3.13.10 \(\int \frac {(a+b \tan (e+f x))^3}{c+d \tan (e+f x)} \, dx\) [1210]

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

[Out]

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

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Rubi [A]
time = 0.18, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3647, 3707, 3698, 31, 3556} \begin {gather*} -\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 \left (c^2+d^2\right )}+\frac {x \left (a^3 c+3 a^2 b d-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 )} \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)/(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 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 3556

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

Rule 3647

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 3698

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 3707

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]

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 \text {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*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.77, size = 126, normalized size = 0.88 \begin {gather*} \frac {\frac {(a+i b)^3 \log (i-\tan (e+f x))}{i c-d}-\frac {(i a+b)^3 \log (i+\tan (e+f x))}{c-i d}+\frac {2 (-b c+a d)^3 \log (c+d \tan (e+f x))}{d^2 \left (c^2+d^2\right )}+\frac {2 b^2 (a+b \tan (e+f x))}{d}}{2 f} \end {gather*}

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 [A]
time = 0.23, size = 164, normalized size = 1.14

method	result	size
derivativedivides	\(\frac {\frac {b^{3} \tan \left (f x +e \right )}{d}+\frac {\left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \ln \left (c +d \tan \left (f x +e \right )\right )}{d^{2} \left (c^{2}+d^{2}\right )}+\frac {\frac {\left (-a^{3} d +3 a^{2} b c +3 a \,b^{2} d -b^{3} c \right ) \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2}+\left (a^{3} c +3 a^{2} b d -3 a \,b^{2} c -b^{3} d \right ) \arctan \left (\tan \left (f x +e \right )\right )}{c^{2}+d^{2}}}{f}\)	\(164\)
default	\(\frac {\frac {b^{3} \tan \left (f x +e \right )}{d}+\frac {\left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \ln \left (c +d \tan \left (f x +e \right )\right )}{d^{2} \left (c^{2}+d^{2}\right )}+\frac {\frac {\left (-a^{3} d +3 a^{2} b c +3 a \,b^{2} d -b^{3} c \right ) \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2}+\left (a^{3} c +3 a^{2} b d -3 a \,b^{2} c -b^{3} d \right ) \arctan \left (\tan \left (f x +e \right )\right )}{c^{2}+d^{2}}}{f}\)	\(164\)
norman	\(\frac {\left (a^{3} c +3 a^{2} b d -3 a \,b^{2} c -b^{3} d \right ) x}{c^{2}+d^{2}}+\frac {b^{3} \tan \left (f x +e \right )}{d f}+\frac {\left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \ln \left (c +d \tan \left (f x +e \right )\right )}{\left (c^{2}+d^{2}\right ) d^{2} f}-\frac {\left (a^{3} d -3 a^{2} b c -3 a \,b^{2} d +b^{3} c \right ) \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2 f \left (c^{2}+d^{2}\right )}\)	\(168\)
risch	\(\frac {6 i a^{2} b c e}{\left (c^{2}+d^{2}\right ) f}+\frac {3 i x \,a^{2} b}{i d -c}-\frac {a^{3} x}{i d -c}+\frac {3 x a \,b^{2}}{i d -c}-\frac {2 i b^{3} c x}{d^{2}}+\frac {6 i b^{2} a x}{d}-\frac {i x \,b^{3}}{i d -c}-\frac {2 i b^{3} c e}{d^{2} f}-\frac {2 i d \,a^{3} e}{\left (c^{2}+d^{2}\right ) f}+\frac {6 i b^{2} a e}{d f}+\frac {2 i b^{3} c^{3} e}{\left (c^{2}+d^{2}\right ) d^{2} f}-\frac {6 i a \,b^{2} c^{2} e}{\left (c^{2}+d^{2}\right ) d f}+\frac {6 i a^{2} b c x}{c^{2}+d^{2}}-\frac {6 i a \,b^{2} c^{2} x}{\left (c^{2}+d^{2}\right ) d}-\frac {2 i d \,a^{3} x}{c^{2}+d^{2}}+\frac {2 i b^{3} c^{3} x}{\left (c^{2}+d^{2}\right ) d^{2}}+\frac {2 i b^{3}}{f d \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}-\frac {3 b^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) a}{d f}+\frac {b^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) c}{d^{2} f}+\frac {d \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {i d +c}{i d -c}\right ) a^{3}}{\left (c^{2}+d^{2}\right ) f}-\frac {3 \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {i d +c}{i d -c}\right ) a^{2} b c}{\left (c^{2}+d^{2}\right ) f}+\frac {3 \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {i d +c}{i d -c}\right ) a \,b^{2} c^{2}}{\left (c^{2}+d^{2}\right ) d f}-\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {i d +c}{i d -c}\right ) b^{3} c^{3}}{\left (c^{2}+d^{2}\right ) d^{2} f}\)	\(563\)

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*(b^3/d*tan(f*x+e)+1/d^2*(a^3*d^3-3*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3*c^3)/(c^2+d^2)*ln(c+d*tan(f*x+e))+1/(c^2+
d^2)*(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 = 0.57, size = 178, normalized size = 1.24 \begin {gather*} \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 {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]

1/2*(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 = 1.41, size = 211, normalized size = 1.47 \begin {gather*} \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 {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]

1/2*(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 [C] Result contains complex when optimal does not.
time = 0.87, size = 1712, normalized size = 11.89 \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)

[Out]

Piecewise((zoo*x*(a + b*tan(e))**3/tan(e), Eq(c, 0) & Eq(d, 0) & Eq(f, 0)), (I*a**3*f*x*tan(e + f*x)/(2*d*f*ta
n(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(ta
n(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*ta
n(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)), ((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)), (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 = 0.74, size = 177, normalized size = 1.23 \begin {gather*} \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 {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]

1/2*(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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Mupad [B]
time = 5.49, size = 177, normalized size = 1.23 \begin {gather*} \frac {b^3\,\mathrm {tan}\left (e+f\,x\right )}{d\,f}+\frac {\ln \left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )\,\left (a^3\,d^3-3\,a^2\,b\,c\,d^2+3\,a\,b^2\,c^2\,d-b^3\,c^3\right )}{f\,\left (c^2\,d^2+d^4\right )}+\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )-\mathrm {i}\right )\,\left (-a^3\,1{}\mathrm {i}+3\,a^2\,b+a\,b^2\,3{}\mathrm {i}-b^3\right )}{2\,f\,\left (c+d\,1{}\mathrm {i}\right )}+\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,\left (-a^3+a^2\,b\,3{}\mathrm {i}+3\,a\,b^2-b^3\,1{}\mathrm {i}\right )}{2\,f\,\left (d+c\,1{}\mathrm {i}\right )} \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]

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

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