\(\int \frac {(A+B x^2) \sqrt {b x^2+c x^4}}{x^{11/2}} \, dx\) [232]

Optimal result

Integrand size = 28, antiderivative size = 167 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^{11/2}} \, dx=-\frac {2 (7 b B-A c) \sqrt {b x^2+c x^4}}{21 b x^{5/2}}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{7 b x^{13/2}}+\frac {2 c^{3/4} (7 b B-A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{21 b^{5/4} \sqrt {b x^2+c x^4}} \] Output:

-2/21*(-A*c+7*B*b)*(c*x^4+b*x^2)^(1/2)/b/x^(5/2)-2/7*A*(c*x^4+b*x^2)^(3/2) 
/b/x^(13/2)+2/21*c^(3/4)*(-A*c+7*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))/b^(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 = 98, normalized size of antiderivative = 0.59 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^{11/2}} \, dx=-\frac {2 \sqrt {x^2 \left (b+c x^2\right )} \left (3 A \left (b+c x^2\right ) \sqrt {1+\frac {c x^2}{b}}+(7 b B-A c) x^2 \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},-\frac {1}{2},\frac {1}{4},-\frac {c x^2}{b}\right )\right )}{21 b x^{9/2} \sqrt {1+\frac {c x^2}{b}}} \] Input:

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

Output:

(-2*Sqrt[x^2*(b + c*x^2)]*(3*A*(b + c*x^2)*Sqrt[1 + (c*x^2)/b] + (7*b*B - 
A*c)*x^2*Hypergeometric2F1[-3/4, -1/2, 1/4, -((c*x^2)/b)]))/(21*b*x^(9/2)* 
Sqrt[1 + (c*x^2)/b])
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 163, 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 = {1944, 1425, 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.

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

\(\Big \downarrow \) 1944

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

\(\Big \downarrow \) 1425

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

\(\Big \downarrow \) 1431

\(\displaystyle \frac {(7 b B-A c) \left (\frac {2 c x \sqrt {b+c x^2} \int \frac {1}{\sqrt {x} \sqrt {c x^2+b}}dx}{3 \sqrt {b x^2+c x^4}}-\frac {2 \sqrt {b x^2+c x^4}}{3 x^{5/2}}\right )}{7 b}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{7 b x^{13/2}}\)

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 761

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

Input:

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

Output:

(-2*A*(b*x^2 + c*x^4)^(3/2))/(7*b*x^(13/2)) + ((7*b*B - A*c)*((-2*Sqrt[b*x 
^2 + c*x^4])/(3*x^(5/2)) + (2*c^(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*b^(1/4)*Sqrt[b*x^2 + c*x^4])))/(7*b)
 

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 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 1944

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + 
(d_.)*(x_)^(n_.)), x_Symbol] :> Simp[c*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j 
+ b*x^(j + n))^(p + 1)/(a*(m + j*p + 1))), x] + Simp[(a*d*(m + j*p + 1) - b 
*c*(m + n + p*(j + n) + 1))/(a*e^n*(m + j*p + 1))   Int[(e*x)^(m + n)*(a*x^ 
j + b*x^(j + n))^p, x], x] /; FreeQ[{a, b, c, d, e, j, p}, x] && EqQ[jn, j 
+ n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && (LtQ[m + j*p, -1 
] || (IntegersQ[m - 1/2, p - 1/2] && LtQ[p, 0] && LtQ[m, (-n)*p - 1])) && ( 
GtQ[e, 0] || IntegersQ[j, n]) && NeQ[m + j*p + 1, 0] && NeQ[m - n + j*p + 1 
, 0]
 

Maple [A] (verified)

