3.3.22 \(\int \frac {\sqrt {c+d x^3}}{x^7 (8 c-d x^3)} \, dx\)

Optimal. Leaf size=107 \[ \frac {d^2 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{256 c^{5/2}}+\frac {d^2 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{256 c^{5/2}}-\frac {d \sqrt {c+d x^3}}{64 c^2 x^3}-\frac {\sqrt {c+d x^3}}{48 c x^6} \]

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Rubi [A]  time = 0.09, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {446, 99, 151, 156, 63, 208, 206} \begin {gather*} \frac {d^2 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{256 c^{5/2}}+\frac {d^2 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{256 c^{5/2}}-\frac {d \sqrt {c+d x^3}}{64 c^2 x^3}-\frac {\sqrt {c+d x^3}}{48 c x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[c + d*x^3]/(x^7*(8*c - d*x^3)),x]

[Out]

-Sqrt[c + d*x^3]/(48*c*x^6) - (d*Sqrt[c + d*x^3])/(64*c^2*x^3) + (d^2*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(2
56*c^(5/2)) + (d^2*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(256*c^(5/2))

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 99

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/((m + 1)*(b*e - a*f)), x] - Dist[1/((m + 1)*(b*e - a*f)), Int[(a +
b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[d*e*n + c*f*(m + p + 2) + d*f*(m + n + p + 2)*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

Rule 151

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegerQ[m]

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, 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 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {\sqrt {c+d x^3}}{x^7 \left (8 c-d x^3\right )} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {\sqrt {c+d x}}{x^3 (8 c-d x)} \, dx,x,x^3\right )\\ &=-\frac {\sqrt {c+d x^3}}{48 c x^6}+\frac {\operatorname {Subst}\left (\int \frac {6 c d+\frac {3 d^2 x}{2}}{x^2 (8 c-d x) \sqrt {c+d x}} \, dx,x,x^3\right )}{48 c}\\ &=-\frac {\sqrt {c+d x^3}}{48 c x^6}-\frac {d \sqrt {c+d x^3}}{64 c^2 x^3}-\frac {\operatorname {Subst}\left (\int \frac {6 c^2 d^2-3 c d^3 x}{x (8 c-d x) \sqrt {c+d x}} \, dx,x,x^3\right )}{384 c^3}\\ &=-\frac {\sqrt {c+d x^3}}{48 c x^6}-\frac {d \sqrt {c+d x^3}}{64 c^2 x^3}-\frac {d^2 \operatorname {Subst}\left (\int \frac {1}{x \sqrt {c+d x}} \, dx,x,x^3\right )}{512 c^2}+\frac {\left (3 d^3\right ) \operatorname {Subst}\left (\int \frac {1}{(8 c-d x) \sqrt {c+d x}} \, dx,x,x^3\right )}{512 c^2}\\ &=-\frac {\sqrt {c+d x^3}}{48 c x^6}-\frac {d \sqrt {c+d x^3}}{64 c^2 x^3}-\frac {d \operatorname {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x^3}\right )}{256 c^2}+\frac {\left (3 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{9 c-x^2} \, dx,x,\sqrt {c+d x^3}\right )}{256 c^2}\\ &=-\frac {\sqrt {c+d x^3}}{48 c x^6}-\frac {d \sqrt {c+d x^3}}{64 c^2 x^3}+\frac {d^2 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{256 c^{5/2}}+\frac {d^2 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{256 c^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 96, normalized size = 0.90 \begin {gather*} \frac {3 d^2 x^6 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )+3 d^2 x^6 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )-4 \sqrt {c} \sqrt {c+d x^3} \left (4 c+3 d x^3\right )}{768 c^{5/2} x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c + d*x^3]/(x^7*(8*c - d*x^3)),x]

[Out]

(-4*Sqrt[c]*Sqrt[c + d*x^3]*(4*c + 3*d*x^3) + 3*d^2*x^6*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])] + 3*d^2*x^6*ArcTa
nh[Sqrt[c + d*x^3]/Sqrt[c]])/(768*c^(5/2)*x^6)

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IntegrateAlgebraic [A]  time = 0.14, size = 95, normalized size = 0.89 \begin {gather*} \frac {d^2 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{256 c^{5/2}}+\frac {d^2 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{256 c^{5/2}}+\frac {\left (-4 c-3 d x^3\right ) \sqrt {c+d x^3}}{192 c^2 x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[c + d*x^3]/(x^7*(8*c - d*x^3)),x]

[Out]

((-4*c - 3*d*x^3)*Sqrt[c + d*x^3])/(192*c^2*x^6) + (d^2*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(256*c^(5/2)) +
(d^2*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(256*c^(5/2))

