3.13 \(\int (b \tanh (c+d x))^{7/2} \, dx\)
Optimal. Leaf size=97 \[ \frac{b^{7/2} \tanh ^{-1}\left (\frac{\sqrt{b \tanh (c+d x)}}{\sqrt{b}}\right )}{d}-\frac{2 b^3 \sqrt{b \tanh (c+d x)}}{d}+\frac{b^{7/2} \tan ^{-1}\left (\frac{\sqrt{b \tanh (c+d x)}}{\sqrt{b}}\right )}{d}-\frac{2 b (b \tanh (c+d x))^{5/2}}{5 d} \]
[Out]
(b^(7/2)*ArcTan[Sqrt[b*Tanh[c + d*x]]/Sqrt[b]])/d + (b^(7/2)*ArcTanh[Sqrt[b*Tanh[c + d*x]]/Sqrt[b]])/d - (2*b^
3*Sqrt[b*Tanh[c + d*x]])/d - (2*b*(b*Tanh[c + d*x])^(5/2))/(5*d)
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Rubi [A] time = 0.0698837, antiderivative size = 97, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 6, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used =
{3473, 3476, 329, 212, 206, 203} \[ \frac{b^{7/2} \tanh ^{-1}\left (\frac{\sqrt{b \tanh (c+d x)}}{\sqrt{b}}\right )}{d}-\frac{2 b^3 \sqrt{b \tanh (c+d x)}}{d}+\frac{b^{7/2} \tan ^{-1}\left (\frac{\sqrt{b \tanh (c+d x)}}{\sqrt{b}}\right )}{d}-\frac{2 b (b \tanh (c+d x))^{5/2}}{5 d} \]
Antiderivative was successfully verified.
[In]
Int[(b*Tanh[c + d*x])^(7/2),x]
[Out]
(b^(7/2)*ArcTan[Sqrt[b*Tanh[c + d*x]]/Sqrt[b]])/d + (b^(7/2)*ArcTanh[Sqrt[b*Tanh[c + d*x]]/Sqrt[b]])/d - (2*b^
3*Sqrt[b*Tanh[c + d*x]])/d - (2*b*(b*Tanh[c + d*x])^(5/2))/(5*d)
Rule 3473
Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[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]
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 212
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/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 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 (b \tanh (c+d x))^{7/2} \, dx &=-\frac{2 b (b \tanh (c+d x))^{5/2}}{5 d}+b^2 \int (b \tanh (c+d x))^{3/2} \, dx\\ &=-\frac{2 b^3 \sqrt{b \tanh (c+d x)}}{d}-\frac{2 b (b \tanh (c+d x))^{5/2}}{5 d}+b^4 \int \frac{1}{\sqrt{b \tanh (c+d x)}} \, dx\\ &=-\frac{2 b^3 \sqrt{b \tanh (c+d x)}}{d}-\frac{2 b (b \tanh (c+d x))^{5/2}}{5 d}-\frac{b^5 \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \left (-b^2+x^2\right )} \, dx,x,b \tanh (c+d x)\right )}{d}\\ &=-\frac{2 b^3 \sqrt{b \tanh (c+d x)}}{d}-\frac{2 b (b \tanh (c+d x))^{5/2}}{5 d}-\frac{\left (2 b^5\right ) \operatorname{Subst}\left (\int \frac{1}{-b^2+x^4} \, dx,x,\sqrt{b \tanh (c+d x)}\right )}{d}\\ &=-\frac{2 b^3 \sqrt{b \tanh (c+d x)}}{d}-\frac{2 b (b \tanh (c+d x))^{5/2}}{5 d}+\frac{b^4 \operatorname{Subst}\left (\int \frac{1}{b-x^2} \, dx,x,\sqrt{b \tanh (c+d x)}\right )}{d}+\frac{b^4 \operatorname{Subst}\left (\int \frac{1}{b+x^2} \, dx,x,\sqrt{b \tanh (c+d x)}\right )}{d}\\ &=\frac{b^{7/2} \tan ^{-1}\left (\frac{\sqrt{b \tanh (c+d x)}}{\sqrt{b}}\right )}{d}+\frac{b^{7/2} \tanh ^{-1}\left (\frac{\sqrt{b \tanh (c+d x)}}{\sqrt{b}}\right )}{d}-\frac{2 b^3 \sqrt{b \tanh (c+d x)}}{d}-\frac{2 b (b \tanh (c+d x))^{5/2}}{5 d}\\ \end{align*}
Mathematica [A] time = 0.231609, size = 83, normalized size = 0.86 \[ \frac{b^3 \sqrt{b \tanh (c+d x)} \left (-2 \tanh ^{\frac{5}{2}}(c+d x)+5 \tanh ^{-1}\left (\sqrt{\tanh (c+d x)}\right )-10 \sqrt{\tanh (c+d x)}+5 \tan ^{-1}\left (\sqrt{\tanh (c+d x)}\right )\right )}{5 d \sqrt{\tanh (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[(b*Tanh[c + d*x])^(7/2),x]
[Out]
(b^3*Sqrt[b*Tanh[c + d*x]]*(5*ArcTan[Sqrt[Tanh[c + d*x]]] + 5*ArcTanh[Sqrt[Tanh[c + d*x]]] - 10*Sqrt[Tanh[c +
d*x]] - 2*Tanh[c + d*x]^(5/2)))/(5*d*Sqrt[Tanh[c + d*x]])
