3.511 \(\int (a+b \tan (d+e x)) (b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x)) \, dx\)

Optimal. Leaf size=72 \[ -\frac {b \left (a^2+b^2\right ) \log (\cos (d+e x))}{e}-a x \left (a^2+b^2\right )+\frac {a^2 (a+b \tan (d+e x))^2}{2 b e}+\frac {2 a b^2 \tan (d+e x)}{e} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.081, Rules used = {3630, 3525, 3475} \[ -\frac {b \left (a^2+b^2\right ) \log (\cos (d+e x))}{e}-a x \left (a^2+b^2\right )+\frac {a^2 (a+b \tan (d+e x))^2}{2 b e}+\frac {2 a b^2 \tan (d+e x)}{e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [C]  time = 0.33, size = 88, normalized size = 1.22 \[ \frac {2 a \left (a^2+2 b^2\right ) \tan (d+e x)+\left (a^2+b^2\right ) ((b+i a) \log (-\tan (d+e x)+i)+(b-i a) \log (\tan (d+e x)+i))+a^2 b \tan ^2(d+e x)}{2 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.12, size = 74, normalized size = 1.03 \[ \frac {a^{2} b \tan \left (e x + d\right )^{2} - 2 \, {\left (a^{3} + a b^{2}\right )} e x - {\left (a^{2} b + b^{3}\right )} \log \left (\frac {1}{\tan \left (e x + d\right )^{2} + 1}\right ) + 2 \, {\left (a^{3} + 2 \, a b^{2}\right )} \tan \left (e x + d\right )}{2 \, e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 1.02, size = 709, normalized size = 9.85 \[ -\frac {2 \, a^{3} x e \tan \left (x e\right )^{2} \tan \relax (d)^{2} + 2 \, a b^{2} x e \tan \left (x e\right )^{2} \tan \relax (d)^{2} + a^{2} b \log \left (\frac {4 \, {\left (\tan \left (x e\right )^{4} \tan \relax (d)^{2} - 2 \, \tan \left (x e\right )^{3} \tan \relax (d) + \tan \left (x e\right )^{2} \tan \relax (d)^{2} + \tan \left (x e\right )^{2} - 2 \, \tan \left (x e\right ) \tan \relax (d) + 1\right )}}{\tan \relax (d)^{2} + 1}\right ) \tan \left (x e\right )^{2} \tan \relax (d)^{2} + b^{3} \log \left (\frac {4 \, {\left (\tan \left (x e\right )^{4} \tan \relax (d)^{2} - 2 \, \tan \left (x e\right )^{3} \tan \relax (d) + \tan \left (x e\right )^{2} \tan \relax (d)^{2} + \tan \left (x e\right )^{2} - 2 \, \tan \left (x e\right ) \tan \relax (d) + 1\right )}}{\tan \relax (d)^{2} + 1}\right ) \tan \left (x e\right )^{2} \tan \relax (d)^{2} - 4 \, a^{3} x e \tan \left (x e\right ) \tan \relax (d) - 4 \, a b^{2} x e \tan \left (x e\right ) \tan \relax (d) - a^{2} b \tan \left (x e\right )^{2} \tan \relax (d)^{2} - 2 \, a^{2} b \log \left (\frac {4 \, {\left (\tan \left (x e\right )^{4} \tan \relax (d)^{2} - 2 \, \tan \left (x e\right )^{3} \tan \relax (d) + \tan \left (x e\right )^{2} \tan \relax (d)^{2} + \tan \left (x e\right )^{2} - 2 \, \tan \left (x e\right ) \tan \relax (d) + 1\right )}}{\tan \relax (d)^{2} + 1}\right ) \tan \left (x e\right ) \tan \relax (d) - 2 \, b^{3} \log \left (\frac {4 \, {\left (\tan \left (x e\right )^{4} \tan \relax (d)^{2} - 2 \, \tan \left (x e\right )^{3} \tan \relax (d) + \tan \left (x e\right )^{2} \tan \relax (d)^{2} + \tan \left (x e\right )^{2} - 2 \, \tan \left (x e\right ) \tan \relax (d) + 1\right )}}{\tan \relax (d)^{2} + 1}\right ) \tan \left (x e\right ) \tan \relax (d) + 2 \, a^{3} \tan \left (x e\right )^{2} \tan \relax (d) + 4 \, a b^{2} \tan \left (x e\right )^{2} \tan \relax (d) + 2 \, a^{3} \tan \left (x e\right ) \tan \relax (d)^{2} + 4 \, a b^{2} \tan \left (x e\right ) \tan \relax (d)^{2} + 2 \, a^{3} x e + 2 \, a b^{2} x e - a^{2} b \tan \left (x e\right )^{2} - a^{2} b \tan \relax (d)^{2} + a^{2} b \log \left (\frac {4 \, {\left (\tan \left (x e\right )^{4} \tan \relax (d)^{2} - 2 \, \tan \left (x e\right )^{3} \tan \relax (d) + \tan \left (x e\right )^{2} \tan \relax (d)^{2} + \tan \left (x e\right )^{2} - 2 \, \tan \left (x e\right ) \tan \relax (d) + 1\right )}}{\tan \relax (d)^{2} + 1}\right ) + b^{3} \log \left (\frac {4 \, {\left (\tan \left (x e\right )^{4} \tan \relax (d)^{2} - 2 \, \tan \left (x e\right )^{3} \tan \relax (d) + \tan \left (x e\right )^{2} \tan \relax (d)^{2} + \tan \left (x e\right )^{2} - 2 \, \tan \left (x e\right ) \tan \relax (d) + 1\right )}}{\tan \relax (d)^{2} + 1}\right ) - 2 \, a^{3} \tan \left (x e\right ) - 4 \, a b^{2} \tan \left (x e\right ) - 2 \, a^{3} \tan \relax (d) - 4 \, a b^{2} \tan \relax (d) - a^{2} b}{2 \, {\left (e \tan \left (x e\right )^{2} \tan \relax (d)^{2} - 2 \, e \tan \left (x e\right ) \tan \relax (d) + e\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 117, normalized size = 1.62 \[ \frac {a^{2} b \left (\tan ^{2}\left (e x +d \right )\right )}{2 e}+\frac {a^{3} \tan \left (e x +d \right )}{e}+\frac {2 a \,b^{2} \tan \left (e x +d \right )}{e}+\frac {\ln \left (1+\tan ^{2}\left (e x +d \right )\right ) a^{2} b}{2 e}+\frac {\ln \left (1+\tan ^{2}\left (e x +d \right )\right ) b^{3}}{2 e}-\frac {\arctan \left (\tan \left (e x +d \right )\right ) a^{3}}{e}-\frac {\arctan \left (\tan \left (e x +d \right )\right ) a \,b^{2}}{e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.41, size = 74, normalized size = 1.03 \[ \frac {a^{2} b \tan \left (e x + d\right )^{2} - 2 \, {\left (a^{3} + a b^{2}\right )} {\left (e x + d\right )} + {\left (a^{2} b + b^{3}\right )} \log \left (\tan \left (e x + d\right )^{2} + 1\right ) + 2 \, {\left (a^{3} + 2 \, a b^{2}\right )} \tan \left (e x + d\right )}{2 \, e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.77, size = 105, normalized size = 1.46 \[ \frac {\mathrm {tan}\left (d+e\,x\right )\,\left (a^3+2\,a\,b^2\right )}{e}+\frac {\ln \left ({\mathrm {tan}\left (d+e\,x\right )}^2+1\right )\,\left (\frac {a^2\,b}{2}+\frac {b^3}{2}\right )}{e}+\frac {a^2\,b\,{\mathrm {tan}\left (d+e\,x\right )}^2}{2\,e}-\frac {a\,\mathrm {atan}\left (\frac {a\,\mathrm {tan}\left (d+e\,x\right )\,\left (a^2+b^2\right )}{a^3+a\,b^2}\right )\,\left (a^2+b^2\right )}{e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.25, size = 122, normalized size = 1.69 \[ \begin {cases} - a^{3} x + \frac {a^{3} \tan {\left (d + e x \right )}}{e} + \frac {a^{2} b \log {\left (\tan ^{2}{\left (d + e x \right )} + 1 \right )}}{2 e} + \frac {a^{2} b \tan ^{2}{\left (d + e x \right )}}{2 e} - a b^{2} x + \frac {2 a b^{2} \tan {\left (d + e x \right )}}{e} + \frac {b^{3} \log {\left (\tan ^{2}{\left (d + e x \right )} + 1 \right )}}{2 e} & \text {for}\: e \neq 0 \\x \left (a + b \tan {\relax (d )}\right ) \left (a^{2} \tan ^{2}{\relax (d )} + 2 a b \tan {\relax (d )} + b^{2}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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