Integrand size = 16, antiderivative size = 200 \[ \int \frac {\coth ^{-1}(a x)}{\left (c+d x^2\right )^{7/2}} \, dx=\frac {a}{15 c \left (a^2 c+d\right ) \left (c+d x^2\right )^{3/2}}+\frac {a \left (7 a^2 c+4 d\right )}{15 c^2 \left (a^2 c+d\right )^2 \sqrt {c+d x^2}}+\frac {x \coth ^{-1}(a x)}{5 c \left (c+d x^2\right )^{5/2}}+\frac {4 x \coth ^{-1}(a x)}{15 c^2 \left (c+d x^2\right )^{3/2}}+\frac {8 x \coth ^{-1}(a x)}{15 c^3 \sqrt {c+d x^2}}-\frac {\left (15 a^4 c^2+20 a^2 c d+8 d^2\right ) \text {arctanh}\left (\frac {a \sqrt {c+d x^2}}{\sqrt {a^2 c+d}}\right )}{15 c^3 \left (a^2 c+d\right )^{5/2}} \]
1/15*a/c/(a^2*c+d)/(d*x^2+c)^(3/2)+1/5*x*arccoth(a*x)/c/(d*x^2+c)^(5/2)+4/ 15*x*arccoth(a*x)/c^2/(d*x^2+c)^(3/2)-1/15*(15*a^4*c^2+20*a^2*c*d+8*d^2)*a rctanh(a*(d*x^2+c)^(1/2)/(a^2*c+d)^(1/2))/c^3/(a^2*c+d)^(5/2)+1/15*a*(7*a^ 2*c+4*d)/c^2/(a^2*c+d)^2/(d*x^2+c)^(1/2)+8/15*x*arccoth(a*x)/c^3/(d*x^2+c) ^(1/2)
Time = 0.39 (sec) , antiderivative size = 329, normalized size of antiderivative = 1.64 \[ \int \frac {\coth ^{-1}(a x)}{\left (c+d x^2\right )^{7/2}} \, dx=\frac {2 a c \sqrt {a^2 c+d} \left (c+d x^2\right ) \left (d \left (5 c+4 d x^2\right )+a^2 c \left (8 c+7 d x^2\right )\right )+2 \left (a^2 c+d\right )^{5/2} x \left (15 c^2+20 c d x^2+8 d^2 x^4\right ) \coth ^{-1}(a x)+\left (15 a^4 c^2+20 a^2 c d+8 d^2\right ) \left (c+d x^2\right )^{5/2} \log (1-a x)+\left (15 a^4 c^2+20 a^2 c d+8 d^2\right ) \left (c+d x^2\right )^{5/2} \log (1+a x)-\left (15 a^4 c^2+20 a^2 c d+8 d^2\right ) \left (c+d x^2\right )^{5/2} \log \left (a c-d x+\sqrt {a^2 c+d} \sqrt {c+d x^2}\right )-\left (15 a^4 c^2+20 a^2 c d+8 d^2\right ) \left (c+d x^2\right )^{5/2} \log \left (a c+d x+\sqrt {a^2 c+d} \sqrt {c+d x^2}\right )}{30 c^3 \left (a^2 c+d\right )^{5/2} \left (c+d x^2\right )^{5/2}} \]
(2*a*c*Sqrt[a^2*c + d]*(c + d*x^2)*(d*(5*c + 4*d*x^2) + a^2*c*(8*c + 7*d*x ^2)) + 2*(a^2*c + d)^(5/2)*x*(15*c^2 + 20*c*d*x^2 + 8*d^2*x^4)*ArcCoth[a*x ] + (15*a^4*c^2 + 20*a^2*c*d + 8*d^2)*(c + d*x^2)^(5/2)*Log[1 - a*x] + (15 *a^4*c^2 + 20*a^2*c*d + 8*d^2)*(c + d*x^2)^(5/2)*Log[1 + a*x] - (15*a^4*c^ 2 + 20*a^2*c*d + 8*d^2)*(c + d*x^2)^(5/2)*Log[a*c - d*x + Sqrt[a^2*c + d]* Sqrt[c + d*x^2]] - (15*a^4*c^2 + 20*a^2*c*d + 8*d^2)*(c + d*x^2)^(5/2)*Log [a*c + d*x + Sqrt[a^2*c + d]*Sqrt[c + d*x^2]])/(30*c^3*(a^2*c + d)^(5/2)*( c + d*x^2)^(5/2))
