∫ (a+b tan(e+fx))3 _________________________

3.1210 \(\int \frac{(a+b \tan (e+f x))^3}{c+d \tan (e+f x)} \, dx\)

Optimal. Leaf size=144 \[ -\frac{\left (3 a^2 b c+a^3 (-d)+3 a b^2 d-b^3 c\right ) \log (\cos (e+f x))}{f \left (c^2+d^2\right )}+\frac{x \left (3 a^2 b d+a^3 c-3 a b^2 c-b^3 d\right )}{c^2+d^2}+\frac{b^2 (a+b \tan (e+f x))}{d f}-\frac{(b c-a d)^3 \log (c+d \tan (e+f x))}{d^2 f \left (c^2+d^2\right )} \]

[Out]

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

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

x]))/(d*f)

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Rubi [A] time = 0.268188, antiderivative size = 144, 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} \[ -\frac{\left (3 a^2 b c+a^3 (-d)+3 a b^2 d-b^3 c\right ) \log (\cos (e+f x))}{f \left (c^2+d^2\right )}+\frac{x \left (3 a^2 b d+a^3 c-3 a b^2 c-b^3 d\right )}{c^2+d^2}+\frac{b^2 (a+b \tan (e+f x))}{d f}-\frac{(b c-a d)^3 \log (c+d \tan (e+f x))}{d^2 f \left (c^2+d^2\right )} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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

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

Mathematica [C] time = 0.718178, size = 126, normalized size = 0.88 \[ \frac{\frac{2 b^2 (a+b \tan (e+f x))}{d}+\frac{2 (a d-b c)^3 \log (c+d \tan (e+f x))}{d^2 \left (c^2+d^2\right )}+\frac{(a+i b)^3 \log (-\tan (e+f x)+i)}{-d+i c}-\frac{(b+i a)^3 \log (\tan (e+f x)+i)}{c-i d}}{2 f} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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

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Maple [B] time = 0.025, size = 364, normalized size = 2.5 \begin{align*}{\frac{{b}^{3}\tan \left ( fx+e \right ) }{fd}}-{\frac{{a}^{3}\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) d}{2\,f \left ({c}^{2}+{d}^{2} \right ) }}+{\frac{3\,\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ){a}^{2}bc}{2\,f \left ({c}^{2}+{d}^{2} \right ) }}+{\frac{3\,\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) a{b}^{2}d}{2\,f \left ({c}^{2}+{d}^{2} \right ) }}-{\frac{\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ){b}^{3}c}{2\,f \left ({c}^{2}+{d}^{2} \right ) }}+{\frac{{a}^{3}\arctan \left ( \tan \left ( fx+e \right ) \right ) c}{f \left ({c}^{2}+{d}^{2} \right ) }}+3\,{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ){a}^{2}bd}{f \left ({c}^{2}+{d}^{2} \right ) }}-3\,{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ) a{b}^{2}c}{f \left ({c}^{2}+{d}^{2} \right ) }}-{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ){b}^{3}d}{f \left ({c}^{2}+{d}^{2} \right ) }}+{\frac{{a}^{3}d\ln \left ( c+d\tan \left ( fx+e \right ) \right ) }{f \left ({c}^{2}+{d}^{2} \right ) }}-3\,{\frac{\ln \left ( c+d\tan \left ( fx+e \right ) \right ){a}^{2}bc}{f \left ({c}^{2}+{d}^{2} \right ) }}+3\,{\frac{\ln \left ( c+d\tan \left ( fx+e \right ) \right ){c}^{2}a{b}^{2}}{fd \left ({c}^{2}+{d}^{2} \right ) }}-{\frac{\ln \left ( c+d\tan \left ( fx+e \right ) \right ){c}^{3}{b}^{3}}{{d}^{2}f \left ({c}^{2}+{d}^{2} \right ) }} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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

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

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

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Maxima [A] time = 2.10474, size = 235, normalized size = 1.63 \begin{align*} \frac{\frac{2 \, b^{3} \tan \left (f x + e\right )}{d} + \frac{2 \,{\left ({\left (a^{3} - 3 \, a b^{2}\right )} c +{\left (3 \, a^{2} b - b^{3}\right )} d\right )}{\left (f x + e\right )}}{c^{2} + d^{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 (d \tan \left (f x + e\right ) + c\right )}{c^{2} d^{2} + d^{4}} + \frac{{\left ({\left (3 \, a^{2} b - b^{3}\right )} c -{\left (a^{3} - 3 \, a b^{2}\right )} d\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{c^{2} + d^{2}}}{2 \, f} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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Fricas [A] time = 1.91054, size = 443, normalized size = 3.08 \begin{align*} \frac{2 \,{\left ({\left (a^{3} - 3 \, a b^{2}\right )} c d^{2} +{\left (3 \, a^{2} b - b^{3}\right )} 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{d^{2} \tan \left (f x + e\right )^{2} + 2 \, c d \tan \left (f x + e\right ) + c^{2}}{\tan \left (f x + e\right )^{2} + 1}\right ) +{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + b^{3} c d^{2} - 3 \, a b^{2} d^{3}\right )} \log \left (\frac{1}{\tan \left (f x + e\right )^{2} + 1}\right ) + 2 \,{\left (b^{3} c^{2} d + b^{3} d^{3}\right )} \tan \left (f x + e\right )}{2 \,{\left (c^{2} d^{2} + d^{4}\right )} f} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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Sympy [A] time = 10.0734, size = 1712, normalized size = 11.89 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x*(a + b*tan(e))**3/tan(e), Eq(c, 0) & Eq(d, 0) & Eq(f, 0)), ((a**3*x + 3*a**2*b*log(tan(e + f*

