\(\int x^{5/2} (A+B x^2) \sqrt {b x^2+c x^4} \, dx\) [224]

Optimal result

Integrand size = 28, antiderivative size = 243 \[ \int x^{5/2} \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {4 b^2 (3 b B-5 A c) \sqrt {b x^2+c x^4}}{231 c^3 \sqrt {x}}-\frac {4 b (3 b B-5 A c) x^{3/2} \sqrt {b x^2+c x^4}}{385 c^2}-\frac {2 (3 b B-5 A c) x^{7/2} \sqrt {b x^2+c x^4}}{55 c}+\frac {2 B x^{3/2} \left (b x^2+c x^4\right )^{3/2}}{15 c}-\frac {2 b^{11/4} (3 b B-5 A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{231 c^{13/4} \sqrt {b x^2+c x^4}} \] Output:

4/231*b^2*(-5*A*c+3*B*b)*(c*x^4+b*x^2)^(1/2)/c^3/x^(1/2)-4/385*b*(-5*A*c+3 
*B*b)*x^(3/2)*(c*x^4+b*x^2)^(1/2)/c^2-2/55*(-5*A*c+3*B*b)*x^(7/2)*(c*x^4+b 
*x^2)^(1/2)/c+2/15*B*x^(3/2)*(c*x^4+b*x^2)^(3/2)/c-2/231*b^(11/4)*(-5*A*c+ 
3*B*b)*x*(b^(1/2)+c^(1/2)*x)*((c*x^2+b)/(b^(1/2)+c^(1/2)*x)^2)^(1/2)*Inver 
seJacobiAM(2*arctan(c^(1/4)*x^(1/2)/b^(1/4)),1/2*2^(1/2))/c^(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 = 10.13 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.56 \[ \int x^{5/2} \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {2 \sqrt {x^2 \left (b+c x^2\right )} \left (\left (b+c x^2\right ) \sqrt {1+\frac {c x^2}{b}} \left (45 b^2 B+7 c^2 x^2 \left (15 A+11 B x^2\right )-3 b c \left (25 A+21 B x^2\right )\right )+15 b^2 (-3 b B+5 A c) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {5}{4},-\frac {c x^2}{b}\right )\right )}{1155 c^3 \sqrt {x} \sqrt {1+\frac {c x^2}{b}}} \] Input:

Integrate[x^(5/2)*(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]*(45*b^2*B + 7*c^ 
2*x^2*(15*A + 11*B*x^2) - 3*b*c*(25*A + 21*B*x^2)) + 15*b^2*(-3*b*B + 5*A* 
c)*Hypergeometric2F1[-1/2, 1/4, 5/4, -((c*x^2)/b)]))/(1155*c^3*Sqrt[x]*Sqr 
t[1 + (c*x^2)/b])
 

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 232, normalized size of antiderivative = 0.95, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1945, 1426, 1429, 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.

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

\(\Big \downarrow \) 1945

\(\displaystyle \frac {2 B x^{3/2} \left (b x^2+c x^4\right )^{3/2}}{15 c}-\frac {(3 b B-5 A c) \int x^{5/2} \sqrt {c x^4+b x^2}dx}{5 c}\)

\(\Big \downarrow \) 1426

\(\displaystyle \frac {2 B x^{3/2} \left (b x^2+c x^4\right )^{3/2}}{15 c}-\frac {(3 b B-5 A c) \left (\frac {2}{11} b \int \frac {x^{9/2}}{\sqrt {c x^4+b x^2}}dx+\frac {2}{11} x^{7/2} \sqrt {b x^2+c x^4}\right )}{5 c}\)

\(\Big \downarrow \) 1429

\(\displaystyle \frac {2 B x^{3/2} \left (b x^2+c x^4\right )^{3/2}}{15 c}-\frac {(3 b B-5 A c) \left (\frac {2}{11} b \left (\frac {2 x^{3/2} \sqrt {b x^2+c x^4}}{7 c}-\frac {5 b \int \frac {x^{5/2}}{\sqrt {c x^4+b x^2}}dx}{7 c}\right )+\frac {2}{11} x^{7/2} \sqrt {b x^2+c x^4}\right )}{5 c}\)

