519, size = 617, normalized size = 3.98 \begin{align*} -\frac{a^{4} b^{2} -{\left (a^{3} b^{3} - a b^{5}\right )} d x -{\left (a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6}\right )} \tan \left (d x + c\right )^{2} +{\left (a^{6} + 2 \, a^{4} b^{2} +{\left (a^{5} b + 2 \, a^{3} 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 ) -{\left (a^{6} + 2 \, a^{4} b^{2} + a^{2} b^{4} +{\left (a^{5} b + 2 \, a^{3} b^{3} + a b^{5}\right )} \tan \left (d x + c\right )\right )} \log \left (\frac{1}{\tan \left (d x + c\right )^{2} + 1}\right ) -{\left (2 \, a^{5} b + 2 \, a^{3} b^{3} + a b^{5} +{\left (a^{2} b^{4} - b^{6}\right )} d x\right )} \tan \left (d x + c\right )}{{\left (a^{4} b^{4} + 2 \, a^{2} b^{6} + b^{8}\right )} d \tan \left (d x + c\right ) +{\left (a^{5} b^{3} + 2 \, a^{3} b^{5} + a b^{7}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a^4*b^2 - (a^3*b^3 - a*b^5)*d*x - (a^4*b^2 + 2*a^2*b^4 + b^6)*tan(d*x + c)^2 + (a^6 + 2*a^4*b^2 + (a^5*b + 2
*a^3*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)) - (a^6 + 2*a
^4*b^2 + a^2*b^4 + (a^5*b + 2*a^3*b^3 + a*b^5)*tan(d*x + c))*log(1/(tan(d*x + c)^2 + 1)) - (2*a^5*b + 2*a^3*b^
3 + a*b^5 + (a^2*b^4 - b^6)*d*x)*tan(d*x + c))/((a^4*b^4 + 2*a^2*b^6 + b^8)*d*tan(d*x + c) + (a^5*b^3 + 2*a^3*
b^5 + a*b^7)*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)**4/(a+b*tan(d*x+c))**2,x)

[Out]

Exception raised: AttributeError

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Giac [A]  time = 2.43756, size = 271, normalized size = 1.75 \begin{align*} -\frac{\frac{a b \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac{{\left (a^{2} - b^{2}\right )}{\left (d x + c\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{2 \,{\left (a^{5} + 2 \, a^{3} b^{2}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{4} b^{3} + 2 \, a^{2} b^{5} + b^{7}} - \frac{2 \, a^{5} b \tan \left (d x + c\right ) + 4 \, a^{3} b^{3} \tan \left (d x + c\right ) + a^{6} + 3 \, a^{4} b^{2}}{{\left (a^{4} b^{3} + 2 \, a^{2} b^{5} + b^{7}\right )}{\left (b \tan \left (d x + c\right ) + a\right )}} - \frac{\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)^4/(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-(a*b*log(tan(d*x + c)^2 + 1)/(a^4 + 2*a^2*b^2 + b^4) - (a^2 - b^2)*(d*x + c)/(a^4 + 2*a^2*b^2 + b^4) + 2*(a^5
+ 2*a^3*b^2)*log(abs(b*tan(d*x + c) + a))/(a^4*b^3 + 2*a^2*b^5 + b^7) - (2*a^5*b*tan(d*x + c) + 4*a^3*b^3*tan
(d*x + c) + a^6 + 3*a^4*b^2)/((a^4*b^3 + 2*a^2*b^5 + b^7)*(b*tan(d*x + c) + a)) - tan(d*x + c)/b^2)/d