3.4.66 \(\int \frac {(b x^2+c x^4)^{3/2}}{\sqrt {x}} \, dx\) [366]

3.4.66.1 Optimal result

Integrand size = 21, antiderivative size = 320 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{\sqrt {x}} \, dx=-\frac {8 b^3 x^{3/2} \left (b+c x^2\right )}{65 c^{3/2} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}+\frac {8 b^2 \sqrt {x} \sqrt {b x^2+c x^4}}{195 c}+\frac {4}{39} b x^{5/2} \sqrt {b x^2+c x^4}+\frac {2}{13} \sqrt {x} \left (b x^2+c x^4\right )^{3/2}+\frac {8 b^{13/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{65 c^{7/4} \sqrt {b x^2+c x^4}}-\frac {4 b^{13/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{65 c^{7/4} \sqrt {b x^2+c x^4}} \]

output

2/13*(c*x^4+b*x^2)^(3/2)*x^(1/2)-8/65*b^3*x^(3/2)*(c*x^2+b)/c^(3/2)/(b^(1/ 
2)+x*c^(1/2))/(c*x^4+b*x^2)^(1/2)+4/39*b*x^(5/2)*(c*x^4+b*x^2)^(1/2)+8/195 
*b^2*x^(1/2)*(c*x^4+b*x^2)^(1/2)/c+8/65*b^(13/4)*x*(cos(2*arctan(c^(1/4)*x 
^(1/2)/b^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x^(1/2)/b^(1/4)))*EllipticE 
(sin(2*arctan(c^(1/4)*x^(1/2)/b^(1/4))),1/2*2^(1/2))*(b^(1/2)+x*c^(1/2))*( 
(c*x^2+b)/(b^(1/2)+x*c^(1/2))^2)^(1/2)/c^(7/4)/(c*x^4+b*x^2)^(1/2)-4/65*b^ 
(13/4)*x*(cos(2*arctan(c^(1/4)*x^(1/2)/b^(1/4)))^2)^(1/2)/cos(2*arctan(c^( 
1/4)*x^(1/2)/b^(1/4)))*EllipticF(sin(2*arctan(c^(1/4)*x^(1/2)/b^(1/4))),1/ 
2*2^(1/2))*(b^(1/2)+x*c^(1/2))*((c*x^2+b)/(b^(1/2)+x*c^(1/2))^2)^(1/2)/c^( 
7/4)/(c*x^4+b*x^2)^(1/2)
 

3.4.66.2 Mathematica [C] (verified)

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

Time = 10.03 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.28 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{\sqrt {x}} \, dx=\frac {2 \sqrt {x} \sqrt {x^2 \left (b+c x^2\right )} \left (\left (b+c x^2\right )^2 \sqrt {1+\frac {c x^2}{b}}-b^2 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {3}{4},\frac {7}{4},-\frac {c x^2}{b}\right )\right )}{13 c \sqrt {1+\frac {c x^2}{b}}} \]

input

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

output

(2*Sqrt[x]*Sqrt[x^2*(b + c*x^2)]*((b + c*x^2)^2*Sqrt[1 + (c*x^2)/b] - b^2* 
Hypergeometric2F1[-3/2, 3/4, 7/4, -((c*x^2)/b)]))/(13*c*Sqrt[1 + (c*x^2)/b 
])
 

3.4.66.3 Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 338, normalized size of antiderivative = 1.06, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {1426, 1426, 1429, 1431, 266, 834, 27, 761, 1510}

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.

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

\(\Big \downarrow \) 1426

\(\displaystyle \frac {6}{13} b \int x^{3/2} \sqrt {c x^4+b x^2}dx+\frac {2}{13} \sqrt {x} \left (b x^2+c x^4\right )^{3/2}\)

\(\Big \downarrow \) 1426

\(\displaystyle \frac {6}{13} b \left (\frac {2}{9} b \int \frac {x^{7/2}}{\sqrt {c x^4+b x^2}}dx+\frac {2}{9} x^{5/2} \sqrt {b x^2+c x^4}\right )+\frac {2}{13} \sqrt {x} \left (b x^2+c x^4\right )^{3/2}\)

\(\Big \downarrow \) 1429

\(\displaystyle \frac {6}{13} b \left (\frac {2}{9} b \left (\frac {2 \sqrt {x} \sqrt {b x^2+c x^4}}{5 c}-\frac {3 b \int \frac {x^{3/2}}{\sqrt {c x^4+b x^2}}dx}{5 c}\right )+\frac {2}{9} x^{5/2} \sqrt {b x^2+c x^4}\right )+\frac {2}{13} \sqrt {x} \left (b x^2+c x^4\right )^{3/2}\)

