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

Optimal. Leaf size=173 \[ \frac{a^2 b x \sqrt{a^2 \sec ^2(d+e x)+2 a b \sec (d+e x)+b^2}}{a^2 \sec (d+e x)+a b}+\frac{a^2 b \tan (d+e x) \sqrt{a^2 \sec ^2(d+e x)+2 a b \sec (d+e x)+b^2}}{e \left (a^2 \sec (d+e x)+a b\right )}+\frac{\left (a^2+b^2\right ) \tanh ^{-1}(\sin (d+e x)) \sqrt{a^2 \sec ^2(d+e x)+2 a b \sec (d+e x)+b^2}}{e (a \sec (d+e x)+b)} \]

[Out]

((a^2 + b^2)*ArcTanh[Sin[d + e*x]]*Sqrt[b^2 + 2*a*b*Sec[d + e*x] + a^2*Sec[d + e*x]^2])/(e*(b + a*Sec[d + e*x]
)) + (a^2*b*x*Sqrt[b^2 + 2*a*b*Sec[d + e*x] + a^2*Sec[d + e*x]^2])/(a*b + a^2*Sec[d + e*x]) + (a^2*b*Sqrt[b^2
+ 2*a*b*Sec[d + e*x] + a^2*Sec[d + e*x]^2]*Tan[d + e*x])/(e*(a*b + a^2*Sec[d + e*x]))

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Rubi [A]  time = 0.118901, antiderivative size = 173, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.122, Rules used = {4174, 3914, 3767, 8, 3770} \[ \frac{a^2 b x \sqrt{a^2 \sec ^2(d+e x)+2 a b \sec (d+e x)+b^2}}{a^2 \sec (d+e x)+a b}+\frac{a^2 b \tan (d+e x) \sqrt{a^2 \sec ^2(d+e x)+2 a b \sec (d+e x)+b^2}}{e \left (a^2 \sec (d+e x)+a b\right )}+\frac{\left (a^2+b^2\right ) \tanh ^{-1}(\sin (d+e x)) \sqrt{a^2 \sec ^2(d+e x)+2 a b \sec (d+e x)+b^2}}{e (a \sec (d+e x)+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a^2 + b^2)*ArcTanh[Sin[d + e*x]]*Sqrt[b^2 + 2*a*b*Sec[d + e*x] + a^2*Sec[d + e*x]^2])/(e*(b + a*Sec[d + e*x]
)) + (a^2*b*x*Sqrt[b^2 + 2*a*b*Sec[d + e*x] + a^2*Sec[d + e*x]^2])/(a*b + a^2*Sec[d + e*x]) + (a^2*b*Sqrt[b^2
+ 2*a*b*Sec[d + e*x] + a^2*Sec[d + e*x]^2]*Tan[d + e*x])/(e*(a*b + a^2*Sec[d + e*x]))

Rule 4174

Int[((A_) + (B_.)*sec[(d_.) + (e_.)*(x_)])*((a_) + (b_.)*sec[(d_.) + (e_.)*(x_)] + (c_.)*sec[(d_.) + (e_.)*(x_
)]^2)^(n_), x_Symbol] :> Dist[(a + b*Sec[d + e*x] + c*Sec[d + e*x]^2)^n/(b + 2*c*Sec[d + e*x])^(2*n), Int[(A +
B*Sec[d + e*x])*(b + 2*c*Sec[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 3914

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Simp[a*c*x, x]
+ (Dist[b*d, Int[Csc[e + f*x]^2, x], x] + Dist[b*c + a*d, Int[Csc[e + f*x], x], x]) /; FreeQ[{a, b, c, d, e,
f}, x] && NeQ[b*c - a*d, 0] && NeQ[b*c + a*d, 0]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3770

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

Rubi steps

\begin{align*} \int (a+b \sec (d+e x)) \sqrt{b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)} \, dx &=\frac{\sqrt{b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)} \int \left (2 a b+2 a^2 \sec (d+e x)\right ) (a+b \sec (d+e x)) \, dx}{2 a b+2 a^2 \sec (d+e x)}\\ &=\frac{a^2 b x \sqrt{b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)}}{a b+a^2 \sec (d+e x)}+\frac{\left (2 a^2 b \sqrt{b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)}\right ) \int \sec ^2(d+e x) \, dx}{2 a b+2 a^2 \sec (d+e x)}+\frac{\left (2 a \left (a^2+b^2\right ) \sqrt{b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)}\right ) \int \sec (d+e x) \, dx}{2 a b+2 a^2 \sec (d+e x)}\\ &=\frac{\left (a^2+b^2\right ) \tanh ^{-1}(\sin (d+e x)) \sqrt{b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)}}{e (b+a \sec (d+e x))}+\frac{a^2 b x \sqrt{b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)}}{a b+a^2 \sec (d+e x)}-\frac{\left (2 a^2 b \sqrt{b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)}\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (d+e x))}{e \left (2 a b+2 a^2 \sec (d+e x)\right )}\\ &=\frac{\left (a^2+b^2\right ) \tanh ^{-1}(\sin (d+e x)) \sqrt{b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)}}{e (b+a \sec (d+e x))}+\frac{a^2 b x \sqrt{b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)}}{a b+a^2 \sec (d+e x)}+\frac{a^2 b \sqrt{b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)} \tan (d+e x)}{e \left (a b+a^2 \sec (d+e x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.263301, size = 67, normalized size = 0.39 \[ \frac{\cos (d+e x) \sqrt{(a \sec (d+e x)+b)^2} \left (\left (a^2+b^2\right ) \tanh ^{-1}(\sin (d+e x))+a b (\tan (d+e x)+e x)\right )}{e (a+b \cos (d+e x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[d + e*x]*Sqrt[(b + a*Sec[d + e*x])^2]*((a^2 + b^2)*ArcTanh[Sin[d + e*x]] + a*b*(e*x + Tan[d + e*x])))/(e*
(a + b*Cos[d + e*x]))

