3.220 \(\int \frac{1}{(a+b \text{sech}^2(c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=183 \[ -\frac{b \left (33 a^2+40 a b+15 b^2\right ) \tanh (c+d x)}{15 a^3 d (a+b)^3 \sqrt{a-b \tanh ^2(c+d x)+b}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a} \tanh (c+d x)}{\sqrt{a-b \tanh ^2(c+d x)+b}}\right )}{a^{7/2} d}-\frac{b (9 a+5 b) \tanh (c+d x)}{15 a^2 d (a+b)^2 \left (a-b \tanh ^2(c+d x)+b\right )^{3/2}}-\frac{b \tanh (c+d x)}{5 a d (a+b) \left (a-b \tanh ^2(c+d x)+b\right )^{5/2}} \]

[Out]

ArcTanh[(Sqrt[a]*Tanh[c + d*x])/Sqrt[a + b - b*Tanh[c + d*x]^2]]/(a^(7/2)*d) - (b*Tanh[c + d*x])/(5*a*(a + b)*
d*(a + b - b*Tanh[c + d*x]^2)^(5/2)) - (b*(9*a + 5*b)*Tanh[c + d*x])/(15*a^2*(a + b)^2*d*(a + b - b*Tanh[c + d
*x]^2)^(3/2)) - (b*(33*a^2 + 40*a*b + 15*b^2)*Tanh[c + d*x])/(15*a^3*(a + b)^3*d*Sqrt[a + b - b*Tanh[c + d*x]^
2])

________________________________________________________________________________________

Rubi [A]  time = 0.191089, antiderivative size = 183, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {4128, 414, 527, 12, 377, 206} \[ -\frac{b \left (33 a^2+40 a b+15 b^2\right ) \tanh (c+d x)}{15 a^3 d (a+b)^3 \sqrt{a-b \tanh ^2(c+d x)+b}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a} \tanh (c+d x)}{\sqrt{a-b \tanh ^2(c+d x)+b}}\right )}{a^{7/2} d}-\frac{b (9 a+5 b) \tanh (c+d x)}{15 a^2 d (a+b)^2 \left (a-b \tanh ^2(c+d x)+b\right )^{3/2}}-\frac{b \tanh (c+d x)}{5 a d (a+b) \left (a-b \tanh ^2(c+d x)+b\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sech[c + d*x]^2)^(-7/2),x]

[Out]

ArcTanh[(Sqrt[a]*Tanh[c + d*x])/Sqrt[a + b - b*Tanh[c + d*x]^2]]/(a^(7/2)*d) - (b*Tanh[c + d*x])/(5*a*(a + b)*
d*(a + b - b*Tanh[c + d*x]^2)^(5/2)) - (b*(9*a + 5*b)*Tanh[c + d*x])/(15*a^2*(a + b)^2*d*(a + b - b*Tanh[c + d
*x]^2)^(3/2)) - (b*(33*a^2 + 40*a*b + 15*b^2)*Tanh[c + d*x])/(15*a^3*(a + b)^3*d*Sqrt[a + b - b*Tanh[c + d*x]^
2])

Rule 4128

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[ff/f, Subst[Int[(a + b + b*ff^2*x^2)^p/(1 + ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p},
x] && NeQ[a + b, 0] && NeQ[p, -1]

Rule 414

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

Rule 527

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[
((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 206

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

Rubi steps

\begin{align*} \int \frac{1}{\left (a+b \text{sech}^2(c+d x)\right )^{7/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right ) \left (a+b-b x^2\right )^{7/2}} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=-\frac{b \tanh (c+d x)}{5 a (a+b) d \left (a+b-b \tanh ^2(c+d x)\right )^{5/2}}-\frac{\operatorname{Subst}\left (\int \frac{-5 a-b-4 b x^2}{\left (1-x^2\right ) \left (a+b-b x^2\right )^{5/2}} \, dx,x,\tanh (c+d x)\right )}{5 a (a+b) d}\\ &=-\frac{b \tanh (c+d x)}{5 a (a+b) d \left (a+b-b \tanh ^2(c+d x)\right )^{5/2}}-\frac{b (9 a+5 b) \tanh (c+d x)}{15 a^2 (a+b)^2 d \left (a+b-b \tanh ^2(c+d x)\right )^{3/2}}+\frac{\operatorname{Subst}\left (\int \frac{15 a^2+12 a b+5 b^2+2 b (9 a+5 b) x^2}{\left (1-x^2\right ) \left (a+b-b x^2\right )^{3/2}} \, dx,x,\tanh (c+d x)\right )}{15 a^2 (a+b)^2 d}\\ &=-\frac{b \tanh (c+d x)}{5 a (a+b) d \left (a+b-b \tanh ^2(c+d x)\right )^{5/2}}-\frac{b (9 a+5 b) \tanh (c+d x)}{15 a^2 (a+b)^2 d \left (a+b-b \tanh ^2(c+d x)\right )^{3/2}}-\frac{b \left (33 a^2+40 a b+15 b^2\right ) \tanh (c+d x)}{15 a^3 (a+b)^3 d \sqrt{a+b-b \tanh ^2(c+d x)}}-\frac{\operatorname{Subst}\left (\int -\frac{15 (a+b)^3}{\left (1-x^2\right ) \sqrt{a+b-b x^2}} \, dx,x,\tanh (c+d x)\right )}{15 a^3 (a+b)^3 d}\\ &=-\frac{b \tanh (c+d x)}{5 a (a+b) d \left (a+b-b \tanh ^2(c+d x)\right )^{5/2}}-\frac{b (9 a+5 b) \tanh (c+d x)}{15 a^2 (a+b)^2 d \left (a+b-b \tanh ^2(c+d x)\right )^{3/2}}-\frac{b \left (33 a^2+40 a b+15 b^2\right ) \tanh (c+d x)}{15 a^3 (a+b)^3 d \sqrt{a+b-b \tanh ^2(c+d x)}}+\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right ) \sqrt{a+b-b x^2}} \, dx,x,\tanh (c+d x)\right )}{a^3 d}\\ &=-\frac{b \tanh (c+d x)}{5 a (a+b) d \left (a+b-b \tanh ^2(c+d x)\right )^{5/2}}-\frac{b (9 a+5 b) \tanh (c+d x)}{15 a^2 (a+b)^2 d \left (a+b-b \tanh ^2(c+d x)\right )^{3/2}}-\frac{b \left (33 a^2+40 a b+15 b^2\right ) \tanh (c+d x)}{15 a^3 (a+b)^3 d \sqrt{a+b-b \tanh ^2(c+d x)}}+\frac{\operatorname{Subst}\left (\int \frac{1}{1-a x^2} \, dx,x,\frac{\tanh (c+d x)}{\sqrt{a+b-b \tanh ^2(c+d x)}}\right )}{a^3 d}\\ &=\frac{\tanh ^{-1}\left (\frac{\sqrt{a} \tanh (c+d x)}{\sqrt{a+b-b \tanh ^2(c+d x)}}\right )}{a^{7/2} d}-\frac{b \tanh (c+d x)}{5 a (a+b) d \left (a+b-b \tanh ^2(c+d x)\right )^{5/2}}-\frac{b (9 a+5 b) \tanh (c+d x)}{15 a^2 (a+b)^2 d \left (a+b-b \tanh ^2(c+d x)\right )^{3/2}}-\frac{b \left (33 a^2+40 a b+15 b^2\right ) \tanh (c+d x)}{15 a^3 (a+b)^3 d \sqrt{a+b-b \tanh ^2(c+d x)}}\\ \end{align*}

