3.78 \(\int (a+a \text {sech}(c+d x))^{5/2} \, dx\)

Optimal. Leaf size=98 \[ \frac {2 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \tanh (c+d x)}{\sqrt {a \text {sech}(c+d x)+a}}\right )}{d}+\frac {14 a^3 \tanh (c+d x)}{3 d \sqrt {a \text {sech}(c+d x)+a}}+\frac {2 a^2 \tanh (c+d x) \sqrt {a \text {sech}(c+d x)+a}}{3 d} \]

[Out]

2*a^(5/2)*arctanh(a^(1/2)*tanh(d*x+c)/(a+a*sech(d*x+c))^(1/2))/d+14/3*a^3*tanh(d*x+c)/d/(a+a*sech(d*x+c))^(1/2
)+2/3*a^2*(a+a*sech(d*x+c))^(1/2)*tanh(d*x+c)/d

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Rubi [A]  time = 0.12, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3775, 3915, 3774, 203, 3792} \[ \frac {2 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \tanh (c+d x)}{\sqrt {a \text {sech}(c+d x)+a}}\right )}{d}+\frac {14 a^3 \tanh (c+d x)}{3 d \sqrt {a \text {sech}(c+d x)+a}}+\frac {2 a^2 \tanh (c+d x) \sqrt {a \text {sech}(c+d x)+a}}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sech[c + d*x])^(5/2),x]

[Out]

(2*a^(5/2)*ArcTanh[(Sqrt[a]*Tanh[c + d*x])/Sqrt[a + a*Sech[c + d*x]]])/d + (14*a^3*Tanh[c + d*x])/(3*d*Sqrt[a
+ a*Sech[c + d*x]]) + (2*a^2*Sqrt[a + a*Sech[c + d*x]]*Tanh[c + d*x])/(3*d)

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])

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 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 3792

Int[csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*b*Cot[e + f*x])/
(f*Sqrt[a + b*Csc[e + f*x]]), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

Rule 3915

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

Rubi steps

\begin {align*} \int (a+a \text {sech}(c+d x))^{5/2} \, dx &=\frac {2 a^2 \sqrt {a+a \text {sech}(c+d x)} \tanh (c+d x)}{3 d}+\frac {1}{3} (2 a) \int \sqrt {a+a \text {sech}(c+d x)} \left (\frac {3 a}{2}+\frac {7}{2} a \text {sech}(c+d x)\right ) \, dx\\ &=\frac {2 a^2 \sqrt {a+a \text {sech}(c+d x)} \tanh (c+d x)}{3 d}+a^2 \int \sqrt {a+a \text {sech}(c+d x)} \, dx+\frac {1}{3} \left (7 a^2\right ) \int \text {sech}(c+d x) \sqrt {a+a \text {sech}(c+d x)} \, dx\\ &=\frac {14 a^3 \tanh (c+d x)}{3 d \sqrt {a+a \text {sech}(c+d x)}}+\frac {2 a^2 \sqrt {a+a \text {sech}(c+d x)} \tanh (c+d x)}{3 d}+\frac {\left (2 i a^3\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^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \tanh (c+d x)}{\sqrt {a+a \text {sech}(c+d x)}}\right )}{d}+\frac {14 a^3 \tanh (c+d x)}{3 d \sqrt {a+a \text {sech}(c+d x)}}+\frac {2 a^2 \sqrt {a+a \text {sech}(c+d x)} \tanh (c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]  time = 0.34, size = 99, normalized size = 1.01 \[ \frac {a^2 \text {sech}\left (\frac {1}{2} (c+d x)\right ) \text {sech}(c+d x) \sqrt {a (\text {sech}(c+d x)+1)} \left (-6 \sinh \left (\frac {1}{2} (c+d x)\right )+8 \sinh \left (\frac {3}{2} (c+d x)\right )+3 \sqrt {2} \sinh ^{-1}\left (\sqrt {2} \sinh \left (\frac {1}{2} (c+d x)\right )\right ) \cosh ^{\frac {3}{2}}(c+d x)\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sech[c + d*x])^(5/2),x]

[Out]

