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

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

[Out]

arctanh(a^(1/2)*tanh(d*x+c)/(a+b-b*tanh(d*x+c)^2)^(1/2))/a^(7/2)/d-1/15*b*(33*a^2+40*a*b+15*b^2)*tanh(d*x+c)/a
^3/(a+b)^3/d/(a+b-b*tanh(d*x+c)^2)^(1/2)-1/5*b*tanh(d*x+c)/a/(a+b)/d/(a+b-b*tanh(d*x+c)^2)^(5/2)-1/15*b*(9*a+5
*b)*tanh(d*x+c)/a^2/(a+b)^2/d/(a+b-b*tanh(d*x+c)^2)^(3/2)

________________________________________________________________________________________

Rubi [A]  time = 0.19, antiderivative size = 183, normalized size of antiderivative = 1.00, 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 12

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

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])

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 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 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]

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 = 7.98, size = 330, normalized size = 1.80 \[ \frac {\text {sech}^7(c+d x) \left (\frac {15}{4} e^{-7 (c+d x)} \left (a \left (e^{2 (c+d x)}+1\right )^2+4 b e^{2 (c+d x)}\right )^{7/2} \left (\tanh ^{-1}\left (\frac {a e^{2 (c+d x)}+a+2 b}{\sqrt {a} \sqrt {a \left (e^{2 (c+d x)}+1\right )^2+4 b e^{2 (c+d x)}}}\right )-\tanh ^{-1}\left (\frac {a e^{2 (c+d x)}+a+2 b e^{2 (c+d x)}}{\sqrt {a} \sqrt {a \left (e^{2 (c+d x)}+1\right )^2+4 b e^{2 (c+d x)}}}\right )\right )-\frac {4 \sqrt {a} b \sinh (c+d x) (a \cosh (2 (c+d x))+a+2 b) \left (135 a^4+480 a^3 b+a^2 \left (45 a^2+60 a b+23 b^2\right ) \cosh (4 (c+d x))+709 a^2 b^2+4 a \left (45 a^3+135 a^2 b+117 a b^2+35 b^3\right ) \cosh (2 (c+d x))+460 a b^3+120 b^4\right )}{(a+b)^3}\right )}{960 a^{7/2} d \left (a+b \text {sech}^2(c+d x)\right )^{7/2}} \]

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))

________________________________________________________________________________________

fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

________________________________________________________________________________________

giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Eval
uation time: 1.44Error: Bad Argument Type

________________________________________________________________________________________

maple [F]  time = 0.60, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a +b \mathrm {sech}\left (d x +c \right )^{2}\right )^{\frac {7}{2}}}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \operatorname {sech}\left (d x + c\right )^{2} + a\right )}^{\frac {7}{2}}}\,{d x} \]

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)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (a+\frac {b}{{\mathrm {cosh}\left (c+d\,x\right )}^2}\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b \operatorname {sech}^{2}{\left (c + d x \right )}\right )^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________