\(\int \frac {x \sqrt {a-b x^2}}{\sqrt [3]{c+d x}} \, dx\) [1616]

Optimal result

Integrand size = 23, antiderivative size = 174 \[ \int \frac {x \sqrt {a-b x^2}}{\sqrt [3]{c+d x}} \, dx=-\frac {\left (a-b x^2\right )^{3/2}}{3 b \sqrt [3]{c+d x}}+\frac {\left (a-b x^2\right )^{3/2} \operatorname {AppellF1}\left (-\frac {1}{3},-\frac {3}{2},-\frac {3}{2},\frac {2}{3},\frac {c+d x}{c-\frac {\sqrt {a} d}{\sqrt {b}}},\frac {c+d x}{c+\frac {\sqrt {a} d}{\sqrt {b}}}\right )}{3 b \sqrt [3]{c+d x} \left (1-\frac {c+d x}{c-\frac {\sqrt {a} d}{\sqrt {b}}}\right )^{3/2} \left (1-\frac {c+d x}{c+\frac {\sqrt {a} d}{\sqrt {b}}}\right )^{3/2}} \] Output:

-1/3*(-b*x^2+a)^(3/2)/b/(d*x+c)^(1/3)+1/3*(-b*x^2+a)^(3/2)*AppellF1(-1/3,- 
3/2,-3/2,2/3,(d*x+c)/(c-a^(1/2)*d/b^(1/2)),(d*x+c)/(c+a^(1/2)*d/b^(1/2)))/ 
b/(d*x+c)^(1/3)/(1-(d*x+c)/(c-a^(1/2)*d/b^(1/2)))^(3/2)/(1-(d*x+c)/(c+a^(1 
/2)*d/b^(1/2)))^(3/2)
                                                                                    
                                                                                    
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(561\) vs. \(2(174)=348\).

Time = 22.75 (sec) , antiderivative size = 561, normalized size of antiderivative = 3.22 \[ \int \frac {x \sqrt {a-b x^2}}{\sqrt [3]{c+d x}} \, dx=\frac {3 \sqrt {a-b x^2} \left (-8 (6 c-5 d x) (c+d x)-\frac {3 \left (4 b c \left (6 b c^2-7 a d^2\right ) (c+d x) \sqrt {\frac {-a d^2+d \sqrt {a b d^2} x+c \left (\sqrt {a b d^2}-b d x\right )}{\left (-b c+\sqrt {a b d^2}\right ) (c+d x)}} \sqrt {\frac {a d^2+d \sqrt {a b d^2} x+c \left (\sqrt {a b d^2}+b d x\right )}{\left (b c+\sqrt {a b d^2}\right ) (c+d x)}} \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},\frac {1}{2},\frac {4}{3},\frac {b c^2-a d^2}{\left (b c+\sqrt {a b d^2}\right ) (c+d x)},\frac {-b c^2+a d^2}{\left (-b c+\sqrt {a b d^2}\right ) (c+d x)}\right )-\left (6 b c^2-5 a d^2\right ) \left (4 d^2 \left (a-b x^2\right )+\left (b c^2-a d^2\right ) \sqrt {\frac {-a d^2+c \sqrt {a b d^2}-b c d x+d \sqrt {a b d^2} x}{\left (-b c+\sqrt {a b d^2}\right ) (c+d x)}} \sqrt {\frac {a d^2+c \sqrt {a b d^2}+b c d x+d \sqrt {a b d^2} x}{\left (b c+\sqrt {a b d^2}\right ) (c+d x)}} \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},\frac {1}{2},\frac {7}{3},\frac {b c^2-a d^2}{\left (b c+\sqrt {a b d^2}\right ) (c+d x)},\frac {-b c^2+a d^2}{\left (-b c+\sqrt {a b d^2}\right ) (c+d x)}\right )\right )\right )}{b d^2 \left (a-b x^2\right )}\right )}{320 d^2 \sqrt [3]{c+d x}} \] Input:

Integrate[(x*Sqrt[a - b*x^2])/(c + d*x)^(1/3),x]
 

