\(\int \frac {(c (a+b x^2)^3)^{3/2}}{x^4} \, dx\) [19]

Optimal result

Integrand size = 19, antiderivative size = 236 \[ \int \frac {\left (c \left (a+b x^2\right )^3\right )^{3/2}}{x^4} \, dx=-\frac {a^4 c \sqrt {c \left (a+b x^2\right )^3}}{3 x^3 \left (a+b x^2\right )}-\frac {13 a^3 b c \sqrt {c \left (a+b x^2\right )^3}}{3 x \left (a+b x^2\right )}+\frac {55 a^2 b^2 c x \sqrt {c \left (a+b x^2\right )^3}}{16 \left (a+b x^2\right )}+\frac {25 a b^3 c x^3 \sqrt {c \left (a+b x^2\right )^3}}{24 \left (a+b x^2\right )}+\frac {b^4 c x^5 \sqrt {c \left (a+b x^2\right )^3}}{6 \left (a+b x^2\right )}+\frac {105 a^{3/2} b^{3/2} c \sqrt {c \left (a+b x^2\right )^3} \text {arcsinh}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{16 \left (1+\frac {b x^2}{a}\right )^{3/2}} \] Output:

-1/3*a^4*c*(c*(b*x^2+a)^3)^(1/2)/x^3/(b*x^2+a)-13/3*a^3*b*c*(c*(b*x^2+a)^3 
)^(1/2)/x/(b*x^2+a)+55*a^2*b^2*c*x*(c*(b*x^2+a)^3)^(1/2)/(16*b*x^2+16*a)+2 
5*a*b^3*c*x^3*(c*(b*x^2+a)^3)^(1/2)/(24*b*x^2+24*a)+b^4*c*x^5*(c*(b*x^2+a) 
^3)^(1/2)/(6*b*x^2+6*a)+105/16*a^(3/2)*b^(3/2)*c*(c*(b*x^2+a)^3)^(1/2)*arc 
sinh(b^(1/2)*x/a^(1/2))/(1+b*x^2/a)^(3/2)
 

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.52 \[ \int \frac {\left (c \left (a+b x^2\right )^3\right )^{3/2}}{x^4} \, dx=-\frac {\left (c \left (a+b x^2\right )^3\right )^{3/2} \left (\sqrt {a+b x^2} \left (16 a^4+208 a^3 b x^2-165 a^2 b^2 x^4-50 a b^3 x^6-8 b^4 x^8\right )+315 a^3 b^{3/2} x^3 \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )\right )}{48 x^3 \left (a+b x^2\right )^{9/2}} \] Input:

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

Output:

-1/48*((c*(a + b*x^2)^3)^(3/2)*(Sqrt[a + b*x^2]*(16*a^4 + 208*a^3*b*x^2 - 
165*a^2*b^2*x^4 - 50*a*b^3*x^6 - 8*b^4*x^8) + 315*a^3*b^(3/2)*x^3*Log[-(Sq 
rt[b]*x) + Sqrt[a + b*x^2]]))/(x^3*(a + b*x^2)^(9/2))
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.78, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {2045, 247, 247, 211, 211, 211, 222}

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.

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

\(\Big \downarrow \) 2045

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

\(\Big \downarrow \) 247

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

\(\Big \downarrow \) 247

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

\(\Big \downarrow \) 211

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

\(\Big \downarrow \) 211

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

\(\Big \downarrow \) 211

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

\(\Big \downarrow \) 222

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

Input:

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

Output:

(a^3*c*Sqrt[c*(a + b*x^2)^3]*(-1/3*(1 + (b*x^2)/a)^(9/2)/x^3 + (3*b*(-((1 
+ (b*x^2)/a)^(7/2)/x) + (7*b*((x*(1 + (b*x^2)/a)^(5/2))/6 + (5*((x*(1 + (b 
*x^2)/a)^(3/2))/4 + (3*((x*Sqrt[1 + (b*x^2)/a])/2 + (Sqrt[a]*ArcSinh[(Sqrt 
[b]*x)/Sqrt[a]])/(2*Sqrt[b])))/4))/6))/a))/a))/(1 + (b*x^2)/a)^(3/2)
 

Defintions of rubi rules used

rule 211

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 
)), x] + Simp[2*a*(p/(2*p + 1))   Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ 
{a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
 

rule 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt 
[a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
 

rule 247

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ 
(m + 1)*((a + b*x^2)^p/(c*(m + 1))), x] - Simp[2*b*(p/(c^2*(m + 1)))   Int[ 
(c*x)^(m + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 
0] && LtQ[m, -1] &&  !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomialQ[a, b, c, 2, 
m, p, x]
 

rule 2045

Int[(u_.)*((c_.)*((a_) + (b_.)*(x_)^(n_.))^(q_))^(p_), x_Symbol] :> Simp[Si 
mp[(c*(a + b*x^n)^q)^p/(1 + b*(x^n/a))^(p*q)]   Int[u*(1 + b*(x^n/a))^(p*q) 
, x], x] /; FreeQ[{a, b, c, n, p, q}, x] &&  !GeQ[a, 0]
 

Maple [A] (verified)