\(\Big \downarrow \) 1431

\(\displaystyle \frac {6}{13} b \left (\frac {2}{9} b \left (\frac {2 \sqrt {x} \sqrt {b x^2+c x^4}}{5 c}-\frac {3 b x \sqrt {b+c x^2} \int \frac {\sqrt {x}}{\sqrt {c x^2+b}}dx}{5 c \sqrt {b x^2+c x^4}}\right )+\frac {2}{9} x^{5/2} \sqrt {b x^2+c x^4}\right )+\frac {2}{13} \sqrt {x} \left (b x^2+c x^4\right )^{3/2}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {6}{13} b \left (\frac {2}{9} b \left (\frac {2 \sqrt {x} \sqrt {b x^2+c x^4}}{5 c}-\frac {6 b x \sqrt {b+c x^2} \int \frac {x}{\sqrt {c x^2+b}}d\sqrt {x}}{5 c \sqrt {b x^2+c x^4}}\right )+\frac {2}{9} x^{5/2} \sqrt {b x^2+c x^4}\right )+\frac {2}{13} \sqrt {x} \left (b x^2+c x^4\right )^{3/2}\)

\(\Big \downarrow \) 834

\(\displaystyle \frac {6}{13} b \left (\frac {2}{9} b \left (\frac {2 \sqrt {x} \sqrt {b x^2+c x^4}}{5 c}-\frac {6 b x \sqrt {b+c x^2} \left (\frac {\sqrt {b} \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}-\frac {\sqrt {b} \int \frac {\sqrt {b}-\sqrt {c} x}{\sqrt {b} \sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}\right )}{5 c \sqrt {b x^2+c x^4}}\right )+\frac {2}{9} x^{5/2} \sqrt {b x^2+c x^4}\right )+\frac {2}{13} \sqrt {x} \left (b x^2+c x^4\right )^{3/2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {6}{13} b \left (\frac {2}{9} b \left (\frac {2 \sqrt {x} \sqrt {b x^2+c x^4}}{5 c}-\frac {6 b x \sqrt {b+c x^2} \left (\frac {\sqrt {b} \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}\right )}{5 c \sqrt {b x^2+c x^4}}\right )+\frac {2}{9} x^{5/2} \sqrt {b x^2+c x^4}\right )+\frac {2}{13} \sqrt {x} \left (b x^2+c x^4\right )^{3/2}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {6}{13} b \left (\frac {2}{9} b \left (\frac {2 \sqrt {x} \sqrt {b x^2+c x^4}}{5 c}-\frac {6 b x \sqrt {b+c x^2} \left (\frac {\sqrt [4]{b} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{2 c^{3/4} \sqrt {b+c x^2}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}\right )}{5 c \sqrt {b x^2+c x^4}}\right )+\frac {2}{9} x^{5/2} \sqrt {b x^2+c x^4}\right )+\frac {2}{13} \sqrt {x} \left (b x^2+c x^4\right )^{3/2}\)

\(\Big \downarrow \) 1510

\(\displaystyle \frac {6}{13} b \left (\frac {2}{9} b \left (\frac {2 \sqrt {x} \sqrt {b x^2+c x^4}}{5 c}-\frac {6 b x \sqrt {b+c x^2} \left (\frac {\sqrt [4]{b} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{2 c^{3/4} \sqrt {b+c x^2}}-\frac {\frac {\sqrt [4]{b} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{\sqrt [4]{c} \sqrt {b+c x^2}}-\frac {\sqrt {x} \sqrt {b+c x^2}}{\sqrt {b}+\sqrt {c} x}}{\sqrt {c}}\right )}{5 c \sqrt {b x^2+c x^4}}\right )+\frac {2}{9} x^{5/2} \sqrt {b x^2+c x^4}\right )+\frac {2}{13} \sqrt {x} \left (b x^2+c x^4\right )^{3/2}\)

input

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

output

(2*Sqrt[x]*(b*x^2 + c*x^4)^(3/2))/13 + (6*b*((2*x^(5/2)*Sqrt[b*x^2 + c*x^4 
])/9 + (2*b*((2*Sqrt[x]*Sqrt[b*x^2 + c*x^4])/(5*c) - (6*b*x*Sqrt[b + c*x^2 
]*(-((-((Sqrt[x]*Sqrt[b + c*x^2])/(Sqrt[b] + Sqrt[c]*x)) + (b^(1/4)*(Sqrt[ 
b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticE[2*ArcT 
an[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(c^(1/4)*Sqrt[b + c*x^2]))/Sqrt[c]) + 
 (b^(1/4)*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]* 
EllipticF[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(2*c^(3/4)*Sqrt[b + c 
*x^2])))/(5*c*Sqrt[b*x^2 + c*x^4])))/9))/13
 

