\(\int \frac {(A+B x^2) (b x^2+c x^4)^{3/2}}{x^7} \, dx\) [177]

Optimal result

Integrand size = 26, antiderivative size = 133 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^7} \, dx=\frac {c (3 b B+2 A c) \sqrt {b x^2+c x^4}}{6 b}-\frac {(3 b B+2 A c) \sqrt {b x^2+c x^4}}{3 x^2}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{3 b x^8}+\frac {1}{2} \sqrt {c} (3 b B+2 A c) \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right ) \] Output:

1/6*c*(2*A*c+3*B*b)*(c*x^4+b*x^2)^(1/2)/b-1/3*(2*A*c+3*B*b)*(c*x^4+b*x^2)^ 
(1/2)/x^2-1/3*A*(c*x^4+b*x^2)^(5/2)/b/x^8+1/2*c^(1/2)*(2*A*c+3*B*b)*arctan 
h(c^(1/2)*x^2/(c*x^4+b*x^2)^(1/2))
 

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.90 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^7} \, dx=\frac {\sqrt {x^2 \left (b+c x^2\right )} \left (\sqrt {b+c x^2} \left (-6 b B x^2+3 B c x^4-2 A \left (b+4 c x^2\right )\right )+6 \sqrt {c} (3 b B+2 A c) x^3 \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {b}+\sqrt {b+c x^2}}\right )\right )}{6 x^4 \sqrt {b+c x^2}} \] Input:

Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^7,x]
 

Output:

(Sqrt[x^2*(b + c*x^2)]*(Sqrt[b + c*x^2]*(-6*b*B*x^2 + 3*B*c*x^4 - 2*A*(b + 
 4*c*x^2)) + 6*Sqrt[c]*(3*b*B + 2*A*c)*x^3*ArcTanh[(Sqrt[c]*x)/(-Sqrt[b] + 
 Sqrt[b + c*x^2])]))/(6*x^4*Sqrt[b + c*x^2])
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.90, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {1940, 1220, 1125, 25, 27, 1160, 1091, 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.

\(\displaystyle \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^7} \, dx\)

\(\Big \downarrow \) 1940

\(\displaystyle \frac {1}{2} \int \frac {\left (B x^2+A\right ) \left (c x^4+b x^2\right )^{3/2}}{x^8}dx^2\)

\(\Big \downarrow \) 1220

\(\displaystyle \frac {1}{2} \left (\frac {(2 A c+3 b B) \int \frac {\left (c x^4+b x^2\right )^{3/2}}{x^6}dx^2}{3 b}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{3 b x^8}\right )\)

\(\Big \downarrow \) 1125

\(\displaystyle \frac {1}{2} \left (\frac {(2 A c+3 b B) \left (-\int -\frac {c \left (c x^2+2 b\right )}{\sqrt {c x^4+b x^2}}dx^2-\frac {2 b \sqrt {b x^2+c x^4}}{x^2}\right )}{3 b}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{3 b x^8}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{2} \left (\frac {(2 A c+3 b B) \left (\int \frac {c \left (c x^2+2 b\right )}{\sqrt {c x^4+b x^2}}dx^2-\frac {2 b \sqrt {b x^2+c x^4}}{x^2}\right )}{3 b}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{3 b x^8}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \left (\frac {(2 A c+3 b B) \left (c \int \frac {c x^2+2 b}{\sqrt {c x^4+b x^2}}dx^2-\frac {2 b \sqrt {b x^2+c x^4}}{x^2}\right )}{3 b}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{3 b x^8}\right )\)

\(\Big \downarrow \) 1160

\(\displaystyle \frac {1}{2} \left (\frac {(2 A c+3 b B) \left (c \left (\frac {3}{2} b \int \frac {1}{\sqrt {c x^4+b x^2}}dx^2+\sqrt {b x^2+c x^4}\right )-\frac {2 b \sqrt {b x^2+c x^4}}{x^2}\right )}{3 b}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{3 b x^8}\right )\)

