\(\int \frac {\sqrt {a+b (c x^3)^{3/2}}}{x^9} \, dx\) [2977]

Optimal result

Integrand size = 21, antiderivative size = 139 \[ \int \frac {\sqrt {a+b \left (c x^3\right )^{3/2}}}{x^9} \, dx=-\frac {\sqrt {a+b \left (c x^3\right )^{3/2}}}{8 x^8}-\frac {9 b c^3 x \sqrt {a+b \left (c x^3\right )^{3/2}}}{112 a \left (c x^3\right )^{3/2}}-\frac {45 b^2 c^3 x \sqrt {1+\frac {b \left (c x^3\right )^{3/2}}{a}} \operatorname {Hypergeometric2F1}\left (\frac {2}{9},\frac {1}{2},\frac {11}{9},-\frac {b \left (c x^3\right )^{3/2}}{a}\right )}{448 a \sqrt {a+b \left (c x^3\right )^{3/2}}} \]

[Out]

-1/8*(a+b*(c*x^3)^(3/2))^(1/2)/x^8-9/112*b*c^3*x*(a+b*(c*x^3)^(3/2))^(1/2)/a/(c*x^3)^(3/2)-45/448*b^2*c^3*x*hy
pergeom([2/9, 1/2],[11/9],-b*(c*x^3)^(3/2)/a)*(1+b*(c*x^3)^(3/2)/a)^(1/2)/a/(a+b*(c*x^3)^(3/2))^(1/2)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {376, 348, 283, 331, 372, 371} \[ \int \frac {\sqrt {a+b \left (c x^3\right )^{3/2}}}{x^9} \, dx=-\frac {45 b^2 c^3 x \sqrt {\frac {b \left (c x^3\right )^{3/2}}{a}+1} \operatorname {Hypergeometric2F1}\left (\frac {2}{9},\frac {1}{2},\frac {11}{9},-\frac {b \left (c x^3\right )^{3/2}}{a}\right )}{448 a \sqrt {a+b \left (c x^3\right )^{3/2}}}-\frac {9 b c^5 x^7 \sqrt {a+b \left (c x^3\right )^{3/2}}}{112 a \left (c x^3\right )^{7/2}}-\frac {\sqrt {a+b \left (c x^3\right )^{3/2}}}{8 x^8} \]

[In]

Int[Sqrt[a + b*(c*x^3)^(3/2)]/x^9,x]

[Out]

-1/8*Sqrt[a + b*(c*x^3)^(3/2)]/x^8 - (9*b*c^5*x^7*Sqrt[a + b*(c*x^3)^(3/2)])/(112*a*(c*x^3)^(7/2)) - (45*b^2*c
^3*x*Sqrt[1 + (b*(c*x^3)^(3/2))/a]*Hypergeometric2F1[2/9, 1/2, 11/9, -((b*(c*x^3)^(3/2))/a)])/(448*a*Sqrt[a +
b*(c*x^3)^(3/2)])

Rule 283

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^p/(c*(m + 1
))), x] - Dist[b*n*(p/(c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 348

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

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/
(1 + b*(x^n/a))^FracPart[p]), Int[(c*x)^m*(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 376

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*} \text {integral}& = \text {Subst}\left (\int \frac {\sqrt {a+b c^{3/2} x^{9/2}}}{x^9} \, dx,\sqrt {x},\frac {\sqrt {c x^3}}{\sqrt {c} x}\right ) \\ & = \text {Subst}\left (2 \text {Subst}\left (\int \frac {\sqrt {a+b c^{3/2} x^9}}{x^{17}} \, dx,x,\sqrt {x}\right ),\sqrt {x},\frac {\sqrt {c x^3}}{\sqrt {c} x}\right ) \\ & = -\frac {\sqrt {a+b \left (c x^3\right )^{3/2}}}{8 x^8}+\text {Subst}\left (\frac {1}{16} \left (9 b c^{3/2}\right ) \text {Subst}\left (\int \frac {1}{x^8 \sqrt {a+b c^{3/2} x^9}} \, dx,x,\sqrt {x}\right ),\sqrt {x},\frac {\sqrt {c x^3}}{\sqrt {c} x}\right ) \\ & = -\frac {\sqrt {a+b \left (c x^3\right )^{3/2}}}{8 x^8}-\frac {9 b c^5 x^7 \sqrt {a+b \left (c x^3\right )^{3/2}}}{112 a \left (c x^3\right )^{7/2}}-\text {Subst}\left (\frac {\left (45 b^2 c^3\right ) \text {Subst}\left (\int \frac {x}{\sqrt {a+b c^{3/2} x^9}} \, dx,x,\sqrt {x}\right )}{224 a},\sqrt {x},\frac {\sqrt {c x^3}}{\sqrt {c} x}\right ) \\ & = -\frac {\sqrt {a+b \left (c x^3\right )^{3/2}}}{8 x^8}-\frac {9 b c^5 x^7 \sqrt {a+b \left (c x^3\right )^{3/2}}}{112 a \left (c x^3\right )^{7/2}}-\text {Subst}\left (\frac {\left (45 b^2 c^3 \sqrt {1+\frac {b c^{3/2} x^{9/2}}{a}}\right ) \text {Subst}\left (\int \frac {x}{\sqrt {1+\frac {b c^{3/2} x^9}{a}}} \, dx,x,\sqrt {x}\right )}{224 a \sqrt {a+b c^{3/2} x^{9/2}}},\sqrt {x},\frac {\sqrt {c x^3}}{\sqrt {c} x}\right ) \\ & = -\frac {\sqrt {a+b \left (c x^3\right )^{3/2}}}{8 x^8}-\frac {9 b c^5 x^7 \sqrt {a+b \left (c x^3\right )^{3/2}}}{112 a \left (c x^3\right )^{7/2}}-\frac {45 b^2 c^3 x \sqrt {1+\frac {b \left (c x^3\right )^{3/2}}{a}} \, _2F_1\left (\frac {2}{9},\frac {1}{2};\frac {11}{9};-\frac {b \left (c x^3\right )^{3/2}}{a}\right )}{448 a \sqrt {a+b \left (c x^3\right )^{3/2}}} \\ \end{align*}