3.4.66.3.1 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 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 834

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

rule 1426

Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp 
[(d*x)^(m + 1)*((b*x^2 + c*x^4)^p/(d*(m + 4*p + 1))), x] + Simp[2*b*(p/(d^2 
*(m + 4*p + 1)))   Int[(d*x)^(m + 2)*(b*x^2 + c*x^4)^(p - 1), x], x] /; Fre 
eQ[{b, c, d, m, p}, x] &&  !IntegerQ[p] && GtQ[p, 0] && NeQ[m + 4*p + 1, 0]
 

rule 1429

Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp 
[d^3*(d*x)^(m - 3)*((b*x^2 + c*x^4)^(p + 1)/(c*(m + 4*p + 1))), x] - Simp[b 
*d^2*((m + 2*p - 1)/(c*(m + 4*p + 1)))   Int[(d*x)^(m - 2)*(b*x^2 + c*x^4)^ 
p, x], x] /; FreeQ[{b, c, d, m, p}, x] &&  !IntegerQ[p] && GtQ[m + 2*p - 1, 
 0] && NeQ[m + 4*p + 1, 0]
 

rule 1431

Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp 
[(b*x^2 + c*x^4)^p/((d*x)^(2*p)*(b + c*x^2)^p)   Int[(d*x)^(m + 2*p)*(b + c 
*x^2)^p, x], x] /; FreeQ[{b, c, d, m, p}, x] &&  !IntegerQ[p]
 

rule 1510

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = 
 Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* 
(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E 
llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e 
}, x] && PosQ[c/a]
 

3.4.66.4 Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 237, normalized size of antiderivative = 0.74

method	result	size
default	\(-\frac {2 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (-15 c^{4} x^{8}-40 c^{3} x^{6} b +12 b^{4} \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, E\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-6 b^{4} \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, F\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-29 b^{2} c^{2} x^{4}-4 x^{2} c \,b^{3}\right )}{195 x^{\frac {7}{2}} \left (c \,x^{2}+b \right )^{2} c^{2}}\)	\(237\)
risch	\(\frac {2 \sqrt {x}\, \left (15 c^{2} x^{4}+25 b c \,x^{2}+4 b^{2}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{195 c}-\frac {4 b^{3} \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \left (-\frac {2 \sqrt {-b c}\, E\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c}+\frac {\sqrt {-b c}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {x \left (c \,x^{2}+b \right )}}{65 c^{2} \sqrt {c \,x^{3}+b x}\, x^{\frac {3}{2}} \left (c \,x^{2}+b \right )}\)	\(241\)

input

int((c*x^4+b*x^2)^(3/2)/x^(1/2),x,method=_RETURNVERBOSE)
 

output

-2/195*(c*x^4+b*x^2)^(3/2)/x^(7/2)/(c*x^2+b)^2/c^2*(-15*c^4*x^8-40*c^3*x^6 
*b+12*b^4*((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-b*c)^(1 
/2))/(-b*c)^(1/2))^(1/2)*(-x*c/(-b*c)^(1/2))^(1/2)*EllipticE(((c*x+(-b*c)^ 
(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))-6*b^4*((c*x+(-b*c)^(1/2))/(-b*c)^( 
1/2))^(1/2)*2^(1/2)*((-c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*(-x*c/(-b*c)^ 
(1/2))^(1/2)*EllipticF(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2) 
)-29*b^2*c^2*x^4-4*x^2*c*b^3)
 

3.4.66.5 Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.09 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.22 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{\sqrt {x}} \, dx=\frac {2 \, {\left (12 \, b^{3} \sqrt {c} {\rm weierstrassZeta}\left (-\frac {4 \, b}{c}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right )\right ) + {\left (15 \, c^{3} x^{4} + 25 \, b c^{2} x^{2} + 4 \, b^{2} c\right )} \sqrt {c x^{4} + b x^{2}} \sqrt {x}\right )}}{195 \, c^{2}} \]

input

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

output

2/195*(12*b^3*sqrt(c)*weierstrassZeta(-4*b/c, 0, weierstrassPInverse(-4*b/ 
c, 0, x)) + (15*c^3*x^4 + 25*b*c^2*x^2 + 4*b^2*c)*sqrt(c*x^4 + b*x^2)*sqrt 
(x))/c^2
 

3.4.66.6 Sympy [F]

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

input

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

output

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

3.4.66.7 Maxima [F]

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

input

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

output

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

3.4.66.8 Giac [F]

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

input

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

output

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

3.4.66.9 Mupad [F(-1)]

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

input

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

output

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