3.1206 \(\int \frac{(c+d \tan (e+f x))^3}{a+b \tan (e+f x)} \, dx\)

Optimal. Leaf size=140 \[ \frac{\left (-3 a c^2 d+a d^3+b c^3-3 b c d^2\right ) \log (\cos (e+f x))}{f \left (a^2+b^2\right )}+\frac{x \left (a c^3-3 a c d^2+3 b c^2 d-b d^3\right )}{a^2+b^2}+\frac{(b c-a d)^3 \log (a+b \tan (e+f x))}{b^2 f \left (a^2+b^2\right )}+\frac{d^2 (c+d \tan (e+f x))}{b f} \]

[Out]

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

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Rubi [A]  time = 0.281758, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3566, 3626, 3617, 31, 3475} \[ \frac{\left (-3 a c^2 d+a d^3+b c^3-3 b c d^2\right ) \log (\cos (e+f x))}{f \left (a^2+b^2\right )}+\frac{x \left (a c^3-3 a c d^2+3 b c^2 d-b d^3\right )}{a^2+b^2}+\frac{(b c-a d)^3 \log (a+b \tan (e+f x))}{b^2 f \left (a^2+b^2\right )}+\frac{d^2 (c+d \tan (e+f x))}{b f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [C]  time = 0.723913, size = 126, normalized size = 0.9 \[ \frac{\frac{2 (b c-a d)^3 \log (a+b \tan (e+f x))}{b^2 \left (a^2+b^2\right )}-\frac{(c-i d)^3 \log (\tan (e+f x)+i)}{b+i a}+\frac{(c+i d)^3 \log (-\tan (e+f x)+i)}{-b+i a}+\frac{2 d^2 (c+d \tan (e+f x))}{b}}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.026, size = 364, normalized size = 2.6 \begin{align*}{\frac{{d}^{3}\tan \left ( fx+e \right ) }{fb}}+{\frac{3\,\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) a{c}^{2}d}{2\,f \left ({a}^{2}+{b}^{2} \right ) }}-{\frac{\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) a{d}^{3}}{2\,f \left ({a}^{2}+{b}^{2} \right ) }}-{\frac{\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) b{c}^{3}}{2\,f \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{3\,\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) bc{d}^{2}}{2\,f \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ) a{c}^{3}}{f \left ({a}^{2}+{b}^{2} \right ) }}-3\,{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ) ac{d}^{2}}{f \left ({a}^{2}+{b}^{2} \right ) }}+3\,{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ) b{c}^{2}d}{f \left ({a}^{2}+{b}^{2} \right ) }}-{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ) b{d}^{3}}{f \left ({a}^{2}+{b}^{2} \right ) }}-{\frac{\ln \left ( a+b\tan \left ( fx+e \right ) \right ){a}^{3}{d}^{3}}{f{b}^{2} \left ({a}^{2}+{b}^{2} \right ) }}+3\,{\frac{\ln \left ( a+b\tan \left ( fx+e \right ) \right ){a}^{2}c{d}^{2}}{fb \left ({a}^{2}+{b}^{2} \right ) }}-3\,{\frac{\ln \left ( a+b\tan \left ( fx+e \right ) \right ) a{c}^{2}d}{f \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{b\ln \left ( a+b\tan \left ( fx+e \right ) \right ){c}^{3}}{f \left ({a}^{2}+{b}^{2} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c+d*tan(f*x+e))^3/(a+b*tan(f*x+e)),x)

[Out]