Time = 3.19 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.61

method	result	size
risch	\(-\frac {\left (-8 b^{4} x^{8}-50 a \,b^{3} x^{6}-165 a^{2} b^{2} x^{4}+208 a^{3} b \,x^{2}+16 a^{4}\right ) c \sqrt {c \left (b \,x^{2}+a \right )^{3}}}{48 \left (b \,x^{2}+a \right ) x^{3}}+\frac {105 a^{3} b^{2} \ln \left (\frac {b c x}{\sqrt {b c}}+\sqrt {x^{2} b c +a c}\right ) c \sqrt {c \left (b \,x^{2}+a \right )^{3}}\, \sqrt {\left (b \,x^{2}+a \right ) c}}{16 \sqrt {b c}\, \left (b \,x^{2}+a \right )^{2}}\)	\(143\)
default	\(-\frac {{\left (c \left (b \,x^{2}+a \right )^{3}\right )}^{\frac {3}{2}} \left (-8 \left (x^{2} b c +a c \right )^{\frac {5}{2}} \sqrt {b c}\, b^{2} x^{4}-210 \left (x^{2} b c +a c \right )^{\frac {3}{2}} \sqrt {b c}\, a \,b^{2} c \,x^{4}-315 \sqrt {x^{2} b c +a c}\, \sqrt {b c}\, a^{2} b^{2} c^{2} x^{4}-315 \ln \left (\frac {b c x +\sqrt {x^{2} b c +a c}\, \sqrt {b c}}{\sqrt {b c}}\right ) a^{3} b^{2} c^{3} x^{3}+176 \left (x^{2} b c +a c \right )^{\frac {5}{2}} \sqrt {b c}\, a b \,x^{2}+16 \left (x^{2} b c +a c \right )^{\frac {5}{2}} \sqrt {b c}\, a^{2}\right )}{48 \left (b \,x^{2}+a \right )^{3} {\left (\left (b \,x^{2}+a \right ) c \right )}^{\frac {3}{2}} x^{3} c \sqrt {b c}}\)	\(221\)

Input:

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

Output:

-1/48/(b*x^2+a)*(-8*b^4*x^8-50*a*b^3*x^6-165*a^2*b^2*x^4+208*a^3*b*x^2+16* 
a^4)/x^3*c*(c*(b*x^2+a)^3)^(1/2)+105/16*a^3*b^2*ln(b*c*x/(b*c)^(1/2)+(b*c* 
x^2+a*c)^(1/2))/(b*c)^(1/2)*c/(b*x^2+a)^2*(c*(b*x^2+a)^3)^(1/2)*((b*x^2+a) 
*c)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 410, normalized size of antiderivative = 1.74 \[ \int \frac {\left (c \left (a+b x^2\right )^3\right )^{3/2}}{x^4} \, dx=\left [\frac {315 \, {\left (a^{3} b^{2} c x^{5} + a^{4} b c x^{3}\right )} \sqrt {b c} \log \left (-\frac {2 \, b^{2} c x^{4} + 3 \, a b c x^{2} + a^{2} c + 2 \, \sqrt {b^{3} c x^{6} + 3 \, a b^{2} c x^{4} + 3 \, a^{2} b c x^{2} + a^{3} c} \sqrt {b c} x}{b x^{2} + a}\right ) + 2 \, {\left (8 \, b^{4} c x^{8} + 50 \, a b^{3} c x^{6} + 165 \, a^{2} b^{2} c x^{4} - 208 \, a^{3} b c x^{2} - 16 \, a^{4} c\right )} \sqrt {b^{3} c x^{6} + 3 \, a b^{2} c x^{4} + 3 \, a^{2} b c x^{2} + a^{3} c}}{96 \, {\left (b x^{5} + a x^{3}\right )}}, -\frac {315 \, {\left (a^{3} b^{2} c x^{5} + a^{4} b c x^{3}\right )} \sqrt {-b c} \arctan \left (\frac {\sqrt {b^{3} c x^{6} + 3 \, a b^{2} c x^{4} + 3 \, a^{2} b c x^{2} + a^{3} c} \sqrt {-b c} x}{b^{2} c x^{4} + 2 \, a b c x^{2} + a^{2} c}\right ) - {\left (8 \, b^{4} c x^{8} + 50 \, a b^{3} c x^{6} + 165 \, a^{2} b^{2} c x^{4} - 208 \, a^{3} b c x^{2} - 16 \, a^{4} c\right )} \sqrt {b^{3} c x^{6} + 3 \, a b^{2} c x^{4} + 3 \, a^{2} b c x^{2} + a^{3} c}}{48 \, {\left (b x^{5} + a x^{3}\right )}}\right ] \] Input:

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

Output:

[1/96*(315*(a^3*b^2*c*x^5 + a^4*b*c*x^3)*sqrt(b*c)*log(-(2*b^2*c*x^4 + 3*a 
*b*c*x^2 + a^2*c + 2*sqrt(b^3*c*x^6 + 3*a*b^2*c*x^4 + 3*a^2*b*c*x^2 + a^3* 
c)*sqrt(b*c)*x)/(b*x^2 + a)) + 2*(8*b^4*c*x^8 + 50*a*b^3*c*x^6 + 165*a^2*b 
^2*c*x^4 - 208*a^3*b*c*x^2 - 16*a^4*c)*sqrt(b^3*c*x^6 + 3*a*b^2*c*x^4 + 3* 
a^2*b*c*x^2 + a^3*c))/(b*x^5 + a*x^3), -1/48*(315*(a^3*b^2*c*x^5 + a^4*b*c 
*x^3)*sqrt(-b*c)*arctan(sqrt(b^3*c*x^6 + 3*a*b^2*c*x^4 + 3*a^2*b*c*x^2 + a 
^3*c)*sqrt(-b*c)*x/(b^2*c*x^4 + 2*a*b*c*x^2 + a^2*c)) - (8*b^4*c*x^8 + 50* 
a*b^3*c*x^6 + 165*a^2*b^2*c*x^4 - 208*a^3*b*c*x^2 - 16*a^4*c)*sqrt(b^3*c*x 
^6 + 3*a*b^2*c*x^4 + 3*a^2*b*c*x^2 + a^3*c))/(b*x^5 + a*x^3)]
 

Sympy [F]

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

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

Output:

Integral((c*(a + b*x**2)**3)**(3/2)/x**4, x)
 

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.11 \[ \int \frac {\left (c \left (a+b x^2\right )^3\right )^{3/2}}{x^4} \, dx=-\frac {1}{96} \, {\left (315 \, \sqrt {b c} a^{3} b \log \left ({\left (\sqrt {b c} x - \sqrt {b c x^{2} + a c}\right )}^{2}\right ) \mathrm {sgn}\left (b x^{2} + a\right ) - 2 \, {\left (165 \, a^{2} b^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 2 \, {\left (4 \, b^{4} x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 25 \, a b^{3} \mathrm {sgn}\left (b x^{2} + a\right )\right )} x^{2}\right )} \sqrt {b c x^{2} + a c} x - \frac {64 \, {\left (15 \, \sqrt {b c} {\left (\sqrt {b c} x - \sqrt {b c x^{2} + a c}\right )}^{4} a^{4} b c \mathrm {sgn}\left (b x^{2} + a\right ) - 24 \, \sqrt {b c} {\left (\sqrt {b c} x - \sqrt {b c x^{2} + a c}\right )}^{2} a^{5} b c^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 13 \, \sqrt {b c} a^{6} b c^{3} \mathrm {sgn}\left (b x^{2} + a\right )\right )}}{{\left ({\left (\sqrt {b c} x - \sqrt {b c x^{2} + a c}\right )}^{2} - a c\right )}^{3}}\right )} c \] Input:

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

Output:

-1/96*(315*sqrt(b*c)*a^3*b*log((sqrt(b*c)*x - sqrt(b*c*x^2 + a*c))^2)*sgn( 
b*x^2 + a) - 2*(165*a^2*b^2*sgn(b*x^2 + a) + 2*(4*b^4*x^2*sgn(b*x^2 + a) + 
 25*a*b^3*sgn(b*x^2 + a))*x^2)*sqrt(b*c*x^2 + a*c)*x - 64*(15*sqrt(b*c)*(s 
qrt(b*c)*x - sqrt(b*c*x^2 + a*c))^4*a^4*b*c*sgn(b*x^2 + a) - 24*sqrt(b*c)* 
(sqrt(b*c)*x - sqrt(b*c*x^2 + a*c))^2*a^5*b*c^2*sgn(b*x^2 + a) + 13*sqrt(b 
*c)*a^6*b*c^3*sgn(b*x^2 + a))/((sqrt(b*c)*x - sqrt(b*c*x^2 + a*c))^2 - a*c 
)^3)*c
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.56 \[ \int \frac {\left (c \left (a+b x^2\right )^3\right )^{3/2}}{x^4} \, dx=\frac {\sqrt {c}\, c \left (-128 \sqrt {b \,x^{2}+a}\, a^{4}-1664 \sqrt {b \,x^{2}+a}\, a^{3} b \,x^{2}+1320 \sqrt {b \,x^{2}+a}\, a^{2} b^{2} x^{4}+400 \sqrt {b \,x^{2}+a}\, a \,b^{3} x^{6}+64 \sqrt {b \,x^{2}+a}\, b^{4} x^{8}+2520 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} b \,x^{3}+567 \sqrt {b}\, a^{3} b \,x^{3}\right )}{384 x^{3}} \] Input:

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

Output:

(sqrt(c)*c*( - 128*sqrt(a + b*x**2)*a**4 - 1664*sqrt(a + b*x**2)*a**3*b*x* 
*2 + 1320*sqrt(a + b*x**2)*a**2*b**2*x**4 + 400*sqrt(a + b*x**2)*a*b**3*x* 
*6 + 64*sqrt(a + b*x**2)*b**4*x**8 + 2520*sqrt(b)*log((sqrt(a + b*x**2) + 
sqrt(b)*x)/sqrt(a))*a**3*b*x**3 + 567*sqrt(b)*a**3*b*x**3))/(384*x**3)