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Maple [A]  time = 0.238, size = 208, normalized size = 1.2 \begin{align*}{\frac{1}{e \left ( b\cos \left ( ex+d \right ) +a \right ) } \left ( \cos \left ( ex+d \right ) \ln \left ({\frac{\sin \left ( ex+d \right ) +1-\cos \left ( ex+d \right ) }{\sin \left ( ex+d \right ) }} \right ){a}^{2}+\cos \left ( ex+d \right ) \ln \left ({\frac{\sin \left ( ex+d \right ) +1-\cos \left ( ex+d \right ) }{\sin \left ( ex+d \right ) }} \right ){b}^{2}-\cos \left ( ex+d \right ) \ln \left ( -{\frac{\cos \left ( ex+d \right ) -1+\sin \left ( ex+d \right ) }{\sin \left ( ex+d \right ) }} \right ){a}^{2}-\cos \left ( ex+d \right ) \ln \left ( -{\frac{\cos \left ( ex+d \right ) -1+\sin \left ( ex+d \right ) }{\sin \left ( ex+d \right ) }} \right ){b}^{2}+\cos \left ( ex+d \right ) \left ( ex+d \right ) ab+ab\sin \left ( ex+d \right ) \right ) \sqrt{{\frac{ \left ( b\cos \left ( ex+d \right ) +a \right ) ^{2}}{ \left ( \cos \left ( ex+d \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.5922, size = 221, normalized size = 1.28 \begin{align*} \frac{{\left (2 \, b \arctan \left (\frac{\sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}\right ) + a \log \left (\frac{\sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} + 1\right ) - a \log \left (\frac{\sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} - 1\right )\right )} a +{\left (b \log \left (\frac{\sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} + 1\right ) - b \log \left (\frac{\sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} - 1\right ) - \frac{2 \, a \sin \left (e x + d\right )}{{\left (\frac{\sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}} - 1\right )}{\left (\cos \left (e x + d\right ) + 1\right )}}\right )} b}{e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*b*arctan(sin(e*x + d)/(cos(e*x + d) + 1)) + a*log(sin(e*x + d)/(cos(e*x + d) + 1) + 1) - a*log(sin(e*x + d
)/(cos(e*x + d) + 1) - 1))*a + (b*log(sin(e*x + d)/(cos(e*x + d) + 1) + 1) - b*log(sin(e*x + d)/(cos(e*x + d)
+ 1) - 1) - 2*a*sin(e*x + d)/((sin(e*x + d)^2/(cos(e*x + d) + 1)^2 - 1)*(cos(e*x + d) + 1)))*b)/e

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Fricas [A]  time = 2.0364, size = 225, normalized size = 1.3 \begin{align*} \frac{2 \, a b e x \cos \left (e x + d\right ) +{\left (a^{2} + b^{2}\right )} \cos \left (e x + d\right ) \log \left (\sin \left (e x + d\right ) + 1\right ) -{\left (a^{2} + b^{2}\right )} \cos \left (e x + d\right ) \log \left (-\sin \left (e x + d\right ) + 1\right ) + 2 \, a b \sin \left (e x + d\right )}{2 \, e \cos \left (e x + d\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*a*b*e*x*cos(e*x + d) + (a^2 + b^2)*cos(e*x + d)*log(sin(e*x + d) + 1) - (a^2 + b^2)*cos(e*x + d)*log(-s
in(e*x + d) + 1) + 2*a*b*sin(e*x + d))/(e*cos(e*x + d))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.36005, size = 302, normalized size = 1.75 \begin{align*}{\left ({\left (x e + d\right )} a b \mathrm{sgn}\left (b \cos \left (x e + d\right )^{2} + a \cos \left (x e + d\right )\right ) - \frac{2 \, a b \mathrm{sgn}\left (b \cos \left (x e + d\right )^{2} + a \cos \left (x e + d\right )\right ) \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )}{\tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{2} - 1} +{\left (a^{2} \mathrm{sgn}\left (b \cos \left (x e + d\right )^{2} + a \cos \left (x e + d\right )\right ) + b^{2} \mathrm{sgn}\left (b \cos \left (x e + d\right )^{2} + a \cos \left (x e + d\right )\right )\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) + 1 \right |}\right ) -{\left (a^{2} \mathrm{sgn}\left (b \cos \left (x e + d\right )^{2} + a \cos \left (x e + d\right )\right ) + b^{2} \mathrm{sgn}\left (b \cos \left (x e + d\right )^{2} + a \cos \left (x e + d\right )\right )\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) - 1 \right |}\right )\right )} e^{\left (-1\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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