\(\int \frac {(b x^2+c x^4)^{3/2}}{x^{14}} \, dx\) [191]

Optimal result

Integrand size = 19, antiderivative size = 165 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{14}} \, dx=-\frac {3 c \sqrt {b x^2+c x^4}}{80 x^9}-\frac {c^2 \sqrt {b x^2+c x^4}}{160 b x^7}+\frac {c^3 \sqrt {b x^2+c x^4}}{128 b^2 x^5}-\frac {3 c^4 \sqrt {b x^2+c x^4}}{256 b^3 x^3}-\frac {\left (b x^2+c x^4\right )^{3/2}}{10 x^{13}}+\frac {3 c^5 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{256 b^{7/2}} \] Output:

-3/80*c*(c*x^4+b*x^2)^(1/2)/x^9-1/160*c^2*(c*x^4+b*x^2)^(1/2)/b/x^7+1/128* 
c^3*(c*x^4+b*x^2)^(1/2)/b^2/x^5-3/256*c^4*(c*x^4+b*x^2)^(1/2)/b^3/x^3-1/10 
*(c*x^4+b*x^2)^(3/2)/x^13+3/256*c^5*arctanh(b^(1/2)*x/(c*x^4+b*x^2)^(1/2)) 
/b^(7/2)
 

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.76 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{14}} \, dx=\frac {\sqrt {b+c x^2} \left (-\sqrt {b} \sqrt {b+c x^2} \left (128 b^4+176 b^3 c x^2+8 b^2 c^2 x^4-10 b c^3 x^6+15 c^4 x^8\right )+15 c^5 x^{10} \text {arctanh}\left (\frac {\sqrt {b+c x^2}}{\sqrt {b}}\right )\right )}{1280 b^{7/2} x^9 \sqrt {x^2 \left (b+c x^2\right )}} \] Input:

Integrate[(b*x^2 + c*x^4)^(3/2)/x^14,x]
 

Output:

(Sqrt[b + c*x^2]*(-(Sqrt[b]*Sqrt[b + c*x^2]*(128*b^4 + 176*b^3*c*x^2 + 8*b 
^2*c^2*x^4 - 10*b*c^3*x^6 + 15*c^4*x^8)) + 15*c^5*x^10*ArcTanh[Sqrt[b + c* 
x^2]/Sqrt[b]]))/(1280*b^(7/2)*x^9*Sqrt[x^2*(b + c*x^2)])
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {1425, 1425, 1430, 1430, 1430, 1400, 219}

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.

\(\displaystyle \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{14}} \, dx\)

\(\Big \downarrow \) 1425

\(\displaystyle \frac {3}{10} c \int \frac {\sqrt {c x^4+b x^2}}{x^{10}}dx-\frac {\left (b x^2+c x^4\right )^{3/2}}{10 x^{13}}\)

\(\Big \downarrow \) 1425

\(\displaystyle \frac {3}{10} c \left (\frac {1}{8} c \int \frac {1}{x^6 \sqrt {c x^4+b x^2}}dx-\frac {\sqrt {b x^2+c x^4}}{8 x^9}\right )-\frac {\left (b x^2+c x^4\right )^{3/2}}{10 x^{13}}\)

\(\Big \downarrow \) 1430

\(\displaystyle \frac {3}{10} c \left (\frac {1}{8} c \left (-\frac {5 c \int \frac {1}{x^4 \sqrt {c x^4+b x^2}}dx}{6 b}-\frac {\sqrt {b x^2+c x^4}}{6 b x^7}\right )-\frac {\sqrt {b x^2+c x^4}}{8 x^9}\right )-\frac {\left (b x^2+c x^4\right )^{3/2}}{10 x^{13}}\)

\(\Big \downarrow \) 1430

\(\displaystyle \frac {3}{10} c \left (\frac {1}{8} c \left (-\frac {5 c \left (-\frac {3 c \int \frac {1}{x^2 \sqrt {c x^4+b x^2}}dx}{4 b}-\frac {\sqrt {b x^2+c x^4}}{4 b x^5}\right )}{6 b}-\frac {\sqrt {b x^2+c x^4}}{6 b x^7}\right )-\frac {\sqrt {b x^2+c x^4}}{8 x^9}\right )-\frac {\left (b x^2+c x^4\right )^{3/2}}{10 x^{13}}\)

\(\Big \downarrow \) 1430

\(\displaystyle \frac {3}{10} c \left (\frac {1}{8} c \left (-\frac {5 c \left (-\frac {3 c \left (-\frac {c \int \frac {1}{\sqrt {c x^4+b x^2}}dx}{2 b}-\frac {\sqrt {b x^2+c x^4}}{2 b x^3}\right )}{4 b}-\frac {\sqrt {b x^2+c x^4}}{4 b x^5}\right )}{6 b}-\frac {\sqrt {b x^2+c x^4}}{6 b x^7}\right )-\frac {\sqrt {b x^2+c x^4}}{8 x^9}\right )-\frac {\left (b x^2+c x^4\right )^{3/2}}{10 x^{13}}\)

