\(\int (a x+b x^3)^{3/2} \, dx\) [55]

Optimal result

Integrand size = 13, antiderivative size = 158 \[ \int \left (a x+b x^3\right )^{3/2} \, dx=\frac {8 a^2 \sqrt {a x+b x^3}}{77 b}+\frac {12}{77} a x^2 \sqrt {a x+b x^3}+\frac {2}{11} x \left (a x+b x^3\right )^{3/2}-\frac {4 a^{11/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{77 b^{5/4} \sqrt {a x+b x^3}} \] Output:

8/77*a^2*(b*x^3+a*x)^(1/2)/b+12/77*a*x^2*(b*x^3+a*x)^(1/2)+2/11*x*(b*x^3+a 
*x)^(3/2)-4/77*a^(11/4)*x^(1/2)*(a^(1/2)+b^(1/2)*x)*((b*x^2+a)/(a^(1/2)+b^ 
(1/2)*x)^2)^(1/2)*InverseJacobiAM(2*arctan(b^(1/4)*x^(1/2)/a^(1/4)),1/2*2^ 
(1/2))/b^(5/4)/(b*x^3+a*x)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.03 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.53 \[ \int \left (a x+b x^3\right )^{3/2} \, dx=\frac {2 \sqrt {x \left (a+b x^2\right )} \left (\left (a+b x^2\right )^2 \sqrt {1+\frac {b x^2}{a}}-a^2 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{4},\frac {5}{4},-\frac {b x^2}{a}\right )\right )}{11 b \sqrt {1+\frac {b x^2}{a}}} \] Input:

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

Output:

(2*Sqrt[x*(a + b*x^2)]*((a + b*x^2)^2*Sqrt[1 + (b*x^2)/a] - a^2*Hypergeome 
tric2F1[-3/2, 1/4, 5/4, -((b*x^2)/a)]))/(11*b*Sqrt[1 + (b*x^2)/a])
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {1910, 1927, 1930, 1917, 266, 761}

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.

\(\displaystyle \int \left (a x+b x^3\right )^{3/2} \, dx\)

\(\Big \downarrow \) 1910

\(\displaystyle \frac {6}{11} a \int x \sqrt {b x^3+a x}dx+\frac {2}{11} x \left (a x+b x^3\right )^{3/2}\)

\(\Big \downarrow \) 1927

\(\displaystyle \frac {6}{11} a \left (\frac {2}{7} a \int \frac {x^2}{\sqrt {b x^3+a x}}dx+\frac {2}{7} x^2 \sqrt {a x+b x^3}\right )+\frac {2}{11} x \left (a x+b x^3\right )^{3/2}\)

\(\Big \downarrow \) 1930

\(\displaystyle \frac {6}{11} a \left (\frac {2}{7} a \left (\frac {2 \sqrt {a x+b x^3}}{3 b}-\frac {a \int \frac {1}{\sqrt {b x^3+a x}}dx}{3 b}\right )+\frac {2}{7} x^2 \sqrt {a x+b x^3}\right )+\frac {2}{11} x \left (a x+b x^3\right )^{3/2}\)

\(\Big \downarrow \) 1917

\(\displaystyle \frac {6}{11} a \left (\frac {2}{7} a \left (\frac {2 \sqrt {a x+b x^3}}{3 b}-\frac {a \sqrt {x} \sqrt {a+b x^2} \int \frac {1}{\sqrt {x} \sqrt {b x^2+a}}dx}{3 b \sqrt {a x+b x^3}}\right )+\frac {2}{7} x^2 \sqrt {a x+b x^3}\right )+\frac {2}{11} x \left (a x+b x^3\right )^{3/2}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {6}{11} a \left (\frac {2}{7} a \left (\frac {2 \sqrt {a x+b x^3}}{3 b}-\frac {2 a \sqrt {x} \sqrt {a+b x^2} \int \frac {1}{\sqrt {b x^2+a}}d\sqrt {x}}{3 b \sqrt {a x+b x^3}}\right )+\frac {2}{7} x^2 \sqrt {a x+b x^3}\right )+\frac {2}{11} x \left (a x+b x^3\right )^{3/2}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {6}{11} a \left (\frac {2}{7} a \left (\frac {2 \sqrt {a x+b x^3}}{3 b}-\frac {a^{3/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{3 b^{5/4} \sqrt {a x+b x^3}}\right )+\frac {2}{7} x^2 \sqrt {a x+b x^3}\right )+\frac {2}{11} x \left (a x+b x^3\right )^{3/2}\)

Input:

Int[(a*x + b*x^3)^(3/2),x]
 

Output:

(2*x*(a*x + b*x^3)^(3/2))/11 + (6*a*((2*x^2*Sqrt[a*x + b*x^3])/7 + (2*a*(( 
2*Sqrt[a*x + b*x^3])/(3*b) - (a^(3/4)*Sqrt[x]*(Sqrt[a] + Sqrt[b]*x)*Sqrt[( 
a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticF[2*ArcTan[(b^(1/4)*Sqrt[x])/a 
^(1/4)], 1/2])/(3*b^(5/4)*Sqrt[a*x + b*x^3])))/7))/11
 

