3.79 \(\int (a+a \text{sech}(c+d x))^{3/2} \, dx\)

Optimal. Leaf size=66 \[ \frac{2 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a} \tanh (c+d x)}{\sqrt{a \text{sech}(c+d x)+a}}\right )}{d}+\frac{2 a^2 \tanh (c+d x)}{d \sqrt{a \text{sech}(c+d x)+a}} \]

[Out]

(2*a^(3/2)*ArcTanh[(Sqrt[a]*Tanh[c + d*x])/Sqrt[a + a*Sech[c + d*x]]])/d + (2*a^2*Tanh[c + d*x])/(d*Sqrt[a + a
*Sech[c + d*x]])

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Rubi [A]  time = 0.0420447, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3775, 21, 3774, 203} \[ \frac{2 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a} \tanh (c+d x)}{\sqrt{a \text{sech}(c+d x)+a}}\right )}{d}+\frac{2 a^2 \tanh (c+d x)}{d \sqrt{a \text{sech}(c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sech[c + d*x])^(3/2),x]

[Out]

(2*a^(3/2)*ArcTanh[(Sqrt[a]*Tanh[c + d*x])/Sqrt[a + a*Sech[c + d*x]]])/d + (2*a^2*Tanh[c + d*x])/(d*Sqrt[a + a
*Sech[c + d*x]])

Rule 3775

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Simp[(b^2*Cot[c + d*x]*(a + b*Csc[c + d*x])^(n
- 2))/(d*(n - 1)), x] + Dist[a/(n - 1), Int[(a + b*Csc[c + d*x])^(n - 2)*(a*(n - 1) + b*(3*n - 4)*Csc[c + d*x]
), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && GtQ[n, 1] && IntegerQ[2*n]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 3774

Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(-2*b)/d, Subst[Int[1/(a + x^2), x], x, (b*C
ot[c + d*x])/Sqrt[a + b*Csc[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int (a+a \text{sech}(c+d x))^{3/2} \, dx &=\frac{2 a^2 \tanh (c+d x)}{d \sqrt{a+a \text{sech}(c+d x)}}+(2 a) \int \frac{\frac{a}{2}+\frac{1}{2} a \text{sech}(c+d x)}{\sqrt{a+a \text{sech}(c+d x)}} \, dx\\ &=\frac{2 a^2 \tanh (c+d x)}{d \sqrt{a+a \text{sech}(c+d x)}}+a \int \sqrt{a+a \text{sech}(c+d x)} \, dx\\ &=\frac{2 a^2 \tanh (c+d x)}{d \sqrt{a+a \text{sech}(c+d x)}}+\frac{\left (2 i a^2\right ) \operatorname{Subst}\left (\int \frac{1}{a+x^2} \, dx,x,-\frac{i a \tanh (c+d x)}{\sqrt{a+a \text{sech}(c+d x)}}\right )}{d}\\ &=\frac{2 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a} \tanh (c+d x)}{\sqrt{a+a \text{sech}(c+d x)}}\right )}{d}+\frac{2 a^2 \tanh (c+d x)}{d \sqrt{a+a \text{sech}(c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.19536, size = 75, normalized size = 1.14 \[ \frac{a \text{sech}\left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\text{sech}(c+d x)+1)} \left (2 \sinh \left (\frac{1}{2} (c+d x)\right )+\sqrt{2} \sinh ^{-1}\left (\sqrt{2} \sinh \left (\frac{1}{2} (c+d x)\right )\right ) \sqrt{\cosh (c+d x)}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sech[c + d*x])^(3/2),x]

[Out]

(a*Sech[(c + d*x)/2]*Sqrt[a*(1 + Sech[c + d*x])]*(Sqrt[2]*ArcSinh[Sqrt[2]*Sinh[(c + d*x)/2]]*Sqrt[Cosh[c + d*x
]] + 2*Sinh[(c + d*x)/2]))/d

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Maple [F]  time = 0.131, size = 0, normalized size = 0. \begin{align*} \int \left ( a+a{\rm sech} \left (dx+c\right ) \right ) ^{{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sech(d*x+c))^(3/2),x)

[Out]

int((a+a*sech(d*x+c))^(3/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \operatorname{sech}\left (d x + c\right ) + a\right )}^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sech(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate((a*sech(d*x + c) + a)^(3/2), x)

