3.13.5 \(\int (a+b \tan (e+f x)) (c+d \tan (e+f x))^3 \, dx\) [1205]

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

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 144, 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 {d \left (2 a c d+b \left (c^2-d^2\right )\right ) \tan (e+f x)}{f}-x \left (b d \left (3 c^2-d^2\right )-a \left (c^3-3 c d^2\right )\right )-\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}+\frac {(a d+b c) (c+d \tan (e+f x))^2}{2 f}+\frac {b (c+d \tan (e+f x))^3}{3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [C] Result contains complex when optimal does not.
time = 1.03, size = 130, normalized size = 0.90 \begin {gather*} \frac {3 (-i a+b) (c+i d)^3 \log (i-\tan (e+f x))+3 (i a+b) (c-i d)^3 \log (i+\tan (e+f x))+6 d \left (3 b c^2+3 a c d-b d^2\right ) \tan (e+f x)+3 d^2 (3 b c+a d) \tan ^2(e+f x)+2 b d^3 \tan ^3(e+f x)}{6 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.09, size = 159, normalized size = 1.10

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.71, size = 148, normalized size = 1.03 \begin {gather*} \frac {2 \, b d^{3} \tan \left (f x + e\right )^{3} + 3 \, {\left (3 \, b c d^{2} + a d^{3}\right )} \tan \left (f x + e\right )^{2} + 6 \, {\left (a c^{3} - 3 \, b c^{2} d - 3 \, a c d^{2} + b d^{3}\right )} {\left (f x + e\right )} + 3 \, {\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 ) + 6 \, {\left (3 \, b c^{2} d + 3 \, a c d^{2} - b d^{3}\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))*(c+d*tan(f*x+e))^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 1.19, size = 146, normalized size = 1.01 \begin {gather*} \frac {2 \, b d^{3} \tan \left (f x + e\right )^{3} + 6 \, {\left (a c^{3} - 3 \, b c^{2} d - 3 \, a c d^{2} + b d^{3}\right )} f x + 3 \, {\left (3 \, b c d^{2} + a d^{3}\right )} \tan \left (f x + e\right )^{2} - 3 \, {\left (b c^{3} + 3 \, a c^{2} d - 3 \, b c d^{2} - a d^{3}\right )} \log \left (\frac {1}{\tan \left (f x + e\right )^{2} + 1}\right ) + 6 \, {\left (3 \, b c^{2} d + 3 \, a c d^{2} - b d^{3}\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))*(c+d*tan(f*x+e))^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.15, size = 240, normalized size = 1.67 \begin {gather*} \begin {cases} a c^{3} x + \frac {3 a c^{2} d \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} - 3 a c d^{2} x + \frac {3 a c d^{2} \tan {\left (e + f x \right )}}{f} - \frac {a d^{3} \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} + \frac {a d^{3} \tan ^{2}{\left (e + f x \right )}}{2 f} + \frac {b c^{3} \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} - 3 b c^{2} d x + \frac {3 b c^{2} d \tan {\left (e + f x \right )}}{f} - \frac {3 b c d^{2} \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} + \frac {3 b c d^{2} \tan ^{2}{\left (e + f x \right )}}{2 f} + b d^{3} x + \frac {b d^{3} \tan ^{3}{\left (e + f x \right )}}{3 f} - \frac {b d^{3} \tan {\left (e + f x \right )}}{f} & \text {for}\: f \neq 0 \\x \left (a + b \tan {\left (e \right )}\right ) \left (c + d \tan {\left (e \right )}\right )^{3} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*c**3*x + 3*a*c**2*d*log(tan(e + f*x)**2 + 1)/(2*f) - 3*a*c*d**2*x + 3*a*c*d**2*tan(e + f*x)/f - a
*d**3*log(tan(e + f*x)**2 + 1)/(2*f) + a*d**3*tan(e + f*x)**2/(2*f) + b*c**3*log(tan(e + f*x)**2 + 1)/(2*f) -
3*b*c**2*d*x + 3*b*c**2*d*tan(e + f*x)/f - 3*b*c*d**2*log(tan(e + f*x)**2 + 1)/(2*f) + 3*b*c*d**2*tan(e + f*x)
**2/(2*f) + b*d**3*x + b*d**3*tan(e + f*x)**3/(3*f) - b*d**3*tan(e + f*x)/f, Ne(f, 0)), (x*(a + b*tan(e))*(c +
d*tan(e))**3, True))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 2046 vs. \(2 (144) = 288\).
time = 1.34, size = 2046, normalized size = 14.21 \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))*(c+d*tan(f*x+e))^3,x, algorithm="giac")

