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)*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)

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Rubi [A]  time = 0.0755105, antiderivative size = 72, normalized size of antiderivative = 1., 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 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]

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 3475

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

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*}

Mathematica [C]  time = 0.333865, 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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Maple [A]  time = 0.005, size = 117, normalized size = 1.6 \begin{align*}{\frac{{a}^{2}b \left ( \tan \left ( ex+d \right ) \right ) ^{2}}{2\,e}}+{\frac{{a}^{3}\tan \left ( ex+d \right ) }{e}}+2\,{\frac{a{b}^{2}\tan \left ( ex+d \right ) }{e}}+{\frac{\ln \left ( 1+ \left ( \tan \left ( ex+d \right ) \right ) ^{2} \right ){a}^{2}b}{2\,e}}+{\frac{\ln \left ( 1+ \left ( \tan \left ( ex+d \right ) \right ) ^{2} \right ){b}^{3}}{2\,e}}-{\frac{\arctan \left ( \tan \left ( ex+d \right ) \right ){a}^{3}}{e}}-{\frac{\arctan \left ( \tan \left ( ex+d \right ) \right ) a{b}^{2}}{e}} \end{align*}

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 = 1.50719, size = 100, normalized size = 1.39 \begin{align*} \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} \end{align*}

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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Fricas [A]  time = 1.79282, size = 174, normalized size = 2.42 \begin{align*} \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} \end{align*}

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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Sympy [A]  time = 0.325562, size = 122, normalized size = 1.69 \begin{align*} \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{\left (d \right )}\right ) \left (a^{2} \tan ^{2}{\left (d \right )} + 2 a b \tan{\left (d \right )} + b^{2}\right ) & \text{otherwise} \end{cases} \end{align*}

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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Giac [B]  time = 2.07487, size = 957, normalized size = 13.29 \begin{align*} \text{result too large to display} \end{align*}

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