\(\int \frac {(b x^2+c x^4)^{3/2}}{x^{11/2}} \, dx\) [247]

Optimal result

Integrand size = 21, antiderivative size = 143 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{11/2}} \, dx=\frac {4 c \sqrt {b x^2+c x^4}}{3 \sqrt {x}}-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{3 x^{9/2}}+\frac {4 b^{3/4} c^{3/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{3 \sqrt {b x^2+c x^4}} \] Output:

4/3*c*(c*x^4+b*x^2)^(1/2)/x^(1/2)-2/3*(c*x^4+b*x^2)^(3/2)/x^(9/2)+4/3*b^(3 
/4)*c^(3/4)*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*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 = 10.01 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.41 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{11/2}} \, dx=-\frac {2 b \sqrt {x^2 \left (b+c x^2\right )} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {3}{4},\frac {1}{4},-\frac {c x^2}{b}\right )}{3 x^{5/2} \sqrt {1+\frac {c x^2}{b}}} \] Input:

Integrate[(b*x^2 + c*x^4)^(3/2)/x^(11/2),x]
 

Output:

(-2*b*Sqrt[x^2*(b + c*x^2)]*Hypergeometric2F1[-3/2, -3/4, 1/4, -((c*x^2)/b 
)])/(3*x^(5/2)*Sqrt[1 + (c*x^2)/b])
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {1425, 1426, 1431, 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.

\(\displaystyle \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{11/2}} \, dx\)

\(\Big \downarrow \) 1425

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

\(\Big \downarrow \) 1426

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

\(\Big \downarrow \) 1431

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 761

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

Input:

Int[(b*x^2 + c*x^4)^(3/2)/x^(11/2),x]
 

Output:

(-2*(b*x^2 + c*x^4)^(3/2))/(3*x^(9/2)) + 2*c*((2*Sqrt[b*x^2 + c*x^4])/(3*S 
qrt[x]) + (2*b^(3/4)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + S 
qrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(3*c^(1/ 
4)*Sqrt[b*x^2 + c*x^4]))
 

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 1425

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 + 2*p + 1))), x] - Simp[2*c*(p/(d^4 
*(m + 2*p + 1)))   Int[(d*x)^(m + 4)*(b*x^2 + c*x^4)^(p - 1), x], x] /; Fre 
eQ[{b, c, d, m, p}, x] &&  !IntegerQ[p] && GtQ[p, 0] && LtQ[m + 2*p + 1, 0]
 

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 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]
 

Maple [A] (verified)

Time = 0.80 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.91

method	result	size
default	\(\frac {2 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (2 \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-b c}\, b x +c^{2} x^{4}-b^{2}\right )}{3 x^{\frac {9}{2}} \left (c \,x^{2}+b \right )^{2}}\)	\(130\)
risch	\(-\frac {2 \left (-c \,x^{2}+b \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{3 x^{\frac {5}{2}}}+\frac {4 b \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {x \left (c \,x^{2}+b \right )}}{3 \sqrt {c \,x^{3}+b x}\, x^{\frac {3}{2}} \left (c \,x^{2}+b \right )}\)	\(170\)

Input:

int((c*x^4+b*x^2)^(3/2)/x^(11/2),x,method=_RETURNVERBOSE)
 

Output:

2/3*(c*x^4+b*x^2)^(3/2)/x^(9/2)/(c*x^2+b)^2*(2*((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 
))*(-b*c)^(1/2)*b*x+c^2*x^4-b^2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.35 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{11/2}} \, dx=\frac {2 \, {\left (4 \, b \sqrt {c} x^{3} {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right ) + \sqrt {c x^{4} + b x^{2}} {\left (c x^{2} - b\right )} \sqrt {x}\right )}}{3 \, x^{3}} \] Input:

integrate((c*x^4+b*x^2)^(3/2)/x^(11/2),x, algorithm="fricas")
 

Output:

2/3*(4*b*sqrt(c)*x^3*weierstrassPInverse(-4*b/c, 0, x) + sqrt(c*x^4 + b*x^ 
2)*(c*x^2 - b)*sqrt(x))/x^3
 

Sympy [F]

\[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{11/2}} \, dx=\int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}{x^{\frac {11}{2}}}\, dx \] Input:

integrate((c*x**4+b*x**2)**(3/2)/x**(11/2),x)
 

Output:

Integral((x**2*(b + c*x**2))**(3/2)/x**(11/2), x)
 

Maxima [F]

\[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{11/2}} \, dx=\int { \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{\frac {11}{2}}} \,d x } \] Input:

integrate((c*x^4+b*x^2)^(3/2)/x^(11/2),x, algorithm="maxima")
 

Output:

integrate((c*x^4 + b*x^2)^(3/2)/x^(11/2), x)
 

Giac [F]

\[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{11/2}} \, dx=\int { \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{\frac {11}{2}}} \,d x } \] Input:

integrate((c*x^4+b*x^2)^(3/2)/x^(11/2),x, algorithm="giac")
 

Output:

integrate((c*x^4 + b*x^2)^(3/2)/x^(11/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{11/2}} \, dx=\int \frac {{\left (c\,x^4+b\,x^2\right )}^{3/2}}{x^{11/2}} \,d x \] Input:

int((b*x^2 + c*x^4)^(3/2)/x^(11/2),x)
 

Output:

int((b*x^2 + c*x^4)^(3/2)/x^(11/2), x)
 

Reduce [F]

\[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{11/2}} \, dx=\frac {-\frac {10 \sqrt {c \,x^{2}+b}\, b}{3}+\frac {2 \sqrt {c \,x^{2}+b}\, c \,x^{2}}{3}-4 \sqrt {x}\, \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c \,x^{5}+b \,x^{3}}d x \right ) b^{2} x}{\sqrt {x}\, x} \] Input:

int((c*x^4+b*x^2)^(3/2)/x^(11/2),x)
 

Output:

(2*( - 5*sqrt(b + c*x**2)*b + sqrt(b + c*x**2)*c*x**2 - 6*sqrt(x)*int((sqr 
t(x)*sqrt(b + c*x**2))/(b*x**3 + c*x**5),x)*b**2*x))/(3*sqrt(x)*x)