Time = 0.81 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.16

method	result	size
risch	\(-\frac {2 \left (2 A c \,x^{2}+7 B b \,x^{2}+3 A b \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{21 x^{\frac {9}{2}} b}-\frac {2 \left (A c -7 B b \right ) \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {x \left (c \,x^{2}+b \right )}}{21 b \sqrt {c \,x^{3}+b x}\, x^{\frac {3}{2}} \left (c \,x^{2}+b \right )}\)	\(194\)
default	\(-\frac {2 \sqrt {c \,x^{4}+b \,x^{2}}\, \left (A \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) c \,x^{3}-7 B \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b \,x^{3}+2 x^{4} A \,c^{2}+7 x^{4} B b c +5 A b c \,x^{2}+7 x^{2} B \,b^{2}+3 b^{2} A \right )}{21 x^{\frac {9}{2}} \left (c \,x^{2}+b \right ) b}\)	\(255\)

Input:

int((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^(11/2),x,method=_RETURNVERBOSE)
 

Output:

-2/21*(2*A*c*x^2+7*B*b*x^2+3*A*b)/x^(9/2)/b*(x^2*(c*x^2+b))^(1/2)-2/21*(A* 
c-7*B*b)/b*(-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 = 71, normalized size of antiderivative = 0.43 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^{11/2}} \, dx=\frac {2 \, {\left (2 \, {\left (7 \, B b - A c\right )} \sqrt {c} x^{5} {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right ) - \sqrt {c x^{4} + b x^{2}} {\left ({\left (7 \, B b + 2 \, A c\right )} x^{2} + 3 \, A b\right )} \sqrt {x}\right )}}{21 \, b x^{5}} \] Input:

integrate((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^(11/2),x, algorithm="fricas")
 

Output:

2/21*(2*(7*B*b - A*c)*sqrt(c)*x^5*weierstrassPInverse(-4*b/c, 0, x) - sqrt 
(c*x^4 + b*x^2)*((7*B*b + 2*A*c)*x^2 + 3*A*b)*sqrt(x))/(b*x^5)
 

Sympy [F]

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

integrate((B*x**2+A)*(c*x**4+b*x**2)**(1/2)/x**(11/2),x)
 

Output:

Integral(sqrt(x**2*(b + c*x**2))*(A + B*x**2)/x**(11/2), x)
 

Maxima [F]

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

integrate((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^(11/2),x, algorithm="maxima")
 

Output:

integrate(sqrt(c*x^4 + b*x^2)*(B*x^2 + A)/x^(11/2), x)
 

Giac [F]

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

integrate((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^(11/2),x, algorithm="giac")
 

Output:

integrate(sqrt(c*x^4 + b*x^2)*(B*x^2 + A)/x^(11/2), x)
 

Mupad [F(-1)]

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

int(((A + B*x^2)*(b*x^2 + c*x^4)^(1/2))/x^(11/2),x)
 

Output:

int(((A + B*x^2)*(b*x^2 + c*x^4)^(1/2))/x^(11/2), x)
 

Reduce [F]

\[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^{11/2}} \, dx=\frac {-\frac {2 \sqrt {c \,x^{2}+b}\, a c}{5}+\frac {4 \sqrt {c \,x^{2}+b}\, b^{2}}{5}-2 \sqrt {c \,x^{2}+b}\, b c \,x^{2}-\frac {2 \sqrt {x}\, \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c \,x^{7}+b \,x^{5}}d x \right ) a b c \,x^{3}}{5}+\frac {14 \sqrt {x}\, \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c \,x^{7}+b \,x^{5}}d x \right ) b^{3} x^{3}}{5}}{\sqrt {x}\, c \,x^{3}} \] Input:

int((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^(11/2),x)
 

Output:

(2*( - sqrt(b + c*x**2)*a*c + 2*sqrt(b + c*x**2)*b**2 - 5*sqrt(b + c*x**2) 
*b*c*x**2 - sqrt(x)*int((sqrt(x)*sqrt(b + c*x**2))/(b*x**5 + c*x**7),x)*a* 
b*c*x**3 + 7*sqrt(x)*int((sqrt(x)*sqrt(b + c*x**2))/(b*x**5 + c*x**7),x)*b 
**3*x**3))/(5*sqrt(x)*c*x**3)