\(\Big \downarrow \) 1429

\(\displaystyle \frac {2 B x^{3/2} \left (b x^2+c x^4\right )^{3/2}}{15 c}-\frac {(3 b B-5 A c) \left (\frac {2}{11} b \left (\frac {2 x^{3/2} \sqrt {b x^2+c x^4}}{7 c}-\frac {5 b \left (\frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {b \int \frac {\sqrt {x}}{\sqrt {c x^4+b x^2}}dx}{3 c}\right )}{7 c}\right )+\frac {2}{11} x^{7/2} \sqrt {b x^2+c x^4}\right )}{5 c}\)

\(\Big \downarrow \) 1431

\(\displaystyle \frac {2 B x^{3/2} \left (b x^2+c x^4\right )^{3/2}}{15 c}-\frac {(3 b B-5 A c) \left (\frac {2}{11} b \left (\frac {2 x^{3/2} \sqrt {b x^2+c x^4}}{7 c}-\frac {5 b \left (\frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {b x \sqrt {b+c x^2} \int \frac {1}{\sqrt {x} \sqrt {c x^2+b}}dx}{3 c \sqrt {b x^2+c x^4}}\right )}{7 c}\right )+\frac {2}{11} x^{7/2} \sqrt {b x^2+c x^4}\right )}{5 c}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {2 B x^{3/2} \left (b x^2+c x^4\right )^{3/2}}{15 c}-\frac {(3 b B-5 A c) \left (\frac {2}{11} b \left (\frac {2 x^{3/2} \sqrt {b x^2+c x^4}}{7 c}-\frac {5 b \left (\frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {2 b x \sqrt {b+c x^2} \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{3 c \sqrt {b x^2+c x^4}}\right )}{7 c}\right )+\frac {2}{11} x^{7/2} \sqrt {b x^2+c x^4}\right )}{5 c}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {2 B x^{3/2} \left (b x^2+c x^4\right )^{3/2}}{15 c}-\frac {(3 b B-5 A c) \left (\frac {2}{11} b \left (\frac {2 x^{3/2} \sqrt {b x^2+c x^4}}{7 c}-\frac {5 b \left (\frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {b^{3/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{3 c^{5/4} \sqrt {b x^2+c x^4}}\right )}{7 c}\right )+\frac {2}{11} x^{7/2} \sqrt {b x^2+c x^4}\right )}{5 c}\)

Input:

Int[x^(5/2)*(A + B*x^2)*Sqrt[b*x^2 + c*x^4],x]
 

Output:

(2*B*x^(3/2)*(b*x^2 + c*x^4)^(3/2))/(15*c) - ((3*b*B - 5*A*c)*((2*x^(7/2)* 
Sqrt[b*x^2 + c*x^4])/11 + (2*b*((2*x^(3/2)*Sqrt[b*x^2 + c*x^4])/(7*c) - (5 
*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)*S 
qrt[x])/b^(1/4)], 1/2])/(3*c^(5/4)*Sqrt[b*x^2 + c*x^4])))/(7*c)))/11))/(5* 
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 = 0.94 (sec) , antiderivative size = 241, normalized size of antiderivative = 0.99

method	result	size
risch	\(-\frac {2 \left (-77 B \,c^{3} x^{6}-105 A \,c^{3} x^{4}-14 B b \,c^{2} x^{4}-30 A b \,c^{2} x^{2}+18 x^{2} B \,b^{2} c +50 A \,b^{2} c -30 B \,b^{3}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{1155 c^{3} \sqrt {x}}+\frac {2 b^{3} \left (5 A c -3 B b \right ) \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {x \left (c \,x^{2}+b \right )}}{231 c^{4} \sqrt {c \,x^{3}+b x}\, x^{\frac {3}{2}} \left (c \,x^{2}+b \right )}\)	\(241\)
default	\(\frac {2 \sqrt {c \,x^{4}+b \,x^{2}}\, \left (77 B \,c^{5} x^{9}+105 A \,c^{5} x^{7}+91 B b \,c^{4} x^{7}+25 A \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b^{3} c +135 A b \,c^{4} x^{5}-15 B \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b^{4}-4 B \,b^{2} c^{3} x^{5}-20 A \,b^{2} c^{3} x^{3}+12 B \,b^{3} c^{2} x^{3}-50 A \,b^{3} c^{2} x +30 B \,b^{4} c x \right )}{1155 x^{\frac {3}{2}} \left (c \,x^{2}+b \right ) c^{4}}\)	\(307\)

