3.2.28 \(\int \frac {(A+B x^2) (b x^2+c x^4)^{3/2}}{x^{14}} \, dx\) [128]

Optimal. Leaf size=214 \[ -\frac {c (2 b B-A c) \sqrt {b x^2+c x^4}}{32 b x^7}-\frac {c^2 (2 b B-A c) \sqrt {b x^2+c x^4}}{128 b^2 x^5}+\frac {3 c^3 (2 b B-A c) \sqrt {b x^2+c x^4}}{256 b^3 x^3}-\frac {(2 b B-A c) \left (b x^2+c x^4\right )^{3/2}}{16 b x^{11}}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{10 b x^{15}}-\frac {3 c^4 (2 b B-A c) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{256 b^{7/2}} \]

[Out]

-1/16*(-A*c+2*B*b)*(c*x^4+b*x^2)^(3/2)/b/x^11-1/10*A*(c*x^4+b*x^2)^(5/2)/b/x^15-3/256*c^4*(-A*c+2*B*b)*arctanh
(x*b^(1/2)/(c*x^4+b*x^2)^(1/2))/b^(7/2)-1/32*c*(-A*c+2*B*b)*(c*x^4+b*x^2)^(1/2)/b/x^7-1/128*c^2*(-A*c+2*B*b)*(
c*x^4+b*x^2)^(1/2)/b^2/x^5+3/256*c^3*(-A*c+2*B*b)*(c*x^4+b*x^2)^(1/2)/b^3/x^3

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Rubi [A]
time = 0.23, antiderivative size = 214, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {2063, 2045, 2050, 2033, 212} \begin {gather*} -\frac {3 c^4 (2 b B-A c) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{256 b^{7/2}}+\frac {3 c^3 \sqrt {b x^2+c x^4} (2 b B-A c)}{256 b^3 x^3}-\frac {c^2 \sqrt {b x^2+c x^4} (2 b B-A c)}{128 b^2 x^5}-\frac {\left (b x^2+c x^4\right )^{3/2} (2 b B-A c)}{16 b x^{11}}-\frac {c \sqrt {b x^2+c x^4} (2 b B-A c)}{32 b x^7}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{10 b x^{15}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/32*(c*(2*b*B - A*c)*Sqrt[b*x^2 + c*x^4])/(b*x^7) - (c^2*(2*b*B - A*c)*Sqrt[b*x^2 + c*x^4])/(128*b^2*x^5) +
(3*c^3*(2*b*B - A*c)*Sqrt[b*x^2 + c*x^4])/(256*b^3*x^3) - ((2*b*B - A*c)*(b*x^2 + c*x^4)^(3/2))/(16*b*x^11) -
(A*(b*x^2 + c*x^4)^(5/2))/(10*b*x^15) - (3*c^4*(2*b*B - A*c)*ArcTanh[(Sqrt[b]*x)/Sqrt[b*x^2 + c*x^4]])/(256*b^
(7/2))

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2033

Int[1/Sqrt[(a_.)*(x_)^2 + (b_.)*(x_)^(n_.)], x_Symbol] :> Dist[2/(2 - n), Subst[Int[1/(1 - a*x^2), x], x, x/Sq
rt[a*x^2 + b*x^n]], x] /; FreeQ[{a, b, n}, x] && NeQ[n, 2]

Rule 2045

Int[((c_.)*(x_))^(m_)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a*x^j + b*
x^n)^p/(c*(m + j*p + 1))), x] - Dist[b*p*((n - j)/(c^n*(m + j*p + 1))), Int[(c*x)^(m + n)*(a*x^j + b*x^n)^(p -
1), x], x] /; FreeQ[{a, b, c}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && GtQ[p
, 0] && LtQ[m + j*p + 1, 0]

Rule 2050

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[c^(j - 1)*(c*x)^(m - j +
1)*((a*x^j + b*x^n)^(p + 1)/(a*(m + j*p + 1))), x] - Dist[b*((m + n*p + n - j + 1)/(a*c^(n - j)*(m + j*p + 1))
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j,
n] && (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[m + j*p + 1, 0]

Rule 2063

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Sim
p[c*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j + b*x^(j + n))^(p + 1)/(a*(m + j*p + 1))), x] + Dist[(a*d*(m + j*p + 1
) - b*c*(m + n + p*(j + n) + 1))/(a*e^n*(m + j*p + 1)), Int[(e*x)^(m + n)*(a*x^j + b*x^(j + n))^p, x], x] /; F
reeQ[{a, b, c, d, e, j, p}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && (LtQ[m
+ j*p, -1] || (IntegersQ[m - 1/2, p - 1/2] && LtQ[p, 0] && LtQ[m, (-n)*p - 1])) && (GtQ[e, 0] || IntegersQ[j,
n]) && NeQ[m + j*p + 1, 0] && NeQ[m - n + j*p + 1, 0]

