Integrand size = 19, antiderivative size = 400 \[ \int x \sqrt {a+b \sqrt {c x^3}} \, dx=\frac {4}{11} x^2 \sqrt {a+b \sqrt {c x^3}}+\frac {12 a x^2 \sqrt {a+b \sqrt {c x^3}}}{55 b \sqrt {c x^3}}-\frac {8\ 3^{3/4} \sqrt {2+\sqrt {3}} a^2 \left (\sqrt [3]{a}+\frac {\sqrt [3]{b} c^{2/3} x^2}{\sqrt {c x^3}}\right ) \sqrt {\frac {a^{2/3}+b^{2/3} \sqrt [3]{c} x-\frac {\sqrt [3]{a} \sqrt [3]{b} c^{2/3} x^2}{\sqrt {c x^3}}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b} c^{2/3} x^2}{\sqrt {c x^3}}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b} c^{2/3} x^2}{\sqrt {c x^3}}}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b} c^{2/3} x^2}{\sqrt {c x^3}}}\right ),-7-4 \sqrt {3}\right )}{55 b^{4/3} c^{2/3} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b} c^{2/3} x^2}{\sqrt {c x^3}}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b} c^{2/3} x^2}{\sqrt {c x^3}}\right )^2}} \sqrt {a+b \sqrt {c x^3}}} \] Output:
4/11*x^2*(a+b*(c*x^3)^(1/2))^(1/2)+12/55*a*x^2*(a+b*(c*x^3)^(1/2))^(1/2)/b /(c*x^3)^(1/2)-8/55*3^(3/4)*(1/2*6^(1/2)+1/2*2^(1/2))*a^2*(a^(1/3)+b^(1/3) *c^(2/3)*x^2/(c*x^3)^(1/2))*((a^(2/3)+b^(2/3)*c^(1/3)*x-a^(1/3)*b^(1/3)*c^ (2/3)*x^2/(c*x^3)^(1/2))/((1+3^(1/2))*a^(1/3)+b^(1/3)*c^(2/3)*x^2/(c*x^3)^ (1/2))^2)^(1/2)*EllipticF(((1-3^(1/2))*a^(1/3)+b^(1/3)*c^(2/3)*x^2/(c*x^3) ^(1/2))/((1+3^(1/2))*a^(1/3)+b^(1/3)*c^(2/3)*x^2/(c*x^3)^(1/2)),I*3^(1/2)+ 2*I)/b^(4/3)/c^(2/3)/(a^(1/3)*(a^(1/3)+b^(1/3)*c^(2/3)*x^2/(c*x^3)^(1/2))/ ((1+3^(1/2))*a^(1/3)+b^(1/3)*c^(2/3)*x^2/(c*x^3)^(1/2))^2)^(1/2)/(a+b*(c*x ^3)^(1/2))^(1/2)
\[ \int x \sqrt {a+b \sqrt {c x^3}} \, dx=\int x \sqrt {a+b \sqrt {c x^3}} \, dx \] Input:
Integrate[x*Sqrt[a + b*Sqrt[c*x^3]],x]
Output:
Integrate[x*Sqrt[a + b*Sqrt[c*x^3]], x]
Time = 0.62 (sec) , antiderivative size = 426, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {893, 864, 811, 843, 759}
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.
\(\displaystyle \int x \sqrt {a+b \sqrt {c x^3}} \, dx\) |
\(\Big \downarrow \) 893 |
\(\displaystyle \int x \sqrt {a+b \sqrt {c} x^{3/2}}dx\) |
\(\Big \downarrow \) 864 |
\(\displaystyle 2 \int \frac {\left (c x^3\right )^{3/2} \sqrt {\frac {b \left (c x^3\right )^{3/2}}{c x^3}+a}}{c^{3/2} x^3}d\frac {\sqrt {c x^3}}{\sqrt {c} x}\) |
\(\Big \downarrow \) 811 |
\(\displaystyle 2 \left (\frac {3}{11} a \int \frac {\left (c x^3\right )^{3/2}}{c^{3/2} x^3 \sqrt {\frac {b \left (c x^3\right )^{3/2}}{c x^3}+a}}d\frac {\sqrt {c x^3}}{\sqrt {c} x}+\frac {2}{11} x^2 \sqrt {a+\frac {b \left (c x^3\right )^{3/2}}{c x^3}}\right )\) |
\(\Big \downarrow \) 843 |
\(\displaystyle 2 \left (\frac {3}{11} a \left (\frac {2 \sqrt {c x^3} \sqrt {a+\frac {b \left (c x^3\right )^{3/2}}{c x^3}}}{5 b c x}-\frac {2 a \int \frac {1}{\sqrt {\frac {b \left (c x^3\right )^{3/2}}{c x^3}+a}}d\frac {\sqrt {c x^3}}{\sqrt {c} x}}{5 b \sqrt {c}}\right )+\frac {2}{11} x^2 \sqrt {a+\frac {b \left (c x^3\right )^{3/2}}{c x^3}}\right )\) |