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fricas [A]  time = 0.80, size = 188, normalized size = 1.76 \begin {gather*} \left [\frac {3 \, \sqrt {c} d^{2} x^{6} \log \left (\frac {d^{2} x^{6} + 24 \, c d x^{3} + 8 \, {\left (d x^{3} + 4 \, c\right )} \sqrt {d x^{3} + c} \sqrt {c} + 32 \, c^{2}}{d x^{6} - 8 \, c x^{3}}\right ) - 8 \, {\left (3 \, c d x^{3} + 4 \, c^{2}\right )} \sqrt {d x^{3} + c}}{1536 \, c^{3} x^{6}}, -\frac {3 \, \sqrt {-c} d^{2} x^{6} \arctan \left (\frac {{\left (d x^{3} + 4 \, c\right )} \sqrt {d x^{3} + c} \sqrt {-c}}{4 \, {\left (c d x^{3} + c^{2}\right )}}\right ) + 4 \, {\left (3 \, c d x^{3} + 4 \, c^{2}\right )} \sqrt {d x^{3} + c}}{768 \, c^{3} x^{6}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^3+c)^(1/2)/x^7/(-d*x^3+8*c),x, algorithm="fricas")

[Out]

[1/1536*(3*sqrt(c)*d^2*x^6*log((d^2*x^6 + 24*c*d*x^3 + 8*(d*x^3 + 4*c)*sqrt(d*x^3 + c)*sqrt(c) + 32*c^2)/(d*x^
6 - 8*c*x^3)) - 8*(3*c*d*x^3 + 4*c^2)*sqrt(d*x^3 + c))/(c^3*x^6), -1/768*(3*sqrt(-c)*d^2*x^6*arctan(1/4*(d*x^3
+ 4*c)*sqrt(d*x^3 + c)*sqrt(-c)/(c*d*x^3 + c^2)) + 4*(3*c*d*x^3 + 4*c^2)*sqrt(d*x^3 + c))/(c^3*x^6)]

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giac [A]  time = 0.18, size = 100, normalized size = 0.93 \begin {gather*} -\frac {d^{2} \arctan \left (\frac {\sqrt {d x^{3} + c}}{\sqrt {-c}}\right )}{256 \, \sqrt {-c} c^{2}} - \frac {d^{2} \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{256 \, \sqrt {-c} c^{2}} - \frac {3 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} d^{2} + \sqrt {d x^{3} + c} c d^{2}}{192 \, c^{2} d^{2} x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^3+c)^(1/2)/x^7/(-d*x^3+8*c),x, algorithm="giac")

[Out]

-1/256*d^2*arctan(sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*c^2) - 1/256*d^2*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(s
qrt(-c)*c^2) - 1/192*(3*(d*x^3 + c)^(3/2)*d^2 + sqrt(d*x^3 + c)*c*d^2)/(c^2*d^2*x^6)

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maple [C]  time = 0.19, size = 574, normalized size = 5.36 \begin {gather*} -\frac {\left (\frac {2 \sqrt {d \,x^{3}+c}}{3 d}+\frac {i \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {\frac {i \left (2 x +\frac {-i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}+\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right ) d}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {\left (x -\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right ) d}{-3 \left (-c \,d^{2}\right )^{\frac {1}{3}}+i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {-\frac {i \left (2 x +\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}+\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right ) d}{2 \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \left (2 \RootOf \left (d \,\textit {\_Z}^{3}-8 c \right )^{2} d^{2}+i \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {3}\, \RootOf \left (d \,\textit {\_Z}^{3}-8 c \right ) d -\left (-c \,d^{2}\right )^{\frac {1}{3}} \RootOf \left (d \,\textit {\_Z}^{3}-8 c \right ) d -i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {2}{3}}-\left (-c \,d^{2}\right )^{\frac {2}{3}}\right ) \EllipticPi \left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}-\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right ) \sqrt {3}\, d}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}}{3}, -\frac {2 i \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {3}\, \RootOf \left (d \,\textit {\_Z}^{3}-8 c \right )^{2} d +i \sqrt {3}\, c d -3 c d -i \left (-c \,d^{2}\right )^{\frac {2}{3}} \sqrt {3}\, \RootOf \left (d \,\textit {\_Z}^{3}-8 c \right )-3 \left (-c \,d^{2}\right )^{\frac {2}{3}} \RootOf \left (d \,\textit {\_Z}^{3}-8 c \right )}{18 c d}, \sqrt {\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{\left (-\frac {3 \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}+\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right ) d}}\right )}{3 d^{3} \sqrt {d \,x^{3}+c}}\right ) d^{3}}{512 c^{3}}+\frac {\frac {d^{2} \arctanh \left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{12 c^{\frac {3}{2}}}-\frac {\sqrt {d \,x^{3}+c}\, d}{12 c \,x^{3}}-\frac {\sqrt {d \,x^{3}+c}}{6 x^{6}}}{8 c}+\frac {\left (-\frac {d \arctanh \left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3 \sqrt {c}}-\frac {\sqrt {d \,x^{3}+c}}{3 x^{3}}\right ) d}{64 c^{2}}+\frac {\left (-\frac {2 \sqrt {c}\, \arctanh \left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3}+\frac {2 \sqrt {d \,x^{3}+c}}{3}\right ) d^{2}}{512 c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^3+c)^(1/2)/x^7/(-d*x^3+8*c),x)

