\(\int \frac {(b x^2+c x^4)^{3/2}}{x^{17/2}} \, dx\) [250]

Optimal result

Integrand size = 21, antiderivative size = 320 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{17/2}} \, dx=\frac {8 c^{5/2} x^{3/2} \left (b+c x^2\right )}{15 b \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}-\frac {4 c \sqrt {b x^2+c x^4}}{15 x^{7/2}}-\frac {8 c^2 \sqrt {b x^2+c x^4}}{15 b x^{3/2}}-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{9 x^{15/2}}-\frac {8 c^{9/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{15 b^{3/4} \sqrt {b x^2+c x^4}}+\frac {4 c^{9/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{15 b^{3/4} \sqrt {b x^2+c x^4}} \] Output:

8/15*c^(5/2)*x^(3/2)*(c*x^2+b)/b/(b^(1/2)+c^(1/2)*x)/(c*x^4+b*x^2)^(1/2)-4 
/15*c*(c*x^4+b*x^2)^(1/2)/x^(7/2)-8/15*c^2*(c*x^4+b*x^2)^(1/2)/b/x^(3/2)-2 
/9*(c*x^4+b*x^2)^(3/2)/x^(15/2)-8/15*c^(9/4)*x*(b^(1/2)+c^(1/2)*x)*((c*x^2 
+b)/(b^(1/2)+c^(1/2)*x)^2)^(1/2)*EllipticE(sin(2*arctan(c^(1/4)*x^(1/2)/b^ 
(1/4))),1/2*2^(1/2))/b^(3/4)/(c*x^4+b*x^2)^(1/2)+4/15*c^(9/4)*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^(3/4)/(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 = 58, normalized size of antiderivative = 0.18 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{17/2}} \, dx=-\frac {2 b \sqrt {x^2 \left (b+c x^2\right )} \operatorname {Hypergeometric2F1}\left (-\frac {9}{4},-\frac {3}{2},-\frac {5}{4},-\frac {c x^2}{b}\right )}{9 x^{11/2} \sqrt {1+\frac {c x^2}{b}}} \] Input:

Integrate[(b*x^2 + c*x^4)^(3/2)/x^(17/2),x]
 

Output:

(-2*b*Sqrt[x^2*(b + c*x^2)]*Hypergeometric2F1[-9/4, -3/2, -5/4, -((c*x^2)/ 
b)])/(9*x^(11/2)*Sqrt[1 + (c*x^2)/b])
 

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {1425, 1425, 1430, 1431, 266, 834, 27, 761, 1510}

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.

\(\displaystyle \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{17/2}} \, dx\)

\(\Big \downarrow \) 1425

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

\(\Big \downarrow \) 1425

\(\displaystyle \frac {2}{3} c \left (\frac {2}{5} c \int \frac {1}{\sqrt {x} \sqrt {c x^4+b x^2}}dx-\frac {2 \sqrt {b x^2+c x^4}}{5 x^{7/2}}\right )-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{9 x^{15/2}}\)

\(\Big \downarrow \) 1430

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

\(\Big \downarrow \) 1431

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 834

\(\displaystyle \frac {2}{3} c \left (\frac {2}{5} c \left (\frac {2 c x \sqrt {b+c x^2} \left (\frac {\sqrt {b} \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}-\frac {\sqrt {b} \int \frac {\sqrt {b}-\sqrt {c} x}{\sqrt {b} \sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}\right )}{b \sqrt {b x^2+c x^4}}-\frac {2 \sqrt {b x^2+c x^4}}{b x^{3/2}}\right )-\frac {2 \sqrt {b x^2+c x^4}}{5 x^{7/2}}\right )-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{9 x^{15/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2}{3} c \left (\frac {2}{5} c \left (\frac {2 c x \sqrt {b+c x^2} \left (\frac {\sqrt {b} \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}\right )}{b \sqrt {b x^2+c x^4}}-\frac {2 \sqrt {b x^2+c x^4}}{b x^{3/2}}\right )-\frac {2 \sqrt {b x^2+c x^4}}{5 x^{7/2}}\right )-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{9 x^{15/2}}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {2}{3} c \left (\frac {2}{5} c \left (\frac {2 c x \sqrt {b+c x^2} \left (\frac {\sqrt [4]{b} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{2 c^{3/4} \sqrt {b+c x^2}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}\right )}{b \sqrt {b x^2+c x^4}}-\frac {2 \sqrt {b x^2+c x^4}}{b x^{3/2}}\right )-\frac {2 \sqrt {b x^2+c x^4}}{5 x^{7/2}}\right )-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{9 x^{15/2}}\)

\(\Big \downarrow \) 1510

\(\displaystyle \frac {2}{3} c \left (\frac {2}{5} c \left (\frac {2 c x \sqrt {b+c x^2} \left (\frac {\sqrt [4]{b} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{2 c^{3/4} \sqrt {b+c x^2}}-\frac {\frac {\sqrt [4]{b} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{\sqrt [4]{c} \sqrt {b+c x^2}}-\frac {\sqrt {x} \sqrt {b+c x^2}}{\sqrt {b}+\sqrt {c} x}}{\sqrt {c}}\right )}{b \sqrt {b x^2+c x^4}}-\frac {2 \sqrt {b x^2+c x^4}}{b x^{3/2}}\right )-\frac {2 \sqrt {b x^2+c x^4}}{5 x^{7/2}}\right )-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{9 x^{15/2}}\)

Input:

Int[(b*x^2 + c*x^4)^(3/2)/x^(17/2),x]
 

Output:

