3.1191 \(\int (a+b \tan (e+f x))^3 (c+d \tan (e+f x)) \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.162, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3528, 3525, 3475} \[ \frac{b \left (a^2 d+2 a b c-b^2 d\right ) \tan (e+f x)}{f}-\frac{\left (3 a^2 b c+a^3 d-3 a b^2 d-b^3 c\right ) \log (\cos (e+f x))}{f}+x \left (-3 a^2 b d+a^3 c-3 a b^2 c+b^3 d\right )+\frac{(a d+b c) (a+b \tan (e+f x))^2}{2 f}+\frac{d (a+b \tan (e+f x))^3}{3 f} \]

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

Rule 3528

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(d
*(a + b*Tan[e + f*x])^m)/(f*m), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]

Rule 3525

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(a*c - b
*d)*x, x] + (Dist[b*c + a*d, Int[Tan[e + f*x], x], x] + Simp[(b*d*Tan[e + f*x])/f, x]) /; FreeQ[{a, b, c, d, e
, f}, x] && NeQ[b*c - a*d, 0] && NeQ[b*c + a*d, 0]

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

Mathematica [C]  time = 1.07232, size = 130, normalized size = 0.93 \[ \frac{6 b \left (3 a^2 d+3 a b c-b^2 d\right ) \tan (e+f x)+3 b^2 (3 a d+b c) \tan ^2(e+f x)+3 (a-i b)^3 (d+i c) \log (\tan (e+f x)+i)+3 (a+i b)^3 (d-i c) \log (-\tan (e+f x)+i)+2 b^3 d \tan ^3(e+f x)}{6 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(a + I*b)^3*((-I)*c + d)*Log[I - Tan[e + f*x]] + 3*(a - I*b)^3*(I*c + d)*Log[I + Tan[e + f*x]] + 6*b*(3*a*b
*c + 3*a^2*d - b^2*d)*Tan[e + f*x] + 3*b^2*(b*c + 3*a*d)*Tan[e + f*x]^2 + 2*b^3*d*Tan[e + f*x]^3)/(6*f)

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Maple [A]  time = 0.01, size = 247, normalized size = 1.8 \begin{align*}{\frac{{b}^{3}d \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{3\,f}}+{\frac{3\, \left ( \tan \left ( fx+e \right ) \right ) ^{2}a{b}^{2}d}{2\,f}}+{\frac{ \left ( \tan \left ( fx+e \right ) \right ) ^{2}{b}^{3}c}{2\,f}}+3\,{\frac{{a}^{2}bd\tan \left ( fx+e \right ) }{f}}+3\,{\frac{a{b}^{2}c\tan \left ( fx+e \right ) }{f}}-{\frac{{b}^{3}d\tan \left ( fx+e \right ) }{f}}+{\frac{{a}^{3}\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) d}{2\,f}}+{\frac{3\,\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ){a}^{2}bc}{2\,f}}-{\frac{3\,\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) a{b}^{2}d}{2\,f}}-{\frac{\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ){b}^{3}c}{2\,f}}+{\frac{{a}^{3}\arctan \left ( \tan \left ( fx+e \right ) \right ) c}{f}}-3\,{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ){a}^{2}bd}{f}}-3\,{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ) a{b}^{2}c}{f}}+{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ){b}^{3}d}{f}} \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]

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

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Maxima [A]  time = 1.75803, size = 200, normalized size = 1.43 \begin{align*} \frac{2 \, b^{3} d \tan \left (f x + e\right )^{3} + 3 \,{\left (b^{3} c + 3 \, a b^{2} d\right )} \tan \left (f x + e\right )^{2} + 6 \,{\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 )} + 3 \,{\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 ) + 6 \,{\left (3 \, a b^{2} c +{\left (3 \, a^{2} b - b^{3}\right )} d\right )} \tan \left (f x + e\right )}{6 \, 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]

