\(\int \frac {(A+B x^2) \sqrt {b x^2+c x^4}}{x^{15/2}} \, dx\) [234]

Optimal result

Integrand size = 28, antiderivative size = 204 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^{15/2}} \, dx=-\frac {2 (11 b B-5 A c) \sqrt {b x^2+c x^4}}{77 b x^{9/2}}-\frac {4 c (11 b B-5 A c) \sqrt {b x^2+c x^4}}{231 b^2 x^{5/2}}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{11 b x^{17/2}}-\frac {2 c^{7/4} (11 b B-5 A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{231 b^{9/4} \sqrt {b x^2+c x^4}} \] Output:

-2/77*(-5*A*c+11*B*b)*(c*x^4+b*x^2)^(1/2)/b/x^(9/2)-4/231*c*(-5*A*c+11*B*b 
)*(c*x^4+b*x^2)^(1/2)/b^2/x^(5/2)-2/11*A*(c*x^4+b*x^2)^(3/2)/b/x^(17/2)-2/ 
231*c^(7/4)*(-5*A*c+11*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^(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 = 10.05 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.48 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^{15/2}} \, dx=-\frac {2 \sqrt {x^2 \left (b+c x^2\right )} \left (7 A \left (b+c x^2\right ) \sqrt {1+\frac {c x^2}{b}}+(11 b B-5 A c) x^2 \operatorname {Hypergeometric2F1}\left (-\frac {7}{4},-\frac {1}{2},-\frac {3}{4},-\frac {c x^2}{b}\right )\right )}{77 b x^{13/2} \sqrt {1+\frac {c x^2}{b}}} \] Input:

Integrate[((A + B*x^2)*Sqrt[b*x^2 + c*x^4])/x^(15/2),x]
 

Output:

(-2*Sqrt[x^2*(b + c*x^2)]*(7*A*(b + c*x^2)*Sqrt[1 + (c*x^2)/b] + (11*b*B - 
 5*A*c)*x^2*Hypergeometric2F1[-7/4, -1/2, -3/4, -((c*x^2)/b)]))/(77*b*x^(1 
3/2)*Sqrt[1 + (c*x^2)/b])
 

Rubi [A] (verified)

Time = 0.61 (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 = {1944, 1425, 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.

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

\(\Big \downarrow \) 1944

\(\displaystyle \frac {(11 b B-5 A c) \int \frac {\sqrt {c x^4+b x^2}}{x^{11/2}}dx}{11 b}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{11 b x^{17/2}}\)

\(\Big \downarrow \) 1425

\(\displaystyle \frac {(11 b B-5 A c) \left (\frac {2}{7} c \int \frac {1}{x^{3/2} \sqrt {c x^4+b x^2}}dx-\frac {2 \sqrt {b x^2+c x^4}}{7 x^{9/2}}\right )}{11 b}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{11 b x^{17/2}}\)

\(\Big \downarrow \) 1430

\(\displaystyle \frac {(11 b B-5 A c) \left (\frac {2}{7} c \left (-\frac {c \int \frac {\sqrt {x}}{\sqrt {c x^4+b x^2}}dx}{3 b}-\frac {2 \sqrt {b x^2+c x^4}}{3 b x^{5/2}}\right )-\frac {2 \sqrt {b x^2+c x^4}}{7 x^{9/2}}\right )}{11 b}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{11 b x^{17/2}}\)

\(\Big \downarrow \) 1431

\(\displaystyle \frac {(11 b B-5 A c) \left (\frac {2}{7} c \left (-\frac {c x \sqrt {b+c x^2} \int \frac {1}{\sqrt {x} \sqrt {c x^2+b}}dx}{3 b \sqrt {b x^2+c x^4}}-\frac {2 \sqrt {b x^2+c x^4}}{3 b x^{5/2}}\right )-\frac {2 \sqrt {b x^2+c x^4}}{7 x^{9/2}}\right )}{11 b}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{11 b x^{17/2}}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {(11 b B-5 A c) \left (\frac {2}{7} c \left (-\frac {2 c x \sqrt {b+c x^2} \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{3 b \sqrt {b x^2+c x^4}}-\frac {2 \sqrt {b x^2+c x^4}}{3 b x^{5/2}}\right )-\frac {2 \sqrt {b x^2+c x^4}}{7 x^{9/2}}\right )}{11 b}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{11 b x^{17/2}}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {(11 b B-5 A c) \left (\frac {2}{7} c \left (-\frac {c^{3/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{3 b^{5/4} \sqrt {b x^2+c x^4}}-\frac {2 \sqrt {b x^2+c x^4}}{3 b x^{5/2}}\right )-\frac {2 \sqrt {b x^2+c x^4}}{7 x^{9/2}}\right )}{11 b}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{11 b x^{17/2}}\)

Input:

Int[((A + B*x^2)*Sqrt[b*x^2 + c*x^4])/x^(15/2),x]
 

Output:

