Integrand size = 21, antiderivative size = 170 \[ \int x^9 \sqrt {a+b \left (c x^3\right )^{3/2}} \, dx=-\frac {792 a^2 x \sqrt {a+b \left (c x^3\right )^{3/2}}}{19747 b^2 c^3}+\frac {4}{49} x^{10} \sqrt {a+b \left (c x^3\right )^{3/2}}+\frac {36 a x \left (c x^3\right )^{3/2} \sqrt {a+b \left (c x^3\right )^{3/2}}}{1519 b c^3}+\frac {792 a^3 x \sqrt {1+\frac {b \left (c x^3\right )^{3/2}}{a}} \operatorname {Hypergeometric2F1}\left (\frac {2}{9},\frac {1}{2},\frac {11}{9},-\frac {b \left (c x^3\right )^{3/2}}{a}\right )}{19747 b^2 c^3 \sqrt {a+b \left (c x^3\right )^{3/2}}} \]
-792/19747*a^2*x*(a+b*(c*x^3)^(3/2))^(1/2)/b^2/c^3+4/49*x^10*(a+b*(c*x^3)^ (3/2))^(1/2)+36/1519*a*x*(c*x^3)^(3/2)*(a+b*(c*x^3)^(3/2))^(1/2)/b/c^3+792 /19747*a^3*x*hypergeom([2/9, 1/2],[11/9],-b*(c*x^3)^(3/2)/a)*(1+b*(c*x^3)^ (3/2)/a)^(1/2)/b^2/c^3/(a+b*(c*x^3)^(3/2))^(1/2)
\[ \int x^9 \sqrt {a+b \left (c x^3\right )^{3/2}} \, dx=\int x^9 \sqrt {a+b \left (c x^3\right )^{3/2}} \, dx \]
Time = 0.35 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.34, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {893, 864, 811, 843, 843, 889, 888}
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.
\(\displaystyle \int x^9 \sqrt {a+b \left (c x^3\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 893 |
\(\displaystyle \int x^9 \sqrt {a+b c^{3/2} x^{9/2}}dx\) |
\(\Big \downarrow \) 864 |
\(\displaystyle 2 \int \frac {\left (c x^3\right )^{19/2} \sqrt {\frac {b \left (c x^3\right )^{9/2}}{c^3 x^9}+a}}{c^{19/2} x^{19}}d\frac {\sqrt {c x^3}}{\sqrt {c} x}\) |
\(\Big \downarrow \) 811 |
\(\displaystyle 2 \left (\frac {9}{49} a \int \frac {\left (c x^3\right )^{19/2}}{c^{19/2} x^{19} \sqrt {\frac {b \left (c x^3\right )^{9/2}}{c^3 x^9}+a}}d\frac {\sqrt {c x^3}}{\sqrt {c} x}+\frac {2}{49} x^{10} \sqrt {a+\frac {b \left (c x^3\right )^{9/2}}{c^3 x^9}}\right )\) |
\(\Big \downarrow \) 843 |
\(\displaystyle 2 \left (\frac {9}{49} a \left (\frac {2 \left (c x^3\right )^{11/2} \sqrt {a+\frac {b \left (c x^3\right )^{9/2}}{c^3 x^9}}}{31 b c^7 x^{11}}-\frac {22 a \int \frac {x^5}{\sqrt {\frac {b \left (c x^3\right )^{9/2}}{c^3 x^9}+a}}d\frac {\sqrt {c x^3}}{\sqrt {c} x}}{31 b c^{3/2}}\right )+\frac {2}{49} x^{10} \sqrt {a+\frac {b \left (c x^3\right )^{9/2}}{c^3 x^9}}\right )\) |
\(\Big \downarrow \) 843 |
\(\displaystyle 2 \left (\frac {9}{49} a \left (\frac {2 \left (c x^3\right )^{11/2} \sqrt {a+\frac {b \left (c x^3\right )^{9/2}}{c^3 x^9}}}{31 b c^7 x^{11}}-\frac {22 a \left (\frac {2 x \sqrt {a+\frac {b \left (c x^3\right )^{9/2}}{c^3 x^9}}}{13 b c^{3/2}}-\frac {4 a \int \frac {\sqrt {c x^3}}{\sqrt {c} x \sqrt {\frac {b \left (c x^3\right )^{9/2}}{c^3 x^9}+a}}d\frac {\sqrt {c x^3}}{\sqrt {c} x}}{13 b c^{3/2}}\right )}{31 b c^{3/2}}\right )+\frac {2}{49} x^{10} \sqrt {a+\frac {b \left (c x^3\right )^{9/2}}{c^3 x^9}}\right )\) |
\(\Big \downarrow \) 889 |
\(\displaystyle 2 \left (\frac {9}{49} a \left (\frac {2 \left (c x^3\right )^{11/2} \sqrt {a+\frac {b \left (c x^3\right )^{9/2}}{c^3 x^9}}}{31 b c^7 x^{11}}-\frac {22 a \left (\frac {2 x \sqrt {a+\frac {b \left (c x^3\right )^{9/2}}{c^3 x^9}}}{13 b c^{3/2}}-\frac {4 a \sqrt {\frac {b \left (c x^3\right )^{9/2}}{a c^3 x^9}+1} \int \frac {\sqrt {c x^3}}{\sqrt {c} x \sqrt {\frac {b \left (c x^3\right )^{9/2}}{a c^3 x^9}+1}}d\frac {\sqrt {c x^3}}{\sqrt {c} x}}{13 b c^{3/2} \sqrt {a+\frac {b \left (c x^3\right )^{9/2}}{c^3 x^9}}}\right )}{31 b c^{3/2}}\right )+\frac {2}{49} x^{10} \sqrt {a+\frac {b \left (c x^3\right )^{9/2}}{c^3 x^9}}\right )\) |