Output:

(3*Sqrt[a - b*x^2]*(-8*(6*c - 5*d*x)*(c + d*x) - (3*(4*b*c*(6*b*c^2 - 7*a* 
d^2)*(c + d*x)*Sqrt[(-(a*d^2) + d*Sqrt[a*b*d^2]*x + c*(Sqrt[a*b*d^2] - b*d 
*x))/((-(b*c) + Sqrt[a*b*d^2])*(c + d*x))]*Sqrt[(a*d^2 + d*Sqrt[a*b*d^2]*x 
 + c*(Sqrt[a*b*d^2] + b*d*x))/((b*c + Sqrt[a*b*d^2])*(c + d*x))]*AppellF1[ 
1/3, 1/2, 1/2, 4/3, (b*c^2 - a*d^2)/((b*c + Sqrt[a*b*d^2])*(c + d*x)), (-( 
b*c^2) + a*d^2)/((-(b*c) + Sqrt[a*b*d^2])*(c + d*x))] - (6*b*c^2 - 5*a*d^2 
)*(4*d^2*(a - b*x^2) + (b*c^2 - a*d^2)*Sqrt[(-(a*d^2) + c*Sqrt[a*b*d^2] - 
b*c*d*x + d*Sqrt[a*b*d^2]*x)/((-(b*c) + Sqrt[a*b*d^2])*(c + d*x))]*Sqrt[(a 
*d^2 + c*Sqrt[a*b*d^2] + b*c*d*x + d*Sqrt[a*b*d^2]*x)/((b*c + Sqrt[a*b*d^2 
])*(c + d*x))]*AppellF1[4/3, 1/2, 1/2, 7/3, (b*c^2 - a*d^2)/((b*c + Sqrt[a 
*b*d^2])*(c + d*x)), (-(b*c^2) + a*d^2)/((-(b*c) + Sqrt[a*b*d^2])*(c + d*x 
))])))/(b*d^2*(a - b*x^2))))/(320*d^2*(c + d*x)^(1/3))
 

Rubi [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(359\) vs. \(2(174)=348\).

Time = 0.48 (sec) , antiderivative size = 359, normalized size of antiderivative = 2.06, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {591, 27, 719, 514, 150}

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.

\(\displaystyle \int \frac {x \sqrt {a-b x^2}}{\sqrt [3]{c+d x}} \, dx\)

\(\Big \downarrow \) 591

\(\displaystyle \frac {9 \int -\frac {a c d+\left (6 b c^2-5 a d^2\right ) x}{3 \sqrt [3]{c+d x} \sqrt {a-b x^2}}dx}{40 d^2}-\frac {3 \sqrt {a-b x^2} (6 c-5 d x) (c+d x)^{2/3}}{40 d^2}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {3 \int \frac {a c d+\left (6 b c^2-5 a d^2\right ) x}{\sqrt [3]{c+d x} \sqrt {a-b x^2}}dx}{40 d^2}-\frac {3 \sqrt {a-b x^2} (c+d x)^{2/3} (6 c-5 d x)}{40 d^2}\)

\(\Big \downarrow \) 719

\(\displaystyle -\frac {3 \left (\frac {\left (6 b c^2-5 a d^2\right ) \int \frac {(c+d x)^{2/3}}{\sqrt {a-b x^2}}dx}{d}-\frac {6 c \left (b c^2-a d^2\right ) \int \frac {1}{\sqrt [3]{c+d x} \sqrt {a-b x^2}}dx}{d}\right )}{40 d^2}-\frac {3 \sqrt {a-b x^2} (c+d x)^{2/3} (6 c-5 d x)}{40 d^2}\)