Time = 1.10 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.97, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {6539, 27, 7266, 1192, 1584, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\coth ^{-1}(a x)}{\left (c+d x^2\right )^{7/2}} \, dx\) |
| \(\Big \downarrow \) 6539 |
| \(\displaystyle -a \int \frac {x \left (8 d^2 x^4+20 c d x^2+15 c^2\right )}{15 c^3 \left (1-a^2 x^2\right ) \left (d x^2+c\right )^{5/2}}dx+\frac {8 x \coth ^{-1}(a x)}{15 c^3 \sqrt {c+d x^2}}+\frac {4 x \coth ^{-1}(a x)}{15 c^2 \left (c+d x^2\right )^{3/2}}+\frac {x \coth ^{-1}(a x)}{5 c \left (c+d x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {a \int \frac {x \left (8 d^2 x^4+20 c d x^2+15 c^2\right )}{\left (1-a^2 x^2\right ) \left (d x^2+c\right )^{5/2}}dx}{15 c^3}+\frac {8 x \coth ^{-1}(a x)}{15 c^3 \sqrt {c+d x^2}}+\frac {4 x \coth ^{-1}(a x)}{15 c^2 \left (c+d x^2\right )^{3/2}}+\frac {x \coth ^{-1}(a x)}{5 c \left (c+d x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 7266 |
| \(\displaystyle -\frac {a \int \frac {8 d^2 x^4+20 c d x^2+15 c^2}{\left (1-a^2 x^2\right ) \left (d x^2+c\right )^{5/2}}dx^2}{30 c^3}+\frac {8 x \coth ^{-1}(a x)}{15 c^3 \sqrt {c+d x^2}}+\frac {4 x \coth ^{-1}(a x)}{15 c^2 \left (c+d x^2\right )^{3/2}}+\frac {x \coth ^{-1}(a x)}{5 c \left (c+d x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 1192 |
| \(\displaystyle -\frac {a \int \frac {8 d^2 x^8+4 c d^2 x^4+3 c^2 d^2}{x^8 \left (-a^2 x^4+a^2 c+d\right )}d\sqrt {d x^2+c}}{15 c^3 d^2}+\frac {8 x \coth ^{-1}(a x)}{15 c^3 \sqrt {c+d x^2}}+\frac {4 x \coth ^{-1}(a x)}{15 c^2 \left (c+d x^2\right )^{3/2}}+\frac {x \coth ^{-1}(a x)}{5 c \left (c+d x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 1584 |
| \(\displaystyle -\frac {a \int \left (\frac {\left (15 c^2 a^4+20 c d a^2+8 d^2\right ) d^2}{\left (c a^2+d\right )^2 \left (-a^2 x^4+a^2 c+d\right )}+\frac {c \left (7 c a^2+4 d\right ) d^2}{\left (c a^2+d\right )^2 x^4}+\frac {3 c^2 d^2}{\left (c a^2+d\right ) x^8}\right )d\sqrt {d x^2+c}}{15 c^3 d^2}+\frac {8 x \coth ^{-1}(a x)}{15 c^3 \sqrt {c+d x^2}}+\frac {4 x \coth ^{-1}(a x)}{15 c^2 \left (c+d x^2\right )^{3/2}}+\frac {x \coth ^{-1}(a x)}{5 c \left (c+d x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {a \left (-\frac {c^2 d^2}{x^6 \left (a^2 c+d\right )}-\frac {c d^2 \left (7 a^2 c+4 d\right )}{x^2 \left (a^2 c+d\right )^2}+\frac {d^2 \left (15 a^4 c^2+20 a^2 c d+8 d^2\right ) \text {arctanh}\left (\frac {a \sqrt {c+d x^2}}{\sqrt {a^2 c+d}}\right )}{a \left (a^2 c+d\right )^{5/2}}\right )}{15 c^3 d^2}+\frac {8 x \coth ^{-1}(a x)}{15 c^3 \sqrt {c+d x^2}}+\frac {4 x \coth ^{-1}(a x)}{15 c^2 \left (c+d x^2\right )^{3/2}}+\frac {x \coth ^{-1}(a x)}{5 c \left (c+d x^2\right )^{5/2}}\) |