\(\Big \downarrow \) 1091

\(\displaystyle \frac {1}{2} \left (\frac {(2 A c+3 b B) \left (c \left (3 b \int \frac {1}{1-c x^4}d\frac {x^2}{\sqrt {c x^4+b x^2}}+\sqrt {b x^2+c x^4}\right )-\frac {2 b \sqrt {b x^2+c x^4}}{x^2}\right )}{3 b}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{3 b x^8}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{2} \left (\frac {(2 A c+3 b B) \left (c \left (\frac {3 b \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{\sqrt {c}}+\sqrt {b x^2+c x^4}\right )-\frac {2 b \sqrt {b x^2+c x^4}}{x^2}\right )}{3 b}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{3 b x^8}\right )\)

Input:

Int[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^7,x]
 

Output:

((-2*A*(b*x^2 + c*x^4)^(5/2))/(3*b*x^8) + ((3*b*B + 2*A*c)*((-2*b*Sqrt[b*x 
^2 + c*x^4])/x^2 + c*(Sqrt[b*x^2 + c*x^4] + (3*b*ArcTanh[(Sqrt[c]*x^2)/Sqr 
t[b*x^2 + c*x^4]])/Sqrt[c])))/(3*b))/2
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 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 1091

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

rule 1125

Int[((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_ 
Symbol] :> Simp[-2*e^(2*m + 3)*(Sqrt[a + b*x + c*x^2]/((-2*c*d + b*e)^(m + 
2)*(d + e*x))), x] - Simp[e^(2*m + 2)   Int[(1/Sqrt[a + b*x + c*x^2])*Expan 
dToSum[((-2*c*d + b*e)^(-m - 1) - ((-c)*d + b*e + c*e*x)^(-m - 1))/(d + e*x 
), x], x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] 
 && ILtQ[m, 0] && EqQ[m + p, -3/2]
 

rule 1160

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol 
] :> Simp[e*((a + b*x + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + Simp[(2*c*d - b 
*e)/(2*c)   Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] 
 && NeQ[p, -1]
 

rule 1220

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d*g - e*f)*(d + e*x)^m*((a + b*x + c*x 
^2)^(p + 1)/((2*c*d - b*e)*(m + p + 1))), x] + Simp[(m*(g*(c*d - b*e) + c*e 
*f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1))   Int[(d + e*x 
)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, 
 x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ((LtQ[m, -1] &&  !IGtQ[m + p + 1, 0 
]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0 
]
 

rule 1940

Int[(x_)^(m_.)*((b_.)*(x_)^(k_.) + (a_.)*(x_)^(j_))^(p_)*((c_) + (d_.)*(x_) 
^(n_))^(q_.), x_Symbol] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1) 
*(a*x^Simplify[j/n] + b*x^Simplify[k/n])^p*(c + d*x)^q, x], x, x^n], x] /; 
FreeQ[{a, b, c, d, j, k, m, n, p, q}, x] &&  !IntegerQ[p] && NeQ[k, j] && I 
ntegerQ[Simplify[j/n]] && IntegerQ[Simplify[k/n]] && IntegerQ[Simplify[(m + 
 1)/n]] && NeQ[n^2, 1]
 

