\(\int x^{3/2} (b x^2+c x^4)^{3/2} \, dx\) [240]

Optimal result

Integrand size = 21, antiderivative size = 350 \[ \int x^{3/2} \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {56 b^4 x^{3/2} \left (b+c x^2\right )}{1105 c^{5/2} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}-\frac {56 b^3 \sqrt {x} \sqrt {b x^2+c x^4}}{3315 c^2}+\frac {8 b^2 x^{5/2} \sqrt {b x^2+c x^4}}{663 c}+\frac {12}{221} b x^{9/2} \sqrt {b x^2+c x^4}+\frac {2}{17} x^{5/2} \left (b x^2+c x^4\right )^{3/2}-\frac {56 b^{17/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{1105 c^{11/4} \sqrt {b x^2+c x^4}}+\frac {28 b^{17/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{1105 c^{11/4} \sqrt {b x^2+c x^4}} \] Output:

56/1105*b^4*x^(3/2)*(c*x^2+b)/c^(5/2)/(b^(1/2)+c^(1/2)*x)/(c*x^4+b*x^2)^(1 
/2)-56/3315*b^3*x^(1/2)*(c*x^4+b*x^2)^(1/2)/c^2+8/663*b^2*x^(5/2)*(c*x^4+b 
*x^2)^(1/2)/c+12/221*b*x^(9/2)*(c*x^4+b*x^2)^(1/2)+2/17*x^(5/2)*(c*x^4+b*x 
^2)^(3/2)-56/1105*b^(17/4)*x*(b^(1/2)+c^(1/2)*x)*((c*x^2+b)/(b^(1/2)+c^(1/ 
2)*x)^2)^(1/2)*EllipticE(sin(2*arctan(c^(1/4)*x^(1/2)/b^(1/4))),1/2*2^(1/2 
))/c^(11/4)/(c*x^4+b*x^2)^(1/2)+28/1105*b^(17/4)*x*(b^(1/2)+c^(1/2)*x)*((c 
*x^2+b)/(b^(1/2)+c^(1/2)*x)^2)^(1/2)*InverseJacobiAM(2*arctan(c^(1/4)*x^(1 
/2)/b^(1/4)),1/2*2^(1/2))/c^(11/4)/(c*x^4+b*x^2)^(1/2)
 

Mathematica [C] (verified)

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

Time = 10.07 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.29 \[ \int x^{3/2} \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {2 \sqrt {x} \sqrt {x^2 \left (b+c x^2\right )} \left (-\left (\left (7 b-13 c x^2\right ) \left (b+c x^2\right )^2 \sqrt {1+\frac {c x^2}{b}}\right )+7 b^3 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {3}{4},\frac {7}{4},-\frac {c x^2}{b}\right )\right )}{221 c^2 \sqrt {1+\frac {c x^2}{b}}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.84 (sec) , antiderivative size = 374, normalized size of antiderivative = 1.07, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {1426, 1426, 1429, 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.

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

\(\Big \downarrow \) 1426

\(\displaystyle \frac {6}{17} b \int x^{7/2} \sqrt {c x^4+b x^2}dx+\frac {2}{17} x^{5/2} \left (b x^2+c x^4\right )^{3/2}\)

\(\Big \downarrow \) 1426

\(\displaystyle \frac {6}{17} b \left (\frac {2}{13} b \int \frac {x^{11/2}}{\sqrt {c x^4+b x^2}}dx+\frac {2}{13} x^{9/2} \sqrt {b x^2+c x^4}\right )+\frac {2}{17} x^{5/2} \left (b x^2+c x^4\right )^{3/2}\)

\(\Big \downarrow \) 1429

\(\displaystyle \frac {6}{17} b \left (\frac {2}{13} b \left (\frac {2 x^{5/2} \sqrt {b x^2+c x^4}}{9 c}-\frac {7 b \int \frac {x^{7/2}}{\sqrt {c x^4+b x^2}}dx}{9 c}\right )+\frac {2}{13} x^{9/2} \sqrt {b x^2+c x^4}\right )+\frac {2}{17} x^{5/2} \left (b x^2+c x^4\right )^{3/2}\)

\(\Big \downarrow \) 1429

\(\displaystyle \frac {6}{17} b \left (\frac {2}{13} b \left (\frac {2 x^{5/2} \sqrt {b x^2+c x^4}}{9 c}-\frac {7 b \left (\frac {2 \sqrt {x} \sqrt {b x^2+c x^4}}{5 c}-\frac {3 b \int \frac {x^{3/2}}{\sqrt {c x^4+b x^2}}dx}{5 c}\right )}{9 c}\right )+\frac {2}{13} x^{9/2} \sqrt {b x^2+c x^4}\right )+\frac {2}{17} x^{5/2} \left (b x^2+c x^4\right )^{3/2}\)