(x*ArcCoth[a*x])/(5*c*(c + d*x^2)^(5/2)) + (4*x*ArcCoth[a*x])/(15*c^2*(c + d*x^2)^(3/2)) + (8*x*ArcCoth[a*x])/(15*c^3*Sqrt[c + d*x^2]) - (a*(-((c^2* d^2)/((a^2*c + d)*x^6)) - (c*d^2*(7*a^2*c + 4*d))/((a^2*c + d)^2*x^2) + (d ^2*(15*a^4*c^2 + 20*a^2*c*d + 8*d^2)*ArcTanh[(a*Sqrt[c + d*x^2])/Sqrt[a^2* c + d]])/(a*(a^2*c + d)^(5/2))))/(15*c^3*d^2)
3.1.46.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[2/e^(n + 2*p + 1) Subst[Int[x^( 2*m + 1)*(e*f - d*g + g*x^2)^n*(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4)^p, x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && IGtQ[p, 0] && ILtQ[n, 0] && IntegerQ[m + 1/2]
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + ( c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q* (a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && NeQ[ b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]
Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))*((d_.) + (e_.)*(x_)^2)^(q_.), x_Sym bol] :> With[{u = IntHide[(d + e*x^2)^q, x]}, Simp[(a + b*ArcCoth[c*x]) u , x] - Simp[b*c Int[SimplifyIntegrand[u/(1 - c^2*x^2), x], x], x]] /; Fre eQ[{a, b, c, d, e}, x] && (IntegerQ[q] || ILtQ[q + 1/2, 0])
Int[(u_)*(x_)^(m_.), x_Symbol] :> Simp[1/(m + 1) Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /; FreeQ[m, x] && NeQ[m, -1] && Function OfQ[x^(m + 1), u, x]
\[\int \frac {\operatorname {arccoth}\left (a x \right )}{\left (d \,x^{2}+c \right )^{\frac {7}{2}}}d x\]
Leaf count of result is larger than twice the leaf count of optimal. 620 vs. \(2 (172) = 344\).
Time = 0.34 (sec) , antiderivative size = 1278, normalized size of antiderivative = 6.39 \[ \int \frac {\coth ^{-1}(a x)}{\left (c+d x^2\right )^{7/2}} \, dx=\text {Too large to display} \]
[1/60*((15*a^4*c^5 + 20*a^2*c^4*d + (15*a^4*c^2*d^3 + 20*a^2*c*d^4 + 8*d^5 )*x^6 + 8*c^3*d^2 + 3*(15*a^4*c^3*d^2 + 20*a^2*c^2*d^3 + 8*c*d^4)*x^4 + 3* (15*a^4*c^4*d + 20*a^2*c^3*d^2 + 8*c^2*d^3)*x^2)*sqrt(a^2*c + d)*log((a^4* d^2*x^4 + 8*a^4*c^2 + 8*a^2*c*d + 2*(4*a^4*c*d + 3*a^2*d^2)*x^2 - 4*(a^3*d *x^2 + 2*a^3*c + a*d)*sqrt(a^2*c + d)*sqrt(d*x^2 + c) + d^2)/(a^4*x^4 - 2* a^2*x^2 + 1)) + 2*(16*a^5*c^5 + 26*a^3*c^4*d + 10*a*c^3*d^2 + 2*(7*a^5*c^3 *d^2 + 11*a^3*c^2*d^3 + 4*a*c*d^4)*x^4 + 6*(5*a^5*c^4*d + 8*a^3*c^3*d^2 + 3*a*c^2*d^3)*x^2 + (8*(a^6*c^3*d^2 + 3*a^4*c^2*d^3 + 3*a^2*c*d^4 + d^5)*x^ 5 + 20*(a^6*c^4*d + 3*a^4*c^3*d^2 + 3*a^2*c^2*d^3 + c*d^4)*x^3 + 15*(a^6*c ^5 + 3*a^4*c^4*d + 3*a^2*c^3*d^2 + c^2*d^3)*x)*log((a*x + 1)/(a*x - 1)))*s qrt(d*x^2 + c))/(a^6*c^9 + 3*a^4*c^8*d + 3*a^2*c^7*d^2 + c^6*d^3 + (a^6*c^ 6*d^3 + 3*a^4*c^5*d^4 + 3*a^2*c^4*d^5 + c^3*d^6)*x^6 + 3*(a^6*c^7*d^2 + 3* a^4*c^6*d^3 + 3*a^2*c^5*d^4 + c^4*d^5)*x^4 + 3*(a^6*c^8*d + 3*a^4*c^7*d^2 + 3*a^2*c^6*d^3 + c^5*d^4)*x^2), 1/30*((15*a^4*c^5 + 20*a^2*c^4*d + (15*a^ 4*c^2*d^3 + 20*a^2*c*d^4 + 8*d^5)*x^6 + 8*c^3*d^2 + 3*(15*a^4*c^3*d^2 + 20 *a^2*c^2*d^3 + 8*c*d^4)*x^4 + 3*(15*a^4*c^4*d + 20*a^2*c^3*d^2 + 8*c^2*d^3 )*x^2)*sqrt(-a^2*c - d)*arctan(1/2*(a^2*d*x^2 + 2*a^2*c + d)*sqrt(-a^2*c - d)*sqrt(d*x^2 + c)/(a^3*c^2 + a*c*d + (a^3*c*d + a*d^2)*x^2)) + (16*a^5*c ^5 + 26*a^3*c^4*d + 10*a*c^3*d^2 + 2*(7*a^5*c^3*d^2 + 11*a^3*c^2*d^3 + 4*a *c*d^4)*x^4 + 6*(5*a^5*c^4*d + 8*a^3*c^3*d^2 + 3*a*c^2*d^3)*x^2 + (8*(a...
\[ \int \frac {\coth ^{-1}(a x)}{\left (c+d x^2\right )^{7/2}} \, dx=\int \frac {\operatorname {acoth}{\left (a x \right )}}{\left (c + d x^{2}\right )^{\frac {7}{2}}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 401 vs. \(2 (172) = 344\).
Time = 0.30 (sec) , antiderivative size = 401, normalized size of antiderivative = 2.00 \[ \int \frac {\coth ^{-1}(a x)}{\left (c+d x^2\right )^{7/2}} \, dx=\frac {1}{30} \, a {\left (\frac {\frac {3 \, a^{3} d \log \left (\frac {\sqrt {d x^{2} + c} a^{2} - \sqrt {a^{2} c + d} a}{\sqrt {d x^{2} + c} a^{2} + \sqrt {a^{2} c + d} a}\right )}{{\left (a^{4} c^{3} + 2 \, a^{2} c^{2} d + c d^{2}\right )} \sqrt {a^{2} c + d}} + \frac {2 \, {\left (3 \, {\left (d x^{2} + c\right )} a^{2} d + a^{2} c d + d^{2}\right )}}{{\left (a^{4} c^{3} + 2 \, a^{2} c^{2} d + c d^{2}\right )} {\left (d x^{2} + c\right )}^{\frac {3}{2}}}}{d} + \frac {4 \, {\left (\frac {a d \log \left (\frac {\sqrt {d x^{2} + c} a^{2} - \sqrt {a^{2} c + d} a}{\sqrt {d x^{2} + c} a^{2} + \sqrt {a^{2} c + d} a}\right )}{{\left (a^{2} c^{3} + c^{2} d\right )} \sqrt {a^{2} c + d}} + \frac {2 \, d}{{\left (a^{2} c^{3} + c^{2} d\right )} \sqrt {d x^{2} + c}}\right )}}{d} + \frac {8 \, \log \left (\frac {\sqrt {d x^{2} + c} a^{2} - \sqrt {a^{2} c + d} a}{\sqrt {d x^{2} + c} a^{2} + \sqrt {a^{2} c + d} a}\right )}{\sqrt {a^{2} c + d} a c^{3}}\right )} + \frac {1}{15} \, {\left (\frac {8 \, x}{\sqrt {d x^{2} + c} c^{3}} + \frac {4 \, x}{{\left (d x^{2} + c\right )}^{\frac {3}{2}} c^{2}} + \frac {3 \, x}{{\left (d x^{2} + c\right )}^{\frac {5}{2}} c}\right )} \operatorname {arcoth}\left (a x\right ) \]