(a^2*Sech[(c + d*x)/2]*Sech[c + d*x]*Sqrt[a*(1 + Sech[c + d*x])]*(3*Sqrt[2]*ArcSinh[Sqrt[2]*Sinh[(c + d*x)/2]]
*Cosh[c + d*x]^(3/2) - 6*Sinh[(c + d*x)/2] + 8*Sinh[(3*(c + d*x))/2]))/(3*d)

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fricas [B]  time = 0.44, size = 924, normalized size = 9.43 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sech(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/6*(3*(a^2*cosh(d*x + c)^2 + 2*a^2*cosh(d*x + c)*sinh(d*x + c) + a^2*sinh(d*x + c)^2 + a^2)*sqrt(a)*log(-(a*c
osh(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)*sinh(d*x + c)^3 + 5*a*cos
h(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*cosh(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)) + 3*(a^2*cosh(d*x + c)^2 + 2*a^2*cosh(d*x + c)*sinh(d*x + c) + a^2*sinh(d*x + c)^2 + a^2
)*sqrt(a)*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*cosh(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) + sinh(d*x + c))) + 8*(4*a^2*cosh(d*x + c)^3 +
4*a^2*sinh(d*x + c)^3 - 3*a^2*cosh(d*x + c)^2 + 3*a^2*cosh(d*x + c) + 3*(4*a^2*cosh(d*x + c) - a^2)*sinh(d*x +
c)^2 - 4*a^2 + 3*(4*a^2*cosh(d*x + c)^2 - 2*a^2*cosh(d*x + c) + a^2)*sinh(d*x + c))*sqrt(a/(cosh(d*x + c)^2 +
2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2 + 1)))/(d*cosh(d*x + c)^2 + 2*d*cosh(d*x + c)*sinh(d*x + c) +
d*sinh(d*x + c)^2 + d)

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giac [A]  time = 0.32, size = 151, normalized size = 1.54 \[ \frac {\frac {6 \, a^{3} \arctan \left (-\frac {\sqrt {a} e^{\left (d x + c\right )} - \sqrt {a e^{\left (2 \, d x + 2 \, c\right )} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - 3 \, a^{\frac {5}{2}} \log \left ({\left | -\sqrt {a} e^{\left (d x + c\right )} + \sqrt {a e^{\left (2 \, d x + 2 \, c\right )} + a} \right |}\right ) - \frac {4 \, {\left (4 \, a^{4} - {\left (3 \, a^{4} e^{c} + {\left (4 \, a^{4} e^{\left (d x + 3 \, c\right )} - 3 \, a^{4} e^{\left (2 \, c\right )}\right )} e^{\left (d x\right )}\right )} e^{\left (d x\right )}\right )}}{{\left (a e^{\left (2 \, d x + 2 \, c\right )} + a\right )}^{\frac {3}{2}}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sech(d*x+c))^(5/2),x, algorithm="giac")

[Out]

1/3*(6*a^3*arctan(-(sqrt(a)*e^(d*x + c) - sqrt(a*e^(2*d*x + 2*c) + a))/sqrt(-a))/sqrt(-a) - 3*a^(5/2)*log(abs(
-sqrt(a)*e^(d*x + c) + sqrt(a*e^(2*d*x + 2*c) + a))) - 4*(4*a^4 - (3*a^4*e^c + (4*a^4*e^(d*x + 3*c) - 3*a^4*e^
(2*c))*e^(d*x))*e^(d*x))/(a*e^(2*d*x + 2*c) + a)^(3/2))/d

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maple [F]  time = 0.51, size = 0, normalized size = 0.00 \[ \int \left (a +a \,\mathrm {sech}\left (d x +c \right )\right )^{\frac {5}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sech(d*x+c))^(5/2),x)

[Out]

int((a+a*sech(d*x+c))^(5/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \operatorname {sech}\left (d x + c\right ) + a\right )}^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sech(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((a*sech(d*x + c) + a)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a+\frac {a}{\mathrm {cosh}\left (c+d\,x\right )}\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cosh(c + d*x))^(5/2),x)

[Out]

int((a + a/cosh(c + d*x))^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \operatorname {sech}{\left (c + d x \right )} + a\right )^{\frac {5}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sech(d*x+c))**(5/2),x)

[Out]

Integral((a*sech(c + d*x) + a)**(5/2), x)

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