\(\Big \downarrow \) 759 |
\(\displaystyle 2 \left (\frac {3}{11} a \left (\frac {2 \sqrt {c x^3} \sqrt {a+\frac {b \left (c x^3\right )^{3/2}}{c x^3}}}{5 b c x}-\frac {4 \sqrt {2+\sqrt {3}} a \left (\sqrt [3]{a}+\frac {\sqrt [3]{b} \sqrt {c x^3}}{\sqrt [3]{c} x}\right ) \sqrt {\frac {a^{2/3}-\frac {\sqrt [3]{a} \sqrt [3]{b} \sqrt {c x^3}}{\sqrt [3]{c} x}+b^{2/3} \sqrt [3]{c} x}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b} \sqrt {c x^3}}{\sqrt [3]{c} x}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\frac {\sqrt [3]{b} \sqrt {c x^3}}{\sqrt [3]{c} x}+\left (1-\sqrt {3}\right ) \sqrt [3]{a}}{\frac {\sqrt [3]{b} \sqrt {c x^3}}{\sqrt [3]{c} x}+\left (1+\sqrt {3}\right ) \sqrt [3]{a}}\right ),-7-4 \sqrt {3}\right )}{5 \sqrt [4]{3} b^{4/3} c^{2/3} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b} \sqrt {c x^3}}{\sqrt [3]{c} x}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b} \sqrt {c x^3}}{\sqrt [3]{c} x}\right )^2}} \sqrt {a+\frac {b \left (c x^3\right )^{3/2}}{c x^3}}}\right )+\frac {2}{11} x^2 \sqrt {a+\frac {b \left (c x^3\right )^{3/2}}{c x^3}}\right )\) |
Input:
Int[x*Sqrt[a + b*Sqrt[c*x^3]],x]
Output:
2*((2*x^2*Sqrt[a + (b*(c*x^3)^(3/2))/(c*x^3)])/11 + (3*a*((2*Sqrt[c*x^3]*S qrt[a + (b*(c*x^3)^(3/2))/(c*x^3)])/(5*b*c*x) - (4*Sqrt[2 + Sqrt[3]]*a*(a^ (1/3) + (b^(1/3)*Sqrt[c*x^3])/(c^(1/3)*x))*Sqrt[(a^(2/3) + b^(2/3)*c^(1/3) *x - (a^(1/3)*b^(1/3)*Sqrt[c*x^3])/(c^(1/3)*x))/((1 + Sqrt[3])*a^(1/3) + ( b^(1/3)*Sqrt[c*x^3])/(c^(1/3)*x))^2]*EllipticF[ArcSin[((1 - Sqrt[3])*a^(1/ 3) + (b^(1/3)*Sqrt[c*x^3])/(c^(1/3)*x))/((1 + Sqrt[3])*a^(1/3) + (b^(1/3)* Sqrt[c*x^3])/(c^(1/3)*x))], -7 - 4*Sqrt[3]])/(5*3^(1/4)*b^(4/3)*c^(2/3)*Sq rt[(a^(1/3)*(a^(1/3) + (b^(1/3)*Sqrt[c*x^3])/(c^(1/3)*x)))/((1 + Sqrt[3])* a^(1/3) + (b^(1/3)*Sqrt[c*x^3])/(c^(1/3)*x))^2]*Sqrt[a + (b*(c*x^3)^(3/2)) /(c*x^3)])))/11)
Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt[2 + Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s *x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[s* ((s + r*x)/((1 + Sqrt[3])*s + r*x)^2)]))*EllipticF[ArcSin[((1 - Sqrt[3])*s + r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]], x]] /; FreeQ[{a, b}, x] & & PosQ[a]
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[((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]
Time = 1.48 (sec) , antiderivative size = 348, normalized size of antiderivative = 0.87
method | result | size |
