\(\int \sqrt {x} (A+B x^2) \sqrt {b x^2+c x^4} \, dx\) [226]

Optimal result

Integrand size = 28, antiderivative size = 204 \[ \int \sqrt {x} \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=-\frac {4 b (5 b B-11 A c) \sqrt {b x^2+c x^4}}{231 c^2 \sqrt {x}}-\frac {2 (5 b B-11 A c) x^{3/2} \sqrt {b x^2+c x^4}}{77 c}+\frac {2 B \left (b x^2+c x^4\right )^{3/2}}{11 c \sqrt {x}}+\frac {2 b^{7/4} (5 b B-11 A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{231 c^{9/4} \sqrt {b x^2+c x^4}} \] Output:

-4/231*b*(-11*A*c+5*B*b)*(c*x^4+b*x^2)^(1/2)/c^2/x^(1/2)-2/77*(-11*A*c+5*B 
*b)*x^(3/2)*(c*x^4+b*x^2)^(1/2)/c+2/11*B*(c*x^4+b*x^2)^(3/2)/c/x^(1/2)+2/2 
31*b^(7/4)*(-11*A*c+5*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^(9/4)/(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 = 10.11 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.54 \[ \int \sqrt {x} \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {2 \sqrt {x^2 \left (b+c x^2\right )} \left (-\left (\left (b+c x^2\right ) \sqrt {1+\frac {c x^2}{b}} \left (5 b B-11 A c-7 B c x^2\right )\right )+b (5 b B-11 A c) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {5}{4},-\frac {c x^2}{b}\right )\right )}{77 c^2 \sqrt {x} \sqrt {1+\frac {c x^2}{b}}} \] Input:

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

Output:

(2*Sqrt[x^2*(b + c*x^2)]*(-((b + c*x^2)*Sqrt[1 + (c*x^2)/b]*(5*b*B - 11*A* 
c - 7*B*c*x^2)) + b*(5*b*B - 11*A*c)*Hypergeometric2F1[-1/2, 1/4, 5/4, -(( 
c*x^2)/b)]))/(77*c^2*Sqrt[x]*Sqrt[1 + (c*x^2)/b])
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {1945, 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.

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

\(\Big \downarrow \) 1945

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

\(\Big \downarrow \) 1426

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

\(\Big \downarrow \) 1429

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

\(\Big \downarrow \) 1431

\(\displaystyle \frac {2 B \left (b x^2+c x^4\right )^{3/2}}{11 c \sqrt {x}}-\frac {(5 b B-11 A c) \left (\frac {2}{7} b \left (\frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {b x \sqrt {b+c x^2} \int \frac {1}{\sqrt {x} \sqrt {c x^2+b}}dx}{3 c \sqrt {b x^2+c x^4}}\right )+\frac {2}{7} x^{3/2} \sqrt {b x^2+c x^4}\right )}{11 c}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {2 B \left (b x^2+c x^4\right )^{3/2}}{11 c \sqrt {x}}-\frac {(5 b B-11 A c) \left (\frac {2}{7} b \left (\frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {2 b x \sqrt {b+c x^2} \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{3 c \sqrt {b x^2+c x^4}}\right )+\frac {2}{7} x^{3/2} \sqrt {b x^2+c x^4}\right )}{11 c}\)

\(\Big \downarrow \) 761

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

Input:

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

Output:

(2*B*(b*x^2 + c*x^4)^(3/2))/(11*c*Sqrt[x]) - ((5*b*B - 11*A*c)*((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]*E 
llipticF[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(3*c^(5/4)*Sqrt[b*x^2 
+ c*x^4])))/7))/(11*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 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]
 

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 = 0.56 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.06

method	result	size
risch	\(\frac {2 \left (21 B \,c^{2} x^{4}+33 A \,c^{2} x^{2}+6 x^{2} B b c +22 A b c -10 B \,b^{2}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{231 c^{2} \sqrt {x}}-\frac {2 b^{2} \left (11 A c -5 B b \right ) \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {x \left (c \,x^{2}+b \right )}}{231 c^{3} \sqrt {c \,x^{3}+b x}\, x^{\frac {3}{2}} \left (c \,x^{2}+b \right )}\)	\(217\)
default	\(-\frac {2 \sqrt {c \,x^{4}+b \,x^{2}}\, \left (-21 B \,c^{4} x^{7}+11 A \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b^{2} c -33 A \,c^{4} x^{5}-5 B \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b^{3}-27 B b \,c^{3} x^{5}-55 A b \,c^{3} x^{3}+4 B \,b^{2} c^{2} x^{3}-22 A \,b^{2} c^{2} x +10 B \,b^{3} c x \right )}{231 x^{\frac {3}{2}} \left (c \,x^{2}+b \right ) c^{3}}\)	\(283\)

Input:

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

Output:

2/231*(21*B*c^2*x^4+33*A*c^2*x^2+6*B*b*c*x^2+22*A*b*c-10*B*b^2)/c^2/x^(1/2 
)*(x^2*(c*x^2+b))^(1/2)-2/231*b^2*(11*A*c-5*B*b)/c^3*(-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 = 0.10 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.48 \[ \int \sqrt {x} \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {2 \, {\left (2 \, {\left (5 \, B b^{3} - 11 \, A b^{2} c\right )} \sqrt {c} x {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right ) + {\left (21 \, B c^{3} x^{4} - 10 \, B b^{2} c + 22 \, A b c^{2} + 3 \, {\left (2 \, B b c^{2} + 11 \, A c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}} \sqrt {x}\right )}}{231 \, c^{3} x} \] Input:

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

Output:

2/231*(2*(5*B*b^3 - 11*A*b^2*c)*sqrt(c)*x*weierstrassPInverse(-4*b/c, 0, x 
) + (21*B*c^3*x^4 - 10*B*b^2*c + 22*A*b*c^2 + 3*(2*B*b*c^2 + 11*A*c^3)*x^2 
)*sqrt(c*x^4 + b*x^2)*sqrt(x))/(c^3*x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \sqrt {x} \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {\frac {4 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, a b c}{21}+\frac {2 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, a \,c^{2} x^{2}}{7}-\frac {20 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, b^{3}}{231}+\frac {4 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, b^{2} c \,x^{2}}{77}+\frac {2 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, b \,c^{2} x^{4}}{11}-\frac {2 \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c \,x^{3}+b x}d x \right ) a \,b^{2} c}{21}+\frac {10 \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c \,x^{3}+b x}d x \right ) b^{4}}{231}}{c^{2}} \] Input:

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

Output:

(2*(22*sqrt(x)*sqrt(b + c*x**2)*a*b*c + 33*sqrt(x)*sqrt(b + c*x**2)*a*c**2 
*x**2 - 10*sqrt(x)*sqrt(b + c*x**2)*b**3 + 6*sqrt(x)*sqrt(b + c*x**2)*b**2 
*c*x**2 + 21*sqrt(x)*sqrt(b + c*x**2)*b*c**2*x**4 - 11*int((sqrt(x)*sqrt(b 
 + c*x**2))/(b*x + c*x**3),x)*a*b**2*c + 5*int((sqrt(x)*sqrt(b + c*x**2))/ 
(b*x + c*x**3),x)*b**4))/(231*c**2)