3.263 \(\int \frac {\tanh (x)}{(a+b \tanh ^4(x))^{5/2}} \, dx\)

Optimal. Leaf size=118 \[ -\frac {3 a^2-b (5 a+2 b) \tanh ^2(x)}{6 a^2 (a+b)^2 \sqrt {a+b \tanh ^4(x)}}-\frac {a-b \tanh ^2(x)}{6 a (a+b) \left (a+b \tanh ^4(x)\right )^{3/2}}+\frac {\tanh ^{-1}\left (\frac {a+b \tanh ^2(x)}{\sqrt {a+b} \sqrt {a+b \tanh ^4(x)}}\right )}{2 (a+b)^{5/2}} \]

[Out]

1/2*arctanh((a+b*tanh(x)^2)/(a+b)^(1/2)/(a+b*tanh(x)^4)^(1/2))/(a+b)^(5/2)+1/6*(-3*a^2+b*(5*a+2*b)*tanh(x)^2)/
a^2/(a+b)^2/(a+b*tanh(x)^4)^(1/2)+1/6*(-a+b*tanh(x)^2)/a/(a+b)/(a+b*tanh(x)^4)^(3/2)

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Rubi [A]  time = 0.20, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {3670, 1248, 741, 823, 12, 725, 206} \[ -\frac {3 a^2-b (5 a+2 b) \tanh ^2(x)}{6 a^2 (a+b)^2 \sqrt {a+b \tanh ^4(x)}}+\frac {\tanh ^{-1}\left (\frac {a+b \tanh ^2(x)}{\sqrt {a+b} \sqrt {a+b \tanh ^4(x)}}\right )}{2 (a+b)^{5/2}}-\frac {a-b \tanh ^2(x)}{6 a (a+b) \left (a+b \tanh ^4(x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]/(a + b*Tanh[x]^4)^(5/2),x]

[Out]

ArcTanh[(a + b*Tanh[x]^2)/(Sqrt[a + b]*Sqrt[a + b*Tanh[x]^4])]/(2*(a + b)^(5/2)) - (a - b*Tanh[x]^2)/(6*a*(a +
 b)*(a + b*Tanh[x]^4)^(3/2)) - (3*a^2 - b*(5*a + 2*b)*Tanh[x]^2)/(6*a^2*(a + b)^2*Sqrt[a + b*Tanh[x]^4])

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 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
 (a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 741

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*(a*e + c*d*x)*(
a + c*x^2)^(p + 1))/(2*a*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*
Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a
, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
 && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 1248

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[(d + e*x)^q
*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x]

Rule 3670

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
 :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff)/f, Subst[Int[(((d*ff*x)/c)^m*(a + b*(ff*x)^n)^p)/(c^
2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

Rubi steps

\begin {align*} \int \frac {\tanh (x)}{\left (a+b \tanh ^4(x)\right )^{5/2}} \, dx &=\operatorname {Subst}\left (\int \frac {x}{\left (1-x^2\right ) \left (a+b x^4\right )^{5/2}} \, dx,x,\tanh (x)\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{(1-x) \left (a+b x^2\right )^{5/2}} \, dx,x,\tanh ^2(x)\right )\\ &=-\frac {a-b \tanh ^2(x)}{6 a (a+b) \left (a+b \tanh ^4(x)\right )^{3/2}}-\frac {\operatorname {Subst}\left (\int \frac {-3 a-2 b+2 b x}{(1-x) \left (a+b x^2\right )^{3/2}} \, dx,x,\tanh ^2(x)\right )}{6 a (a+b)}\\ &=-\frac {a-b \tanh ^2(x)}{6 a (a+b) \left (a+b \tanh ^4(x)\right )^{3/2}}-\frac {3 a^2-b (5 a+2 b) \tanh ^2(x)}{6 a^2 (a+b)^2 \sqrt {a+b \tanh ^4(x)}}+\frac {\operatorname {Subst}\left (\int \frac {3 a^2 b}{(1-x) \sqrt {a+b x^2}} \, dx,x,\tanh ^2(x)\right )}{6 a^2 b (a+b)^2}\\ &=-\frac {a-b \tanh ^2(x)}{6 a (a+b) \left (a+b \tanh ^4(x)\right )^{3/2}}-\frac {3 a^2-b (5 a+2 b) \tanh ^2(x)}{6 a^2 (a+b)^2 \sqrt {a+b \tanh ^4(x)}}+\frac {\operatorname {Subst}\left (\int \frac {1}{(1-x) \sqrt {a+b x^2}} \, dx,x,\tanh ^2(x)\right )}{2 (a+b)^2}\\ &=-\frac {a-b \tanh ^2(x)}{6 a (a+b) \left (a+b \tanh ^4(x)\right )^{3/2}}-\frac {3 a^2-b (5 a+2 b) \tanh ^2(x)}{6 a^2 (a+b)^2 \sqrt {a+b \tanh ^4(x)}}-\frac {\operatorname {Subst}\left (\int \frac {1}{a+b-x^2} \, dx,x,\frac {-a-b \tanh ^2(x)}{\sqrt {a+b \tanh ^4(x)}}\right )}{2 (a+b)^2}\\ &=\frac {\tanh ^{-1}\left (\frac {a+b \tanh ^2(x)}{\sqrt {a+b} \sqrt {a+b \tanh ^4(x)}}\right )}{2 (a+b)^{5/2}}-\frac {a-b \tanh ^2(x)}{6 a (a+b) \left (a+b \tanh ^4(x)\right )^{3/2}}-\frac {3 a^2-b (5 a+2 b) \tanh ^2(x)}{6 a^2 (a+b)^2 \sqrt {a+b \tanh ^4(x)}}\\ \end {align*}