\(\Big \downarrow \) 514

\(\displaystyle -\frac {3 \left (\frac {\left (6 b c^2-5 a d^2\right ) \sqrt {1-\frac {c+d x}{c-\frac {\sqrt {a} d}{\sqrt {b}}}} \sqrt {1-\frac {c+d x}{\frac {\sqrt {a} d}{\sqrt {b}}+c}} \int \frac {(c+d x)^{2/3}}{\sqrt {1-\frac {c+d x}{c-\frac {\sqrt {a} d}{\sqrt {b}}}} \sqrt {1-\frac {c+d x}{c+\frac {\sqrt {a} d}{\sqrt {b}}}}}d(c+d x)}{d^2 \sqrt {a-b x^2}}-\frac {6 c \left (b c^2-a d^2\right ) \sqrt {1-\frac {c+d x}{c-\frac {\sqrt {a} d}{\sqrt {b}}}} \sqrt {1-\frac {c+d x}{\frac {\sqrt {a} d}{\sqrt {b}}+c}} \int \frac {1}{\sqrt [3]{c+d x} \sqrt {1-\frac {c+d x}{c-\frac {\sqrt {a} d}{\sqrt {b}}}} \sqrt {1-\frac {c+d x}{c+\frac {\sqrt {a} d}{\sqrt {b}}}}}d(c+d x)}{d^2 \sqrt {a-b x^2}}\right )}{40 d^2}-\frac {3 \sqrt {a-b x^2} (c+d x)^{2/3} (6 c-5 d x)}{40 d^2}\)

\(\Big \downarrow \) 150

\(\displaystyle -\frac {3 \left (\frac {3 (c+d x)^{5/3} \left (6 b c^2-5 a d^2\right ) \sqrt {1-\frac {c+d x}{c-\frac {\sqrt {a} d}{\sqrt {b}}}} \sqrt {1-\frac {c+d x}{\frac {\sqrt {a} d}{\sqrt {b}}+c}} \operatorname {AppellF1}\left (\frac {5}{3},\frac {1}{2},\frac {1}{2},\frac {8}{3},\frac {c+d x}{c-\frac {\sqrt {a} d}{\sqrt {b}}},\frac {c+d x}{c+\frac {\sqrt {a} d}{\sqrt {b}}}\right )}{5 d^2 \sqrt {a-b x^2}}-\frac {9 c (c+d x)^{2/3} \left (b c^2-a d^2\right ) \sqrt {1-\frac {c+d x}{c-\frac {\sqrt {a} d}{\sqrt {b}}}} \sqrt {1-\frac {c+d x}{\frac {\sqrt {a} d}{\sqrt {b}}+c}} \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{2},\frac {1}{2},\frac {5}{3},\frac {c+d x}{c-\frac {\sqrt {a} d}{\sqrt {b}}},\frac {c+d x}{c+\frac {\sqrt {a} d}{\sqrt {b}}}\right )}{d^2 \sqrt {a-b x^2}}\right )}{40 d^2}-\frac {3 \sqrt {a-b x^2} (c+d x)^{2/3} (6 c-5 d x)}{40 d^2}\)

Input:

Int[(x*Sqrt[a - b*x^2])/(c + d*x)^(1/3),x]
 

Output:

(-3*(6*c - 5*d*x)*(c + d*x)^(2/3)*Sqrt[a - b*x^2])/(40*d^2) - (3*((-9*c*(b 
*c^2 - a*d^2)*(c + d*x)^(2/3)*Sqrt[1 - (c + d*x)/(c - (Sqrt[a]*d)/Sqrt[b]) 
]*Sqrt[1 - (c + d*x)/(c + (Sqrt[a]*d)/Sqrt[b])]*AppellF1[2/3, 1/2, 1/2, 5/ 
3, (c + d*x)/(c - (Sqrt[a]*d)/Sqrt[b]), (c + d*x)/(c + (Sqrt[a]*d)/Sqrt[b] 
)])/(d^2*Sqrt[a - b*x^2]) + (3*(6*b*c^2 - 5*a*d^2)*(c + d*x)^(5/3)*Sqrt[1 
- (c + d*x)/(c - (Sqrt[a]*d)/Sqrt[b])]*Sqrt[1 - (c + d*x)/(c + (Sqrt[a]*d) 
/Sqrt[b])]*AppellF1[5/3, 1/2, 1/2, 8/3, (c + d*x)/(c - (Sqrt[a]*d)/Sqrt[b] 
), (c + d*x)/(c + (Sqrt[a]*d)/Sqrt[b])])/(5*d^2*Sqrt[a - b*x^2])))/(40*d^2 
)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 150