\(\Big \downarrow \) 1400

\(\displaystyle \frac {3}{10} c \left (\frac {1}{8} c \left (-\frac {5 c \left (-\frac {3 c \left (\frac {c \int \frac {1}{1-\frac {b x^2}{c x^4+b x^2}}d\frac {x}{\sqrt {c x^4+b x^2}}}{2 b}-\frac {\sqrt {b x^2+c x^4}}{2 b x^3}\right )}{4 b}-\frac {\sqrt {b x^2+c x^4}}{4 b x^5}\right )}{6 b}-\frac {\sqrt {b x^2+c x^4}}{6 b x^7}\right )-\frac {\sqrt {b x^2+c x^4}}{8 x^9}\right )-\frac {\left (b x^2+c x^4\right )^{3/2}}{10 x^{13}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {3}{10} c \left (\frac {1}{8} c \left (-\frac {5 c \left (-\frac {3 c \left (\frac {c \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{2 b^{3/2}}-\frac {\sqrt {b x^2+c x^4}}{2 b x^3}\right )}{4 b}-\frac {\sqrt {b x^2+c x^4}}{4 b x^5}\right )}{6 b}-\frac {\sqrt {b x^2+c x^4}}{6 b x^7}\right )-\frac {\sqrt {b x^2+c x^4}}{8 x^9}\right )-\frac {\left (b x^2+c x^4\right )^{3/2}}{10 x^{13}}\)

Input:

Int[(b*x^2 + c*x^4)^(3/2)/x^14,x]
 

Output:

-1/10*(b*x^2 + c*x^4)^(3/2)/x^13 + (3*c*(-1/8*Sqrt[b*x^2 + c*x^4]/x^9 + (c 
*(-1/6*Sqrt[b*x^2 + c*x^4]/(b*x^7) - (5*c*(-1/4*Sqrt[b*x^2 + c*x^4]/(b*x^5 
) - (3*c*(-1/2*Sqrt[b*x^2 + c*x^4]/(b*x^3) + (c*ArcTanh[(Sqrt[b]*x)/Sqrt[b 
*x^2 + c*x^4]])/(2*b^(3/2))))/(4*b)))/(6*b)))/8))/10
 

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1400

Int[1/Sqrt[(b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> -Subst[Int[1/(1 - b*x 
^2), x], x, x/Sqrt[b*x^2 + c*x^4]] /; FreeQ[{b, c}, x]
 

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 
]
 

Maple [A] (verified)

Time = 1.83 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.74

method	result	size
risch	\(-\frac {\left (15 c^{4} x^{8}-10 b \,c^{3} x^{6}+8 b^{2} c^{2} x^{4}+176 x^{2} b^{3} c +128 b^{4}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{1280 x^{11} b^{3}}+\frac {3 c^{5} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{256 b^{\frac {7}{2}} x \sqrt {c \,x^{2}+b}}\)	\(122\)
default	\(\frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (15 b^{\frac {3}{2}} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) c^{5} x^{10}-5 \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{5} x^{10}+5 \left (c \,x^{2}+b \right )^{\frac {5}{2}} c^{4} x^{8}-15 \sqrt {c \,x^{2}+b}\, b \,c^{5} x^{10}+10 \left (c \,x^{2}+b \right )^{\frac {5}{2}} b \,c^{3} x^{6}-40 \left (c \,x^{2}+b \right )^{\frac {5}{2}} b^{2} c^{2} x^{4}+80 \left (c \,x^{2}+b \right )^{\frac {5}{2}} b^{3} c \,x^{2}-128 \left (c \,x^{2}+b \right )^{\frac {5}{2}} b^{4}\right )}{1280 x^{13} \left (c \,x^{2}+b \right )^{\frac {3}{2}} b^{5}}\)	\(186\)

Input:

int((c*x^4+b*x^2)^(3/2)/x^14,x,method=_RETURNVERBOSE)
 

Output:

-1/1280*(15*c^4*x^8-10*b*c^3*x^6+8*b^2*c^2*x^4+176*b^3*c*x^2+128*b^4)/x^11 
/b^3*(x^2*(c*x^2+b))^(1/2)+3/256*c^5/b^(7/2)*ln((2*b+2*b^(1/2)*(c*x^2+b)^( 
1/2))/x)*(x^2*(c*x^2+b))^(1/2)/x/(c*x^2+b)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.36 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{14}} \, dx=\left [\frac {15 \, \sqrt {b} c^{5} x^{11} \log \left (-\frac {c x^{3} + 2 \, b x + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {b}}{x^{3}}\right ) - 2 \, {\left (15 \, b c^{4} x^{8} - 10 \, b^{2} c^{3} x^{6} + 8 \, b^{3} c^{2} x^{4} + 176 \, b^{4} c x^{2} + 128 \, b^{5}\right )} \sqrt {c x^{4} + b x^{2}}}{2560 \, b^{4} x^{11}}, -\frac {15 \, \sqrt {-b} c^{5} x^{11} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-b}}{b x}\right ) + {\left (15 \, b c^{4} x^{8} - 10 \, b^{2} c^{3} x^{6} + 8 \, b^{3} c^{2} x^{4} + 176 \, b^{4} c x^{2} + 128 \, b^{5}\right )} \sqrt {c x^{4} + b x^{2}}}{1280 \, b^{4} x^{11}}\right ] \] Input:

integrate((c*x^4+b*x^2)^(3/2)/x^14,x, algorithm="fricas")
 

