3.20 \(\int \frac {1}{(b \tanh (c+d x))^{7/2}} \, dx\)
Optimal. Leaf size=100 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {b \tanh (c+d x)}}{\sqrt {b}}\right )}{b^{7/2} d}-\frac {\tan ^{-1}\left (\frac {\sqrt {b \tanh (c+d x)}}{\sqrt {b}}\right )}{b^{7/2} d}-\frac {2}{b^3 d \sqrt {b \tanh (c+d x)}}-\frac {2}{5 b d (b \tanh (c+d x))^{5/2}} \]
[Out]
-arctan((b*tanh(d*x+c))^(1/2)/b^(1/2))/b^(7/2)/d+arctanh((b*tanh(d*x+c))^(1/2)/b^(1/2))/b^(7/2)/d-2/b^3/d/(b*t
anh(d*x+c))^(1/2)-2/5/b/d/(b*tanh(d*x+c))^(5/2)
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Rubi [A] time = 0.07, antiderivative size = 100, normalized size of antiderivative = 1.00,
number of steps used = 7, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used =
{3474, 3476, 329, 298, 203, 206} \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {b \tanh (c+d x)}}{\sqrt {b}}\right )}{b^{7/2} d}-\frac {2}{b^3 d \sqrt {b \tanh (c+d x)}}-\frac {\tan ^{-1}\left (\frac {\sqrt {b \tanh (c+d x)}}{\sqrt {b}}\right )}{b^{7/2} d}-\frac {2}{5 b d (b \tanh (c+d x))^{5/2}} \]
Antiderivative was successfully verified.
[In]
Int[(b*Tanh[c + d*x])^(-7/2),x]
[Out]
-(ArcTan[Sqrt[b*Tanh[c + d*x]]/Sqrt[b]]/(b^(7/2)*d)) + ArcTanh[Sqrt[b*Tanh[c + d*x]]/Sqrt[b]]/(b^(7/2)*d) - 2/
(5*b*d*(b*Tanh[c + d*x])^(5/2)) - 2/(b^3*d*Sqrt[b*Tanh[c + d*x]])
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 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 298
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
& !GtQ[a/b, 0]
Rule 329
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]
Rule 3474
Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Tan[c + d*x])^(n + 1)/(b*d*(n + 1)), x] - Dist[
1/b^2, Int[(b*Tan[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1]
Rule 3476
Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] && !IntegerQ[n]
Rubi steps
\begin {align*} \int \frac {1}{(b \tanh (c+d x))^{7/2}} \, dx &=-\frac {2}{5 b d (b \tanh (c+d x))^{5/2}}+\frac {\int \frac {1}{(b \tanh (c+d x))^{3/2}} \, dx}{b^2}\\ &=-\frac {2}{5 b d (b \tanh (c+d x))^{5/2}}-\frac {2}{b^3 d \sqrt {b \tanh (c+d x)}}+\frac {\int \sqrt {b \tanh (c+d x)} \, dx}{b^4}\\ &=-\frac {2}{5 b d (b \tanh (c+d x))^{5/2}}-\frac {2}{b^3 d \sqrt {b \tanh (c+d x)}}-\frac {\operatorname {Subst}\left (\int \frac {\sqrt {x}}{-b^2+x^2} \, dx,x,b \tanh (c+d x)\right )}{b^3 d}\\ &=-\frac {2}{5 b d (b \tanh (c+d x))^{5/2}}-\frac {2}{b^3 d \sqrt {b \tanh (c+d x)}}-\frac {2 \operatorname {Subst}\left (\int \frac {x^2}{-b^2+x^4} \, dx,x,\sqrt {b \tanh (c+d x)}\right )}{b^3 d}\\ &=-\frac {2}{5 b d (b \tanh (c+d x))^{5/2}}-\frac {2}{b^3 d \sqrt {b \tanh (c+d x)}}+\frac {\operatorname {Subst}\left (\int \frac {1}{b-x^2} \, dx,x,\sqrt {b \tanh (c+d x)}\right )}{b^3 d}-\frac {\operatorname {Subst}\left (\int \frac {1}{b+x^2} \, dx,x,\sqrt {b \tanh (c+d x)}\right )}{b^3 d}\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt {b \tanh (c+d x)}}{\sqrt {b}}\right )}{b^{7/2} d}+\frac {\tanh ^{-1}\left (\frac {\sqrt {b \tanh (c+d x)}}{\sqrt {b}}\right )}{b^{7/2} d}-\frac {2}{5 b d (b \tanh (c+d x))^{5/2}}-\frac {2}{b^3 d \sqrt {b \tanh (c+d x)}}\\ \end {align*}
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Mathematica [C] time = 0.12, size = 38, normalized size = 0.38 \[ -\frac {2 \, _2F_1\left (-\frac {5}{4},1;-\frac {1}{4};\tanh ^2(c+d x)\right )}{5 b d (b \tanh (c+d x))^{5/2}} \]
Antiderivative was successfully verified.