\(\Big \downarrow \) 1431

\(\displaystyle \frac {6}{17} b \left (\frac {2}{13} b \left (\frac {2 x^{5/2} \sqrt {b x^2+c x^4}}{9 c}-\frac {7 b \left (\frac {2 \sqrt {x} \sqrt {b x^2+c x^4}}{5 c}-\frac {3 b x \sqrt {b+c x^2} \int \frac {\sqrt {x}}{\sqrt {c x^2+b}}dx}{5 c \sqrt {b x^2+c x^4}}\right )}{9 c}\right )+\frac {2}{13} x^{9/2} \sqrt {b x^2+c x^4}\right )+\frac {2}{17} x^{5/2} \left (b x^2+c x^4\right )^{3/2}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {6}{17} b \left (\frac {2}{13} b \left (\frac {2 x^{5/2} \sqrt {b x^2+c x^4}}{9 c}-\frac {7 b \left (\frac {2 \sqrt {x} \sqrt {b x^2+c x^4}}{5 c}-\frac {6 b x \sqrt {b+c x^2} \int \frac {x}{\sqrt {c x^2+b}}d\sqrt {x}}{5 c \sqrt {b x^2+c x^4}}\right )}{9 c}\right )+\frac {2}{13} x^{9/2} \sqrt {b x^2+c x^4}\right )+\frac {2}{17} x^{5/2} \left (b x^2+c x^4\right )^{3/2}\)

\(\Big \downarrow \) 834

\(\displaystyle \frac {6}{17} b \left (\frac {2}{13} b \left (\frac {2 x^{5/2} \sqrt {b x^2+c x^4}}{9 c}-\frac {7 b \left (\frac {2 \sqrt {x} \sqrt {b x^2+c x^4}}{5 c}-\frac {6 b x \sqrt {b+c x^2} \left (\frac {\sqrt {b} \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}-\frac {\sqrt {b} \int \frac {\sqrt {b}-\sqrt {c} x}{\sqrt {b} \sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}\right )}{5 c \sqrt {b x^2+c x^4}}\right )}{9 c}\right )+\frac {2}{13} x^{9/2} \sqrt {b x^2+c x^4}\right )+\frac {2}{17} x^{5/2} \left (b x^2+c x^4\right )^{3/2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {6}{17} b \left (\frac {2}{13} b \left (\frac {2 x^{5/2} \sqrt {b x^2+c x^4}}{9 c}-\frac {7 b \left (\frac {2 \sqrt {x} \sqrt {b x^2+c x^4}}{5 c}-\frac {6 b x \sqrt {b+c x^2} \left (\frac {\sqrt {b} \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}\right )}{5 c \sqrt {b x^2+c x^4}}\right )}{9 c}\right )+\frac {2}{13} x^{9/2} \sqrt {b x^2+c x^4}\right )+\frac {2}{17} x^{5/2} \left (b x^2+c x^4\right )^{3/2}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {6}{17} b \left (\frac {2}{13} b \left (\frac {2 x^{5/2} \sqrt {b x^2+c x^4}}{9 c}-\frac {7 b \left (\frac {2 \sqrt {x} \sqrt {b x^2+c x^4}}{5 c}-\frac {6 b x \sqrt {b+c x^2} \left (\frac {\sqrt [4]{b} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{2 c^{3/4} \sqrt {b+c x^2}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}\right )}{5 c \sqrt {b x^2+c x^4}}\right )}{9 c}\right )+\frac {2}{13} x^{9/2} \sqrt {b x^2+c x^4}\right )+\frac {2}{17} x^{5/2} \left (b x^2+c x^4\right )^{3/2}\)

\(\Big \downarrow \) 1510

\(\displaystyle \frac {6}{17} b \left (\frac {2}{13} b \left (\frac {2 x^{5/2} \sqrt {b x^2+c x^4}}{9 c}-\frac {7 b \left (\frac {2 \sqrt {x} \sqrt {b x^2+c x^4}}{5 c}-\frac {6 b x \sqrt {b+c x^2} \left (\frac {\sqrt [4]{b} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{2 c^{3/4} \sqrt {b+c x^2}}-\frac {\frac {\sqrt [4]{b} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{\sqrt [4]{c} \sqrt {b+c x^2}}-\frac {\sqrt {x} \sqrt {b+c x^2}}{\sqrt {b}+\sqrt {c} x}}{\sqrt {c}}\right )}{5 c \sqrt {b x^2+c x^4}}\right )}{9 c}\right )+\frac {2}{13} x^{9/2} \sqrt {b x^2+c x^4}\right )+\frac {2}{17} x^{5/2} \left (b x^2+c x^4\right )^{3/2}\)