(-2*(b*x^2 + c*x^4)^(3/2))/(9*x^(15/2)) + (2*c*((-2*Sqrt[b*x^2 + c*x^4])/( 
5*x^(7/2)) + (2*c*((-2*Sqrt[b*x^2 + c*x^4])/(b*x^(3/2)) + (2*c*x*Sqrt[b + 
c*x^2]*(-((-((Sqrt[x]*Sqrt[b + c*x^2])/(Sqrt[b] + Sqrt[c]*x)) + (b^(1/4)*( 
Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticE[2 
*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(c^(1/4)*Sqrt[b + c*x^2]))/Sqrt[ 
c]) + (b^(1/4)*(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])/(2*c^(3/4)*Sqrt[ 
b + c*x^2])))/(b*Sqrt[b*x^2 + c*x^4])))/5))/3
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 834

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, S 
imp[1/q   Int[1/Sqrt[a + b*x^4], x], x] - Simp[1/q   Int[(1 - q*x^2)/Sqrt[a 
 + b*x^4], x], 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 1510

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = 
 Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* 
(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E 
llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e 
}, x] && PosQ[c/a]
 

Maple [A] (verified)

Time = 1.53 (sec) , antiderivative size = 239, normalized size of antiderivative = 0.75

method	result	size
default	\(\frac {2 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (12 \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticE}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b \,c^{2} x^{4}-6 \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b \,c^{2} x^{4}-12 c^{3} x^{6}-23 b \,c^{2} x^{4}-16 b^{2} c \,x^{2}-5 b^{3}\right )}{45 x^{\frac {15}{2}} \left (c \,x^{2}+b \right )^{2} b}\)	\(239\)
risch	\(-\frac {2 \left (12 c^{2} x^{4}+11 b c \,x^{2}+5 b^{2}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{45 x^{\frac {11}{2}} b}+\frac {4 c^{2} \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \left (-\frac {2 \sqrt {-b c}\, \operatorname {EllipticE}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c}+\frac {\sqrt {-b c}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {x \left (c \,x^{2}+b \right )}}{15 b \sqrt {c \,x^{3}+b x}\, x^{\frac {3}{2}} \left (c \,x^{2}+b \right )}\)	\(241\)

Input:

int((c*x^4+b*x^2)^(3/2)/x^(17/2),x,method=_RETURNVERBOSE)
 

Output:

2/45*(c*x^4+b*x^2)^(3/2)/x^(15/2)/(c*x^2+b)^2*(12*((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)*EllipticE(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^( 
1/2))*b*c^2*x^4-6*((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))*b*c^2*x^4-12*c^3*x^6-23*b* 
c^2*x^4-16*b^2*c*x^2-5*b^3)/b
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.22 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{17/2}} \, dx=-\frac {2 \, {\left (12 \, c^{\frac {5}{2}} x^{6} {\rm weierstrassZeta}\left (-\frac {4 \, b}{c}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right )\right ) + {\left (12 \, c^{2} x^{4} + 11 \, b c x^{2} + 5 \, b^{2}\right )} \sqrt {c x^{4} + b x^{2}} \sqrt {x}\right )}}{45 \, b x^{6}} \] Input:

integrate((c*x^4+b*x^2)^(3/2)/x^(17/2),x, algorithm="fricas")
 

Output:

-2/45*(12*c^(5/2)*x^6*weierstrassZeta(-4*b/c, 0, weierstrassPInverse(-4*b/ 
c, 0, x)) + (12*c^2*x^4 + 11*b*c*x^2 + 5*b^2)*sqrt(c*x^4 + b*x^2)*sqrt(x)) 
/(b*x^6)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{17/2}} \, dx=\text {Timed out} \] Input:

integrate((c*x**4+b*x**2)**(3/2)/x**(17/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{17/2}} \, dx=\int { \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{\frac {17}{2}}} \,d x } \] Input:

integrate((c*x^4+b*x^2)^(3/2)/x^(17/2),x, algorithm="maxima")
 

Output:

integrate((c*x^4 + b*x^2)^(3/2)/x^(17/2), x)
 

Giac [F]

\[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{17/2}} \, dx=\int { \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{\frac {17}{2}}} \,d x } \] Input:

integrate((c*x^4+b*x^2)^(3/2)/x^(17/2),x, algorithm="giac")
 

Output:

integrate((c*x^4 + b*x^2)^(3/2)/x^(17/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{17/2}} \, dx=\int \frac {{\left (c\,x^4+b\,x^2\right )}^{3/2}}{x^{17/2}} \,d x \] Input:

int((b*x^2 + c*x^4)^(3/2)/x^(17/2),x)
                                                                                    
                                                                                    
 

Output:

int((b*x^2 + c*x^4)^(3/2)/x^(17/2), x)
 

Reduce [F]

\[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{17/2}} \, dx=\frac {-\frac {2 \sqrt {c \,x^{2}+b}\, b}{21}-\frac {2 \sqrt {c \,x^{2}+b}\, c \,x^{2}}{3}+\frac {4 \sqrt {x}\, \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c \,x^{8}+b \,x^{6}}d x \right ) b^{2} x^{4}}{7}}{\sqrt {x}\, x^{4}} \] Input:

int((c*x^4+b*x^2)^(3/2)/x^(17/2),x)
 

Output:

(2*( - sqrt(b + c*x**2)*b - 7*sqrt(b + c*x**2)*c*x**2 + 6*sqrt(x)*int((sqr 
t(x)*sqrt(b + c*x**2))/(b*x**6 + c*x**8),x)*b**2*x**4))/(21*sqrt(x)*x**4)