[In]
Integrate[(b*Tanh[c + d*x])^(-7/2),x]
[Out]
(-2*Hypergeometric2F1[-5/4, 1, -1/4, Tanh[c + d*x]^2])/(5*b*d*(b*Tanh[c + d*x])^(5/2))
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fricas [B] time = 0.70, size = 2144, normalized size = 21.44 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(b*tanh(d*x+c))^(7/2),x, algorithm="fricas")
[Out]
[-1/20*(10*(cosh(d*x + c)^6 + 6*cosh(d*x + c)*sinh(d*x + c)^5 + sinh(d*x + c)^6 + 3*(5*cosh(d*x + c)^2 - 1)*si
nh(d*x + c)^4 - 3*cosh(d*x + c)^4 + 4*(5*cosh(d*x + c)^3 - 3*cosh(d*x + c))*sinh(d*x + c)^3 + 3*(5*cosh(d*x +
c)^4 - 6*cosh(d*x + c)^2 + 1)*sinh(d*x + c)^2 + 3*cosh(d*x + c)^2 + 6*(cosh(d*x + c)^5 - 2*cosh(d*x + c)^3 + c
osh(d*x + c))*sinh(d*x + c) - 1)*sqrt(-b)*arctan((cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x +
c)^2)*sqrt(-b)*sqrt(b*sinh(d*x + c)/cosh(d*x + c))/(b*cosh(d*x + c)^2 + 2*b*cosh(d*x + c)*sinh(d*x + c) + b*s
inh(d*x + c)^2 - b)) + 5*(cosh(d*x + c)^6 + 6*cosh(d*x + c)*sinh(d*x + c)^5 + sinh(d*x + c)^6 + 3*(5*cosh(d*x
+ c)^2 - 1)*sinh(d*x + c)^4 - 3*cosh(d*x + c)^4 + 4*(5*cosh(d*x + c)^3 - 3*cosh(d*x + c))*sinh(d*x + c)^3 + 3*
(5*cosh(d*x + c)^4 - 6*cosh(d*x + c)^2 + 1)*sinh(d*x + c)^2 + 3*cosh(d*x + c)^2 + 6*(cosh(d*x + c)^5 - 2*cosh(
d*x + c)^3 + cosh(d*x + c))*sinh(d*x + c) - 1)*sqrt(-b)*log(-(b*cosh(d*x + c)^4 + 4*b*cosh(d*x + c)^3*sinh(d*x
+ c) + 6*b*cosh(d*x + c)^2*sinh(d*x + c)^2 + 4*b*cosh(d*x + c)*sinh(d*x + c)^3 + b*sinh(d*x + c)^4 + 2*(cosh(
d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2 + 1)*sqrt(-b)*sqrt(b*sinh(d*x + c)/cosh(d*x + c))
- 2*b)/(cosh(d*x + c)^4 + 4*cosh(d*x + c)^3*sinh(d*x + c) + 6*cosh(d*x + c)^2*sinh(d*x + c)^2 + 4*cosh(d*x +
c)*sinh(d*x + c)^3 + sinh(d*x + c)^4)) + 16*(3*cosh(d*x + c)^6 + 18*cosh(d*x + c)*sinh(d*x + c)^5 + 3*sinh(d*x
+ c)^6 + (45*cosh(d*x + c)^2 - 1)*sinh(d*x + c)^4 - cosh(d*x + c)^4 + 4*(15*cosh(d*x + c)^3 - cosh(d*x + c))*
sinh(d*x + c)^3 + (45*cosh(d*x + c)^4 - 6*cosh(d*x + c)^2 - 1)*sinh(d*x + c)^2 - cosh(d*x + c)^2 + 2*(9*cosh(d
*x + c)^5 - 2*cosh(d*x + c)^3 - cosh(d*x + c))*sinh(d*x + c) + 3)*sqrt(b*sinh(d*x + c)/cosh(d*x + c)))/(b^4*d*