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

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Fricas [A]  time = 1.34181, size = 324, normalized size = 2.31 \begin{align*} \frac{2 \, b^{3} d \tan \left (f x + e\right )^{3} + 6 \,{\left ({\left (a^{3} - 3 \, a b^{2}\right )} c -{\left (3 \, a^{2} b - b^{3}\right )} d\right )} f x + 3 \,{\left (b^{3} c + 3 \, a b^{2} d\right )} \tan \left (f x + e\right )^{2} - 3 \,{\left ({\left (3 \, a^{2} b - b^{3}\right )} c +{\left (a^{3} - 3 \, a b^{2}\right )} d\right )} \log \left (\frac{1}{\tan \left (f x + e\right )^{2} + 1}\right ) + 6 \,{\left (3 \, a b^{2} c +{\left (3 \, a^{2} b - b^{3}\right )} d\right )} \tan \left (f x + e\right )}{6 \, 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]

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

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Sympy [A]  time = 0.631693, size = 240, normalized size = 1.71 \begin{align*} \begin{cases} a^{3} c x + \frac{a^{3} d \log{\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} + \frac{3 a^{2} b c \log{\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} - 3 a^{2} b d x + \frac{3 a^{2} b d \tan{\left (e + f x \right )}}{f} - 3 a b^{2} c x + \frac{3 a b^{2} c \tan{\left (e + f x \right )}}{f} - \frac{3 a b^{2} d \log{\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} + \frac{3 a b^{2} d \tan ^{2}{\left (e + f x \right )}}{2 f} - \frac{b^{3} c \log{\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} + \frac{b^{3} c \tan ^{2}{\left (e + f x \right )}}{2 f} + b^{3} d x + \frac{b^{3} d \tan ^{3}{\left (e + f x \right )}}{3 f} - \frac{b^{3} d \tan{\left (e + f x \right )}}{f} & \text{for}\: f \neq 0 \\x \left (a + b \tan{\left (e \right )}\right )^{3} \left (c + d \tan{\left (e \right )}\right ) & \text{otherwise} \end{cases} \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((a**3*c*x + a**3*d*log(tan(e + f*x)**2 + 1)/(2*f) + 3*a**2*b*c*log(tan(e + f*x)**2 + 1)/(2*f) - 3*a*
*2*b*d*x + 3*a**2*b*d*tan(e + f*x)/f - 3*a*b**2*c*x + 3*a*b**2*c*tan(e + f*x)/f - 3*a*b**2*d*log(tan(e + f*x)*
*2 + 1)/(2*f) + 3*a*b**2*d*tan(e + f*x)**2/(2*f) - b**3*c*log(tan(e + f*x)**2 + 1)/(2*f) + b**3*c*tan(e + f*x)
**2/(2*f) + b**3*d*x + b**3*d*tan(e + f*x)**3/(3*f) - b**3*d*tan(e + f*x)/f, Ne(f, 0)), (x*(a + b*tan(e))**3*(
c + d*tan(e)), True))

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Giac [B]  time = 3.467, size = 2762, normalized size = 19.73 \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, algorithm="giac")

[Out]