Mathematica [F]

\[ \int \frac {\sqrt {a+b \left (c x^3\right )^{3/2}}}{x^9} \, dx=\int \frac {\sqrt {a+b \left (c x^3\right )^{3/2}}}{x^9} \, dx \]

[In]

Integrate[Sqrt[a + b*(c*x^3)^(3/2)]/x^9,x]

[Out]

Integrate[Sqrt[a + b*(c*x^3)^(3/2)]/x^9, x]

Maple [F]

\[\int \frac {\sqrt {a +b \left (c \,x^{3}\right )^{\frac {3}{2}}}}{x^{9}}d x\]

[In]

int((a+b*(c*x^3)^(3/2))^(1/2)/x^9,x)

[Out]

int((a+b*(c*x^3)^(3/2))^(1/2)/x^9,x)

Fricas [F]

\[ \int \frac {\sqrt {a+b \left (c x^3\right )^{3/2}}}{x^9} \, dx=\int { \frac {\sqrt {\left (c x^{3}\right )^{\frac {3}{2}} b + a}}{x^{9}} \,d x } \]

[In]

integrate((a+b*(c*x^3)^(3/2))^(1/2)/x^9,x, algorithm="fricas")

[Out]

integral(sqrt(sqrt(c*x^3)*b*c*x^3 + a)/x^9, x)

Sympy [F]

\[ \int \frac {\sqrt {a+b \left (c x^3\right )^{3/2}}}{x^9} \, dx=\int \frac {\sqrt {a + b \left (c x^{3}\right )^{\frac {3}{2}}}}{x^{9}}\, dx \]

[In]

integrate((a+b*(c*x**3)**(3/2))**(1/2)/x**9,x)

[Out]

Integral(sqrt(a + b*(c*x**3)**(3/2))/x**9, x)

Maxima [F]

\[ \int \frac {\sqrt {a+b \left (c x^3\right )^{3/2}}}{x^9} \, dx=\int { \frac {\sqrt {\left (c x^{3}\right )^{\frac {3}{2}} b + a}}{x^{9}} \,d x } \]

[In]

integrate((a+b*(c*x^3)^(3/2))^(1/2)/x^9,x, algorithm="maxima")

[Out]

integrate(sqrt((c*x^3)^(3/2)*b + a)/x^9, x)

Giac [F]

\[ \int \frac {\sqrt {a+b \left (c x^3\right )^{3/2}}}{x^9} \, dx=\int { \frac {\sqrt {\left (c x^{3}\right )^{\frac {3}{2}} b + a}}{x^{9}} \,d x } \]

[In]

integrate((a+b*(c*x^3)^(3/2))^(1/2)/x^9,x, algorithm="giac")

[Out]

integrate(sqrt((c*x^3)^(3/2)*b + a)/x^9, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+b \left (c x^3\right )^{3/2}}}{x^9} \, dx=\int \frac {\sqrt {a+b\,{\left (c\,x^3\right )}^{3/2}}}{x^9} \,d x \]

[In]

int((a + b*(c*x^3)^(3/2))^(1/2)/x^9,x)

[Out]

int((a + b*(c*x^3)^(3/2))^(1/2)/x^9, x)