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ 
] :> Simp[c^n*e^p*((b*x)^(m + 1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2 
, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] &&  !In 
tegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
 

rule 514

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ 
{q = Rt[-a/b, 2]}, Simp[(a + b*x^2)^p/(d*(1 - (c + d*x)/(c - d*q))^p*(1 - ( 
c + d*x)/(c + d*q))^p)   Subst[Int[x^n*Simp[1 - x/(c + d*q), x]^p*Simp[1 - 
x/(c - d*q), x]^p, x], x, c + d*x], x]] /; FreeQ[{a, b, c, d, n, p}, x] && 
NeQ[b*c^2 + a*d^2, 0]
 

rule 591

Int[(x_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
Simp[(-(c + d*x)^(n + 1))*(a + b*x^2)^p*((c*(2*p + 1) - d*(n + 2*p + 1)*x)/ 
(d^2*(n + 2*p + 1)*(n + 2*p + 2))), x] + Simp[2*(p/(d^2*(n + 2*p + 1)*(n + 
2*p + 2)))   Int[(c + d*x)^n*(a + b*x^2)^(p - 1)*Simp[a*c*d*n + (b*c^2*(2*p 
 + 1) + a*d^2*(n + 2*p + 1))*x, x], x], x] /; FreeQ[{a, b, c, d, n}, x] && 
GtQ[p, 0] && LeQ[-1, n, 0] &&  !ILtQ[n + 2*p, 0]
 

rule 719

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p 
_.), x_Symbol] :> Simp[g/e   Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] + 
Simp[(e*f - d*g)/e   Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, 
d, e, f, g, m, p}, x] &&  !IGtQ[m, 0]
 

Maple [F]

\[\int \frac {x \sqrt {-b \,x^{2}+a}}{\left (d x +c \right )^{\frac {1}{3}}}d x\]

Input:

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

Output:

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

Fricas [F]

\[ \int \frac {x \sqrt {a-b x^2}}{\sqrt [3]{c+d x}} \, dx=\int { \frac {\sqrt {-b x^{2} + a} x}{{\left (d x + c\right )}^{\frac {1}{3}}} \,d x } \] Input:

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

Output:

integral(sqrt(-b*x^2 + a)*x/(d*x + c)^(1/3), x)
 

Sympy [F]

\[ \int \frac {x \sqrt {a-b x^2}}{\sqrt [3]{c+d x}} \, dx=\int \frac {x \sqrt {a - b x^{2}}}{\sqrt [3]{c + d x}}\, dx \] Input:

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

Output:

Integral(x*sqrt(a - b*x**2)/(c + d*x)**(1/3), x)
 

Maxima [F]

\[ \int \frac {x \sqrt {a-b x^2}}{\sqrt [3]{c+d x}} \, dx=\int { \frac {\sqrt {-b x^{2} + a} x}{{\left (d x + c\right )}^{\frac {1}{3}}} \,d x } \] Input:

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

Output:

integrate(sqrt(-b*x^2 + a)*x/(d*x + c)^(1/3), x)
 

Giac [F]

\[ \int \frac {x \sqrt {a-b x^2}}{\sqrt [3]{c+d x}} \, dx=\int { \frac {\sqrt {-b x^{2} + a} x}{{\left (d x + c\right )}^{\frac {1}{3}}} \,d x } \] Input:

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

Output:

integrate(sqrt(-b*x^2 + a)*x/(d*x + c)^(1/3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x \sqrt {a-b x^2}}{\sqrt [3]{c+d x}} \, dx=\int \frac {x\,\sqrt {a-b\,x^2}}{{\left (c+d\,x\right )}^{1/3}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {x \sqrt {a-b x^2}}{\sqrt [3]{c+d x}} \, dx=\int \frac {x \sqrt {-b \,x^{2}+a}}{\left (d x +c \right )^{\frac {1}{3}}}d x \] Input:

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

Output:

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