1/6*(6*a^3*c*f*x*tan(f*x)^3*tan(e)^3 - 18*a*b^2*c*f*x*tan(f*x)^3*tan(e)^3 - 18*a^2*b*d*f*x*tan(f*x)^3*tan(e)^3
+ 6*b^3*d*f*x*tan(f*x)^3*tan(e)^3 - 9*a^2*b*c*log(4*(tan(e)^2 + 1)/(tan(f*x)^4*tan(e)^2 - 2*tan(f*x)^3*tan(e)
+ tan(f*x)^2*tan(e)^2 + tan(f*x)^2 - 2*tan(f*x)*tan(e) + 1))*tan(f*x)^3*tan(e)^3 + 3*b^3*c*log(4*(tan(e)^2 +
1)/(tan(f*x)^4*tan(e)^2 - 2*tan(f*x)^3*tan(e) + tan(f*x)^2*tan(e)^2 + tan(f*x)^2 - 2*tan(f*x)*tan(e) + 1))*tan
(f*x)^3*tan(e)^3 - 3*a^3*d*log(4*(tan(e)^2 + 1)/(tan(f*x)^4*tan(e)^2 - 2*tan(f*x)^3*tan(e) + tan(f*x)^2*tan(e)
^2 + tan(f*x)^2 - 2*tan(f*x)*tan(e) + 1))*tan(f*x)^3*tan(e)^3 + 9*a*b^2*d*log(4*(tan(e)^2 + 1)/(tan(f*x)^4*tan
(e)^2 - 2*tan(f*x)^3*tan(e) + tan(f*x)^2*tan(e)^2 + tan(f*x)^2 - 2*tan(f*x)*tan(e) + 1))*tan(f*x)^3*tan(e)^3 -
18*a^3*c*f*x*tan(f*x)^2*tan(e)^2 + 54*a*b^2*c*f*x*tan(f*x)^2*tan(e)^2 + 54*a^2*b*d*f*x*tan(f*x)^2*tan(e)^2 -
18*b^3*d*f*x*tan(f*x)^2*tan(e)^2 + 3*b^3*c*tan(f*x)^3*tan(e)^3 + 9*a*b^2*d*tan(f*x)^3*tan(e)^3 + 27*a^2*b*c*lo
g(4*(tan(e)^2 + 1)/(tan(f*x)^4*tan(e)^2 - 2*tan(f*x)^3*tan(e) + tan(f*x)^2*tan(e)^2 + tan(f*x)^2 - 2*tan(f*x)*
tan(e) + 1))*tan(f*x)^2*tan(e)^2 - 9*b^3*c*log(4*(tan(e)^2 + 1)/(tan(f*x)^4*tan(e)^2 - 2*tan(f*x)^3*tan(e) + t
an(f*x)^2*tan(e)^2 + tan(f*x)^2 - 2*tan(f*x)*tan(e) + 1))*tan(f*x)^2*tan(e)^2 + 9*a^3*d*log(4*(tan(e)^2 + 1)/(
tan(f*x)^4*tan(e)^2 - 2*tan(f*x)^3*tan(e) + tan(f*x)^2*tan(e)^2 + tan(f*x)^2 - 2*tan(f*x)*tan(e) + 1))*tan(f*x
)^2*tan(e)^2 - 27*a*b^2*d*log(4*(tan(e)^2 + 1)/(tan(f*x)^4*tan(e)^2 - 2*tan(f*x)^3*tan(e) + tan(f*x)^2*tan(e)^
2 + tan(f*x)^2 - 2*tan(f*x)*tan(e) + 1))*tan(f*x)^2*tan(e)^2 - 18*a*b^2*c*tan(f*x)^3*tan(e)^2 - 18*a^2*b*d*tan
(f*x)^3*tan(e)^2 + 6*b^3*d*tan(f*x)^3*tan(e)^2 - 18*a*b^2*c*tan(f*x)^2*tan(e)^3 - 18*a^2*b*d*tan(f*x)^2*tan(e)
^3 + 6*b^3*d*tan(f*x)^2*tan(e)^3 + 18*a^3*c*f*x*tan(f*x)*tan(e) - 54*a*b^2*c*f*x*tan(f*x)*tan(e) - 54*a^2*b*d*
f*x*tan(f*x)*tan(e) + 18*b^3*d*f*x*tan(f*x)*tan(e) + 3*b^3*c*tan(f*x)^3*tan(e) + 9*a*b^2*d*tan(f*x)^3*tan(e) -