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Mathematica [A]  time = 0.85, size = 113, normalized size = 0.96 \[ \frac {1}{6} \left (\frac {-3 a^2 b \tanh ^4(x)-a^2 (4 a+b)+b^2 (5 a+2 b) \tanh ^6(x)+3 a b (2 a+b) \tanh ^2(x)}{a^2 (a+b)^2 \left (a+b \tanh ^4(x)\right )^{3/2}}+\frac {3 \tanh ^{-1}\left (\frac {a+b \tanh ^2(x)}{\sqrt {a+b} \sqrt {a+b \tanh ^4(x)}}\right )}{(a+b)^{5/2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[x]/(a + b*Tanh[x]^4)^(5/2),x]

[Out]

((3*ArcTanh[(a + b*Tanh[x]^2)/(Sqrt[a + b]*Sqrt[a + b*Tanh[x]^4])])/(a + b)^(5/2) + (-(a^2*(4*a + b)) + 3*a*b*
(2*a + b)*Tanh[x]^2 - 3*a^2*b*Tanh[x]^4 + b^2*(5*a + 2*b)*Tanh[x]^6)/(a^2*(a + b)^2*(a + b*Tanh[x]^4)^(3/2)))/
6

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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(tanh(x)/(a+b*tanh(x)^4)^(5/2),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tanh \relax (x)}{{\left (b \tanh \relax (x)^{4} + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)/(a+b*tanh(x)^4)^(5/2),x, algorithm="giac")

[Out]

integrate(tanh(x)/(b*tanh(x)^4 + a)^(5/2), x)

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maple [C]  time = 0.14, size = 637, normalized size = 5.40 \[ -\frac {\left (-\frac {\tanh ^{3}\relax (x )}{6 a \left (a +b \right ) b}-\frac {\tanh ^{2}\relax (x )}{6 a \left (a +b \right ) b}-\frac {\tanh \relax (x )}{6 a \left (a +b \right ) b}+\frac {1}{6 \left (a +b \right ) b^{2}}\right ) \sqrt {a +b \left (\tanh ^{4}\relax (x )\right )}}{2 \left (\tanh ^{4}\relax (x )+\frac {a}{b}\right )^{2}}+\frac {b \left (\frac {\left (3 a +b \right ) \left (\tanh ^{3}\relax (x )\right )}{8 a^{2} \left (a +b \right )^{2}}+\frac {\left (5 a +2 b \right ) \left (\tanh ^{2}\relax (x )\right )}{12 a^{2} \left (a +b \right )^{2}}+\frac {\left (11 a +5 b \right ) \tanh \relax (x )}{24 a^{2} \left (a +b \right )^{2}}-\frac {1}{4 \left (a +b \right )^{2} b}\right )}{\sqrt {\left (\tanh ^{4}\relax (x )+\frac {a}{b}\right ) b}}-\frac {-\frac {\arctanh \left (\frac {2 b \left (\tanh ^{2}\relax (x )\right )+2 a}{2 \sqrt {a +b}\, \sqrt {a +b \left (\tanh ^{4}\relax (x )\right )}}\right )}{2 \sqrt {a +b}}-\frac {\sqrt {1-\frac {i \sqrt {b}\, \left (\tanh ^{2}\relax (x )\right )}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, \left (\tanh ^{2}\relax (x )\right )}{\sqrt {a}}}\, \EllipticPi \left (\tanh \relax (x ) \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, -\frac {i \sqrt {a}}{\sqrt {b}}, \frac {\sqrt {-\frac {i \sqrt {b}}{\sqrt {a}}}}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}}\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {a +b \left (\tanh ^{4}\relax (x )\right )}}}{2 \left (a +b \right )^{2}}-\frac {\left (\frac {\tanh ^{3}\relax (x )}{6 a \left (a +b \right ) b}-\frac {\tanh ^{2}\relax (x )}{6 a \left (a +b \right ) b}+\frac {\tanh \relax (x )}{6 a \left (a +b \right ) b}+\frac {1}{6 \left (a +b \right ) b^{2}}\right ) \sqrt {a +b \left (\tanh ^{4}\relax (x )\right )}}{2 \left (\tanh ^{4}\relax (x )+\frac {a}{b}\right )^{2}}+\frac {b \left (-\frac {\left (3 a +b \right ) \left (\tanh ^{3}\relax (x )\right )}{8 a^{2} \left (a +b \right )^{2}}+\frac {\left (5 a +2 b \right ) \left (\tanh ^{2}\relax (x )\right )}{12 a^{2} \left (a +b \right )^{2}}-\frac {\left (11 a +5 b \right ) \tanh \relax (x )}{24 a^{2} \left (a +b \right )^{2}}-\frac {1}{4 \left (a +b \right )^{2} b}\right )}{\sqrt {\left (\tanh ^{4}\relax (x )+\frac {a}{b}\right ) b}}-\frac {-\frac {\arctanh \left (\frac {2 b \left (\tanh ^{2}\relax (x )\right )+2 a}{2 \sqrt {a +b}\, \sqrt {a +b \left (\tanh ^{4}\relax (x )\right )}}\right )}{2 \sqrt {a +b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, \left (\tanh ^{2}\relax (x )\right )}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, \left (\tanh ^{2}\relax (x )\right )}{\sqrt {a}}}\, \EllipticPi \left (\tanh \relax (x ) \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, -\frac {i \sqrt {a}}{\sqrt {b}}, \frac {\sqrt {-\frac {i \sqrt {b}}{\sqrt {a}}}}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}}\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {a +b \left (\tanh ^{4}\relax (x )\right )}}}{2 \left (a +b \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)/(a+b*tanh(x)^4)^(5/2),x)