cosh(d*x + c)^6 + 6*b^4*d*cosh(d*x + c)*sinh(d*x + c)^5 + b^4*d*sinh(d*x + c)^6 - 3*b^4*d*cosh(d*x + c)^4 + 3*
b^4*d*cosh(d*x + c)^2 - b^4*d + 3*(5*b^4*d*cosh(d*x + c)^2 - b^4*d)*sinh(d*x + c)^4 + 4*(5*b^4*d*cosh(d*x + c)
^3 - 3*b^4*d*cosh(d*x + c))*sinh(d*x + c)^3 + 3*(5*b^4*d*cosh(d*x + c)^4 - 6*b^4*d*cosh(d*x + c)^2 + b^4*d)*si
nh(d*x + c)^2 + 6*(b^4*d*cosh(d*x + c)^5 - 2*b^4*d*cosh(d*x + c)^3 + b^4*d*cosh(d*x + c))*sinh(d*x + c)), 1/20
*(10*(cosh(d*x + c)^6 + 6*cosh(d*x + c)*sinh(d*x + c)^5 + sinh(d*x + c)^6 + 3*(5*cosh(d*x + c)^2 - 1)*sinh(d*x
+ c)^4 - 3*cosh(d*x + c)^4 + 4*(5*cosh(d*x + c)^3 - 3*cosh(d*x + c))*sinh(d*x + c)^3 + 3*(5*cosh(d*x + c)^4 -
6*cosh(d*x + c)^2 + 1)*sinh(d*x + c)^2 + 3*cosh(d*x + c)^2 + 6*(cosh(d*x + c)^5 - 2*cosh(d*x + c)^3 + cosh(d*
x + c))*sinh(d*x + c) - 1)*sqrt(b)*arctan(sqrt(b)*sqrt(b*sinh(d*x + c)/cosh(d*x + c))/(b*cosh(d*x + c)^2 + 2*b
*cosh(d*x + c)*sinh(d*x + c) + b*sinh(d*x + c)^2 - b)) + 5*(cosh(d*x + c)^6 + 6*cosh(d*x + c)*sinh(d*x + c)^5
+ sinh(d*x + c)^6 + 3*(5*cosh(d*x + c)^2 - 1)*sinh(d*x + c)^4 - 3*cosh(d*x + c)^4 + 4*(5*cosh(d*x + c)^3 - 3*c
osh(d*x + c))*sinh(d*x + c)^3 + 3*(5*cosh(d*x + c)^4 - 6*cosh(d*x + c)^2 + 1)*sinh(d*x + c)^2 + 3*cosh(d*x + c
)^2 + 6*(cosh(d*x + c)^5 - 2*cosh(d*x + c)^3 + cosh(d*x + c))*sinh(d*x + c) - 1)*sqrt(b)*log(2*b*cosh(d*x + c)
^4 + 8*b*cosh(d*x + c)^3*sinh(d*x + c) + 12*b*cosh(d*x + c)^2*sinh(d*x + c)^2 + 8*b*cosh(d*x + c)*sinh(d*x + c
)^3 + 2*b*sinh(d*x + c)^4 + 2*(cosh(d*x + c)^4 + 4*cosh(d*x + c)*sinh(d*x + c)^3 + sinh(d*x + c)^4 + (6*cosh(d
*x + c)^2 + 1)*sinh(d*x + c)^2 + cosh(d*x + c)^2 + 2*(2*cosh(d*x + c)^3 + cosh(d*x + c))*sinh(d*x + c))*sqrt(b
)*sqrt(b*sinh(d*x + c)/cosh(d*x + c)) - b) - 16*(3*cosh(d*x + c)^6 + 18*cosh(d*x + c)*sinh(d*x + c)^5 + 3*sinh
(d*x + c)^6 + (45*cosh(d*x + c)^2 - 1)*sinh(d*x + c)^4 - cosh(d*x + c)^4 + 4*(15*cosh(d*x + c)^3 - cosh(d*x +
c))*sinh(d*x + c)^3 + (45*cosh(d*x + c)^4 - 6*cosh(d*x + c)^2 - 1)*sinh(d*x + c)^2 - cosh(d*x + c)^2 + 2*(9*co
sh(d*x + c)^5 - 2*cosh(d*x + c)^3 - cosh(d*x + c))*sinh(d*x + c) + 3)*sqrt(b*sinh(d*x + c)/cosh(d*x + c)))/(b^
4*d*cosh(d*x + c)^6 + 6*b^4*d*cosh(d*x + c)*sinh(d*x + c)^5 + b^4*d*sinh(d*x + c)^6 - 3*b^4*d*cosh(d*x + c)^4