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Maple [A] time = 0.032, size = 80, normalized size = 0.8 \begin{align*}{\frac{1}{d}{b}^{{\frac{7}{2}}}\arctan \left ({\sqrt{b\tanh \left ( dx+c \right ) }{\frac{1}{\sqrt{b}}}} \right ) }+{\frac{1}{d}{b}^{{\frac{7}{2}}}{\it Artanh} \left ({\sqrt{b\tanh \left ( dx+c \right ) }{\frac{1}{\sqrt{b}}}} \right ) }-2\,{\frac{{b}^{3}\sqrt{b\tanh \left ( dx+c \right ) }}{d}}-{\frac{2\,b}{5\,d} \left ( b\tanh \left ( dx+c \right ) \right ) ^{{\frac{5}{2}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((b*tanh(d*x+c))^(7/2),x)
[Out]
b^(7/2)*arctan((b*tanh(d*x+c))^(1/2)/b^(1/2))/d+b^(7/2)*arctanh((b*tanh(d*x+c))^(1/2)/b^(1/2))/d-2*b^3*(b*tanh
(d*x+c))^(1/2)/d-2/5*b*(b*tanh(d*x+c))^(5/2)/d
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \tanh \left (d x + c\right )\right )^{\frac{7}{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*tanh(d*x+c))^(7/2),x, algorithm="maxima")
[Out]
integrate((b*tanh(d*x + c))^(7/2), x)
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Fricas [B] time = 2.67289, size = 4136, normalized size = 42.64 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*tanh(d*x+c))^(7/2),x, algorithm="fricas")
[Out]
[-1/20*(10*(b^3*cosh(d*x + c)^4 + 4*b^3*cosh(d*x + c)*sinh(d*x + c)^3 + b^3*sinh(d*x + c)^4 + 2*b^3*cosh(d*x +
c)^2 + b^3 + 2*(3*b^3*cosh(d*x + c)^2 + b^3)*sinh(d*x + c)^2 + 4*(b^3*cosh(d*x + c)^3 + b^3*cosh(d*x + c))*si
nh(d*x + c))*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*sinh(d*x + c)^2 - b))
- 5*(b^3*cosh(d*x + c)^4 + 4*b^3*cosh(d*x + c)*sinh(d*x + c)^3 + b^3*sinh(d*x + c)^4 + 2*b^3*cosh(d*x + c)^2
+ b^3 + 2*(3*b^3*cosh(d*x + c)^2 + b^3)*sinh(d*x + c)^2 + 4*(b^3*cosh(d*x + c)^3 + b^3*cosh(d*x + c))*sinh(d*x
+ c))*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*b^3*cosh(d*x + c)^4 + 12*b^3*cosh(d*x + c)*sinh(d*x + c)^3 + 3*b^3*sinh(d*x + c)^4 + 4*b^3*cosh(d*x + c
)^2 + 3*b^3 + 2*(9*b^3*cosh(d*x + c)^2 + 2*b^3)*sinh(d*x + c)^2 + 4*(3*b^3*cosh(d*x + c)^3 + 2*b^3*cosh(d*x +
c))*sinh(d*x + c))*sqrt(b*sinh(d*x + c)/cosh(d*x + c)))/(d*cosh(d*x + c)^4 + 4*d*cosh(d*x + c)*sinh(d*x + c)^3
+ d*sinh(d*x + c)^4 + 2*d*cosh(d*x + c)^2 + 2*(3*d*cosh(d*x + c)^2 + d)*sinh(d*x + c)^2 + 4*(d*cosh(d*x + c)^
3 + d*cosh(d*x + c))*sinh(d*x + c) + d), -1/20*(10*(b^3*cosh(d*x + c)^4 + 4*b^3*cosh(d*x + c)*sinh(d*x + c)^3
+ b^3*sinh(d*x + c)^4 + 2*b^3*cosh(d*x + c)^2 + b^3 + 2*(3*b^3*cosh(d*x + c)^2 + b^3)*sinh(d*x + c)^2 + 4*(b^3
*cosh(d*x + c)^3 + b^3*cosh(d*x + c))*sinh(d*x + c))*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*(b^3*cosh(d*x + c)^4 + 4*
b^3*cosh(d*x + c)*sinh(d*x + c)^3 + b^3*sinh(d*x + c)^4 + 2*b^3*cosh(d*x + c)^2 + b^3 + 2*(3*b^3*cosh(d*x + c)
^2 + b^3)*sinh(d*x + c)^2 + 4*(b^3*cosh(d*x + c)^3 + b^3*cosh(d*x + c))*sinh(d*x + c))*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*b^3*cosh(d*x + c)^4 + 12*b^3*cosh(d*x + c)*sinh(d*x +
c)^3 + 3*b^3*sinh(d*x + c)^4 + 4*b^3*cosh(d*x + c)^2 + 3*b^3 + 2*(9*b^3*cosh(d*x + c)^2 + 2*b^3)*sinh(d*x + c
)^2 + 4*(3*b^3*cosh(d*x + c)^3 + 2*b^3*cosh(d*x + c))*sinh(d*x + c))*sqrt(b*sinh(d*x + c)/cosh(d*x + c)))/(d*c
osh(d*x + c)^4 + 4*d*cosh(d*x + c)*sinh(d*x + c)^3 + d*sinh(d*x + c)^4 + 2*d*cosh(d*x + c)^2 + 2*(3*d*cosh(d*x
+ c)^2 + d)*sinh(d*x + c)^2 + 4*(d*cosh(d*x + c)^3 + d*cosh(d*x + c))*sinh(d*x + c) + d)]
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \tanh{\left (c + d x \right )}\right )^{\frac{7}{2}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*tanh(d*x+c))**(7/2),x)
[Out]
Integral((b*tanh(c + d*x))**(7/2), x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \tanh \left (d x + c\right )\right )^{\frac{7}{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*tanh(d*x+c))^(7/2),x, algorithm="giac")
[Out]
integrate((b*tanh(d*x + c))^(7/2), x)