[Out]

1/64/c^2*d*(-1/3*(d*x^3+c)^(1/2)/x^3-1/3*d*arctanh((d*x^3+c)^(1/2)/c^(1/2))/c^(1/2))+1/8/c*(-1/6*(d*x^3+c)^(1/
2)/x^6-1/12*d*(d*x^3+c)^(1/2)/c/x^3+1/12*d^2*arctanh((d*x^3+c)^(1/2)/c^(1/2))/c^(3/2))-1/512/c^3*d^3*(2/3*(d*x
^3+c)^(1/2)/d+1/3*I/d^3*2^(1/2)*sum((-c*d^2)^(1/3)*(1/2*I*(2*x+(-I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3))/d)/(
-c*d^2)^(1/3)*d)^(1/2)*((x-(-c*d^2)^(1/3)/d)/(-3*(-c*d^2)^(1/3)+I*3^(1/2)*(-c*d^2)^(1/3))*d)^(1/2)*(-1/2*I*(2*
x+(I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3))/d)/(-c*d^2)^(1/3)*d)^(1/2)/(d*x^3+c)^(1/2)*(2*_alpha^2*d^2+I*(-c*d
^2)^(1/3)*3^(1/2)*_alpha*d-(-c*d^2)^(1/3)*_alpha*d-I*3^(1/2)*(-c*d^2)^(2/3)-(-c*d^2)^(2/3))*EllipticPi(1/3*3^(
1/2)*(I*(x+1/2*(-c*d^2)^(1/3)/d-1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)*3^(1/2)/(-c*d^2)^(1/3)*d)^(1/2),-1/18*(2*I*(-c
*d^2)^(1/3)*3^(1/2)*_alpha^2*d+I*3^(1/2)*c*d-3*c*d-I*(-c*d^2)^(2/3)*3^(1/2)*_alpha-3*(-c*d^2)^(2/3)*_alpha)/c/
d,(I*3^(1/2)*(-c*d^2)^(1/3)/(-3/2*(-c*d^2)^(1/3)/d+1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)/d)^(1/2)),_alpha=RootOf(_Z^
3*d-8*c)))+1/512/c^3*d^2*(2/3*(d*x^3+c)^(1/2)-2/3*arctanh((d*x^3+c)^(1/2)/c^(1/2))*c^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\int \frac {\sqrt {d x^{3} + c}}{{\left (d x^{3} - 8 \, c\right )} x^{7}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^3+c)^(1/2)/x^7/(-d*x^3+8*c),x, algorithm="maxima")

[Out]

-integrate(sqrt(d*x^3 + c)/((d*x^3 - 8*c)*x^7), x)

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mupad [B]  time = 3.91, size = 83, normalized size = 0.78 \begin {gather*} \frac {d^2\,\mathrm {atanh}\left (\frac {d^4\,\sqrt {d\,x^3+c}}{2048\,c^{7/2}\,\left (\frac {d^4}{2048\,c^3}+\frac {d^5\,x^3}{8192\,c^4}\right )}\right )}{256\,c^{5/2}}-\frac {\sqrt {d\,x^3+c}}{192\,c\,x^6}-\frac {{\left (d\,x^3+c\right )}^{3/2}}{64\,c^2\,x^6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x^3)^(1/2)/(x^7*(8*c - d*x^3)),x)

[Out]

(d^2*atanh((d^4*(c + d*x^3)^(1/2))/(2048*c^(7/2)*(d^4/(2048*c^3) + (d^5*x^3)/(8192*c^4)))))/(256*c^(5/2)) - (c
+ d*x^3)^(1/2)/(192*c*x^6) - (c + d*x^3)^(3/2)/(64*c^2*x^6)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {\sqrt {c + d x^{3}}}{- 8 c x^{7} + d x^{10}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**3+c)**(1/2)/x**7/(-d*x**3+8*c),x)

[Out]

-Integral(sqrt(c + d*x**3)/(-8*c*x**7 + d*x**10), x)

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