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.76

method	result	size
risch	\(-\frac {\left (-3 B c \,x^{4}+8 A c \,x^{2}+6 B b \,x^{2}+2 A b \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{6 x^{4}}+\frac {\left (2 A c +3 B b \right ) \sqrt {c}\, \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{2 x \sqrt {c \,x^{2}+b}}\)	\(101\)
pseudoelliptic	\(\frac {3 x^{4} \left (3 B \sqrt {c}\, b +2 A \,c^{\frac {3}{2}}\right ) \ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right )+2 \left (3 B c \,x^{4}+2 \left (-4 A c -3 B b \right ) x^{2}-2 A b \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}-6 \left (\frac {3 B \sqrt {c}\, b}{2}+A \,c^{\frac {3}{2}}\right ) \ln \left (2\right ) x^{4}}{12 x^{4}}\)	\(118\)
default	\(\frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (4 A \,c^{\frac {5}{2}} \left (c \,x^{2}+b \right )^{\frac {3}{2}} x^{4}+6 A \,c^{\frac {5}{2}} \sqrt {c \,x^{2}+b}\, b \,x^{4}+6 B \,c^{\frac {3}{2}} \left (c \,x^{2}+b \right )^{\frac {3}{2}} b \,x^{4}-4 A \,c^{\frac {3}{2}} \left (c \,x^{2}+b \right )^{\frac {5}{2}} x^{2}+9 B \,c^{\frac {3}{2}} \sqrt {c \,x^{2}+b}\, b^{2} x^{4}-6 B \sqrt {c}\, \left (c \,x^{2}+b \right )^{\frac {5}{2}} b \,x^{2}+6 A \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b^{2} c^{2} x^{3}+9 B \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b^{3} c \,x^{3}-2 A \sqrt {c}\, \left (c \,x^{2}+b \right )^{\frac {5}{2}} b \right )}{6 x^{6} \left (c \,x^{2}+b \right )^{\frac {3}{2}} b^{2} \sqrt {c}}\)	\(219\)

Input:

int((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^7,x,method=_RETURNVERBOSE)
 

Output:

-1/6*(-3*B*c*x^4+8*A*c*x^2+6*B*b*x^2+2*A*b)/x^4*(x^2*(c*x^2+b))^(1/2)+1/2* 
(2*A*c+3*B*b)*c^(1/2)*ln(c^(1/2)*x+(c*x^2+b)^(1/2))*(x^2*(c*x^2+b))^(1/2)/ 
x/(c*x^2+b)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.42 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^7} \, dx=\left [\frac {3 \, {\left (3 \, B b + 2 \, A c\right )} \sqrt {c} x^{4} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) + 2 \, {\left (3 \, B c x^{4} - 2 \, {\left (3 \, B b + 4 \, A c\right )} x^{2} - 2 \, A b\right )} \sqrt {c x^{4} + b x^{2}}}{12 \, x^{4}}, -\frac {3 \, {\left (3 \, B b + 2 \, A c\right )} \sqrt {-c} x^{4} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) - {\left (3 \, B c x^{4} - 2 \, {\left (3 \, B b + 4 \, A c\right )} x^{2} - 2 \, A b\right )} \sqrt {c x^{4} + b x^{2}}}{6 \, x^{4}}\right ] \] Input:

integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^7,x, algorithm="fricas")
 

Output:

[1/12*(3*(3*B*b + 2*A*c)*sqrt(c)*x^4*log(-2*c*x^2 - b - 2*sqrt(c*x^4 + b*x 
^2)*sqrt(c)) + 2*(3*B*c*x^4 - 2*(3*B*b + 4*A*c)*x^2 - 2*A*b)*sqrt(c*x^4 + 
b*x^2))/x^4, -1/6*(3*(3*B*b + 2*A*c)*sqrt(-c)*x^4*arctan(sqrt(c*x^4 + b*x^ 
2)*sqrt(-c)/(c*x^2 + b)) - (3*B*c*x^4 - 2*(3*B*b + 4*A*c)*x^2 - 2*A*b)*sqr 
t(c*x^4 + b*x^2))/x^4]
 

Sympy [F]

\[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^7} \, dx=\int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}} \left (A + B x^{2}\right )}{x^{7}}\, dx \] Input:

integrate((B*x**2+A)*(c*x**4+b*x**2)**(3/2)/x**7,x)
 

Output:

Integral((x**2*(b + c*x**2))**(3/2)*(A + B*x**2)/x**7, x)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.26 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^7} \, dx=\frac {1}{6} \, {\left (3 \, c^{\frac {3}{2}} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - \frac {7 \, \sqrt {c x^{4} + b x^{2}} c}{x^{2}} - \frac {\sqrt {c x^{4} + b x^{2}} b}{x^{4}} - \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{6}}\right )} A + \frac {1}{4} \, {\left (3 \, b \sqrt {c} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - \frac {6 \, \sqrt {c x^{4} + b x^{2}} b}{x^{2}} + \frac {2 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{4}}\right )} B \] Input:

integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^7,x, algorithm="maxima")
 

