\(\int \frac {(b x^2+c x^4)^{3/2}}{x^{3/2}} \, dx\) [243]

Optimal result

Integrand size = 21, antiderivative size = 173 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{3/2}} \, dx=\frac {8 b^2 \sqrt {b x^2+c x^4}}{77 c \sqrt {x}}+\frac {12}{77} b x^{3/2} \sqrt {b x^2+c x^4}+\frac {2 \left (b x^2+c x^4\right )^{3/2}}{11 \sqrt {x}}-\frac {4 b^{11/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{77 c^{5/4} \sqrt {b x^2+c x^4}} \] Output:

8/77*b^2*(c*x^4+b*x^2)^(1/2)/c/x^(1/2)+12/77*b*x^(3/2)*(c*x^4+b*x^2)^(1/2) 
+2/11*(c*x^4+b*x^2)^(3/2)/x^(1/2)-4/77*b^(11/4)*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^(5/4)/(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 = 10.04 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.52 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{3/2}} \, dx=\frac {2 \sqrt {x^2 \left (b+c x^2\right )} \left (\left (b+c x^2\right )^2 \sqrt {1+\frac {c x^2}{b}}-b^2 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{4},\frac {5}{4},-\frac {c x^2}{b}\right )\right )}{11 c \sqrt {x} \sqrt {1+\frac {c x^2}{b}}} \] Input:

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

Output:

(2*Sqrt[x^2*(b + c*x^2)]*((b + c*x^2)^2*Sqrt[1 + (c*x^2)/b] - b^2*Hypergeo 
metric2F1[-3/2, 1/4, 5/4, -((c*x^2)/b)]))/(11*c*Sqrt[x]*Sqrt[1 + (c*x^2)/b 
])
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {1426, 1426, 1429, 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.

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

\(\Big \downarrow \) 1426

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

\(\Big \downarrow \) 1426

\(\displaystyle \frac {6}{11} b \left (\frac {2}{7} b \int \frac {x^{5/2}}{\sqrt {c x^4+b x^2}}dx+\frac {2}{7} x^{3/2} \sqrt {b x^2+c x^4}\right )+\frac {2 \left (b x^2+c x^4\right )^{3/2}}{11 \sqrt {x}}\)

\(\Big \downarrow \) 1429

\(\displaystyle \frac {6}{11} b \left (\frac {2}{7} b \left (\frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {b \int \frac {\sqrt {x}}{\sqrt {c x^4+b x^2}}dx}{3 c}\right )+\frac {2}{7} x^{3/2} \sqrt {b x^2+c x^4}\right )+\frac {2 \left (b x^2+c x^4\right )^{3/2}}{11 \sqrt {x}}\)

\(\Big \downarrow \) 1431

\(\displaystyle \frac {6}{11} b \left (\frac {2}{7} b \left (\frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {b x \sqrt {b+c x^2} \int \frac {1}{\sqrt {x} \sqrt {c x^2+b}}dx}{3 c \sqrt {b x^2+c x^4}}\right )+\frac {2}{7} x^{3/2} \sqrt {b x^2+c x^4}\right )+\frac {2 \left (b x^2+c x^4\right )^{3/2}}{11 \sqrt {x}}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {6}{11} b \left (\frac {2}{7} b \left (\frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {2 b x \sqrt {b+c x^2} \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{3 c \sqrt {b x^2+c x^4}}\right )+\frac {2}{7} x^{3/2} \sqrt {b x^2+c x^4}\right )+\frac {2 \left (b x^2+c x^4\right )^{3/2}}{11 \sqrt {x}}\)

\(\Big \downarrow \) 761

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

Input:

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

Output:

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

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 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]
 

Maple [A] (verified)

Time = 0.74 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.91

method	result	size
default	\(-\frac {2 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (-7 c^{4} x^{7}+2 b^{3} \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-20 b \,c^{3} x^{5}-17 b^{2} c^{2} x^{3}-4 b^{3} c x \right )}{77 x^{\frac {7}{2}} \left (c \,x^{2}+b \right )^{2} c^{2}}\)	\(157\)
risch	\(\frac {2 \left (7 c^{2} x^{4}+13 b c \,x^{2}+4 b^{2}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{77 c \sqrt {x}}-\frac {4 b^{3} \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {x \left (c \,x^{2}+b \right )}}{77 c^{2} \sqrt {c \,x^{3}+b x}\, x^{\frac {3}{2}} \left (c \,x^{2}+b \right )}\)	\(191\)

Input:

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

Output:

-2/77*(c*x^4+b*x^2)^(3/2)/x^(7/2)/(c*x^2+b)^2*(-7*c^4*x^7+2*b^3*(-b*c)^(1/ 
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))-20*b*c^3*x^5-17*b^2*c^2*x^3-4*b^3*c*x)/c^ 
2
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.40 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{3/2}} \, dx=-\frac {2 \, {\left (4 \, b^{3} \sqrt {c} x {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right ) - {\left (7 \, c^{3} x^{4} + 13 \, b c^{2} x^{2} + 4 \, b^{2} c\right )} \sqrt {c x^{4} + b x^{2}} \sqrt {x}\right )}}{77 \, c^{2} x} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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