Input:

int(x^(5/2)*(B*x^2+A)*(c*x^4+b*x^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-2/1155*(-77*B*c^3*x^6-105*A*c^3*x^4-14*B*b*c^2*x^4-30*A*b*c^2*x^2+18*B*b^ 
2*c*x^2+50*A*b^2*c-30*B*b^3)/c^3/x^(1/2)*(x^2*(c*x^2+b))^(1/2)+2/231*b^3*( 
5*A*c-3*B*b)/c^4*(-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 = 123, normalized size of antiderivative = 0.51 \[ \int x^{5/2} \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=-\frac {2 \, {\left (10 \, {\left (3 \, B b^{4} - 5 \, A b^{3} c\right )} \sqrt {c} x {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right ) - {\left (77 \, B c^{4} x^{6} + 30 \, B b^{3} c - 50 \, A b^{2} c^{2} + 7 \, {\left (2 \, B b c^{3} + 15 \, A c^{4}\right )} x^{4} - 6 \, {\left (3 \, B b^{2} c^{2} - 5 \, A b c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}} \sqrt {x}\right )}}{1155 \, c^{4} x} \] Input:

integrate(x^(5/2)*(B*x^2+A)*(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")
 

Output:

-2/1155*(10*(3*B*b^4 - 5*A*b^3*c)*sqrt(c)*x*weierstrassPInverse(-4*b/c, 0, 
 x) - (77*B*c^4*x^6 + 30*B*b^3*c - 50*A*b^2*c^2 + 7*(2*B*b*c^3 + 15*A*c^4) 
*x^4 - 6*(3*B*b^2*c^2 - 5*A*b*c^3)*x^2)*sqrt(c*x^4 + b*x^2)*sqrt(x))/(c^4* 
x)
 

Sympy [F]

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

integrate(x**(5/2)*(B*x**2+A)*(c*x**4+b*x**2)**(1/2),x)
 

Output:

Integral(x**(5/2)*sqrt(x**2*(b + c*x**2))*(A + B*x**2), x)
 

Maxima [F]

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

integrate(x^(5/2)*(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)*x^(5/2), x)
 

Giac [F]

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

integrate(x^(5/2)*(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)*x^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int x^{5/2} \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\int x^{5/2}\,\left (B\,x^2+A\right )\,\sqrt {c\,x^4+b\,x^2} \,d x \] Input:

int(x^(5/2)*(A + B*x^2)*(b*x^2 + c*x^4)^(1/2),x)
 

Output:

int(x^(5/2)*(A + B*x^2)*(b*x^2 + c*x^4)^(1/2), x)
 

Reduce [F]

\[ \int x^{5/2} \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {-\frac {20 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, a \,b^{2} c}{231}+\frac {4 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, a b \,c^{2} x^{2}}{77}+\frac {2 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, a \,c^{3} x^{4}}{11}+\frac {4 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, b^{4}}{77}-\frac {12 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, b^{3} c \,x^{2}}{385}+\frac {4 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, b^{2} c^{2} x^{4}}{165}+\frac {2 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, b \,c^{3} x^{6}}{15}+\frac {10 \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c \,x^{3}+b x}d x \right ) a \,b^{3} c}{231}-\frac {2 \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c \,x^{3}+b x}d x \right ) b^{5}}{77}}{c^{3}} \] Input:

int(x^(5/2)*(B*x^2+A)*(c*x^4+b*x^2)^(1/2),x)
 

Output:

(2*( - 50*sqrt(x)*sqrt(b + c*x**2)*a*b**2*c + 30*sqrt(x)*sqrt(b + c*x**2)* 
a*b*c**2*x**2 + 105*sqrt(x)*sqrt(b + c*x**2)*a*c**3*x**4 + 30*sqrt(x)*sqrt 
(b + c*x**2)*b**4 - 18*sqrt(x)*sqrt(b + c*x**2)*b**3*c*x**2 + 14*sqrt(x)*s 
qrt(b + c*x**2)*b**2*c**2*x**4 + 77*sqrt(x)*sqrt(b + c*x**2)*b*c**3*x**6 + 
 25*int((sqrt(x)*sqrt(b + c*x**2))/(b*x + c*x**3),x)*a*b**3*c - 15*int((sq 
rt(x)*sqrt(b + c*x**2))/(b*x + c*x**3),x)*b**5))/(1155*c**3)