[Out]

1/6*(6*a*c^3*f*x*tan(f*x)^3*tan(e)^3 - 18*b*c^2*d*f*x*tan(f*x)^3*tan(e)^3 - 18*a*c*d^2*f*x*tan(f*x)^3*tan(e)^3
+ 6*b*d^3*f*x*tan(f*x)^3*tan(e)^3 - 3*b*c^3*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*c^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 + 9*b*c*d^2*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*d^3*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*c^3*f*x*tan(f*x)^2*tan(e)^2 + 54*b*c^2*d*f*x*tan(f*x)^2*tan(e)^2 + 54*a*c*d^2*f*x*tan(f*x)^2*tan(e)^2 -
18*b*d^3*f*x*tan(f*x)^2*tan(e)^2 + 9*b*c*d^2*tan(f*x)^3*tan(e)^3 + 3*a*d^3*tan(f*x)^3*tan(e)^3 + 9*b*c^3*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*c^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 - 27*b*c*d^2*log(4*(tan(f*x)^4*t
an(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*d^3*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*b*c^2*d*tan(f*x)^3*tan(e)^2 - 18*a*c*d^2*tan
(f*x)^3*tan(e)^2 + 6*b*d^3*tan(f*x)^3*tan(e)^2 - 18*b*c^2*d*tan(f*x)^2*tan(e)^3 - 18*a*c*d^2*tan(f*x)^2*tan(e)
^3 + 6*b*d^3*tan(f*x)^2*tan(e)^3 + 18*a*c^3*f*x*tan(f*x)*tan(e) - 54*b*c^2*d*f*x*tan(f*x)*tan(e) - 54*a*c*d^2*
f*x*tan(f*x)*tan(e) + 18*b*d^3*f*x*tan(f*x)*tan(e) + 9*b*c*d^2*tan(f*x)^3*tan(e) + 3*a*d^3*tan(f*x)^3*tan(e) -
9*b*c*d^2*tan(f*x)^2*tan(e)^2 - 3*a*d^3*tan(f*x)^2*tan(e)^2 + 9*b*c*d^2*tan(f*x)*tan(e)^3 + 3*a*d^3*tan(f*x)*
tan(e)^3 - 2*b*d^3*tan(f*x)^3 - 9*b*c^3*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*c^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)*ta
n(e) + 27*b*c*d^2*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*a*d^3*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*b*c^2*d*tan(f*x
)^2*tan(e) + 36*a*c*d^2*tan(f*x)^2*tan(e) - 18*b*d^3*tan(f*x)^2*tan(e) + 36*b*c^2*d*tan(f*x)*tan(e)^2 + 36*a*c
*d^2*tan(f*x)*tan(e)^2 - 18*b*d^3*tan(f*x)*tan(e)^2 - 2*b*d^3*tan(e)^3 - 6*a*c^3*f*x + 18*b*c^2*d*f*x + 18*a*c
*d^2*f*x - 6*b*d^3*f*x - 9*b*c*d^2*tan(f*x)^2 - 3*a*d^3*tan(f*x)^2 + 9*b*c*d^2*tan(f*x)*tan(e) + 3*a*d^3*tan(f
*x)*tan(e) - 9*b*c*d^2*tan(e)^2 - 3*a*d^3*tan(e)^2 + 3*b*c^3*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*c^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)) - 9*b*c
*d^2*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*d^3*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*b*c^2*d*tan(f*x) - 18*a*c*d^2*tan(f*x) + 6*b*d^3*tan(f*x) -
18*b*c^2*d*tan(e) - 18*a*c*d^2*tan(e) + 6*b*d^3*tan(e) - 9*b*c*d^2 - 3*a*d^3)/(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 = 5.23, size = 142, normalized size = 0.99 \begin {gather*} x\,\left (a\,c^3-3\,b\,c^2\,d-3\,a\,c\,d^2+b\,d^3\right )-\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (b\,d^3-3\,c\,d\,\left (a\,d+b\,c\right )\right )}{f}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (\frac {a\,d^3}{2}+\frac {3\,b\,c\,d^2}{2}\right )}{f}-\frac {\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )\,\left (-\frac {b\,c^3}{2}-\frac {3\,a\,c^2\,d}{2}+\frac {3\,b\,c\,d^2}{2}+\frac {a\,d^3}{2}\right )}{f}+\frac {b\,d^3\,{\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))*(c + d*tan(e + f*x))^3,x)

[Out]

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

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