\(\int \frac {(A+B x^2) \sqrt {b x^2+c x^4}}{x^{3/2}} \, dx\) [228]

Optimal result

Integrand size = 28, antiderivative size = 165 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^{3/2}} \, dx=-\frac {2 (b B-7 A c) \sqrt {b x^2+c x^4}}{21 c \sqrt {x}}+\frac {2 B \left (b x^2+c x^4\right )^{3/2}}{7 c x^{5/2}}-\frac {2 b^{3/4} (b B-7 A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{21 c^{5/4} \sqrt {b x^2+c x^4}} \] Output:

-2/21*(-7*A*c+B*b)*(c*x^4+b*x^2)^(1/2)/c/x^(1/2)+2/7*B*(c*x^4+b*x^2)^(3/2) 
/c/x^(5/2)-2/21*b^(3/4)*(-7*A*c+B*b)*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 = 94, normalized size of antiderivative = 0.57 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^{3/2}} \, dx=\frac {2 \sqrt {x^2 \left (b+c x^2\right )} \left (B \left (b+c x^2\right ) \sqrt {1+\frac {c x^2}{b}}+(-b B+7 A c) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {5}{4},-\frac {c x^2}{b}\right )\right )}{7 c \sqrt {x} \sqrt {1+\frac {c x^2}{b}}} \] Input:

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

Output:

(2*Sqrt[x^2*(b + c*x^2)]*(B*(b + c*x^2)*Sqrt[1 + (c*x^2)/b] + (-(b*B) + 7* 
A*c)*Hypergeometric2F1[-1/2, 1/4, 5/4, -((c*x^2)/b)]))/(7*c*Sqrt[x]*Sqrt[1 
 + (c*x^2)/b])
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {1945, 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.

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

\(\Big \downarrow \) 1945

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

\(\Big \downarrow \) 1426

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

\(\Big \downarrow \) 1431

\(\displaystyle \frac {2 B \left (b x^2+c x^4\right )^{3/2}}{7 c x^{5/2}}-\frac {(b B-7 A c) \left (\frac {2 b x \sqrt {b+c x^2} \int \frac {1}{\sqrt {x} \sqrt {c x^2+b}}dx}{3 \sqrt {b x^2+c x^4}}+\frac {2 \sqrt {b x^2+c x^4}}{3 \sqrt {x}}\right )}{7 c}\)

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 761

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

Input:

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

Output:

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

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 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 1945

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

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.17

method	result	size
risch	\(\frac {2 \left (3 B c \,x^{2}+7 A c +2 B b \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{21 c \sqrt {x}}+\frac {2 b \left (7 A c -B b \right ) \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {x \left (c \,x^{2}+b \right )}}{21 c^{2} \sqrt {c \,x^{3}+b x}\, x^{\frac {3}{2}} \left (c \,x^{2}+b \right )}\)	\(193\)
default	\(\frac {2 \sqrt {c \,x^{4}+b \,x^{2}}\, \left (7 A \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b c -B \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b^{2}+3 B \,c^{3} x^{5}+7 A \,c^{3} x^{3}+5 x^{3} B b \,c^{2}+7 A b \,c^{2} x +2 B \,b^{2} c x \right )}{21 x^{\frac {3}{2}} \left (c \,x^{2}+b \right ) c^{2}}\)	\(257\)

Input:

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

Output:

2/21*(3*B*c*x^2+7*A*c+2*B*b)/c/x^(1/2)*(x^2*(c*x^2+b))^(1/2)+2/21*b*(7*A*c 
-B*b)/c^2*(-b*c)^(1/2)*((x+1/c*(-b*c)^(1/2))*c/(-b*c)^(1/2))^(1/2)*(-2*(x- 
1/c*(-b*c)^(1/2))*c/(-b*c)^(1/2))^(1/2)*(-c/(-b*c)^(1/2)*x)^(1/2)/(c*x^3+b 
*x)^(1/2)*EllipticF(((x+1/c*(-b*c)^(1/2))*c/(-b*c)^(1/2))^(1/2),1/2*2^(1/2 
))*(x^2*(c*x^2+b))^(1/2)/x^(3/2)/(c*x^2+b)*(x*(c*x^2+b))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.45 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^{3/2}} \, dx=-\frac {2 \, {\left (2 \, {\left (B b^{2} - 7 \, A b c\right )} \sqrt {c} x {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right ) - {\left (3 \, B c^{2} x^{2} + 2 \, B b c + 7 \, A c^{2}\right )} \sqrt {c x^{4} + b x^{2}} \sqrt {x}\right )}}{21 \, c^{2} x} \] Input:

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

Output:

-2/21*(2*(B*b^2 - 7*A*b*c)*sqrt(c)*x*weierstrassPInverse(-4*b/c, 0, x) - ( 
3*B*c^2*x^2 + 2*B*b*c + 7*A*c^2)*sqrt(c*x^4 + b*x^2)*sqrt(x))/(c^2*x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^{3/2}} \, dx=\frac {\frac {2 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, a c}{3}+\frac {4 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, b^{2}}{21}+\frac {2 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, b c \,x^{2}}{7}+\frac {2 \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c \,x^{3}+b x}d x \right ) a b c}{3}-\frac {2 \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c \,x^{3}+b x}d x \right ) b^{3}}{21}}{c} \] Input:

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

Output:

(2*(7*sqrt(x)*sqrt(b + c*x**2)*a*c + 2*sqrt(x)*sqrt(b + c*x**2)*b**2 + 3*s 
qrt(x)*sqrt(b + c*x**2)*b*c*x**2 + 7*int((sqrt(x)*sqrt(b + c*x**2))/(b*x + 
 c*x**3),x)*a*b*c - int((sqrt(x)*sqrt(b + c*x**2))/(b*x + c*x**3),x)*b**3) 
)/(21*c)