Defintions of rubi rules used

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 761

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 
1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* 
EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
 

rule 1910

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[x*((a*x^j 
 + b*x^n)^p/(n*p + 1)), x] + Simp[a*(n - j)*(p/(n*p + 1))   Int[x^j*(a*x^j 
+ b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] &&  !IntegerQ[p] && LtQ[0, j, 
n] && GtQ[p, 0] && NeQ[n*p + 1, 0]
 

rule 1917

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + 
b*x^n)^FracPart[p]/(x^(j*FracPart[p])*(a + b*x^(n - j))^FracPart[p])   Int[ 
x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] &&  !Integ 
erQ[p] && NeQ[n, j] && PosQ[n - j]
 

rule 1927

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

rule 1930

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol 
] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a*x^j + b*x^n)^(p + 1)/(b*(m + n*p 
+ 1))), x] - Simp[a*c^(n - j)*((m + j*p - n + j + 1)/(b*(m + n*p + 1)))   I 
nt[(c*x)^(m - (n - j))*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, 
x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && Gt 
Q[m + j*p - n + j + 1, 0] && NeQ[m + n*p + 1, 0]
 

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.00

method	result	size
risch	\(\frac {2 \left (7 b^{2} x^{4}+13 a b \,x^{2}+4 a^{2}\right ) x \left (b \,x^{2}+a \right )}{77 b \sqrt {x \left (b \,x^{2}+a \right )}}-\frac {4 a^{3} \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{77 b^{2} \sqrt {b \,x^{3}+a x}}\)	\(158\)
default	\(\frac {2 b \,x^{4} \sqrt {b \,x^{3}+a x}}{11}+\frac {26 a \,x^{2} \sqrt {b \,x^{3}+a x}}{77}+\frac {8 a^{2} \sqrt {b \,x^{3}+a x}}{77 b}-\frac {4 a^{3} \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{77 b^{2} \sqrt {b \,x^{3}+a x}}\)	\(166\)
elliptic	\(\frac {2 b \,x^{4} \sqrt {b \,x^{3}+a x}}{11}+\frac {26 a \,x^{2} \sqrt {b \,x^{3}+a x}}{77}+\frac {8 a^{2} \sqrt {b \,x^{3}+a x}}{77 b}-\frac {4 a^{3} \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{77 b^{2} \sqrt {b \,x^{3}+a x}}\)	\(166\)

Input:

int((b*x^3+a*x)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

2/77*(7*b^2*x^4+13*a*b*x^2+4*a^2)/b*x*(b*x^2+a)/(x*(b*x^2+a))^(1/2)-4/77*a 
^3/b^2*(-a*b)^(1/2)*((x+(-a*b)^(1/2)/b)/(-a*b)^(1/2)*b)^(1/2)*(-2*(x-(-a*b 
)^(1/2)/b)/(-a*b)^(1/2)*b)^(1/2)*(-1/(-a*b)^(1/2)*b*x)^(1/2)/(b*x^3+a*x)^( 
1/2)*EllipticF(((x+(-a*b)^(1/2)/b)/(-a*b)^(1/2)*b)^(1/2),1/2*2^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.38 \[ \int \left (a x+b x^3\right )^{3/2} \, dx=-\frac {2 \, {\left (4 \, a^{3} \sqrt {b} {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right ) - {\left (7 \, b^{3} x^{4} + 13 \, a b^{2} x^{2} + 4 \, a^{2} b\right )} \sqrt {b x^{3} + a x}\right )}}{77 \, b^{2}} \] Input:

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

Output:

-2/77*(4*a^3*sqrt(b)*weierstrassPInverse(-4*a/b, 0, x) - (7*b^3*x^4 + 13*a 
*b^2*x^2 + 4*a^2*b)*sqrt(b*x^3 + a*x))/b^2
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 8.74 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.25 \[ \int \left (a x+b x^3\right )^{3/2} \, dx=\frac {2\,x\,{\left (b\,x^3+a\,x\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},\frac {5}{4};\ \frac {9}{4};\ -\frac {b\,x^2}{a}\right )}{5\,{\left (\frac {b\,x^2}{a}+1\right )}^{3/2}} \] Input:

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

Output:

(2*x*(a*x + b*x^3)^(3/2)*hypergeom([-3/2, 5/4], 9/4, -(b*x^2)/a))/(5*((b*x 
^2)/a + 1)^(3/2))
 

Reduce [F]

\[ \int \left (a x+b x^3\right )^{3/2} \, dx=\frac {\frac {8 \sqrt {x}\, \sqrt {b \,x^{2}+a}\, a^{2}}{77}+\frac {26 \sqrt {x}\, \sqrt {b \,x^{2}+a}\, a b \,x^{2}}{77}+\frac {2 \sqrt {x}\, \sqrt {b \,x^{2}+a}\, b^{2} x^{4}}{11}-\frac {4 \left (\int \frac {\sqrt {x}\, \sqrt {b \,x^{2}+a}}{b \,x^{3}+a x}d x \right ) a^{3}}{77}}{b} \] Input:

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

Output:

(2*(4*sqrt(x)*sqrt(a + b*x**2)*a**2 + 13*sqrt(x)*sqrt(a + b*x**2)*a*b*x**2 
 + 7*sqrt(x)*sqrt(a + b*x**2)*b**2*x**4 - 2*int((sqrt(x)*sqrt(a + b*x**2)) 
/(a*x + b*x**3),x)*a**3))/(77*b)