1/30*a*((3*a^3*d*log((sqrt(d*x^2 + c)*a^2 - sqrt(a^2*c + d)*a)/(sqrt(d*x^2 + c)*a^2 + sqrt(a^2*c + d)*a))/((a^4*c^3 + 2*a^2*c^2*d + c*d^2)*sqrt(a^2* c + d)) + 2*(3*(d*x^2 + c)*a^2*d + a^2*c*d + d^2)/((a^4*c^3 + 2*a^2*c^2*d + c*d^2)*(d*x^2 + c)^(3/2)))/d + 4*(a*d*log((sqrt(d*x^2 + c)*a^2 - sqrt(a^ 2*c + d)*a)/(sqrt(d*x^2 + c)*a^2 + sqrt(a^2*c + d)*a))/((a^2*c^3 + c^2*d)* sqrt(a^2*c + d)) + 2*d/((a^2*c^3 + c^2*d)*sqrt(d*x^2 + c)))/d + 8*log((sqr t(d*x^2 + c)*a^2 - sqrt(a^2*c + d)*a)/(sqrt(d*x^2 + c)*a^2 + sqrt(a^2*c + d)*a))/(sqrt(a^2*c + d)*a*c^3)) + 1/15*(8*x/(sqrt(d*x^2 + c)*c^3) + 4*x/(( d*x^2 + c)^(3/2)*c^2) + 3*x/((d*x^2 + c)^(5/2)*c))*arccoth(a*x)
Time = 0.31 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.13 \[ \int \frac {\coth ^{-1}(a x)}{\left (c+d x^2\right )^{7/2}} \, dx=\frac {1}{15} \, a {\left (\frac {{\left (15 \, a^{4} c^{2} + 20 \, a^{2} c d + 8 \, d^{2}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c} a}{\sqrt {-a^{2} c - d}}\right )}{{\left (a^{4} c^{5} + 2 \, a^{2} c^{4} d + c^{3} d^{2}\right )} \sqrt {-a^{2} c - d} a} + \frac {7 \, {\left (d x^{2} + c\right )} a^{2} c + a^{2} c^{2} + 4 \, {\left (d x^{2} + c\right )} d + c d}{{\left (a^{4} c^{4} + 2 \, a^{2} c^{3} d + c^{2} d^{2}\right )} {\left (d x^{2} + c\right )}^{\frac {3}{2}}}\right )} + \frac {{\left (4 \, x^{2} {\left (\frac {2 \, d^{2} x^{2}}{c^{3}} + \frac {5 \, d}{c^{2}}\right )} + \frac {15}{c}\right )} x \log \left (-\frac {\frac {1}{a x} + 1}{\frac {1}{a x} - 1}\right )}{30 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}}} \]
1/15*a*((15*a^4*c^2 + 20*a^2*c*d + 8*d^2)*arctan(sqrt(d*x^2 + c)*a/sqrt(-a ^2*c - d))/((a^4*c^5 + 2*a^2*c^4*d + c^3*d^2)*sqrt(-a^2*c - d)*a) + (7*(d* x^2 + c)*a^2*c + a^2*c^2 + 4*(d*x^2 + c)*d + c*d)/((a^4*c^4 + 2*a^2*c^3*d + c^2*d^2)*(d*x^2 + c)^(3/2))) + 1/30*(4*x^2*(2*d^2*x^2/c^3 + 5*d/c^2) + 1 5/c)*x*log(-(1/(a*x) + 1)/(1/(a*x) - 1))/(d*x^2 + c)^(5/2)
Timed out. \[ \int \frac {\coth ^{-1}(a x)}{\left (c+d x^2\right )^{7/2}} \, dx=\int \frac {\mathrm {acoth}\left (a\,x\right )}{{\left (d\,x^2+c\right )}^{7/2}} \,d x \]