Mathematica [A]  time = 7.45835, size = 330, normalized size = 1.8 \[ \frac{\text{sech}^7(c+d x) \left (\frac{15}{4} e^{-7 (c+d x)} \left (a \left (e^{2 (c+d x)}+1\right )^2+4 b e^{2 (c+d x)}\right )^{7/2} \left (\tanh ^{-1}\left (\frac{a e^{2 (c+d x)}+a+2 b}{\sqrt{a} \sqrt{a \left (e^{2 (c+d x)}+1\right )^2+4 b e^{2 (c+d x)}}}\right )-\tanh ^{-1}\left (\frac{a e^{2 (c+d x)}+a+2 b e^{2 (c+d x)}}{\sqrt{a} \sqrt{a \left (e^{2 (c+d x)}+1\right )^2+4 b e^{2 (c+d x)}}}\right )\right )-\frac{4 \sqrt{a} b \sinh (c+d x) (a \cosh (2 (c+d x))+a+2 b) \left (a^2 \left (45 a^2+60 a b+23 b^2\right ) \cosh (4 (c+d x))+4 a \left (135 a^2 b+45 a^3+117 a b^2+35 b^3\right ) \cosh (2 (c+d x))+709 a^2 b^2+480 a^3 b+135 a^4+460 a b^3+120 b^4\right )}{(a+b)^3}\right )}{960 a^{7/2} d \left (a+b \text{sech}^2(c+d x)\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sech[c + d*x]^2)^(-7/2),x]

[Out]

(Sech[c + d*x]^7*((15*(4*b*E^(2*(c + d*x)) + a*(1 + E^(2*(c + d*x)))^2)^(7/2)*(ArcTanh[(a + 2*b + a*E^(2*(c +
d*x)))/(Sqrt[a]*Sqrt[4*b*E^(2*(c + d*x)) + a*(1 + E^(2*(c + d*x)))^2])] - ArcTanh[(a + a*E^(2*(c + d*x)) + 2*b
*E^(2*(c + d*x)))/(Sqrt[a]*Sqrt[4*b*E^(2*(c + d*x)) + a*(1 + E^(2*(c + d*x)))^2])]))/(4*E^(7*(c + d*x))) - (4*
Sqrt[a]*b*(a + 2*b + a*Cosh[2*(c + d*x)])*(135*a^4 + 480*a^3*b + 709*a^2*b^2 + 460*a*b^3 + 120*b^4 + 4*a*(45*a
^3 + 135*a^2*b + 117*a*b^2 + 35*b^3)*Cosh[2*(c + d*x)] + a^2*(45*a^2 + 60*a*b + 23*b^2)*Cosh[4*(c + d*x)])*Sin
h[c + d*x])/(a + b)^3))/(960*a^(7/2)*d*(a + b*Sech[c + d*x]^2)^(7/2))

________________________________________________________________________________________

Maple [F]  time = 0.152, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b \left ({\rm sech} \left (dx+c\right ) \right ) ^{2} \right ) ^{-{\frac{7}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*sech(d*x+c)^2)^(7/2),x)

[Out]

int(1/(a+b*sech(d*x+c)^2)^(7/2),x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \operatorname{sech}\left (d x + c\right )^{2} + a\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sech(d*x+c)^2)^(7/2),x, algorithm="maxima")

[Out]

integrate((b*sech(d*x + c)^2 + a)^(-7/2), x)

________________________________________________________________________________________

Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sech(d*x+c)^2)^(7/2),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sech(d*x+c)**2)**(7/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \operatorname{sech}\left (d x + c\right )^{2} + a\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sech(d*x+c)^2)^(7/2),x, algorithm="giac")

[Out]

integrate((b*sech(d*x + c)^2 + a)^(-7/2), x)