\(\int \frac {(b x^2+c x^4)^{3/2}}{x^{15/2}} \, dx\) [249]

Optimal result

Integrand size = 21, antiderivative size = 143 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{15/2}} \, dx=-\frac {4 c \sqrt {b x^2+c x^4}}{7 x^{5/2}}-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{7 x^{13/2}}+\frac {4 c^{7/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{7 \sqrt [4]{b} \sqrt {b x^2+c x^4}} \] Output:

-4/7*c*(c*x^4+b*x^2)^(1/2)/x^(5/2)-2/7*(c*x^4+b*x^2)^(3/2)/x^(13/2)+4/7*c^ 
(7/4)*x*(b^(1/2)+c^(1/2)*x)*((c*x^2+b)/(b^(1/2)+c^(1/2)*x)^2)^(1/2)*Invers 
eJacobiAM(2*arctan(c^(1/4)*x^(1/2)/b^(1/4)),1/2*2^(1/2))/b^(1/4)/(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 = 10.02 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.41 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{15/2}} \, dx=-\frac {2 b \sqrt {x^2 \left (b+c x^2\right )} \operatorname {Hypergeometric2F1}\left (-\frac {7}{4},-\frac {3}{2},-\frac {3}{4},-\frac {c x^2}{b}\right )}{7 x^{9/2} \sqrt {1+\frac {c x^2}{b}}} \] Input:

Integrate[(b*x^2 + c*x^4)^(3/2)/x^(15/2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {1425, 1425, 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.

\(\displaystyle \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{15/2}} \, dx\)

\(\Big \downarrow \) 1425

\(\displaystyle \frac {6}{7} c \int \frac {\sqrt {c x^4+b x^2}}{x^{7/2}}dx-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{7 x^{13/2}}\)

\(\Big \downarrow \) 1425

\(\displaystyle \frac {6}{7} c \left (\frac {2}{3} c \int \frac {\sqrt {x}}{\sqrt {c x^4+b x^2}}dx-\frac {2 \sqrt {b x^2+c x^4}}{3 x^{5/2}}\right )-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{7 x^{13/2}}\)

\(\Big \downarrow \) 1431

\(\displaystyle \frac {6}{7} c \left (\frac {2 c x \sqrt {b+c x^2} \int \frac {1}{\sqrt {x} \sqrt {c x^2+b}}dx}{3 \sqrt {b x^2+c x^4}}-\frac {2 \sqrt {b x^2+c x^4}}{3 x^{5/2}}\right )-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{7 x^{13/2}}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {6}{7} c \left (\frac {4 c x \sqrt {b+c x^2} \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{3 \sqrt {b x^2+c x^4}}-\frac {2 \sqrt {b x^2+c x^4}}{3 x^{5/2}}\right )-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{7 x^{13/2}}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {6}{7} c \left (\frac {2 c^{3/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{3 \sqrt [4]{b} \sqrt {b x^2+c x^4}}-\frac {2 \sqrt {b x^2+c x^4}}{3 x^{5/2}}\right )-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{7 x^{13/2}}\)

Input:

Int[(b*x^2 + c*x^4)^(3/2)/x^(15/2),x]
 

Output:

(-2*(b*x^2 + c*x^4)^(3/2))/(7*x^(13/2)) + (6*c*((-2*Sqrt[b*x^2 + c*x^4])/( 
3*x^(5/2)) + (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^ 
(1/4)*Sqrt[b*x^2 + c*x^4])))/7
 

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 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 = 1.24 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.98

method	result	size
default	\(\frac {2 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (2 \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) c \,x^{3}-3 c^{2} x^{4}-4 b c \,x^{2}-b^{2}\right )}{7 x^{\frac {13}{2}} \left (c \,x^{2}+b \right )^{2}}\)	\(140\)
risch	\(-\frac {2 \left (3 c \,x^{2}+b \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{7 x^{\frac {9}{2}}}+\frac {4 c \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {x \left (c \,x^{2}+b \right )}}{7 \sqrt {c \,x^{3}+b x}\, x^{\frac {3}{2}} \left (c \,x^{2}+b \right )}\)	\(170\)

Input:

int((c*x^4+b*x^2)^(3/2)/x^(15/2),x,method=_RETURNVERBOSE)
 

Output:

2/7*(c*x^4+b*x^2)^(3/2)/x^(13/2)/(c*x^2+b)^2*(2*(-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^3-3*c^2*x^4-4*b*c*x^2-b^2)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.34 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{15/2}} \, dx=\frac {2 \, {\left (4 \, c^{\frac {3}{2}} x^{5} {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right ) - \sqrt {c x^{4} + b x^{2}} {\left (3 \, c x^{2} + b\right )} \sqrt {x}\right )}}{7 \, x^{5}} \] Input:

integrate((c*x^4+b*x^2)^(3/2)/x^(15/2),x, algorithm="fricas")
 

Output:

2/7*(4*c^(3/2)*x^5*weierstrassPInverse(-4*b/c, 0, x) - sqrt(c*x^4 + b*x^2) 
*(3*c*x^2 + b)*sqrt(x))/x^5
 

Sympy [F]

\[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{15/2}} \, dx=\int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}{x^{\frac {15}{2}}}\, dx \] Input:

integrate((c*x**4+b*x**2)**(3/2)/x**(15/2),x)
 

Output:

Integral((x**2*(b + c*x**2))**(3/2)/x**(15/2), x)
 

Maxima [F]

\[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{15/2}} \, dx=\int { \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{\frac {15}{2}}} \,d x } \] Input:

integrate((c*x^4+b*x^2)^(3/2)/x^(15/2),x, algorithm="maxima")
 

Output:

integrate((c*x^4 + b*x^2)^(3/2)/x^(15/2), x)
 

Giac [F]

\[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{15/2}} \, dx=\int { \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{\frac {15}{2}}} \,d x } \] Input:

integrate((c*x^4+b*x^2)^(3/2)/x^(15/2),x, algorithm="giac")
 

Output:

integrate((c*x^4 + b*x^2)^(3/2)/x^(15/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{15/2}} \, dx=\int \frac {{\left (c\,x^4+b\,x^2\right )}^{3/2}}{x^{15/2}} \,d x \] Input:

int((b*x^2 + c*x^4)^(3/2)/x^(15/2),x)
 

Output:

int((b*x^2 + c*x^4)^(3/2)/x^(15/2), x)
 

Reduce [F]

\[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{15/2}} \, dx=\frac {\frac {2 \sqrt {c \,x^{2}+b}\, b}{5}-2 \sqrt {c \,x^{2}+b}\, c \,x^{2}+\frac {12 \sqrt {x}\, \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c \,x^{7}+b \,x^{5}}d x \right ) b^{2} x^{3}}{5}}{\sqrt {x}\, x^{3}} \] Input:

int((c*x^4+b*x^2)^(3/2)/x^(15/2),x)
                                                                                    
                                                                                    
 

Output:

(2*(sqrt(b + c*x**2)*b - 5*sqrt(b + c*x**2)*c*x**2 + 6*sqrt(x)*int((sqrt(x 
)*sqrt(b + c*x**2))/(b*x**5 + c*x**7),x)*b**2*x**3))/(5*sqrt(x)*x**3)