Rubi steps

\begin {align*} \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{14}} \, dx &=-\frac {A \left (b x^2+c x^4\right )^{5/2}}{10 b x^{15}}-\frac {(-10 b B+5 A c) \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{12}} \, dx}{10 b}\\ &=-\frac {(2 b B-A c) \left (b x^2+c x^4\right )^{3/2}}{16 b x^{11}}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{10 b x^{15}}+\frac {(3 c (2 b B-A c)) \int \frac {\sqrt {b x^2+c x^4}}{x^8} \, dx}{16 b}\\ &=-\frac {c (2 b B-A c) \sqrt {b x^2+c x^4}}{32 b x^7}-\frac {(2 b B-A c) \left (b x^2+c x^4\right )^{3/2}}{16 b x^{11}}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{10 b x^{15}}+\frac {\left (c^2 (2 b B-A c)\right ) \int \frac {1}{x^4 \sqrt {b x^2+c x^4}} \, dx}{32 b}\\ &=-\frac {c (2 b B-A c) \sqrt {b x^2+c x^4}}{32 b x^7}-\frac {c^2 (2 b B-A c) \sqrt {b x^2+c x^4}}{128 b^2 x^5}-\frac {(2 b B-A c) \left (b x^2+c x^4\right )^{3/2}}{16 b x^{11}}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{10 b x^{15}}-\frac {\left (3 c^3 (2 b B-A c)\right ) \int \frac {1}{x^2 \sqrt {b x^2+c x^4}} \, dx}{128 b^2}\\ &=-\frac {c (2 b B-A c) \sqrt {b x^2+c x^4}}{32 b x^7}-\frac {c^2 (2 b B-A c) \sqrt {b x^2+c x^4}}{128 b^2 x^5}+\frac {3 c^3 (2 b B-A c) \sqrt {b x^2+c x^4}}{256 b^3 x^3}-\frac {(2 b B-A c) \left (b x^2+c x^4\right )^{3/2}}{16 b x^{11}}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{10 b x^{15}}+\frac {\left (3 c^4 (2 b B-A c)\right ) \int \frac {1}{\sqrt {b x^2+c x^4}} \, dx}{256 b^3}\\ &=-\frac {c (2 b B-A c) \sqrt {b x^2+c x^4}}{32 b x^7}-\frac {c^2 (2 b B-A c) \sqrt {b x^2+c x^4}}{128 b^2 x^5}+\frac {3 c^3 (2 b B-A c) \sqrt {b x^2+c x^4}}{256 b^3 x^3}-\frac {(2 b B-A c) \left (b x^2+c x^4\right )^{3/2}}{16 b x^{11}}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{10 b x^{15}}-\frac {\left (3 c^4 (2 b B-A c)\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {b x^2+c x^4}}\right )}{256 b^3}\\ &=-\frac {c (2 b B-A c) \sqrt {b x^2+c x^4}}{32 b x^7}-\frac {c^2 (2 b B-A c) \sqrt {b x^2+c x^4}}{128 b^2 x^5}+\frac {3 c^3 (2 b B-A c) \sqrt {b x^2+c x^4}}{256 b^3 x^3}-\frac {(2 b B-A c) \left (b x^2+c x^4\right )^{3/2}}{16 b x^{11}}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{10 b x^{15}}-\frac {3 c^4 (2 b B-A c) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{256 b^{7/2}}\\ \end {align*}

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Mathematica [A]
time = 0.33, size = 172, normalized size = 0.80 \begin {gather*} \frac {-\sqrt {b} \left (b+c x^2\right ) \left (10 b B x^2 \left (16 b^3+24 b^2 c x^2+2 b c^2 x^4-3 c^3 x^6\right )+A \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 )\right )+15 c^4 (-2 b B+A c) x^{10} \sqrt {b+c x^2} \tanh ^{-1}\left (\frac {\sqrt {b+c x^2}}{\sqrt {b}}\right )}{1280 b^{7/2} x^9 \sqrt {x^2 \left (b+c x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-(Sqrt[b]*(b + c*x^2)*(10*b*B*x^2*(16*b^3 + 24*b^2*c*x^2 + 2*b*c^2*x^4 - 3*c^3*x^6) + A*(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^4*(-2*b*B + A*c)*x^10*Sqrt[b + c*x^2]*ArcTanh[Sqrt[b
+ c*x^2]/Sqrt[b]])/(1280*b^(7/2)*x^9*Sqrt[x^2*(b + c*x^2)])