Input:

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

Output:

(2*x^(5/2)*(b*x^2 + c*x^4)^(3/2))/17 + (6*b*((2*x^(9/2)*Sqrt[b*x^2 + c*x^4 
])/13 + (2*b*((2*x^(5/2)*Sqrt[b*x^2 + c*x^4])/(9*c) - (7*b*((2*Sqrt[x]*Sqr 
t[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*ArcTan[(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*c)))/13))/17
 

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]
 

Maple [A] (verified)

Time = 1.46 (sec) , antiderivative size = 248, normalized size of antiderivative = 0.71

method	result	size
default	\(\frac {2 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (195 c^{5} x^{10}+480 c^{4} x^{8} b +305 b^{2} c^{3} x^{6}+84 b^{5} \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticE}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-42 b^{5} \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-8 c^{2} x^{4} b^{3}-28 x^{2} c \,b^{4}\right )}{3315 x^{\frac {7}{2}} \left (c \,x^{2}+b \right )^{2} c^{3}}\)	\(248\)
risch	\(-\frac {2 \sqrt {x}\, \left (-195 c^{3} x^{6}-285 b \,c^{2} x^{4}-20 b^{2} c \,x^{2}+28 b^{3}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{3315 c^{2}}+\frac {28 b^{4} \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \left (-\frac {2 \sqrt {-b c}\, \operatorname {EllipticE}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c}+\frac {\sqrt {-b c}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {x \left (c \,x^{2}+b \right )}}{1105 c^{3} \sqrt {c \,x^{3}+b x}\, x^{\frac {3}{2}} \left (c \,x^{2}+b \right )}\)	\(252\)

Input:

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

Output:

2/3315*(c*x^4+b*x^2)^(3/2)/x^(7/2)/(c*x^2+b)^2/c^3*(195*c^5*x^10+480*c^4*x 
^8*b+305*b^2*c^3*x^6+84*b^5*((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)*(-c/(-b*c)^(1/2)*x)^(1/2)*Ellip 
ticE(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))-42*b^5*((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)*(-c/(-b*c)^(1/2)*x)^(1/2)*EllipticF(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2) 
)^(1/2),1/2*2^(1/2))-8*c^2*x^4*b^3-28*x^2*c*b^4)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.24 \[ \int x^{3/2} \left (b x^2+c x^4\right )^{3/2} \, dx=-\frac {2 \, {\left (84 \, b^{4} \sqrt {c} {\rm weierstrassZeta}\left (-\frac {4 \, b}{c}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right )\right ) - {\left (195 \, c^{4} x^{6} + 285 \, b c^{3} x^{4} + 20 \, b^{2} c^{2} x^{2} - 28 \, b^{3} c\right )} \sqrt {c x^{4} + b x^{2}} \sqrt {x}\right )}}{3315 \, c^{3}} \] Input:

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

Output:

-2/3315*(84*b^4*sqrt(c)*weierstrassZeta(-4*b/c, 0, weierstrassPInverse(-4* 
b/c, 0, x)) - (195*c^4*x^6 + 285*b*c^3*x^4 + 20*b^2*c^2*x^2 - 28*b^3*c)*sq 
rt(c*x^4 + b*x^2)*sqrt(x))/c^3
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int x^{3/2} \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {-\frac {56 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, b^{3} x}{3315}+\frac {8 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, b^{2} c \,x^{3}}{663}+\frac {38 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, b \,c^{2} x^{5}}{221}+\frac {2 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, c^{3} x^{7}}{17}+\frac {28 \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c \,x^{2}+b}d x \right ) b^{4}}{1105}}{c^{2}} \] Input:

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

Output:

(2*( - 28*sqrt(x)*sqrt(b + c*x**2)*b**3*x + 20*sqrt(x)*sqrt(b + c*x**2)*b* 
*2*c*x**3 + 285*sqrt(x)*sqrt(b + c*x**2)*b*c**2*x**5 + 195*sqrt(x)*sqrt(b 
+ c*x**2)*c**3*x**7 + 42*int((sqrt(x)*sqrt(b + c*x**2))/(b + c*x**2),x)*b* 
*4))/(3315*c**2)