\(\int \frac {x}{\sqrt {c+d x} (a+b x^2)^{4/3}} \, dx\) [1641]

Optimal result

Integrand size = 22, antiderivative size = 180 \[ \int \frac {x}{\sqrt {c+d x} \left (a+b x^2\right )^{4/3}} \, dx=-\frac {3}{2 b \sqrt {c+d x} \sqrt [3]{a+b x^2}}+\frac {3 \sqrt [3]{1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}} \sqrt [3]{1-\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}} \operatorname {AppellF1}\left (-\frac {1}{2},\frac {1}{3},\frac {1}{3},\frac {1}{2},\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}},\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}\right )}{2 b \sqrt {c+d x} \sqrt [3]{a+b x^2}} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 21.79 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.42 \[ \int \frac {x}{\sqrt {c+d x} \left (a+b x^2\right )^{4/3}} \, dx=\frac {3 \left (-7+3 \sqrt [3]{\frac {a d^2+c \sqrt {-a b d^2}-b c d x+d \sqrt {-a b d^2} x}{\left (-b c+\sqrt {-a b d^2}\right ) (c+d x)}} \sqrt [3]{\frac {-a d^2+c \sqrt {-a b d^2}+b c d x+d \sqrt {-a b d^2} x}{\left (b c+\sqrt {-a b d^2}\right ) (c+d x)}} \operatorname {AppellF1}\left (\frac {7}{6},\frac {1}{3},\frac {1}{3},\frac {13}{6},\frac {b c^2+a d^2}{\left (b c-\sqrt {-a b d^2}\right ) (c+d x)},\frac {b c^2+a d^2}{\left (b c+\sqrt {-a b d^2}\right ) (c+d x)}\right )\right )}{14 b \sqrt {c+d x} \sqrt [3]{a+b x^2}} \] Input:

Integrate[x/(Sqrt[c + d*x]*(a + b*x^2)^(4/3)),x]
 

Output:

(3*(-7 + 3*((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)))^(1/3)*((-(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) 
))^(1/3)*AppellF1[7/6, 1/3, 1/3, 13/6, (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))])) 
/(14*b*Sqrt[c + d*x]*(a + b*x^2)^(1/3))
 

Rubi [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(366\) vs. \(2(180)=360\).

Time = 0.45 (sec) , antiderivative size = 366, normalized size of antiderivative = 2.03, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {593, 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.

\(\displaystyle \int \frac {x}{\left (a+b x^2\right )^{4/3} \sqrt {c+d x}} \, dx\)

\(\Big \downarrow \) 593

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

\(\Big \downarrow \) 27

\(\displaystyle -\frac {d \int \frac {3 c+5 d x}{\sqrt {c+d x} \sqrt [3]{b x^2+a}}dx}{4 \left (a d^2+b c^2\right )}-\frac {3 \sqrt {c+d x} (c-d x)}{2 \sqrt [3]{a+b x^2} \left (a d^2+b c^2\right )}\)

\(\Big \downarrow \) 719

\(\displaystyle -\frac {d \left (5 \int \frac {\sqrt {c+d x}}{\sqrt [3]{b x^2+a}}dx-2 c \int \frac {1}{\sqrt {c+d x} \sqrt [3]{b x^2+a}}dx\right )}{4 \left (a d^2+b c^2\right )}-\frac {3 \sqrt {c+d x} (c-d x)}{2 \sqrt [3]{a+b x^2} \left (a d^2+b c^2\right )}\)

\(\Big \downarrow \) 514

\(\displaystyle -\frac {d \left (\frac {5 \sqrt [3]{1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}} \sqrt [3]{1-\frac {c+d x}{\frac {\sqrt {-a} d}{\sqrt {b}}+c}} \int \frac {\sqrt {c+d x}}{\sqrt [3]{1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}} \sqrt [3]{1-\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}}}d(c+d x)}{d \sqrt [3]{a+b x^2}}-\frac {2 c \sqrt [3]{1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}} \sqrt [3]{1-\frac {c+d x}{\frac {\sqrt {-a} d}{\sqrt {b}}+c}} \int \frac {1}{\sqrt {c+d x} \sqrt [3]{1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}} \sqrt [3]{1-\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}}}d(c+d x)}{d \sqrt [3]{a+b x^2}}\right )}{4 \left (a d^2+b c^2\right )}-\frac {3 \sqrt {c+d x} (c-d x)}{2 \sqrt [3]{a+b x^2} \left (a d^2+b c^2\right )}\)

\(\Big \downarrow \) 150

\(\displaystyle -\frac {d \left (\frac {10 (c+d x)^{3/2} \sqrt [3]{1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}} \sqrt [3]{1-\frac {c+d x}{\frac {\sqrt {-a} d}{\sqrt {b}}+c}} \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{3},\frac {1}{3},\frac {5}{2},\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}},\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}\right )}{3 d \sqrt [3]{a+b x^2}}-\frac {4 c \sqrt {c+d x} \sqrt [3]{1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}} \sqrt [3]{1-\frac {c+d x}{\frac {\sqrt {-a} d}{\sqrt {b}}+c}} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{3},\frac {1}{3},\frac {3}{2},\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}},\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}\right )}{d \sqrt [3]{a+b x^2}}\right )}{4 \left (a d^2+b c^2\right )}-\frac {3 \sqrt {c+d x} (c-d x)}{2 \sqrt [3]{a+b x^2} \left (a d^2+b c^2\right )}\)

Input:

Int[x/(Sqrt[c + d*x]*(a + b*x^2)^(4/3)),x]
 

Output:

(-3*(c - d*x)*Sqrt[c + d*x])/(2*(b*c^2 + a*d^2)*(a + b*x^2)^(1/3)) - (d*(( 
-4*c*Sqrt[c + d*x]*(1 - (c + d*x)/(c - (Sqrt[-a]*d)/Sqrt[b]))^(1/3)*(1 - ( 
c + d*x)/(c + (Sqrt[-a]*d)/Sqrt[b]))^(1/3)*AppellF1[1/2, 1/3, 1/3, 3/2, (c 
 + d*x)/(c - (Sqrt[-a]*d)/Sqrt[b]), (c + d*x)/(c + (Sqrt[-a]*d)/Sqrt[b])]) 
/(d*(a + b*x^2)^(1/3)) + (10*(c + d*x)^(3/2)*(1 - (c + d*x)/(c - (Sqrt[-a] 
*d)/Sqrt[b]))^(1/3)*(1 - (c + d*x)/(c + (Sqrt[-a]*d)/Sqrt[b]))^(1/3)*Appel 
lF1[3/2, 1/3, 1/3, 5/2, (c + d*x)/(c - (Sqrt[-a]*d)/Sqrt[b]), (c + d*x)/(c 
 + (Sqrt[-a]*d)/Sqrt[b])])/(3*d*(a + b*x^2)^(1/3))))/(4*(b*c^2 + a*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 593

Int[(x_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
 Simp[(c + d*x)^(n + 1)*(c - d*x)*((a + b*x^2)^(p + 1)/(2*(p + 1)*(b*c^2 + 
a*d^2))), x] - Simp[d/(2*(p + 1)*(b*c^2 + a*d^2))   Int[(c + d*x)^n*(a + b* 
x^2)^(p + 1)*(c*n - d*(n + 2*p + 4)*x), x], x] /; FreeQ[{a, b, c, d, n}, x] 
 && LtQ[p, -1] && NeQ[b*c^2 + a*d^2, 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]

\[\int \frac {x}{\sqrt {d x +c}\, \left (b \,x^{2}+a \right )^{\frac {4}{3}}}d x\]

Input:

int(x/(d*x+c)^(1/2)/(b*x^2+a)^(4/3),x)
 

Output:

int(x/(d*x+c)^(1/2)/(b*x^2+a)^(4/3),x)
 

Fricas [F]

\[ \int \frac {x}{\sqrt {c+d x} \left (a+b x^2\right )^{4/3}} \, dx=\int { \frac {x}{{\left (b x^{2} + a\right )}^{\frac {4}{3}} \sqrt {d x + c}} \,d x } \] Input:

integrate(x/(d*x+c)^(1/2)/(b*x^2+a)^(4/3),x, algorithm="fricas")
 

Output:

integral((b*x^2 + a)^(2/3)*sqrt(d*x + c)*x/(b^2*d*x^5 + b^2*c*x^4 + 2*a*b* 
d*x^3 + 2*a*b*c*x^2 + a^2*d*x + a^2*c), x)
 

Sympy [F]

\[ \int \frac {x}{\sqrt {c+d x} \left (a+b x^2\right )^{4/3}} \, dx=\int \frac {x}{\left (a + b x^{2}\right )^{\frac {4}{3}} \sqrt {c + d x}}\, dx \] Input:

integrate(x/(d*x+c)**(1/2)/(b*x**2+a)**(4/3),x)
 

Output:

Integral(x/((a + b*x**2)**(4/3)*sqrt(c + d*x)), x)
 

Maxima [F]

\[ \int \frac {x}{\sqrt {c+d x} \left (a+b x^2\right )^{4/3}} \, dx=\int { \frac {x}{{\left (b x^{2} + a\right )}^{\frac {4}{3}} \sqrt {d x + c}} \,d x } \] Input:

integrate(x/(d*x+c)^(1/2)/(b*x^2+a)^(4/3),x, algorithm="maxima")
 

Output:

integrate(x/((b*x^2 + a)^(4/3)*sqrt(d*x + c)), x)
 

Giac [F]

\[ \int \frac {x}{\sqrt {c+d x} \left (a+b x^2\right )^{4/3}} \, dx=\int { \frac {x}{{\left (b x^{2} + a\right )}^{\frac {4}{3}} \sqrt {d x + c}} \,d x } \] Input:

integrate(x/(d*x+c)^(1/2)/(b*x^2+a)^(4/3),x, algorithm="giac")
 

Output:

integrate(x/((b*x^2 + a)^(4/3)*sqrt(d*x + c)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x}{\sqrt {c+d x} \left (a+b x^2\right )^{4/3}} \, dx=\int \frac {x}{{\left (b\,x^2+a\right )}^{4/3}\,\sqrt {c+d\,x}} \,d x \] Input:

int(x/((a + b*x^2)^(4/3)*(c + d*x)^(1/2)),x)
 

Output:

int(x/((a + b*x^2)^(4/3)*(c + d*x)^(1/2)), x)
 

Reduce [F]

\[ \int \frac {x}{\sqrt {c+d x} \left (a+b x^2\right )^{4/3}} \, dx=\int \frac {\sqrt {d x +c}\, x}{\left (b \,x^{2}+a \right )^{\frac {1}{3}} a c +\left (b \,x^{2}+a \right )^{\frac {1}{3}} a d x +\left (b \,x^{2}+a \right )^{\frac {1}{3}} b c \,x^{2}+\left (b \,x^{2}+a \right )^{\frac {1}{3}} b d \,x^{3}}d x \] Input:

int(x/(d*x+c)^(1/2)/(b*x^2+a)^(4/3),x)
 

Output:

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