3.53 \(\int (c+d \tan (e+f x)) (A+B \tan (e+f x)+C \tan ^2(e+f x)) \, dx\)

Optimal. Leaf size=73 \[ -\frac {(d (A-C)+B c) \log (\cos (e+f x))}{f}+x (A c-B d-c C)+\frac {B d \tan (e+f x)}{f}+\frac {C (c+d \tan (e+f x))^2}{2 d f} \]

[Out]

(A*c-B*d-C*c)*x-(B*c+(A-C)*d)*ln(cos(f*x+e))/f+B*d*tan(f*x+e)/f+1/2*C*(c+d*tan(f*x+e))^2/d/f

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Rubi [A]  time = 0.06, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {3630, 3525, 3475} \[ -\frac {(d (A-C)+B c) \log (\cos (e+f x))}{f}+x (A c-B d-c C)+\frac {B d \tan (e+f x)}{f}+\frac {C (c+d \tan (e+f x))^2}{2 d f} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*Tan[e + f*x])*(A + B*Tan[e + f*x] + C*Tan[e + f*x]^2),x]

[Out]

(A*c - c*C - B*d)*x - ((B*c + (A - C)*d)*Log[Cos[e + f*x]])/f + (B*d*Tan[e + f*x])/f + (C*(c + d*Tan[e + f*x])
^2)/(2*d*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 3630

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(C*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Int[(a + b*Tan[e + f*x])
^m*Simp[A - C + B*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0]
&&  !LeQ[m, -1]

Rubi steps

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

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Mathematica [A]  time = 0.47, size = 76, normalized size = 1.04 \[ \frac {-2 (d (A-C)+B c) \log (\cos (e+f x))+2 A c f x-2 (B d+c C) \tan ^{-1}(\tan (e+f x))+2 (B d+c C) \tan (e+f x)+C d \tan ^2(e+f x)}{2 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*Tan[e + f*x])*(A + B*Tan[e + f*x] + C*Tan[e + f*x]^2),x]

[Out]

(2*A*c*f*x - 2*(c*C + B*d)*ArcTan[Tan[e + f*x]] - 2*(B*c + (A - C)*d)*Log[Cos[e + f*x]] + 2*(c*C + B*d)*Tan[e
+ f*x] + C*d*Tan[e + f*x]^2)/(2*f)

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fricas [A]  time = 1.33, size = 74, normalized size = 1.01 \[ \frac {C d \tan \left (f x + e\right )^{2} + 2 \, {\left ({\left (A - C\right )} c - B d\right )} f x - {\left (B c + {\left (A - C\right )} d\right )} \log \left (\frac {1}{\tan \left (f x + e\right )^{2} + 1}\right ) + 2 \, {\left (C c + B d\right )} \tan \left (f x + e\right )}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))*(A+B*tan(f*x+e)+C*tan(f*x+e)^2),x, algorithm="fricas")

[Out]

1/2*(C*d*tan(f*x + e)^2 + 2*((A - C)*c - B*d)*f*x - (B*c + (A - C)*d)*log(1/(tan(f*x + e)^2 + 1)) + 2*(C*c + B
*d)*tan(f*x + e))/f

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giac [B]  time = 2.93, size = 918, normalized size = 12.58 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))*(A+B*tan(f*x+e)+C*tan(f*x+e)^2),x, algorithm="giac")

[Out]

1/2*(2*A*c*f*x*tan(f*x)^2*tan(e)^2 - 2*C*c*f*x*tan(f*x)^2*tan(e)^2 - 2*B*d*f*x*tan(f*x)^2*tan(e)^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)^2*tan(e)^2 - A*d*log(4*(tan(f*x)^4*tan(e)^2 - 2*tan(f*x)^3*tan(e) + tan(f*x)^2*tan(e)^2 + t
an(f*x)^2 - 2*tan(f*x)*tan(e) + 1)/(tan(e)^2 + 1))*tan(f*x)^2*tan(e)^2 + C*d*log(4*(tan(f*x)^4*tan(e)^2 - 2*ta
n(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*f*x*tan(f*x)*tan(e) + 4*C*c*f*x*tan(f*x)*tan(e) + 4*B*d*f*x*tan(f*x)*tan(e) + C*d*tan(f*x)^2*tan(e)^
2 + 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) + 2*A*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*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*C*c*tan(f*x)^2*tan(e) - 2*B*d*tan(f*x)^2*tan(e) - 2*C*c*tan(f*x)*tan(e)^2 - 2*B*d*tan(f*x)*tan(e)
^2 + 2*A*c*f*x - 2*C*c*f*x - 2*B*d*f*x + C*d*tan(f*x)^2 + C*d*tan(e)^2 - B*c*log(4*(tan(f*x)^4*tan(e)^2 - 2*ta
n(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)) - A*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
)) + 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)) + 2*C*c*tan(f*x) + 2*B*d*tan(f*x) + 2*C*c*tan(e) + 2*B*d*tan(e) + C*d)/(f*tan(f*x)^2*t
an(e)^2 - 2*f*tan(f*x)*tan(e) + f)