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

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Maxima [A]  time = 1.69857, size = 231, normalized size = 1.65 \begin{align*} \frac{\frac{2 \, d^{3} \tan \left (f x + e\right )}{b} + \frac{2 \,{\left (a c^{3} + 3 \, b c^{2} d - 3 \, a c d^{2} - b d^{3}\right )}{\left (f x + e\right )}}{a^{2} + b^{2}} + \frac{2 \,{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \log \left (b \tan \left (f x + e\right ) + a\right )}{a^{2} b^{2} + b^{4}} - \frac{{\left (b c^{3} - 3 \, a c^{2} d - 3 \, b c d^{2} + a d^{3}\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{a^{2} + b^{2}}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))^3/(a+b*tan(f*x+e)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.82528, size = 435, normalized size = 3.11 \begin{align*} \frac{2 \,{\left (a^{2} b + b^{3}\right )} d^{3} \tan \left (f x + e\right ) + 2 \,{\left (a b^{2} c^{3} + 3 \, b^{3} c^{2} d - 3 \, a b^{2} c d^{2} - b^{3} d^{3}\right )} f x +{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \log \left (\frac{b^{2} \tan \left (f x + e\right )^{2} + 2 \, a b \tan \left (f x + e\right ) + a^{2}}{\tan \left (f x + e\right )^{2} + 1}\right ) -{\left (3 \,{\left (a^{2} b + b^{3}\right )} c d^{2} -{\left (a^{3} + a b^{2}\right )} d^{3}\right )} \log \left (\frac{1}{\tan \left (f x + e\right )^{2} + 1}\right )}{2 \,{\left (a^{2} b^{2} + b^{4}\right )} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))^3/(a+b*tan(f*x+e)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 9.95967, size = 1712, normalized size = 12.23 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))**3/(a+b*tan(f*x+e)),x)

[Out]