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Maple [A]
time = 0.40, size = 344, normalized size = 1.61

method	result	size
risch	\(-\frac {\left (15 A \,c^{4} x^{8}-30 B b \,c^{3} x^{8}-10 A b \,c^{3} x^{6}+20 B \,b^{2} c^{2} x^{6}+8 A \,b^{2} c^{2} x^{4}+240 B \,b^{3} c \,x^{4}+176 A \,b^{3} c \,x^{2}+160 B \,b^{4} x^{2}+128 A \,b^{4}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{1280 x^{11} b^{3}}+\frac {\left (\frac {3 c^{5} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) A}{256 b^{\frac {7}{2}}}-\frac {3 c^{4} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) B}{128 b^{\frac {5}{2}}}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{x \sqrt {c \,x^{2}+b}}\)	\(203\)
default	\(-\frac {\left (x^{4} c +b \,x^{2}\right )^{\frac {3}{2}} \left (5 A \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{5} x^{10}-15 A \,b^{\frac {3}{2}} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) c^{5} x^{10}-10 B \left (c \,x^{2}+b \right )^{\frac {3}{2}} b \,c^{4} x^{10}+30 B \,b^{\frac {5}{2}} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) c^{4} x^{10}-5 A \left (c \,x^{2}+b \right )^{\frac {5}{2}} c^{4} x^{8}+15 A \sqrt {c \,x^{2}+b}\, b \,c^{5} x^{10}+10 B \left (c \,x^{2}+b \right )^{\frac {5}{2}} b \,c^{3} x^{8}-30 B \sqrt {c \,x^{2}+b}\, b^{2} c^{4} x^{10}-10 A \left (c \,x^{2}+b \right )^{\frac {5}{2}} b \,c^{3} x^{6}+20 B \left (c \,x^{2}+b \right )^{\frac {5}{2}} b^{2} c^{2} x^{6}+40 A \left (c \,x^{2}+b \right )^{\frac {5}{2}} b^{2} c^{2} x^{4}-80 B \left (c \,x^{2}+b \right )^{\frac {5}{2}} b^{3} c \,x^{4}-80 A \left (c \,x^{2}+b \right )^{\frac {5}{2}} b^{3} c \,x^{2}+160 B \left (c \,x^{2}+b \right )^{\frac {5}{2}} b^{4} x^{2}+128 A \left (c \,x^{2}+b \right )^{\frac {5}{2}} b^{4}\right )}{1280 x^{13} \left (c \,x^{2}+b \right )^{\frac {3}{2}} b^{5}}\)	\(344\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/1280*(c*x^4+b*x^2)^(3/2)*(5*A*(c*x^2+b)^(3/2)*c^5*x^10-15*A*b^(3/2)*ln(2*(b^(1/2)*(c*x^2+b)^(1/2)+b)/x)*c^5
*x^10-10*B*(c*x^2+b)^(3/2)*b*c^4*x^10+30*B*b^(5/2)*ln(2*(b^(1/2)*(c*x^2+b)^(1/2)+b)/x)*c^4*x^10-5*A*(c*x^2+b)^
(5/2)*c^4*x^8+15*A*(c*x^2+b)^(1/2)*b*c^5*x^10+10*B*(c*x^2+b)^(5/2)*b*c^3*x^8-30*B*(c*x^2+b)^(1/2)*b^2*c^4*x^10
-10*A*(c*x^2+b)^(5/2)*b*c^3*x^6+20*B*(c*x^2+b)^(5/2)*b^2*c^2*x^6+40*A*(c*x^2+b)^(5/2)*b^2*c^2*x^4-80*B*(c*x^2+
b)^(5/2)*b^3*c*x^4-80*A*(c*x^2+b)^(5/2)*b^3*c*x^2+160*B*(c*x^2+b)^(5/2)*b^4*x^2+128*A*(c*x^2+b)^(5/2)*b^4)/x^1
3/(c*x^2+b)^(3/2)/b^5