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maple [A]  time = 0.02, size = 136, normalized size = 1.86 \[ \frac {C d \left (\tan ^{2}\left (f x +e \right )\right )}{2 f}+\frac {B d \tan \left (f x +e \right )}{f}+\frac {c C \tan \left (f x +e \right )}{f}+\frac {\ln \left (1+\tan ^{2}\left (f x +e \right )\right ) A d}{2 f}+\frac {\ln \left (1+\tan ^{2}\left (f x +e \right )\right ) B c}{2 f}-\frac {\ln \left (1+\tan ^{2}\left (f x +e \right )\right ) C d}{2 f}+\frac {A \arctan \left (\tan \left (f x +e \right )\right ) c}{f}-\frac {B \arctan \left (\tan \left (f x +e \right )\right ) d}{f}-\frac {C \arctan \left (\tan \left (f x +e \right )\right ) c}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c+d*tan(f*x+e))*(A+B*tan(f*x+e)+C*tan(f*x+e)^2),x)

[Out]

1/2/f*C*d*tan(f*x+e)^2+B*d*tan(f*x+e)/f+1/f*c*C*tan(f*x+e)+1/2/f*ln(1+tan(f*x+e)^2)*A*d+1/2/f*ln(1+tan(f*x+e)^
2)*B*c-1/2/f*ln(1+tan(f*x+e)^2)*C*d+1/f*A*arctan(tan(f*x+e))*c-1/f*B*arctan(tan(f*x+e))*d-1/f*C*arctan(tan(f*x
+e))*c

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maxima [A]  time = 0.51, size = 74, normalized size = 1.01 \[ \frac {C d \tan \left (f x + e\right )^{2} + 2 \, {\left ({\left (A - C\right )} c - B d\right )} {\left (f x + e\right )} + {\left (B c + {\left (A - C\right )} d\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right ) + 2 \, {\left (C c + B d\right )} \tan \left (f x + e\right )}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))*(A+B*tan(f*x+e)+C*tan(f*x+e)^2),x, algorithm="maxima")

[Out]

1/2*(C*d*tan(f*x + e)^2 + 2*((A - C)*c - B*d)*(f*x + e) + (B*c + (A - C)*d)*log(tan(f*x + e)^2 + 1) + 2*(C*c +
B*d)*tan(f*x + e))/f

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mupad [B]  time = 8.68, size = 75, normalized size = 1.03 \[ \frac {\mathrm {tan}\left (e+f\,x\right )\,\left (B\,d+C\,c\right )}{f}-x\,\left (B\,d-A\,c+C\,c\right )+\frac {\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )\,\left (\frac {A\,d}{2}+\frac {B\,c}{2}-\frac {C\,d}{2}\right )}{f}+\frac {C\,d\,{\mathrm {tan}\left (e+f\,x\right )}^2}{2\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*tan(e + f*x))*(A + B*tan(e + f*x) + C*tan(e + f*x)^2),x)

[Out]

(tan(e + f*x)*(B*d + C*c))/f - x*(B*d - A*c + C*c) + (log(tan(e + f*x)^2 + 1)*((A*d)/2 + (B*c)/2 - (C*d)/2))/f
+ (C*d*tan(e + f*x)^2)/(2*f)

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sympy [A]  time = 0.28, size = 131, normalized size = 1.79 \[ \begin {cases} A c x + \frac {A d \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} + \frac {B c \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} - B d x + \frac {B d \tan {\left (e + f x \right )}}{f} - C c x + \frac {C c \tan {\left (e + f x \right )}}{f} - \frac {C d \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} + \frac {C d \tan ^{2}{\left (e + f x \right )}}{2 f} & \text {for}\: f \neq 0 \\x \left (c + d \tan {\relax (e )}\right ) \left (A + B \tan {\relax (e )} + C \tan ^{2}{\relax (e )}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))*(A+B*tan(f*x+e)+C*tan(f*x+e)**2),x)

[Out]

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

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