3.2973 \(\int \frac {\sqrt {a+b (c x^3)^{3/2}}}{x^{10}} \, dx\)

Optimal. Leaf size=101 \[ \frac {b^2 c^3 \tanh ^{-1}\left (\frac {\sqrt {a+b \left (c x^3\right )^{3/2}}}{\sqrt {a}}\right )}{18 a^{3/2}}-\frac {b c^3 \sqrt {a+b \left (c x^3\right )^{3/2}}}{18 a \left (c x^3\right )^{3/2}}-\frac {\sqrt {a+b \left (c x^3\right )^{3/2}}}{9 x^9} \]

[Out]

1/18*b^2*c^3*arctanh((a+b*(c*x^3)^(3/2))^(1/2)/a^(1/2))/a^(3/2)-1/9*(a+b*(c*x^3)^(3/2))^(1/2)/x^9-1/18*b*c^3*(
a+b*(c*x^3)^(3/2))^(1/2)/a/(c*x^3)^(3/2)

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Rubi [A]  time = 0.07, antiderivative size = 104, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {369, 266, 47, 51, 63, 208} \[ \frac {b^2 c^3 \tanh ^{-1}\left (\frac {\sqrt {a+b \left (c x^3\right )^{3/2}}}{\sqrt {a}}\right )}{18 a^{3/2}}-\frac {b c^6 x^9 \sqrt {a+b \left (c x^3\right )^{3/2}}}{18 a \left (c x^3\right )^{9/2}}-\frac {\sqrt {a+b \left (c x^3\right )^{3/2}}}{9 x^9} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*(c*x^3)^(3/2)]/x^10,x]

[Out]

-Sqrt[a + b*(c*x^3)^(3/2)]/(9*x^9) - (b*c^6*x^9*Sqrt[a + b*(c*x^3)^(3/2)])/(18*a*(c*x^3)^(9/2)) + (b^2*c^3*Arc
Tanh[Sqrt[a + b*(c*x^3)^(3/2)]/Sqrt[a]])/(18*a^(3/2))

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

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 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 266

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

Rule 369

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*((c_.)*(x_)^(q_))^(n_))^(p_.), x_Symbol] :> With[{k = Denominator[n]}, Su
bst[Int[(d*x)^m*(a + b*c^n*x^(n*q))^p, x], x^(1/k), (c*x^q)^(1/k)/(c^(1/k)*(x^(1/k))^(q - 1))]] /; FreeQ[{a, b
, c, d, m, p, q}, x] && FractionQ[n]

Rubi steps

\begin {align*} \int \frac {\sqrt {a+b \left (c x^3\right )^{3/2}}}{x^{10}} \, dx &=\operatorname {Subst}\left (\int \frac {\sqrt {a+b c^{3/2} x^{9/2}}}{x^{10}} \, dx,\sqrt {x},\frac {\sqrt {c x^3}}{\sqrt {c} x}\right )\\ &=\operatorname {Subst}\left (\frac {2}{9} \operatorname {Subst}\left (\int \frac {\sqrt {a+b c^{3/2} x}}{x^3} \, dx,x,x^{9/2}\right ),\sqrt {x},\frac {\sqrt {c x^3}}{\sqrt {c} x}\right )\\ &=-\frac {\sqrt {a+b \left (c x^3\right )^{3/2}}}{9 x^9}+\operatorname {Subst}\left (\frac {1}{18} \left (b c^{3/2}\right ) \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {a+b c^{3/2} x}} \, dx,x,x^{9/2}\right ),\sqrt {x},\frac {\sqrt {c x^3}}{\sqrt {c} x}\right )\\ &=-\frac {\sqrt {a+b \left (c x^3\right )^{3/2}}}{9 x^9}-\frac {b c^6 x^9 \sqrt {a+b \left (c x^3\right )^{3/2}}}{18 a \left (c x^3\right )^{9/2}}-\operatorname {Subst}\left (\frac {\left (b^2 c^3\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b c^{3/2} x}} \, dx,x,x^{9/2}\right )}{36 a},\sqrt {x},\frac {\sqrt {c x^3}}{\sqrt {c} x}\right )\\ &=-\frac {\sqrt {a+b \left (c x^3\right )^{3/2}}}{9 x^9}-\frac {b c^6 x^9 \sqrt {a+b \left (c x^3\right )^{3/2}}}{18 a \left (c x^3\right )^{9/2}}-\operatorname {Subst}\left (\frac {\left (b c^{3/2}\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b c^{3/2}}+\frac {x^2}{b c^{3/2}}} \, dx,x,\sqrt {a+b c^{3/2} x^{9/2}}\right )}{18 a},\sqrt {x},\frac {\sqrt {c x^3}}{\sqrt {c} x}\right )\\ &=-\frac {\sqrt {a+b \left (c x^3\right )^{3/2}}}{9 x^9}-\frac {b c^6 x^9 \sqrt {a+b \left (c x^3\right )^{3/2}}}{18 a \left (c x^3\right )^{9/2}}+\frac {b^2 c^3 \tanh ^{-1}\left (\frac {\sqrt {a+b \left (c x^3\right )^{3/2}}}{\sqrt {a}}\right )}{18 a^{3/2}}\\ \end {align*}