Output:

[1/2560*(15*sqrt(b)*c^5*x^11*log(-(c*x^3 + 2*b*x + 2*sqrt(c*x^4 + b*x^2)*s 
qrt(b))/x^3) - 2*(15*b*c^4*x^8 - 10*b^2*c^3*x^6 + 8*b^3*c^2*x^4 + 176*b^4* 
c*x^2 + 128*b^5)*sqrt(c*x^4 + b*x^2))/(b^4*x^11), -1/1280*(15*sqrt(-b)*c^5 
*x^11*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-b)/(b*x)) + (15*b*c^4*x^8 - 10*b^2* 
c^3*x^6 + 8*b^3*c^2*x^4 + 176*b^4*c*x^2 + 128*b^5)*sqrt(c*x^4 + b*x^2))/(b 
^4*x^11)]
 

Sympy [F]

\[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{14}} \, dx=\int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}{x^{14}}\, dx \] Input:

integrate((c*x**4+b*x**2)**(3/2)/x**14,x)
 

Output:

Integral((x**2*(b + c*x**2))**(3/2)/x**14, x)
 

Maxima [F]

\[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{14}} \, dx=\int { \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{14}} \,d x } \] Input:

integrate((c*x^4+b*x^2)^(3/2)/x^14,x, algorithm="maxima")
 

Output:

integrate((c*x^4 + b*x^2)^(3/2)/x^14, x)
 

Giac [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.73 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{14}} \, dx=-\frac {1}{1280} \, c^{5} {\left (\frac {15 \, \arctan \left (\frac {\sqrt {c x^{2} + b}}{\sqrt {-b}}\right ) \mathrm {sgn}\left (x\right )}{\sqrt {-b} b^{3}} + \frac {15 \, {\left (c x^{2} + b\right )}^{\frac {9}{2}} \mathrm {sgn}\left (x\right ) - 70 \, {\left (c x^{2} + b\right )}^{\frac {7}{2}} b \mathrm {sgn}\left (x\right ) + 128 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} b^{2} \mathrm {sgn}\left (x\right ) + 70 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} b^{3} \mathrm {sgn}\left (x\right ) - 15 \, \sqrt {c x^{2} + b} b^{4} \mathrm {sgn}\left (x\right )}{b^{3} c^{5} x^{10}}\right )} \] Input:

integrate((c*x^4+b*x^2)^(3/2)/x^14,x, algorithm="giac")
 

Output:

-1/1280*c^5*(15*arctan(sqrt(c*x^2 + b)/sqrt(-b))*sgn(x)/(sqrt(-b)*b^3) + ( 
15*(c*x^2 + b)^(9/2)*sgn(x) - 70*(c*x^2 + b)^(7/2)*b*sgn(x) + 128*(c*x^2 + 
 b)^(5/2)*b^2*sgn(x) + 70*(c*x^2 + b)^(3/2)*b^3*sgn(x) - 15*sqrt(c*x^2 + b 
)*b^4*sgn(x))/(b^3*c^5*x^10))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{14}} \, dx=\int \frac {{\left (c\,x^4+b\,x^2\right )}^{3/2}}{x^{14}} \,d x \] Input:

int((b*x^2 + c*x^4)^(3/2)/x^14,x)
 

Output:

int((b*x^2 + c*x^4)^(3/2)/x^14, x)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.96 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{14}} \, dx=\frac {-128 \sqrt {c \,x^{2}+b}\, b^{5}-176 \sqrt {c \,x^{2}+b}\, b^{4} c \,x^{2}-8 \sqrt {c \,x^{2}+b}\, b^{3} c^{2} x^{4}+10 \sqrt {c \,x^{2}+b}\, b^{2} c^{3} x^{6}-15 \sqrt {c \,x^{2}+b}\, b \,c^{4} x^{8}-15 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}-\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) c^{5} x^{10}+15 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) c^{5} x^{10}}{1280 b^{4} x^{10}} \] Input:

int((c*x^4+b*x^2)^(3/2)/x^14,x)
 

Output:

( - 128*sqrt(b + c*x**2)*b**5 - 176*sqrt(b + c*x**2)*b**4*c*x**2 - 8*sqrt( 
b + c*x**2)*b**3*c**2*x**4 + 10*sqrt(b + c*x**2)*b**2*c**3*x**6 - 15*sqrt( 
b + c*x**2)*b*c**4*x**8 - 15*sqrt(b)*log((sqrt(b + c*x**2) - sqrt(b) + sqr 
t(c)*x)/sqrt(b))*c**5*x**10 + 15*sqrt(b)*log((sqrt(b + c*x**2) + sqrt(b) + 
 sqrt(c)*x)/sqrt(b))*c**5*x**10)/(1280*b**4*x**10)