Output:

1/6*(3*c^(3/2)*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) - 7*sqrt(c 
*x^4 + b*x^2)*c/x^2 - sqrt(c*x^4 + b*x^2)*b/x^4 - (c*x^4 + b*x^2)^(3/2)/x^ 
6)*A + 1/4*(3*b*sqrt(c)*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) - 
 6*sqrt(c*x^4 + b*x^2)*b/x^2 + 2*(c*x^4 + b*x^2)^(3/2)/x^4)*B
 

Giac [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.69 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^7} \, dx=\frac {1}{2} \, \sqrt {c x^{2} + b} B c x \mathrm {sgn}\left (x\right ) - \frac {1}{4} \, {\left (3 \, B b \sqrt {c} \mathrm {sgn}\left (x\right ) + 2 \, A c^{\frac {3}{2}} \mathrm {sgn}\left (x\right )\right )} \log \left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2}\right ) + \frac {2 \, {\left (3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} B b^{2} \sqrt {c} \mathrm {sgn}\left (x\right ) + 6 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} A b c^{\frac {3}{2}} \mathrm {sgn}\left (x\right ) - 6 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} B b^{3} \sqrt {c} \mathrm {sgn}\left (x\right ) - 6 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} A b^{2} c^{\frac {3}{2}} \mathrm {sgn}\left (x\right ) + 3 \, B b^{4} \sqrt {c} \mathrm {sgn}\left (x\right ) + 4 \, A b^{3} c^{\frac {3}{2}} \mathrm {sgn}\left (x\right )\right )}}{3 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b\right )}^{3}} \] Input:

integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^7,x, algorithm="giac")
 

Output:

1/2*sqrt(c*x^2 + b)*B*c*x*sgn(x) - 1/4*(3*B*b*sqrt(c)*sgn(x) + 2*A*c^(3/2) 
*sgn(x))*log((sqrt(c)*x - sqrt(c*x^2 + b))^2) + 2/3*(3*(sqrt(c)*x - sqrt(c 
*x^2 + b))^4*B*b^2*sqrt(c)*sgn(x) + 6*(sqrt(c)*x - sqrt(c*x^2 + b))^4*A*b* 
c^(3/2)*sgn(x) - 6*(sqrt(c)*x - sqrt(c*x^2 + b))^2*B*b^3*sqrt(c)*sgn(x) - 
6*(sqrt(c)*x - sqrt(c*x^2 + b))^2*A*b^2*c^(3/2)*sgn(x) + 3*B*b^4*sqrt(c)*s 
gn(x) + 4*A*b^3*c^(3/2)*sgn(x))/((sqrt(c)*x - sqrt(c*x^2 + b))^2 - b)^3
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^7} \, dx=\int \frac {\left (B\,x^2+A\right )\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{x^7} \,d x \] Input:

int(((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^7,x)
 

Output:

int(((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^7, x)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.98 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^7} \, dx=\frac {-4 \sqrt {c \,x^{2}+b}\, a b -16 \sqrt {c \,x^{2}+b}\, a c \,x^{2}-12 \sqrt {c \,x^{2}+b}\, b^{2} x^{2}+6 \sqrt {c \,x^{2}+b}\, b c \,x^{4}+12 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a c \,x^{3}+18 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{2} x^{3}+5 \sqrt {c}\, b^{2} x^{3}}{12 x^{3}} \] Input:

int((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^7,x)
 

Output:

( - 4*sqrt(b + c*x**2)*a*b - 16*sqrt(b + c*x**2)*a*c*x**2 - 12*sqrt(b + c* 
x**2)*b**2*x**2 + 6*sqrt(b + c*x**2)*b*c*x**4 + 12*sqrt(c)*log((sqrt(b + c 
*x**2) + sqrt(c)*x)/sqrt(b))*a*c*x**3 + 18*sqrt(c)*log((sqrt(b + c*x**2) + 
 sqrt(c)*x)/sqrt(b))*b**2*x**3 + 5*sqrt(c)*b**2*x**3)/(12*x**3)