(-2*A*(b*x^2 + c*x^4)^(3/2))/(11*b*x^(17/2)) + ((11*b*B - 5*A*c)*((-2*Sqrt 
[b*x^2 + c*x^4])/(7*x^(9/2)) + (2*c*((-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]*EllipticF[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(3*b^(5/4)*Sqrt[ 
b*x^2 + c*x^4])))/7))/(11*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 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 = 1.13 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.07

method	result	size
risch	\(-\frac {2 \left (-10 x^{4} A \,c^{2}+22 x^{4} B b c +6 A b c \,x^{2}+33 x^{2} B \,b^{2}+21 b^{2} A \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{231 x^{\frac {13}{2}} b^{2}}+\frac {2 c \left (5 A c -11 B b \right ) \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {x \left (c \,x^{2}+b \right )}}{231 b^{2} \sqrt {c \,x^{3}+b x}\, x^{\frac {3}{2}} \left (c \,x^{2}+b \right )}\)	\(218\)
default	\(\frac {2 \sqrt {c \,x^{4}+b \,x^{2}}\, \left (5 A \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) c^{2} x^{5}-11 B \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b c \,x^{5}+10 A \,c^{3} x^{6}-22 B b \,c^{2} x^{6}+4 A b \,c^{2} x^{4}-55 x^{4} B \,b^{2} c -27 A \,b^{2} c \,x^{2}-33 x^{2} B \,b^{3}-21 A \,b^{3}\right )}{231 x^{\frac {13}{2}} \left (c \,x^{2}+b \right ) b^{2}}\)	\(283\)

Input:

int((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^(15/2),x,method=_RETURNVERBOSE)
 

Output:

-2/231*(-10*A*c^2*x^4+22*B*b*c*x^4+6*A*b*c*x^2+33*B*b^2*x^2+21*A*b^2)/x^(1 
3/2)/b^2*(x^2*(c*x^2+b))^(1/2)+2/231*c*(5*A*c-11*B*b)/b^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 = 0.10 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.47 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^{15/2}} \, dx=-\frac {2 \, {\left (2 \, {\left (11 \, B b c - 5 \, A c^{2}\right )} \sqrt {c} x^{7} {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right ) + {\left (2 \, {\left (11 \, B b c - 5 \, A c^{2}\right )} x^{4} + 21 \, A b^{2} + 3 \, {\left (11 \, B b^{2} + 2 \, A b c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}} \sqrt {x}\right )}}{231 \, b^{2} x^{7}} \] Input:

integrate((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^(15/2),x, algorithm="fricas")
 

Output:

-2/231*(2*(11*B*b*c - 5*A*c^2)*sqrt(c)*x^7*weierstrassPInverse(-4*b/c, 0, 
x) + (2*(11*B*b*c - 5*A*c^2)*x^4 + 21*A*b^2 + 3*(11*B*b^2 + 2*A*b*c)*x^2)* 
sqrt(c*x^4 + b*x^2)*sqrt(x))/(b^2*x^7)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^{15/2}} \, dx=\text {Timed out} \] Input:

integrate((B*x**2+A)*(c*x**4+b*x**2)**(1/2)/x**(15/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^{15/2}} \, dx=\int { \frac {\sqrt {c x^{4} + b x^{2}} {\left (B x^{2} + A\right )}}{x^{\frac {15}{2}}} \,d x } \] Input:

integrate((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^(15/2),x, algorithm="maxima")
 

Output:

integrate(sqrt(c*x^4 + b*x^2)*(B*x^2 + A)/x^(15/2), x)
 

Giac [F]

\[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^{15/2}} \, dx=\int { \frac {\sqrt {c x^{4} + b x^{2}} {\left (B x^{2} + A\right )}}{x^{\frac {15}{2}}} \,d x } \] Input:

integrate((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^(15/2),x, algorithm="giac")
 

Output:

integrate(sqrt(c*x^4 + b*x^2)*(B*x^2 + A)/x^(15/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^{15/2}} \, dx=\int \frac {\left (B\,x^2+A\right )\,\sqrt {c\,x^4+b\,x^2}}{x^{15/2}} \,d x \] Input:

int(((A + B*x^2)*(b*x^2 + c*x^4)^(1/2))/x^(15/2),x)
 

Output:

int(((A + B*x^2)*(b*x^2 + c*x^4)^(1/2))/x^(15/2), x)
 

Reduce [F]

\[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^{15/2}} \, dx=\frac {-\frac {2 \sqrt {c \,x^{2}+b}\, a c}{9}+\frac {4 \sqrt {c \,x^{2}+b}\, b^{2}}{45}-\frac {2 \sqrt {c \,x^{2}+b}\, b c \,x^{2}}{5}-\frac {2 \sqrt {x}\, \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c \,x^{9}+b \,x^{7}}d x \right ) a b c \,x^{5}}{9}+\frac {22 \sqrt {x}\, \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c \,x^{9}+b \,x^{7}}d x \right ) b^{3} x^{5}}{45}}{\sqrt {x}\, c \,x^{5}} \] Input:

int((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^(15/2),x)
 

Output:

(2*( - 5*sqrt(b + c*x**2)*a*c + 2*sqrt(b + c*x**2)*b**2 - 9*sqrt(b + c*x** 
2)*b*c*x**2 - 5*sqrt(x)*int((sqrt(x)*sqrt(b + c*x**2))/(b*x**7 + c*x**9),x 
)*a*b*c*x**5 + 11*sqrt(x)*int((sqrt(x)*sqrt(b + c*x**2))/(b*x**7 + c*x**9) 
,x)*b**3*x**5))/(45*sqrt(x)*c*x**5)