3.1199 \(\int (a+b \tan (e+f x)) (c+d \tan (e+f x))^2 \, dx\)

Optimal. Leaf size=89 \[ -\frac {\left (2 a c d+b \left (c^2-d^2\right )\right ) \log (\cos (e+f x))}{f}-x \left (2 b c d-a \left (c^2-d^2\right )\right )+\frac {d (a d+b c) \tan (e+f x)}{f}+\frac {b (c+d \tan (e+f x))^2}{2 f} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3528, 3525, 3475} \[ -\frac {\left (2 a c d+b \left (c^2-d^2\right )\right ) \log (\cos (e+f x))}{f}-x \left (2 b c d-a \left (c^2-d^2\right )\right )+\frac {d (a d+b c) \tan (e+f x)}{f}+\frac {b (c+d \tan (e+f x))^2}{2 f} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[e + f*x])*(c + d*Tan[e + f*x])^2,x]

[Out]

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

Rule 3475

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

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 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]

Rubi steps

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

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Mathematica [C]  time = 0.47, size = 96, normalized size = 1.08 \[ \frac {2 d (a d+2 b c) \tan (e+f x)+(b+i a) (c-i d)^2 \log (\tan (e+f x)+i)+(b-i a) (c+i d)^2 \log (-\tan (e+f x)+i)+b d^2 \tan ^2(e+f x)}{2 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[e + f*x])*(c + d*Tan[e + f*x])^2,x]

[Out]

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

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fricas [A]  time = 0.47, size = 91, normalized size = 1.02 \[ \frac {b d^{2} \tan \left (f x + e\right )^{2} + 2 \, {\left (a c^{2} - 2 \, b c d - a d^{2}\right )} f x - {\left (b c^{2} + 2 \, a c d - b d^{2}\right )} \log \left (\frac {1}{\tan \left (f x + e\right )^{2} + 1}\right ) + 2 \, {\left (2 \, b c d + a d^{2}\right )} \tan \left (f x + e\right )}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))*(c+d*tan(f*x+e))^2,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 1.38, size = 968, normalized size = 10.88 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))*(c+d*tan(f*x+e))^2,x, algorithm="giac")

[Out]

1/2*(2*a*c^2*f*x*tan(f*x)^2*tan(e)^2 - 4*b*c*d*f*x*tan(f*x)^2*tan(e)^2 - 2*a*d^2*f*x*tan(f*x)^2*tan(e)^2 - b*c
^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)^2*tan(e)^2 - 2*a*c*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 + b*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 - 4*a*c^2*f*x*tan(f*x)*tan(e) + 8*b*c*d*f*x*tan(f*x)*tan(e) + 4*a*d^2*f*x*tan(f*x)*tan(e) + b*
d^2*tan(f*x)^2*tan(e)^2 + 2*b*c^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) + 4*a*c*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) - 2*
b*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) - 4*b*c*d*tan(f*x)^2*tan(e) - 2*a*d^2*tan(f*x)^2*tan(e) - 4*b*c*d*tan(f*x
)*tan(e)^2 - 2*a*d^2*tan(f*x)*tan(e)^2 + 2*a*c^2*f*x - 4*b*c*d*f*x - 2*a*d^2*f*x + b*d^2*tan(f*x)^2 + b*d^2*ta
n(e)^2 - b*c^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)) - 2*a*c*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)) + b*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)) + 4*b*c*d*tan(f*x) + 2*a*d^2*tan
(f*x) + 4*b*c*d*tan(e) + 2*a*d^2*tan(e) + b*d^2)/(f*tan(f*x)^2*tan(e)^2 - 2*f*tan(f*x)*tan(e) + f)

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maple [A]  time = 0.02, size = 151, normalized size = 1.70 \[ \frac {b \,d^{2} \left (\tan ^{2}\left (f x +e \right )\right )}{2 f}+\frac {a \,d^{2} \tan \left (f x +e \right )}{f}+\frac {2 b c d \tan \left (f x +e \right )}{f}+\frac {\ln \left (1+\tan ^{2}\left (f x +e \right )\right ) a c d}{f}+\frac {\ln \left (1+\tan ^{2}\left (f x +e \right )\right ) c^{2} b}{2 f}-\frac {\ln \left (1+\tan ^{2}\left (f x +e \right )\right ) b \,d^{2}}{2 f}+\frac {\arctan \left (\tan \left (f x +e \right )\right ) a \,c^{2}}{f}-\frac {\arctan \left (\tan \left (f x +e \right )\right ) a \,d^{2}}{f}-\frac {2 \arctan \left (\tan \left (f x +e \right )\right ) b c d}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(f*x+e))*(c+d*tan(f*x+e))^2,x)

[Out]

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

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maxima [A]  time = 0.86, size = 91, normalized size = 1.02 \[ \frac {b d^{2} \tan \left (f x + e\right )^{2} + 2 \, {\left (a c^{2} - 2 \, b c d - a d^{2}\right )} {\left (f x + e\right )} + {\left (b c^{2} + 2 \, a c d - b d^{2}\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right ) + 2 \, {\left (2 \, b c d + a d^{2}\right )} \tan \left (f x + e\right )}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))*(c+d*tan(f*x+e))^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.18, size = 91, normalized size = 1.02 \[ \frac {\mathrm {tan}\left (e+f\,x\right )\,\left (a\,d^2+2\,b\,c\,d\right )}{f}-x\,\left (-a\,c^2+2\,b\,c\,d+a\,d^2\right )+\frac {\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )\,\left (\frac {b\,c^2}{2}+a\,c\,d-\frac {b\,d^2}{2}\right )}{f}+\frac {b\,d^2\,{\mathrm {tan}\left (e+f\,x\right )}^2}{2\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tan(e + f*x))*(c + d*tan(e + f*x))^2,x)

[Out]

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

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sympy [A]  time = 0.30, size = 143, normalized size = 1.61 \[ \begin {cases} a c^{2} x + \frac {a c d \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{f} - a d^{2} x + \frac {a d^{2} \tan {\left (e + f x \right )}}{f} + \frac {b c^{2} \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} - 2 b c d x + \frac {2 b c d \tan {\left (e + f x \right )}}{f} - \frac {b d^{2} \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} + \frac {b d^{2} \tan ^{2}{\left (e + f x \right )}}{2 f} & \text {for}\: f \neq 0 \\x \left (a + b \tan {\relax (e )}\right ) \left (c + d \tan {\relax (e )}\right )^{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))*(c+d*tan(f*x+e))**2,x)

[Out]

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

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