\(\int \frac {A+B x^2}{x^{3/2} (b x^2+c x^4)^{3/2}} \, dx\) [271]

Optimal result

Integrand size = 28, antiderivative size = 203 \[ \int \frac {A+B x^2}{x^{3/2} \left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {2 A}{7 b x^{5/2} \sqrt {b x^2+c x^4}}+\frac {7 b B-9 A c}{7 b^2 \sqrt {x} \sqrt {b x^2+c x^4}}-\frac {5 (7 b B-9 A c) \sqrt {b x^2+c x^4}}{21 b^3 x^{5/2}}-\frac {5 c^{3/4} (7 b B-9 A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{42 b^{13/4} \sqrt {b x^2+c x^4}} \] Output:

-2/7*A/b/x^(5/2)/(c*x^4+b*x^2)^(1/2)+1/7*(-9*A*c+7*B*b)/b^2/x^(1/2)/(c*x^4 
+b*x^2)^(1/2)-5/21*(-9*A*c+7*B*b)*(c*x^4+b*x^2)^(1/2)/b^3/x^(5/2)-5/42*c^( 
3/4)*(-9*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^( 
13/4)/(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 = 79, normalized size of antiderivative = 0.39 \[ \int \frac {A+B x^2}{x^{3/2} \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {-6 A b+2 (-7 b B+9 A c) x^2 \sqrt {1+\frac {c x^2}{b}} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {3}{2},\frac {1}{4},-\frac {c x^2}{b}\right )}{21 b^2 x^{5/2} \sqrt {x^2 \left (b+c x^2\right )}} \] Input:

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

Output:

(-6*A*b + 2*(-7*b*B + 9*A*c)*x^2*Sqrt[1 + (c*x^2)/b]*Hypergeometric2F1[-3/ 
4, 3/2, 1/4, -((c*x^2)/b)])/(21*b^2*x^(5/2)*Sqrt[x^2*(b + c*x^2)])
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.98, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {1944, 1428, 1430, 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.

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

\(\Big \downarrow \) 1944

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

\(\Big \downarrow \) 1428

\(\displaystyle \frac {(7 b B-9 A c) \left (\frac {5 \int \frac {1}{x^{3/2} \sqrt {c x^4+b x^2}}dx}{2 b}+\frac {1}{b \sqrt {x} \sqrt {b x^2+c x^4}}\right )}{7 b}-\frac {2 A}{7 b x^{5/2} \sqrt {b x^2+c x^4}}\)

\(\Big \downarrow \) 1430

\(\displaystyle \frac {(7 b B-9 A c) \left (\frac {5 \left (-\frac {c \int \frac {\sqrt {x}}{\sqrt {c x^4+b x^2}}dx}{3 b}-\frac {2 \sqrt {b x^2+c x^4}}{3 b x^{5/2}}\right )}{2 b}+\frac {1}{b \sqrt {x} \sqrt {b x^2+c x^4}}\right )}{7 b}-\frac {2 A}{7 b x^{5/2} \sqrt {b x^2+c x^4}}\)

\(\Big \downarrow \) 1431

\(\displaystyle \frac {(7 b B-9 A c) \left (\frac {5 \left (-\frac {c x \sqrt {b+c x^2} \int \frac {1}{\sqrt {x} \sqrt {c x^2+b}}dx}{3 b \sqrt {b x^2+c x^4}}-\frac {2 \sqrt {b x^2+c x^4}}{3 b x^{5/2}}\right )}{2 b}+\frac {1}{b \sqrt {x} \sqrt {b x^2+c x^4}}\right )}{7 b}-\frac {2 A}{7 b x^{5/2} \sqrt {b x^2+c x^4}}\)

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 761

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

Input:

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

Output:

(-2*A)/(7*b*x^(5/2)*Sqrt[b*x^2 + c*x^4]) + ((7*b*B - 9*A*c)*(1/(b*Sqrt[x]* 
Sqrt[b*x^2 + c*x^4]) + (5*((-2*Sqrt[b*x^2 + c*x^4])/(3*b*x^(5/2)) - (c^(3/ 
4)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*Ellip 
ticF[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(3*b^(5/4)*Sqrt[b*x^2 + c* 
x^4])))/(2*b)))/(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 1428

Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp 
[(-d)*(d*x)^(m - 1)*((b*x^2 + c*x^4)^(p + 1)/(2*b*(p + 1))), x] + Simp[d^2* 
((m + 4*p + 3)/(2*b*(p + 1)))   Int[(d*x)^(m - 2)*(b*x^2 + c*x^4)^(p + 1), 
x], x] /; FreeQ[{b, c, d, m, p}, x] &&  !IntegerQ[p] && LtQ[p, -1]
 

rule 1430

Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp 
[d*(d*x)^(m - 1)*((b*x^2 + c*x^4)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[c*( 
(m + 4*p + 3)/(b*d^2*(m + 2*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] && 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 = 2.16 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.25

method	result	size
default	\(\frac {\left (c \,x^{2}+b \right ) \left (45 A \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) c \,x^{3}-35 B \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b \,x^{3}+90 x^{4} A \,c^{2}-70 x^{4} B b c +36 A b c \,x^{2}-28 x^{2} B \,b^{2}-12 b^{2} A \right )}{42 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \sqrt {x}\, b^{3}}\)	\(254\)
risch	\(-\frac {2 \left (c \,x^{2}+b \right ) \left (-12 A c \,x^{2}+7 B b \,x^{2}+3 A b \right )}{21 b^{3} x^{\frac {5}{2}} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}+\frac {c \left (\frac {12 A \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{\sqrt {c \,x^{3}+b x}}-\frac {7 B b \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c \sqrt {c \,x^{3}+b x}}+21 b \left (A c -B b \right ) \left (\frac {x}{b \sqrt {\left (x^{2}+\frac {b}{c}\right ) c x}}+\frac {\sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{2 b c \sqrt {c \,x^{3}+b x}}\right )\right ) \sqrt {x}\, \sqrt {x \left (c \,x^{2}+b \right )}}{21 b^{3} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\)	\(441\)

Input:

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

Output:

1/42/(c*x^4+b*x^2)^(3/2)/x^(1/2)*(c*x^2+b)*(45*A*(-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))*c*x^3-35*B*(-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))*b* 
x^3+90*x^4*A*c^2-70*x^4*B*b*c+36*A*b*c*x^2-28*x^2*B*b^2-12*b^2*A)/b^3
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.62 \[ \int \frac {A+B x^2}{x^{3/2} \left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {5 \, {\left ({\left (7 \, B b c - 9 \, A c^{2}\right )} x^{7} + {\left (7 \, B b^{2} - 9 \, A b c\right )} x^{5}\right )} \sqrt {c} {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right ) + {\left (5 \, {\left (7 \, B b c - 9 \, A c^{2}\right )} x^{4} + 6 \, A b^{2} + 2 \, {\left (7 \, B b^{2} - 9 \, A b c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}} \sqrt {x}}{21 \, {\left (b^{3} c x^{7} + b^{4} x^{5}\right )}} \] Input:

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

Output:

-1/21*(5*((7*B*b*c - 9*A*c^2)*x^7 + (7*B*b^2 - 9*A*b*c)*x^5)*sqrt(c)*weier 
strassPInverse(-4*b/c, 0, x) + (5*(7*B*b*c - 9*A*c^2)*x^4 + 6*A*b^2 + 2*(7 
*B*b^2 - 9*A*b*c)*x^2)*sqrt(c*x^4 + b*x^2)*sqrt(x))/(b^3*c*x^7 + b^4*x^5)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {A+B x^2}{x^{3/2} \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {-2 \sqrt {c \,x^{2}+b}\, b +9 \sqrt {x}\, \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c^{2} x^{9}+2 b c \,x^{7}+b^{2} x^{5}}d x \right ) a b c \,x^{3}+9 \sqrt {x}\, \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c^{2} x^{9}+2 b c \,x^{7}+b^{2} x^{5}}d x \right ) a \,c^{2} x^{5}-7 \sqrt {x}\, \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c^{2} x^{9}+2 b c \,x^{7}+b^{2} x^{5}}d x \right ) b^{3} x^{3}-7 \sqrt {x}\, \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c^{2} x^{9}+2 b c \,x^{7}+b^{2} x^{5}}d x \right ) b^{2} c \,x^{5}}{9 \sqrt {x}\, c \,x^{3} \left (c \,x^{2}+b \right )} \] Input:

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

Output:

( - 2*sqrt(b + c*x**2)*b + 9*sqrt(x)*int((sqrt(x)*sqrt(b + c*x**2))/(b**2* 
x**5 + 2*b*c*x**7 + c**2*x**9),x)*a*b*c*x**3 + 9*sqrt(x)*int((sqrt(x)*sqrt 
(b + c*x**2))/(b**2*x**5 + 2*b*c*x**7 + c**2*x**9),x)*a*c**2*x**5 - 7*sqrt 
(x)*int((sqrt(x)*sqrt(b + c*x**2))/(b**2*x**5 + 2*b*c*x**7 + c**2*x**9),x) 
*b**3*x**3 - 7*sqrt(x)*int((sqrt(x)*sqrt(b + c*x**2))/(b**2*x**5 + 2*b*c*x 
**7 + c**2*x**9),x)*b**2*c*x**5)/(9*sqrt(x)*c*x**3*(b + c*x**2))