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

*x)**2/(2*f))/c, Eq(d, 0)), (-I*a**3*f*x*tan(e + f*x)/(-2*d*f*tan(e + f*x) + 2*I*d*f) - a**3*f*x/(-2*d*f*tan(e

+ f*x) + 2*I*d*f) - I*a**3/(-2*d*f*tan(e + f*x) + 2*I*d*f) - 3*a**2*b*f*x*tan(e + f*x)/(-2*d*f*tan(e + f*x) +

2*I*d*f) + 3*I*a**2*b*f*x/(-2*d*f*tan(e + f*x) + 2*I*d*f) + 3*a**2*b/(-2*d*f*tan(e + f*x) + 2*I*d*f) - 3*I*a*

b**2*f*x*tan(e + f*x)/(-2*d*f*tan(e + f*x) + 2*I*d*f) - 3*a*b**2*f*x/(-2*d*f*tan(e + f*x) + 2*I*d*f) - 3*a*b**

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

(-2*d*f*tan(e + f*x) + 2*I*d*f) + 3*I*a*b**2/(-2*d*f*tan(e + f*x) + 2*I*d*f) + 3*b**3*f*x*tan(e + f*x)/(-2*d*f

*tan(e + f*x) + 2*I*d*f) - 3*I*b**3*f*x/(-2*d*f*tan(e + f*x) + 2*I*d*f) - I*b**3*log(tan(e + f*x)**2 + 1)*tan(

e + f*x)/(-2*d*f*tan(e + f*x) + 2*I*d*f) - b**3*log(tan(e + f*x)**2 + 1)/(-2*d*f*tan(e + f*x) + 2*I*d*f) - 2*b

**3*tan(e + f*x)**2/(-2*d*f*tan(e + f*x) + 2*I*d*f) - 3*b**3/(-2*d*f*tan(e + f*x) + 2*I*d*f), Eq(c, -I*d)), (-

I*a**3*f*x*tan(e + f*x)/(2*d*f*tan(e + f*x) + 2*I*d*f) + a**3*f*x/(2*d*f*tan(e + f*x) + 2*I*d*f) - I*a**3/(2*d

*f*tan(e + f*x) + 2*I*d*f) + 3*a**2*b*f*x*tan(e + f*x)/(2*d*f*tan(e + f*x) + 2*I*d*f) + 3*I*a**2*b*f*x/(2*d*f*

tan(e + f*x) + 2*I*d*f) - 3*a**2*b/(2*d*f*tan(e + f*x) + 2*I*d*f) - 3*I*a*b**2*f*x*tan(e + f*x)/(2*d*f*tan(e +

f*x) + 2*I*d*f) + 3*a*b**2*f*x/(2*d*f*tan(e + f*x) + 2*I*d*f) + 3*a*b**2*log(tan(e + f*x)**2 + 1)*tan(e + f*x

)/(2*d*f*tan(e + f*x) + 2*I*d*f) + 3*I*a*b**2*log(tan(e + f*x)**2 + 1)/(2*d*f*tan(e + f*x) + 2*I*d*f) + 3*I*a*

b**2/(2*d*f*tan(e + f*x) + 2*I*d*f) - 3*b**3*f*x*tan(e + f*x)/(2*d*f*tan(e + f*x) + 2*I*d*f) - 3*I*b**3*f*x/(2

*d*f*tan(e + f*x) + 2*I*d*f) - I*b**3*log(tan(e + f*x)**2 + 1)*tan(e + f*x)/(2*d*f*tan(e + f*x) + 2*I*d*f) + b

**3*log(tan(e + f*x)**2 + 1)/(2*d*f*tan(e + f*x) + 2*I*d*f) + 2*b**3*tan(e + f*x)**2/(2*d*f*tan(e + f*x) + 2*I

*d*f) + 3*b**3/(2*d*f*tan(e + f*x) + 2*I*d*f), Eq(c, I*d)), (x*(a + b*tan(e))**3/(c + d*tan(e)), Eq(f, 0)), (2

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

a**3*d**3*log(tan(e + f*x)**2 + 1)/(2*c**2*d**2*f + 2*d**4*f) - 6*a**2*b*c*d**2*log(c/d + tan(e + f*x))/(2*c*

*2*d**2*f + 2*d**4*f) + 3*a**2*b*c*d**2*log(tan(e + f*x)**2 + 1)/(2*c**2*d**2*f + 2*d**4*f) + 6*a**2*b*d**3*f*

x/(2*c**2*d**2*f + 2*d**4*f) + 6*a*b**2*c**2*d*log(c/d + tan(e + f*x))/(2*c**2*d**2*f + 2*d**4*f) - 6*a*b**2*c

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

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

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

**4*f) + 2*b**3*d**3*tan(e + f*x)/(2*c**2*d**2*f + 2*d**4*f), True))

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Giac [A] time = 1.85158, size = 239, normalized size = 1.66 \begin{align*} \frac{\frac{2 \, b^{3} \tan \left (f x + e\right )}{d} + \frac{2 \,{\left (a^{3} c - 3 \, a b^{2} c + 3 \, a^{2} b d - b^{3} d\right )}{\left (f x + e\right )}}{c^{2} + d^{2}} + \frac{{\left (3 \, a^{2} b c - b^{3} c - a^{3} d + 3 \, a b^{2} d\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{c^{2} + d^{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 | d \tan \left (f x + e\right ) + c \right |}\right )}{c^{2} d^{2} + d^{4}}}{2 \, f} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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