+ 3*b^4*d*cosh(d*x + c)^2 - b^4*d + 3*(5*b^4*d*cosh(d*x + c)^2 - b^4*d)*sinh(d*x + c)^4 + 4*(5*b^4*d*cosh(d*x
+ c)^3 - 3*b^4*d*cosh(d*x + c))*sinh(d*x + c)^3 + 3*(5*b^4*d*cosh(d*x + c)^4 - 6*b^4*d*cosh(d*x + c)^2 + b^4*d
)*sinh(d*x + c)^2 + 6*(b^4*d*cosh(d*x + c)^5 - 2*b^4*d*cosh(d*x + c)^3 + b^4*d*cosh(d*x + c))*sinh(d*x + c))]
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giac [B] time = 0.72, size = 307, normalized size = 3.07 \[ \frac {\frac {5 \, \pi + 5 \, \log \left ({\left | b \right |}\right ) + 48}{b^{\frac {7}{2}}} - \frac {20 \, \arctan \left (-\frac {\sqrt {b} e^{\left (2 \, d x + 2 \, c\right )} - \sqrt {b e^{\left (4 \, d x + 4 \, c\right )} - b}}{\sqrt {b}}\right )}{b^{\frac {7}{2}}} - \frac {10 \, \log \left ({\left | -\sqrt {b} e^{\left (2 \, d x + 2 \, c\right )} + \sqrt {b e^{\left (4 \, d x + 4 \, c\right )} - b} \right |}\right )}{b^{\frac {7}{2}}} + \frac {32 \, {\left (5 \, {\left (\sqrt {b} e^{\left (2 \, d x + 2 \, c\right )} - \sqrt {b e^{\left (4 \, d x + 4 \, c\right )} - b}\right )}^{4} - 10 \, {\left (\sqrt {b} e^{\left (2 \, d x + 2 \, c\right )} - \sqrt {b e^{\left (4 \, d x + 4 \, c\right )} - b}\right )}^{3} \sqrt {b} + 20 \, {\left (\sqrt {b} e^{\left (2 \, d x + 2 \, c\right )} - \sqrt {b e^{\left (4 \, d x + 4 \, c\right )} - b}\right )}^{2} b - 10 \, {\left (\sqrt {b} e^{\left (2 \, d x + 2 \, c\right )} - \sqrt {b e^{\left (4 \, d x + 4 \, c\right )} - b}\right )} b^{\frac {3}{2}} + 3 \, b^{2}\right )}}{{\left (\sqrt {b} e^{\left (2 \, d x + 2 \, c\right )} - \sqrt {b e^{\left (4 \, d x + 4 \, c\right )} - b} - \sqrt {b}\right )}^{5} b^{3}}}{20 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(b*tanh(d*x+c))^(7/2),x, algorithm="giac")
[Out]
1/20*((5*pi + 5*log(abs(b)) + 48)/b^(7/2) - 20*arctan(-(sqrt(b)*e^(2*d*x + 2*c) - sqrt(b*e^(4*d*x + 4*c) - b))
/sqrt(b))/b^(7/2) - 10*log(abs(-sqrt(b)*e^(2*d*x + 2*c) + sqrt(b*e^(4*d*x + 4*c) - b)))/b^(7/2) + 32*(5*(sqrt(
b)*e^(2*d*x + 2*c) - sqrt(b*e^(4*d*x + 4*c) - b))^4 - 10*(sqrt(b)*e^(2*d*x + 2*c) - sqrt(b*e^(4*d*x + 4*c) - b
))^3*sqrt(b) + 20*(sqrt(b)*e^(2*d*x + 2*c) - sqrt(b*e^(4*d*x + 4*c) - b))^2*b - 10*(sqrt(b)*e^(2*d*x + 2*c) -
