\(\int \frac {\sqrt {b x^2+c x^4}}{x^{15/2}} \, dx\) [239]

Optimal result

Integrand size = 21, antiderivative size = 176 \[ \int \frac {\sqrt {b x^2+c x^4}}{x^{15/2}} \, dx=-\frac {2 \sqrt {b x^2+c x^4}}{11 x^{13/2}}-\frac {4 c \sqrt {b x^2+c x^4}}{77 b x^{9/2}}+\frac {20 c^2 \sqrt {b x^2+c x^4}}{231 b^2 x^{5/2}}+\frac {10 c^{11/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{231 b^{9/4} \sqrt {b x^2+c x^4}} \] Output:

-2/11*(c*x^4+b*x^2)^(1/2)/x^(13/2)-4/77*c*(c*x^4+b*x^2)^(1/2)/b/x^(9/2)+20 
/231*c^2*(c*x^4+b*x^2)^(1/2)/b^2/x^(5/2)+10/231*c^(11/4)*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 = 10.01 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.32 \[ \int \frac {\sqrt {b x^2+c x^4}}{x^{15/2}} \, dx=-\frac {2 \sqrt {x^2 \left (b+c x^2\right )} \operatorname {Hypergeometric2F1}\left (-\frac {11}{4},-\frac {1}{2},-\frac {7}{4},-\frac {c x^2}{b}\right )}{11 x^{13/2} \sqrt {1+\frac {c x^2}{b}}} \] Input:

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

Output:

(-2*Sqrt[x^2*(b + c*x^2)]*Hypergeometric2F1[-11/4, -1/2, -7/4, -((c*x^2)/b 
)])/(11*x^(13/2)*Sqrt[1 + (c*x^2)/b])
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.06, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {1425, 1430, 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.

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

\(\Big \downarrow \) 1425

\(\displaystyle \frac {2}{11} c \int \frac {1}{x^{7/2} \sqrt {c x^4+b x^2}}dx-\frac {2 \sqrt {b x^2+c x^4}}{11 x^{13/2}}\)

\(\Big \downarrow \) 1430

\(\displaystyle \frac {2}{11} c \left (-\frac {5 c \int \frac {1}{x^{3/2} \sqrt {c x^4+b x^2}}dx}{7 b}-\frac {2 \sqrt {b x^2+c x^4}}{7 b x^{9/2}}\right )-\frac {2 \sqrt {b x^2+c x^4}}{11 x^{13/2}}\)

\(\Big \downarrow \) 1430

\(\displaystyle \frac {2}{11} c \left (-\frac {5 c \left (-\frac {c \int \frac {\sqrt {x}}{\sqrt {c x^4+b x^2}}dx}{3 b}-\frac {2 \sqrt {b x^2+c x^4}}{3 b x^{5/2}}\right )}{7 b}-\frac {2 \sqrt {b x^2+c x^4}}{7 b x^{9/2}}\right )-\frac {2 \sqrt {b x^2+c x^4}}{11 x^{13/2}}\)

\(\Big \downarrow \) 1431

\(\displaystyle \frac {2}{11} c \left (-\frac {5 c \left (-\frac {c x \sqrt {b+c x^2} \int \frac {1}{\sqrt {x} \sqrt {c x^2+b}}dx}{3 b \sqrt {b x^2+c x^4}}-\frac {2 \sqrt {b x^2+c x^4}}{3 b x^{5/2}}\right )}{7 b}-\frac {2 \sqrt {b x^2+c x^4}}{7 b x^{9/2}}\right )-\frac {2 \sqrt {b x^2+c x^4}}{11 x^{13/2}}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {2}{11} c \left (-\frac {5 c \left (-\frac {2 c x \sqrt {b+c x^2} \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{3 b \sqrt {b x^2+c x^4}}-\frac {2 \sqrt {b x^2+c x^4}}{3 b x^{5/2}}\right )}{7 b}-\frac {2 \sqrt {b x^2+c x^4}}{7 b x^{9/2}}\right )-\frac {2 \sqrt {b x^2+c x^4}}{11 x^{13/2}}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {2}{11} c \left (-\frac {5 c \left (-\frac {c^{3/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{3 b^{5/4} \sqrt {b x^2+c x^4}}-\frac {2 \sqrt {b x^2+c x^4}}{3 b x^{5/2}}\right )}{7 b}-\frac {2 \sqrt {b x^2+c x^4}}{7 b x^{9/2}}\right )-\frac {2 \sqrt {b x^2+c x^4}}{11 x^{13/2}}\)

Input:

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

Output:

(-2*Sqrt[b*x^2 + c*x^4])/(11*x^(13/2)) + (2*c*((-2*Sqrt[b*x^2 + c*x^4])/(7 
*b*x^(9/2)) - (5*c*((-2*Sqrt[b*x^2 + c*x^4])/(3*b*x^(5/2)) - (c^(3/4)*x*(S 
qrt[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*b)))/11
 

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]
 

Maple [A] (verified)

Time = 0.62 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.89

method	result	size
default	\(\frac {2 \sqrt {c \,x^{4}+b \,x^{2}}\, \left (5 \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) c^{2} x^{5}+10 c^{3} x^{6}+4 b \,c^{2} x^{4}-27 b^{2} c \,x^{2}-21 b^{3}\right )}{231 x^{\frac {13}{2}} \left (c \,x^{2}+b \right ) b^{2}}\)	\(156\)
risch	\(-\frac {2 \left (-10 c^{2} x^{4}+6 b c \,x^{2}+21 b^{2}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{231 x^{\frac {13}{2}} b^{2}}+\frac {10 c^{2} \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {x \left (c \,x^{2}+b \right )}}{231 b^{2} \sqrt {c \,x^{3}+b x}\, x^{\frac {3}{2}} \left (c \,x^{2}+b \right )}\)	\(191\)

Input:

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

Output:

2/231*(c*x^4+b*x^2)^(1/2)/x^(13/2)/(c*x^2+b)*(5*(-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^2*x^5+10*c^3*x^6+4*b*c^2*x^4-27*b^2*c*x^2-21*b^3)/b^2
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.36 \[ \int \frac {\sqrt {b x^2+c x^4}}{x^{15/2}} \, dx=\frac {2 \, {\left (10 \, c^{\frac {5}{2}} x^{7} {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right ) + {\left (10 \, c^{2} x^{4} - 6 \, b c x^{2} - 21 \, b^{2}\right )} \sqrt {c x^{4} + b x^{2}} \sqrt {x}\right )}}{231 \, b^{2} x^{7}} \] Input:

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

Output:

2/231*(10*c^(5/2)*x^7*weierstrassPInverse(-4*b/c, 0, x) + (10*c^2*x^4 - 6* 
b*c*x^2 - 21*b^2)*sqrt(c*x^4 + b*x^2)*sqrt(x))/(b^2*x^7)
 

Sympy [F]

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

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

Output:

Integral(sqrt(x**2*(b + c*x**2))/x**(15/2), x)
 

Maxima [F]

\[ \int \frac {\sqrt {b x^2+c x^4}}{x^{15/2}} \, dx=\int { \frac {\sqrt {c x^{4} + b x^{2}}}{x^{\frac {15}{2}}} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {\sqrt {b x^2+c x^4}}{x^{15/2}} \, dx=\int { \frac {\sqrt {c x^{4} + b x^{2}}}{x^{\frac {15}{2}}} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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