3.12.91 \(\int (a+b \tan (e+f x))^3 (c+d \tan (e+f x)) \, dx\) [1191]

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

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3609, 3606, 3556} \begin {gather*} \frac {b \left (a^2 d+2 a b c-b^2 d\right ) \tan (e+f x)}{f}-\frac {\left (a^3 d+3 a^2 b c-3 a b^2 d-b^3 c\right ) \log (\cos (e+f x))}{f}+x \left (a^3 c-3 a^2 b d-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} \end {gather*}

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 3556

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3606

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 3609

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]

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*}

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Mathematica [C] Result contains complex when optimal does not.
time = 1.11, size = 130, normalized size = 0.93 \begin {gather*} \frac {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 \left (3 a b c+3 a^2 d-b^2 d\right ) \tan (e+f x)+3 b^2 (b c+3 a d) \tan ^2(e+f x)+2 b^3 d \tan ^3(e+f x)}{6 f} \end {gather*}

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 = 0.09, size = 159, normalized size = 1.14

method	result	size
norman	\(\left (a^{3} c -3 a^{2} b d -3 a \,b^{2} c +b^{3} d \right ) x +\frac {b \left (3 a^{2} d +3 a b c -b^{2} d \right ) \tan \left (f x +e \right )}{f}+\frac {b^{2} \left (3 a d +b c \right ) \left (\tan ^{2}\left (f x +e \right )\right )}{2 f}+\frac {b^{3} d \left (\tan ^{3}\left (f x +e \right )\right )}{3 f}+\frac {\left (a^{3} d +3 a^{2} b c -3 a \,b^{2} d -b^{3} c \right ) \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2 f}\)	\(141\)
derivativedivides	\(\frac {\frac {b^{3} d \left (\tan ^{3}\left (f x +e \right )\right )}{3}+\frac {3 a \,b^{2} d \left (\tan ^{2}\left (f x +e \right )\right )}{2}+\frac {b^{3} c \left (\tan ^{2}\left (f x +e \right )\right )}{2}+3 a^{2} b d \tan \left (f x +e \right )+3 a \,b^{2} c \tan \left (f x +e \right )-b^{3} d \tan \left (f x +e \right )+\frac {\left (a^{3} d +3 a^{2} b c -3 a \,b^{2} d -b^{3} c \right ) \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2}+\left (a^{3} c -3 a^{2} b d -3 a \,b^{2} c +b^{3} d \right ) \arctan \left (\tan \left (f x +e \right )\right )}{f}\)	\(159\)
default	\(\frac {\frac {b^{3} d \left (\tan ^{3}\left (f x +e \right )\right )}{3}+\frac {3 a \,b^{2} d \left (\tan ^{2}\left (f x +e \right )\right )}{2}+\frac {b^{3} c \left (\tan ^{2}\left (f x +e \right )\right )}{2}+3 a^{2} b d \tan \left (f x +e \right )+3 a \,b^{2} c \tan \left (f x +e \right )-b^{3} d \tan \left (f x +e \right )+\frac {\left (a^{3} d +3 a^{2} b c -3 a \,b^{2} d -b^{3} c \right ) \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2}+\left (a^{3} c -3 a^{2} b d -3 a \,b^{2} c +b^{3} d \right ) \arctan \left (\tan \left (f x +e \right )\right )}{f}\)	\(159\)
risch	\(a^{3} c x -3 a^{2} b d x -3 a \,b^{2} c x +b^{3} d x +3 i a^{2} b c x +\frac {2 i a^{3} d e}{f}-3 i a \,b^{2} d x -\frac {2 i b^{3} c e}{f}-i b^{3} c x +i a^{3} d x -\frac {6 i a \,b^{2} d e}{f}+\frac {2 i b \left (9 a^{2} d \,{\mathrm e}^{4 i \left (f x +e \right )}+9 a b c \,{\mathrm e}^{4 i \left (f x +e \right )}-6 b^{2} d \,{\mathrm e}^{4 i \left (f x +e \right )}-9 i a b d \,{\mathrm e}^{4 i \left (f x +e \right )}-3 i b^{2} c \,{\mathrm e}^{4 i \left (f x +e \right )}+18 a^{2} d \,{\mathrm e}^{2 i \left (f x +e \right )}+18 a b c \,{\mathrm e}^{2 i \left (f x +e \right )}-6 b^{2} d \,{\mathrm e}^{2 i \left (f x +e \right )}-9 i a b d \,{\mathrm e}^{2 i \left (f x +e \right )}-3 i b^{2} c \,{\mathrm e}^{2 i \left (f x +e \right )}+9 a^{2} d +9 a b c -4 b^{2} d \right )}{3 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{3}}+\frac {6 i a^{2} b c e}{f}-\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) a^{3} d}{f}-\frac {3 \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) a^{2} b c}{f}+\frac {3 \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) a \,b^{2} d}{f}+\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) b^{3} c}{f}\)	\(383\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.54, size = 153, normalized size = 1.09 \begin {gather*} \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 {gather*}

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 = 1.13, size = 151, normalized size = 1.08 \begin {gather*} \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 {gather*}

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 = 0.15, size = 240, normalized size = 1.71 \begin {gather*} \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 {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 2046 vs. \(2 (140) = 280\).
time = 1.35, size = 2046, normalized size = 14.61 \begin {gather*} \text {Too large to display} \end {gather*}

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

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Mupad [B]
time = 5.27, size = 141, normalized size = 1.01 \begin {gather*} x\,\left (c\,a^3-3\,d\,a^2\,b-3\,c\,a\,b^2+d\,b^3\right )-\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (b^3\,d-3\,a\,b\,\left (a\,d+b\,c\right )\right )}{f}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (\frac {c\,b^3}{2}+\frac {3\,a\,d\,b^2}{2}\right )}{f}+\frac {\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )\,\left (\frac {d\,a^3}{2}+\frac {3\,c\,a^2\,b}{2}-\frac {3\,d\,a\,b^2}{2}-\frac {c\,b^3}{2}\right )}{f}+\frac {b^3\,d\,{\mathrm {tan}\left (e+f\,x\right )}^3}{3\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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