\(\Big \downarrow \) 888 |
\(\displaystyle 2 \left (\frac {2}{49} x^{10} \sqrt {a+\frac {b \left (c x^3\right )^{9/2}}{c^3 x^9}}+\frac {9}{49} a \left (\frac {2 \left (c x^3\right )^{11/2} \sqrt {a+\frac {b \left (c x^3\right )^{9/2}}{c^3 x^9}}}{31 b c^7 x^{11}}-\frac {22 a \left (\frac {2 x \sqrt {a+\frac {b \left (c x^3\right )^{9/2}}{c^3 x^9}}}{13 b c^{3/2}}-\frac {2 a x \sqrt {\frac {b \left (c x^3\right )^{9/2}}{a c^3 x^9}+1} \operatorname {Hypergeometric2F1}\left (\frac {2}{9},\frac {1}{2},\frac {11}{9},-\frac {b \left (c x^3\right )^{9/2}}{a c^3 x^9}\right )}{13 b c^{3/2} \sqrt {a+\frac {b \left (c x^3\right )^{9/2}}{c^3 x^9}}}\right )}{31 b c^{3/2}}\right )\right )\) |
2*((2*x^10*Sqrt[a + (b*(c*x^3)^(9/2))/(c^3*x^9)])/49 + (9*a*((2*(c*x^3)^(1 1/2)*Sqrt[a + (b*(c*x^3)^(9/2))/(c^3*x^9)])/(31*b*c^7*x^11) - (22*a*((2*x* Sqrt[a + (b*(c*x^3)^(9/2))/(c^3*x^9)])/(13*b*c^(3/2)) - (2*a*x*Sqrt[1 + (b *(c*x^3)^(9/2))/(a*c^3*x^9)]*Hypergeometric2F1[2/9, 1/2, 11/9, -((b*(c*x^3 )^(9/2))/(a*c^3*x^9))])/(13*b*c^(3/2)*Sqrt[a + (b*(c*x^3)^(9/2))/(c^3*x^9) ])))/(31*b*c^(3/2))))/49)
3.30.75.3.1 Defintions of rubi rules used
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c* x)^(m + 1)*((a + b*x^n)^p/(c*(m + n*p + 1))), x] + Simp[a*n*(p/(m + n*p + 1 )) Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x] && I GtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m , p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Simp[ a*c^n*((m - n + 1)/(b*(m + n*p + 1))) Int[(c*x)^(m - n)*(a + b*x^n)^p, x] , x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n* p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denomi nator[n]}, Simp[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]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p *((c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/n, (m + 1)/n + 1 , (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] && !IGtQ[p, 0] && (ILt Q[p, 0] || GtQ[a, 0])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^I ntPart[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])
Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*((c_.)*(x_)^(q_))^(n_))^(p_.), x_Symbo l] :> With[{k = Denominator[n]}, Subst[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]
\[\int x^{9} \sqrt {a +b \left (c \,x^{3}\right )^{\frac {3}{2}}}d x\]
\[ \int x^9 \sqrt {a+b \left (c x^3\right )^{3/2}} \, dx=\int { \sqrt {\left (c x^{3}\right )^{\frac {3}{2}} b + a} x^{9} \,d x } \]
\[ \int x^9 \sqrt {a+b \left (c x^3\right )^{3/2}} \, dx=\int x^{9} \sqrt {a + b \left (c x^{3}\right )^{\frac {3}{2}}}\, dx \]
\[ \int x^9 \sqrt {a+b \left (c x^3\right )^{3/2}} \, dx=\int { \sqrt {\left (c x^{3}\right )^{\frac {3}{2}} b + a} x^{9} \,d x } \]
\[ \int x^9 \sqrt {a+b \left (c x^3\right )^{3/2}} \, dx=\int { \sqrt {\left (c x^{3}\right )^{\frac {3}{2}} b + a} x^{9} \,d x } \]
Timed out. \[ \int x^9 \sqrt {a+b \left (c x^3\right )^{3/2}} \, dx=\int x^9\,\sqrt {a+b\,{\left (c\,x^3\right )}^{3/2}} \,d x \]