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Mathematica [F]  time = 0.07, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a+b \left (c x^3\right )^{3/2}}}{x^{10}} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[Sqrt[a + b*(c*x^3)^(3/2)]/x^10,x]

[Out]

Integrate[Sqrt[a + b*(c*x^3)^(3/2)]/x^10, x]

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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((a+b*(c*x^3)^(3/2))^(1/2)/x^10,x, algorithm="fricas")

[Out]

Timed out

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giac [A]  time = 0.25, size = 130, normalized size = 1.29 \[ -\frac {{\left (\frac {b^{3} c^{6} \arctan \left (\frac {\sqrt {\sqrt {c x} b c^{5} x^{4} + a c^{4}}}{\sqrt {-a} c^{2}}\right )}{\sqrt {-a} a} + \frac {\sqrt {\sqrt {c x} b c^{5} x^{4} + a c^{4}} a b^{3} c^{12} + {\left (\sqrt {c x} b c^{5} x^{4} + a c^{4}\right )}^{\frac {3}{2}} b^{3} c^{8}}{a b^{2} c^{11} x^{9}}\right )} {\left | c \right |}^{2}}{18 \, b c^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*(c*x^3)^(3/2))^(1/2)/x^10,x, algorithm="giac")

[Out]

-1/18*(b^3*c^6*arctan(sqrt(sqrt(c*x)*b*c^5*x^4 + a*c^4)/(sqrt(-a)*c^2))/(sqrt(-a)*a) + (sqrt(sqrt(c*x)*b*c^5*x
^4 + a*c^4)*a*b^3*c^12 + (sqrt(c*x)*b*c^5*x^4 + a*c^4)^(3/2)*b^3*c^8)/(a*b^2*c^11*x^9))*abs(c)^2/(b*c^5)

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maple [F]  time = 0.24, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a +\left (c \,x^{3}\right )^{\frac {3}{2}} b}}{x^{10}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+(c*x^3)^(3/2)*b)^(1/2)/x^10,x)

[Out]

int((a+(c*x^3)^(3/2)*b)^(1/2)/x^10,x)

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maxima [A]  time = 1.21, size = 128, normalized size = 1.27 \[ -\frac {1}{36} \, {\left (\frac {b^{2} \log \left (\frac {\sqrt {\left (c x^{3}\right )^{\frac {3}{2}} b + a} - \sqrt {a}}{\sqrt {\left (c x^{3}\right )^{\frac {3}{2}} b + a} + \sqrt {a}}\right )}{a^{\frac {3}{2}}} + \frac {2 \, {\left ({\left (\left (c x^{3}\right )^{\frac {3}{2}} b + a\right )}^{\frac {3}{2}} b^{2} + \sqrt {\left (c x^{3}\right )^{\frac {3}{2}} b + a} a b^{2}\right )}}{{\left (\left (c x^{3}\right )^{\frac {3}{2}} b + a\right )}^{2} a - 2 \, {\left (\left (c x^{3}\right )^{\frac {3}{2}} b + a\right )} a^{2} + a^{3}}\right )} c^{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*(c*x^3)^(3/2))^(1/2)/x^10,x, algorithm="maxima")

[Out]

-1/36*(b^2*log((sqrt((c*x^3)^(3/2)*b + a) - sqrt(a))/(sqrt((c*x^3)^(3/2)*b + a) + sqrt(a)))/a^(3/2) + 2*(((c*x
^3)^(3/2)*b + a)^(3/2)*b^2 + sqrt((c*x^3)^(3/2)*b + a)*a*b^2)/(((c*x^3)^(3/2)*b + a)^2*a - 2*((c*x^3)^(3/2)*b
+ a)*a^2 + a^3))*c^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {a+b\,{\left (c\,x^3\right )}^{3/2}}}{x^{10}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*(c*x^3)^(3/2))^(1/2)/x^10,x)

[Out]

int((a + b*(c*x^3)^(3/2))^(1/2)/x^10, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a + b \left (c x^{3}\right )^{\frac {3}{2}}}}{x^{10}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*(c*x**3)**(3/2))**(1/2)/x**10,x)

[Out]

Integral(sqrt(a + b*(c*x**3)**(3/2))/x**10, x)

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