3*b^3*c*tan(f*x)^2*tan(e)^2 - 9*a*b^2*d*tan(f*x)^2*tan(e)^2 + 3*b^3*c*tan(f*x)*tan(e)^3 + 9*a*b^2*d*tan(f*x)*
tan(e)^3 - 2*b^3*d*tan(f*x)^3 - 27*a^2*b*c*log(4*(tan(e)^2 + 1)/(tan(f*x)^4*tan(e)^2 - 2*tan(f*x)^3*tan(e) + t
an(f*x)^2*tan(e)^2 + tan(f*x)^2 - 2*tan(f*x)*tan(e) + 1))*tan(f*x)*tan(e) + 9*b^3*c*log(4*(tan(e)^2 + 1)/(tan(
f*x)^4*tan(e)^2 - 2*tan(f*x)^3*tan(e) + tan(f*x)^2*tan(e)^2 + tan(f*x)^2 - 2*tan(f*x)*tan(e) + 1))*tan(f*x)*ta
n(e) - 9*a^3*d*log(4*(tan(e)^2 + 1)/(tan(f*x)^4*tan(e)^2 - 2*tan(f*x)^3*tan(e) + tan(f*x)^2*tan(e)^2 + tan(f*x
)^2 - 2*tan(f*x)*tan(e) + 1))*tan(f*x)*tan(e) + 27*a*b^2*d*log(4*(tan(e)^2 + 1)/(tan(f*x)^4*tan(e)^2 - 2*tan(f
*x)^3*tan(e) + tan(f*x)^2*tan(e)^2 + tan(f*x)^2 - 2*tan(f*x)*tan(e) + 1))*tan(f*x)*tan(e) + 36*a*b^2*c*tan(f*x
)^2*tan(e) + 36*a^2*b*d*tan(f*x)^2*tan(e) - 18*b^3*d*tan(f*x)^2*tan(e) + 36*a*b^2*c*tan(f*x)*tan(e)^2 + 36*a^2
*b*d*tan(f*x)*tan(e)^2 - 18*b^3*d*tan(f*x)*tan(e)^2 - 2*b^3*d*tan(e)^3 - 6*a^3*c*f*x + 18*a*b^2*c*f*x + 18*a^2
*b*d*f*x - 6*b^3*d*f*x - 3*b^3*c*tan(f*x)^2 - 9*a*b^2*d*tan(f*x)^2 + 3*b^3*c*tan(f*x)*tan(e) + 9*a*b^2*d*tan(f
*x)*tan(e) - 3*b^3*c*tan(e)^2 - 9*a*b^2*d*tan(e)^2 + 9*a^2*b*c*log(4*(tan(e)^2 + 1)/(tan(f*x)^4*tan(e)^2 - 2*t
an(f*x)^3*tan(e) + tan(f*x)^2*tan(e)^2 + tan(f*x)^2 - 2*tan(f*x)*tan(e) + 1)) - 3*b^3*c*log(4*(tan(e)^2 + 1)/(
tan(f*x)^4*tan(e)^2 - 2*tan(f*x)^3*tan(e) + tan(f*x)^2*tan(e)^2 + tan(f*x)^2 - 2*tan(f*x)*tan(e) + 1)) + 3*a^3
*d*log(4*(tan(e)^2 + 1)/(tan(f*x)^4*tan(e)^2 - 2*tan(f*x)^3*tan(e) + tan(f*x)^2*tan(e)^2 + tan(f*x)^2 - 2*tan(
f*x)*tan(e) + 1)) - 9*a*b^2*d*log(4*(tan(e)^2 + 1)/(tan(f*x)^4*tan(e)^2 - 2*tan(f*x)^3*tan(e) + tan(f*x)^2*tan
(e)^2 + tan(f*x)^2 - 2*tan(f*x)*tan(e) + 1)) - 18*a*b^2*c*tan(f*x) - 18*a^2*b*d*tan(f*x) + 6*b^3*d*tan(f*x) -
18*a*b^2*c*tan(e) - 18*a^2*b*d*tan(e) + 6*b^3*d*tan(e) - 3*b^3*c - 9*a*b^2*d)/(f*tan(f*x)^3*tan(e)^3 - 3*f*tan
(f*x)^2*tan(e)^2 + 3*f*tan(f*x)*tan(e) - f)