default | \(\frac {\frac {4 i a^{2} \sqrt {3}\, \left (-a \,b^{2} c \right )^{\frac {1}{3}} \sqrt {2}\, \sqrt {\frac {i \left (-i \sqrt {3}\, x \left (-a \,b^{2} c \right )^{\frac {1}{3}}+2 b \sqrt {c \,x^{3}}+\left (-a \,b^{2} c \right )^{\frac {1}{3}} x \right ) \sqrt {3}}{\left (-a \,b^{2} c \right )^{\frac {1}{3}} x}}\, \sqrt {\frac {b \sqrt {c \,x^{3}}-\left (-a \,b^{2} c \right )^{\frac {1}{3}} x}{x \left (-a \,b^{2} c \right )^{\frac {1}{3}} \left (i \sqrt {3}-3\right )}}\, \sqrt {-\frac {i \left (i \sqrt {3}\, x \left (-a \,b^{2} c \right )^{\frac {1}{3}}+2 b \sqrt {c \,x^{3}}+\left (-a \,b^{2} c \right )^{\frac {1}{3}} x \right ) \sqrt {3}}{\left (-a \,b^{2} c \right )^{\frac {1}{3}} x}}\, \operatorname {EllipticF}\left (\frac {\sqrt {3}\, \sqrt {2}\, \sqrt {\frac {i \left (-i \sqrt {3}\, x \left (-a \,b^{2} c \right )^{\frac {1}{3}}+2 b \sqrt {c \,x^{3}}+\left (-a \,b^{2} c \right )^{\frac {1}{3}} x \right ) \sqrt {3}}{\left (-a \,b^{2} c \right )^{\frac {1}{3}} x}}}{6}, \sqrt {2}\, \sqrt {\frac {i \sqrt {3}}{i \sqrt {3}-3}}\right ) \sqrt {c \,x^{3}}}{55}+\frac {4 c^{2} x^{5} b^{3}}{11}+\frac {32 \sqrt {c \,x^{3}}\, a \,b^{2} c \,x^{2}}{55}+\frac {12 a^{2} b c \,x^{2}}{55}}{c \,b^{2} \sqrt {c \,x^{3}}\, \sqrt {a +b \sqrt {c \,x^{3}}}}\) | \(348\) |
Input:
int(x*(a+b*(c*x^3)^(1/2))^(1/2),x,method=_RETURNVERBOSE)
Output:
4/55*(I*a^2*3^(1/2)*(-a*b^2*c)^(1/3)*2^(1/2)*(I*(-I*3^(1/2)*x*(-a*b^2*c)^( 1/3)+2*b*(c*x^3)^(1/2)+(-a*b^2*c)^(1/3)*x)*3^(1/2)/(-a*b^2*c)^(1/3)/x)^(1/ 2)*((b*(c*x^3)^(1/2)-(-a*b^2*c)^(1/3)*x)/x/(-a*b^2*c)^(1/3)/(I*3^(1/2)-3)) ^(1/2)*(-I*(I*3^(1/2)*x*(-a*b^2*c)^(1/3)+2*b*(c*x^3)^(1/2)+(-a*b^2*c)^(1/3 )*x)*3^(1/2)/(-a*b^2*c)^(1/3)/x)^(1/2)*EllipticF(1/6*3^(1/2)*2^(1/2)*(I*(- I*3^(1/2)*x*(-a*b^2*c)^(1/3)+2*b*(c*x^3)^(1/2)+(-a*b^2*c)^(1/3)*x)*3^(1/2) /(-a*b^2*c)^(1/3)/x)^(1/2),2^(1/2)*(I*3^(1/2)/(I*3^(1/2)-3))^(1/2))*(c*x^3 )^(1/2)+5*c^2*x^5*b^3+8*(c*x^3)^(1/2)*a*b^2*c*x^2+3*a^2*b*c*x^2)/c/b^2/(c* x^3)^(1/2)/(a+b*(c*x^3)^(1/2))^(1/2)
\[ \int x \sqrt {a+b \sqrt {c x^3}} \, dx=\int { \sqrt {\sqrt {c x^{3}} b + a} x \,d x } \] Input:
integrate(x*(a+b*(c*x^3)^(1/2))^(1/2),x, algorithm="fricas")
Output:
integral(sqrt(sqrt(c*x^3)*b + a)*x, x)
\[ \int x \sqrt {a+b \sqrt {c x^3}} \, dx=\int x \sqrt {a + b \sqrt {c x^{3}}}\, dx \] Input:
integrate(x*(a+b*(c*x**3)**(1/2))**(1/2),x)
Output:
Integral(x*sqrt(a + b*sqrt(c*x**3)), x)
\[ \int x \sqrt {a+b \sqrt {c x^3}} \, dx=\int { \sqrt {\sqrt {c x^{3}} b + a} x \,d x } \] Input:
integrate(x*(a+b*(c*x^3)^(1/2))^(1/2),x, algorithm="maxima")
Output:
integrate(sqrt(sqrt(c*x^3)*b + a)*x, x)
\[ \int x \sqrt {a+b \sqrt {c x^3}} \, dx=\int { \sqrt {\sqrt {c x^{3}} b + a} x \,d x } \] Input:
integrate(x*(a+b*(c*x^3)^(1/2))^(1/2),x, algorithm="giac")
Output:
integrate(sqrt(sqrt(c*x^3)*b + a)*x, x)
Timed out. \[ \int x \sqrt {a+b \sqrt {c x^3}} \, dx=\int x\,\sqrt {a+b\,\sqrt {c\,x^3}} \,d x \] Input:
int(x*(a + b*(c*x^3)^(1/2))^(1/2),x)
Output:
int(x*(a + b*(c*x^3)^(1/2))^(1/2), x)
\[ \int x \sqrt {a+b \sqrt {c x^3}} \, dx=\int x \sqrt {a +b \sqrt {c \,x^{3}}}d x \] Input:
int(x*(a+b*(c*x^3)^(1/2))^(1/2),x)
Output:
int(x*(a+b*(c*x^3)^(1/2))^(1/2),x)