sqrt(b*e^(4*d*x + 4*c) - b))*b^(3/2) + 3*b^2)/((sqrt(b)*e^(2*d*x + 2*c) - sqrt(b*e^(4*d*x + 4*c) - b) - sqrt(b
))^5*b^3))/d
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maple [A] time = 0.08, size = 83, normalized size = 0.83 \[ -\frac {\arctan \left (\frac {\sqrt {b \tanh \left (d x +c \right )}}{\sqrt {b}}\right )}{b^{\frac {7}{2}} d}+\frac {\arctanh \left (\frac {\sqrt {b \tanh \left (d x +c \right )}}{\sqrt {b}}\right )}{b^{\frac {7}{2}} d}-\frac {2}{b^{3} d \sqrt {b \tanh \left (d x +c \right )}}-\frac {2}{5 b d \left (b \tanh \left (d x +c \right )\right )^{\frac {5}{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(b*tanh(d*x+c))^(7/2),x)
[Out]
-arctan((b*tanh(d*x+c))^(1/2)/b^(1/2))/b^(7/2)/d+arctanh((b*tanh(d*x+c))^(1/2)/b^(1/2))/b^(7/2)/d-2/b^3/d/(b*t
anh(d*x+c))^(1/2)-2/5/b/d/(b*tanh(d*x+c))^(5/2)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \tanh \left (d x + c\right )\right )^{\frac {7}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(b*tanh(d*x+c))^(7/2),x, algorithm="maxima")
[Out]
integrate((b*tanh(d*x + c))^(-7/2), x)
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mupad [B] time = 1.51, size = 80, normalized size = 0.80 \[ \frac {\mathrm {atanh}\left (\frac {\sqrt {b\,\mathrm {tanh}\left (c+d\,x\right )}}{\sqrt {b}}\right )}{b^{7/2}\,d}-\frac {\mathrm {atan}\left (\frac {\sqrt {b\,\mathrm {tanh}\left (c+d\,x\right )}}{\sqrt {b}}\right )}{b^{7/2}\,d}-\frac {\frac {2}{5\,b}+\frac {2\,{\mathrm {tanh}\left (c+d\,x\right )}^2}{b}}{d\,{\left (b\,\mathrm {tanh}\left (c+d\,x\right )\right )}^{5/2}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(b*tanh(c + d*x))^(7/2),x)
[Out]
atanh((b*tanh(c + d*x))^(1/2)/b^(1/2))/(b^(7/2)*d) - atan((b*tanh(c + d*x))^(1/2)/b^(1/2))/(b^(7/2)*d) - (2/(5
*b) + (2*tanh(c + d*x)^2)/b)/(d*(b*tanh(c + d*x))^(5/2))
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \tanh {\left (c + d x \right )}\right )^{\frac {7}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(b*tanh(d*x+c))**(7/2),x)
[Out]
Integral((b*tanh(c + d*x))**(-7/2), x)
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