[Out]

-1/2*(-1/6/a/(a+b)/b*tanh(x)^3-1/6/a/(a+b)/b*tanh(x)^2-1/6/a/(a+b)/b*tanh(x)+1/6/(a+b)/b^2)*(a+b*tanh(x)^4)^(1
/2)/(tanh(x)^4+a/b)^2+b*(1/8*(3*a+b)/a^2/(a+b)^2*tanh(x)^3+1/12*(5*a+2*b)/a^2/(a+b)^2*tanh(x)^2+1/24/a^2*(11*a
+5*b)/(a+b)^2*tanh(x)-1/4/(a+b)^2/b)/((tanh(x)^4+a/b)*b)^(1/2)-1/2/(a+b)^2*(-1/2/(a+b)^(1/2)*arctanh(1/2*(2*b*
tanh(x)^2+2*a)/(a+b)^(1/2)/(a+b*tanh(x)^4)^(1/2))-1/(I/a^(1/2)*b^(1/2))^(1/2)*(1-I/a^(1/2)*b^(1/2)*tanh(x)^2)^
(1/2)*(1+I/a^(1/2)*b^(1/2)*tanh(x)^2)^(1/2)/(a+b*tanh(x)^4)^(1/2)*EllipticPi(tanh(x)*(I/a^(1/2)*b^(1/2))^(1/2)
,-I*a^(1/2)/b^(1/2),(-I/a^(1/2)*b^(1/2))^(1/2)/(I/a^(1/2)*b^(1/2))^(1/2)))-1/2*(1/6/a/(a+b)/b*tanh(x)^3-1/6/a/
(a+b)/b*tanh(x)^2+1/6/a/(a+b)/b*tanh(x)+1/6/(a+b)/b^2)*(a+b*tanh(x)^4)^(1/2)/(tanh(x)^4+a/b)^2+b*(-1/8*(3*a+b)
/a^2/(a+b)^2*tanh(x)^3+1/12*(5*a+2*b)/a^2/(a+b)^2*tanh(x)^2-1/24/a^2*(11*a+5*b)/(a+b)^2*tanh(x)-1/4/(a+b)^2/b)
/((tanh(x)^4+a/b)*b)^(1/2)-1/2/(a+b)^2*(-1/2/(a+b)^(1/2)*arctanh(1/2*(2*b*tanh(x)^2+2*a)/(a+b)^(1/2)/(a+b*tanh
(x)^4)^(1/2))+1/(I/a^(1/2)*b^(1/2))^(1/2)*(1-I/a^(1/2)*b^(1/2)*tanh(x)^2)^(1/2)*(1+I/a^(1/2)*b^(1/2)*tanh(x)^2
)^(1/2)/(a+b*tanh(x)^4)^(1/2)*EllipticPi(tanh(x)*(I/a^(1/2)*b^(1/2))^(1/2),-I*a^(1/2)/b^(1/2),(-I/a^(1/2)*b^(1
/2))^(1/2)/(I/a^(1/2)*b^(1/2))^(1/2)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tanh \relax (x)}{{\left (b \tanh \relax (x)^{4} + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)/(a+b*tanh(x)^4)^(5/2),x, algorithm="maxima")

[Out]

integrate(tanh(x)/(b*tanh(x)^4 + a)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {tanh}\relax (x)}{{\left (b\,{\mathrm {tanh}\relax (x)}^4+a\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)/(a + b*tanh(x)^4)^(5/2),x)

[Out]

int(tanh(x)/(a + b*tanh(x)^4)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tanh {\relax (x )}}{\left (a + b \tanh ^{4}{\relax (x )}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)/(a+b*tanh(x)**4)**(5/2),x)

[Out]

Integral(tanh(x)/(a + b*tanh(x)**4)**(5/2), x)

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