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Fricas [B]  time = 2.43816, size = 1975, normalized size = 29.92 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sech(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/2*(a^(3/2)*log(-(a*cosh(d*x + c)^4 + a*sinh(d*x + c)^4 - 3*a*cosh(d*x + c)^3 + (4*a*cosh(d*x + c) - 3*a)*sin
h(d*x + c)^3 + 5*a*cosh(d*x + c)^2 + (6*a*cosh(d*x + c)^2 - 9*a*cosh(d*x + c) + 5*a)*sinh(d*x + c)^2 + (cosh(d
*x + c)^5 + (5*cosh(d*x + c) - 3)*sinh(d*x + c)^4 + sinh(d*x + c)^5 - 3*cosh(d*x + c)^4 + (10*cosh(d*x + c)^2
- 12*cosh(d*x + c) + 5)*sinh(d*x + c)^3 + 5*cosh(d*x + c)^3 + (10*cosh(d*x + c)^3 - 18*cosh(d*x + c)^2 + 15*co
sh(d*x + c) - 7)*sinh(d*x + c)^2 - 7*cosh(d*x + c)^2 + (5*cosh(d*x + c)^4 - 12*cosh(d*x + c)^3 + 15*cosh(d*x +
c)^2 - 14*cosh(d*x + c) + 4)*sinh(d*x + c) + 4*cosh(d*x + c) - 4)*sqrt(a)*sqrt(a/(cosh(d*x + c)^2 + 2*cosh(d*
x + c)*sinh(d*x + c) + sinh(d*x + c)^2 + 1)) - 4*a*cosh(d*x + c) + (4*a*cosh(d*x + c)^3 - 9*a*cosh(d*x + c)^2
+ 10*a*cosh(d*x + c) - 4*a)*sinh(d*x + c) + 4*a)/(cosh(d*x + c)^3 + 3*cosh(d*x + c)^2*sinh(d*x + c) + 3*cosh(d
*x + c)*sinh(d*x + c)^2 + sinh(d*x + c)^3)) + a^(3/2)*log((a*cosh(d*x + c)^2 + a*sinh(d*x + c)^2 + (cosh(d*x +
c)^3 + (3*cosh(d*x + c) + 1)*sinh(d*x + c)^2 + sinh(d*x + c)^3 + cosh(d*x + c)^2 + (3*cosh(d*x + c)^2 + 2*cos
h(d*x + c) + 1)*sinh(d*x + c) + cosh(d*x + c) + 1)*sqrt(a)*sqrt(a/(cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x
+ c) + sinh(d*x + c)^2 + 1)) + a*cosh(d*x + c) + (2*a*cosh(d*x + c) + a)*sinh(d*x + c) + a)/(cosh(d*x + c) + s
inh(d*x + c))) + 4*(a*cosh(d*x + c) + a*sinh(d*x + c) - a)*sqrt(a/(cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x
+ c) + sinh(d*x + c)^2 + 1)))/d

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \operatorname{sech}{\left (c + d x \right )} + a\right )^{\frac{3}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sech(d*x+c))**(3/2),x)

[Out]

Integral((a*sech(c + d*x) + a)**(3/2), x)

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Giac [B]  time = 1.22772, size = 159, normalized size = 2.41 \begin{align*} \frac{\frac{2 \, a^{2} \arctan \left (-\frac{\sqrt{a} e^{\left (d x + c\right )} - \sqrt{a e^{\left (2 \, d x + 2 \, c\right )} + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} - a^{\frac{3}{2}} \log \left ({\left | -\sqrt{a} e^{\left (d x + c\right )} + \sqrt{a e^{\left (2 \, d x + 2 \, c\right )} + a} \right |}\right ) + \frac{2 \,{\left (a^{2} e^{\left (d x + c\right )} - a^{2}\right )}}{\sqrt{a e^{\left (2 \, d x + 2 \, c\right )} + a}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sech(d*x+c))^(3/2),x, algorithm="giac")

[Out]

(2*a^2*arctan(-(sqrt(a)*e^(d*x + c) - sqrt(a*e^(2*d*x + 2*c) + a))/sqrt(-a))/sqrt(-a) - a^(3/2)*log(abs(-sqrt(
a)*e^(d*x + c) + sqrt(a*e^(2*d*x + 2*c) + a))) + 2*(a^2*e^(d*x + c) - a^2)/sqrt(a*e^(2*d*x + 2*c) + a))/d