3.515 \(\int (a+b \tan (d+e x)) \sqrt{b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x)} \, dx\)

Optimal. Leaf size=122 \[ \frac{a^2 b \tan (d+e x) \sqrt{a^2 \tan ^2(d+e x)+2 a b \tan (d+e x)+b^2}}{e \left (a^2 \tan (d+e x)+a b\right )}-\frac{\left (a^2+b^2\right ) \log (\cos (d+e x)) \sqrt{a^2 \tan ^2(d+e x)+2 a b \tan (d+e x)+b^2}}{e (a \tan (d+e x)+b)} \]

[Out]

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

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Rubi [A]  time = 0.100776, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.073, Rules used = {3710, 3525, 3475} \[ \frac{a^2 b \tan (d+e x) \sqrt{a^2 \tan ^2(d+e x)+2 a b \tan (d+e x)+b^2}}{e \left (a^2 \tan (d+e x)+a b\right )}-\frac{\left (a^2+b^2\right ) \log (\cos (d+e x)) \sqrt{a^2 \tan ^2(d+e x)+2 a b \tan (d+e x)+b^2}}{e (a \tan (d+e x)+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3710

Int[((A_) + (B_.)*tan[(d_.) + (e_.)*(x_)])*((a_) + (b_.)*tan[(d_.) + (e_.)*(x_)] + (c_.)*tan[(d_.) + (e_.)*(x_
)]^2)^(n_), x_Symbol] :> Dist[(a + b*Tan[d + e*x] + c*Tan[d + e*x]^2)^n/(b + 2*c*Tan[d + e*x])^(2*n), Int[(A +
B*Tan[d + e*x])*(b + 2*c*Tan[d + e*x])^(2*n), x], x] /; FreeQ[{a, b, c, d, e, A, B}, x] && EqQ[b^2 - 4*a*c, 0
] &&  !IntegerQ[n]

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

Mathematica [A]  time = 0.288498, size = 58, normalized size = 0.48 \[ \frac{\sqrt{(a \tan (d+e x)+b)^2} \left (a b \tan (d+e x)-\left (a^2+b^2\right ) \log (\cos (d+e x))\right )}{e (a \tan (d+e x)+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.095, size = 75, normalized size = 0.6 \begin{align*}{\frac{{\it csgn} \left ( b+a\tan \left ( ex+d \right ) \right ) \left ( \ln \left ({a}^{2} \left ( \tan \left ( ex+d \right ) \right ) ^{2}+{a}^{2} \right ){a}^{2}+\ln \left ({a}^{2} \left ( \tan \left ( ex+d \right ) \right ) ^{2}+{a}^{2} \right ){b}^{2}+2\,ab\tan \left ( ex+d \right ) +2\,{b}^{2} \right ) }{2\,e}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b^2+2*a*b*tan(e*x+d)+a^2*tan(e*x+d)^2)^(1/2)*(a+b*tan(e*x+d)),x)

[Out]

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

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Maxima [A]  time = 1.55801, size = 88, normalized size = 0.72 \begin{align*} \frac{{\left (2 \,{\left (e x + d\right )} b + a \log \left (\tan \left (e x + d\right )^{2} + 1\right )\right )} a -{\left (2 \,{\left (e x + d\right )} a - b \log \left (\tan \left (e x + d\right )^{2} + 1\right ) - 2 \, a \tan \left (e x + d\right )\right )} b}{2 \, e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2+2*a*b*tan(e*x+d)+a^2*tan(e*x+d)^2)^(1/2)*(a+b*tan(e*x+d)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.73161, size = 95, normalized size = 0.78 \begin{align*} \frac{2 \, a b \tan \left (e x + d\right ) -{\left (a^{2} + b^{2}\right )} \log \left (\frac{1}{\tan \left (e x + d\right )^{2} + 1}\right )}{2 \, e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2+2*a*b*tan(e*x+d)+a^2*tan(e*x+d)^2)^(1/2)*(a+b*tan(e*x+d)),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \tan{\left (d + e x \right )}\right ) \sqrt{\left (a \tan{\left (d + e x \right )} + b\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b**2+2*a*b*tan(e*x+d)+a**2*tan(e*x+d)**2)**(1/2)*(a+b*tan(e*x+d)),x)

[Out]

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

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Giac [A]  time = 1.25722, size = 100, normalized size = 0.82 \begin{align*} a b e^{\left (-1\right )} \mathrm{sgn}\left (a \tan \left (x e + d\right ) + b\right ) \tan \left (x e + d\right ) + \frac{1}{2} \,{\left (a^{2} \mathrm{sgn}\left (a \tan \left (x e + d\right ) + b\right ) + b^{2} \mathrm{sgn}\left (a \tan \left (x e + d\right ) + b\right )\right )} e^{\left (-1\right )} \log \left (\tan \left (x e + d\right )^{2} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2+2*a*b*tan(e*x+d)+a^2*tan(e*x+d)^2)^(1/2)*(a+b*tan(e*x+d)),x, algorithm="giac")

[Out]

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