Piecewise((zoo*x*(c + d*tan(e))**3/tan(e), Eq(a, 0) & Eq(b, 0) & Eq(f, 0)), ((c**3*x + 3*c**2*d*log(tan(e + f*
x)**2 + 1)/(2*f) - 3*c*d**2*x + 3*c*d**2*tan(e + f*x)/f - d**3*log(tan(e + f*x)**2 + 1)/(2*f) + d**3*tan(e + f
*x)**2/(2*f))/a, Eq(b, 0)), (-I*c**3*f*x*tan(e + f*x)/(-2*b*f*tan(e + f*x) + 2*I*b*f) - c**3*f*x/(-2*b*f*tan(e
+ f*x) + 2*I*b*f) - I*c**3/(-2*b*f*tan(e + f*x) + 2*I*b*f) - 3*c**2*d*f*x*tan(e + f*x)/(-2*b*f*tan(e + f*x) +
2*I*b*f) + 3*I*c**2*d*f*x/(-2*b*f*tan(e + f*x) + 2*I*b*f) + 3*c**2*d/(-2*b*f*tan(e + f*x) + 2*I*b*f) - 3*I*c*
d**2*f*x*tan(e + f*x)/(-2*b*f*tan(e + f*x) + 2*I*b*f) - 3*c*d**2*f*x/(-2*b*f*tan(e + f*x) + 2*I*b*f) - 3*c*d**
2*log(tan(e + f*x)**2 + 1)*tan(e + f*x)/(-2*b*f*tan(e + f*x) + 2*I*b*f) + 3*I*c*d**2*log(tan(e + f*x)**2 + 1)/
(-2*b*f*tan(e + f*x) + 2*I*b*f) + 3*I*c*d**2/(-2*b*f*tan(e + f*x) + 2*I*b*f) + 3*d**3*f*x*tan(e + f*x)/(-2*b*f
*tan(e + f*x) + 2*I*b*f) - 3*I*d**3*f*x/(-2*b*f*tan(e + f*x) + 2*I*b*f) - I*d**3*log(tan(e + f*x)**2 + 1)*tan(
e + f*x)/(-2*b*f*tan(e + f*x) + 2*I*b*f) - d**3*log(tan(e + f*x)**2 + 1)/(-2*b*f*tan(e + f*x) + 2*I*b*f) - 2*d
**3*tan(e + f*x)**2/(-2*b*f*tan(e + f*x) + 2*I*b*f) - 3*d**3/(-2*b*f*tan(e + f*x) + 2*I*b*f), Eq(a, -I*b)), (-
I*c**3*f*x*tan(e + f*x)/(2*b*f*tan(e + f*x) + 2*I*b*f) + c**3*f*x/(2*b*f*tan(e + f*x) + 2*I*b*f) - I*c**3/(2*b
*f*tan(e + f*x) + 2*I*b*f) + 3*c**2*d*f*x*tan(e + f*x)/(2*b*f*tan(e + f*x) + 2*I*b*f) + 3*I*c**2*d*f*x/(2*b*f*
tan(e + f*x) + 2*I*b*f) - 3*c**2*d/(2*b*f*tan(e + f*x) + 2*I*b*f) - 3*I*c*d**2*f*x*tan(e + f*x)/(2*b*f*tan(e +
f*x) + 2*I*b*f) + 3*c*d**2*f*x/(2*b*f*tan(e + f*x) + 2*I*b*f) + 3*c*d**2*log(tan(e + f*x)**2 + 1)*tan(e + f*x
)/(2*b*f*tan(e + f*x) + 2*I*b*f) + 3*I*c*d**2*log(tan(e + f*x)**2 + 1)/(2*b*f*tan(e + f*x) + 2*I*b*f) + 3*I*c*
d**2/(2*b*f*tan(e + f*x) + 2*I*b*f) - 3*d**3*f*x*tan(e + f*x)/(2*b*f*tan(e + f*x) + 2*I*b*f) - 3*I*d**3*f*x/(2
*b*f*tan(e + f*x) + 2*I*b*f) - I*d**3*log(tan(e + f*x)**2 + 1)*tan(e + f*x)/(2*b*f*tan(e + f*x) + 2*I*b*f) + d
**3*log(tan(e + f*x)**2 + 1)/(2*b*f*tan(e + f*x) + 2*I*b*f) + 2*d**3*tan(e + f*x)**2/(2*b*f*tan(e + f*x) + 2*I
*b*f) + 3*d**3/(2*b*f*tan(e + f*x) + 2*I*b*f), Eq(a, I*b)), (x*(c + d*tan(e))**3/(a + b*tan(e)), Eq(f, 0)), (-
2*a**3*d**3*log(a/b + tan(e + f*x))/(2*a**2*b**2*f + 2*b**4*f) + 6*a**2*b*c*d**2*log(a/b + tan(e + f*x))/(2*a*
*2*b**2*f + 2*b**4*f) + 2*a**2*b*d**3*tan(e + f*x)/(2*a**2*b**2*f + 2*b**4*f) + 2*a*b**2*c**3*f*x/(2*a**2*b**2
*f + 2*b**4*f) - 6*a*b**2*c**2*d*log(a/b + tan(e + f*x))/(2*a**2*b**2*f + 2*b**4*f) + 3*a*b**2*c**2*d*log(tan(
e + f*x)**2 + 1)/(2*a**2*b**2*f + 2*b**4*f) - 6*a*b**2*c*d**2*f*x/(2*a**2*b**2*f + 2*b**4*f) - a*b**2*d**3*log
(tan(e + f*x)**2 + 1)/(2*a**2*b**2*f + 2*b**4*f) + 2*b**3*c**3*log(a/b + tan(e + f*x))/(2*a**2*b**2*f + 2*b**4
*f) - b**3*c**3*log(tan(e + f*x)**2 + 1)/(2*a**2*b**2*f + 2*b**4*f) + 6*b**3*c**2*d*f*x/(2*a**2*b**2*f + 2*b**
4*f) + 3*b**3*c*d**2*log(tan(e + f*x)**2 + 1)/(2*a**2*b**2*f + 2*b**4*f) - 2*b**3*d**3*f*x/(2*a**2*b**2*f + 2*
b**4*f) + 2*b**3*d**3*tan(e + f*x)/(2*a**2*b**2*f + 2*b**4*f), True))

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Giac [A]  time = 1.87252, size = 238, normalized size = 1.7 \begin{align*} \frac{\frac{2 \, d^{3} \tan \left (f x + e\right )}{b} + \frac{2 \,{\left (a c^{3} + 3 \, b c^{2} d - 3 \, a c d^{2} - b d^{3}\right )}{\left (f x + e\right )}}{a^{2} + b^{2}} - \frac{{\left (b c^{3} - 3 \, a c^{2} d - 3 \, b c d^{2} + a d^{3}\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{a^{2} + b^{2}} + \frac{2 \,{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \log \left ({\left | b \tan \left (f x + e\right ) + a \right |}\right )}{a^{2} b^{2} + b^{4}}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))^3/(a+b*tan(f*x+e)),x, algorithm="giac")

[Out]

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