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 1.44, size = 345, normalized size = 1.61 \begin {gather*} \left [-\frac {15 \, {\left (2 \, B b c^{4} - A c^{5}\right )} \sqrt {b} 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 \, {\left (2 \, B b^{2} c^{3} - A b c^{4}\right )} x^{8} - 10 \, {\left (2 \, B b^{3} c^{2} - A b^{2} c^{3}\right )} x^{6} - 128 \, A b^{5} - 8 \, {\left (30 \, B b^{4} c + A b^{3} c^{2}\right )} x^{4} - 16 \, {\left (10 \, B b^{5} + 11 \, A b^{4} c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{2560 \, b^{4} x^{11}}, \frac {15 \, {\left (2 \, B b c^{4} - A c^{5}\right )} \sqrt {-b} x^{11} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-b}}{c x^{3} + b x}\right ) + {\left (15 \, {\left (2 \, B b^{2} c^{3} - A b c^{4}\right )} x^{8} - 10 \, {\left (2 \, B b^{3} c^{2} - A b^{2} c^{3}\right )} x^{6} - 128 \, A b^{5} - 8 \, {\left (30 \, B b^{4} c + A b^{3} c^{2}\right )} x^{4} - 16 \, {\left (10 \, B b^{5} + 11 \, A b^{4} c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{1280 \, b^{4} x^{11}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2560*(15*(2*B*b*c^4 - A*c^5)*sqrt(b)*x^11*log(-(c*x^3 + 2*b*x + 2*sqrt(c*x^4 + b*x^2)*sqrt(b))/x^3) - 2*(1
5*(2*B*b^2*c^3 - A*b*c^4)*x^8 - 10*(2*B*b^3*c^2 - A*b^2*c^3)*x^6 - 128*A*b^5 - 8*(30*B*b^4*c + A*b^3*c^2)*x^4
- 16*(10*B*b^5 + 11*A*b^4*c)*x^2)*sqrt(c*x^4 + b*x^2))/(b^4*x^11), 1/1280*(15*(2*B*b*c^4 - A*c^5)*sqrt(-b)*x^1
1*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-b)/(c*x^3 + b*x)) + (15*(2*B*b^2*c^3 - A*b*c^4)*x^8 - 10*(2*B*b^3*c^2 - A*b
^2*c^3)*x^6 - 128*A*b^5 - 8*(30*B*b^4*c + A*b^3*c^2)*x^4 - 16*(10*B*b^5 + 11*A*b^4*c)*x^2)*sqrt(c*x^4 + b*x^2)
)/(b^4*x^11)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}} \left (A + B x^{2}\right )}{x^{14}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.94, size = 234, normalized size = 1.09 \begin {gather*} \frac {\frac {15 \, {\left (2 \, B b c^{5} \mathrm {sgn}\left (x\right ) - A c^{6} \mathrm {sgn}\left (x\right )\right )} \arctan \left (\frac {\sqrt {c x^{2} + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{3}} + \frac {30 \, {\left (c x^{2} + b\right )}^{\frac {9}{2}} B b c^{5} \mathrm {sgn}\left (x\right ) - 140 \, {\left (c x^{2} + b\right )}^{\frac {7}{2}} B b^{2} c^{5} \mathrm {sgn}\left (x\right ) + 140 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} B b^{4} c^{5} \mathrm {sgn}\left (x\right ) - 30 \, \sqrt {c x^{2} + b} B b^{5} c^{5} \mathrm {sgn}\left (x\right ) - 15 \, {\left (c x^{2} + b\right )}^{\frac {9}{2}} A c^{6} \mathrm {sgn}\left (x\right ) + 70 \, {\left (c x^{2} + b\right )}^{\frac {7}{2}} A b c^{6} \mathrm {sgn}\left (x\right ) - 128 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} A b^{2} c^{6} \mathrm {sgn}\left (x\right ) - 70 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} A b^{3} c^{6} \mathrm {sgn}\left (x\right ) + 15 \, \sqrt {c x^{2} + b} A b^{4} c^{6} \mathrm {sgn}\left (x\right )}{b^{3} c^{5} x^{10}}}{1280 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1280*(15*(2*B*b*c^5*sgn(x) - A*c^6*sgn(x))*arctan(sqrt(c*x^2 + b)/sqrt(-b))/(sqrt(-b)*b^3) + (30*(c*x^2 + b)
^(9/2)*B*b*c^5*sgn(x) - 140*(c*x^2 + b)^(7/2)*B*b^2*c^5*sgn(x) + 140*(c*x^2 + b)^(3/2)*B*b^4*c^5*sgn(x) - 30*s
qrt(c*x^2 + b)*B*b^5*c^5*sgn(x) - 15*(c*x^2 + b)^(9/2)*A*c^6*sgn(x) + 70*(c*x^2 + b)^(7/2)*A*b*c^6*sgn(x) - 12
8*(c*x^2 + b)^(5/2)*A*b^2*c^6*sgn(x) - 70*(c*x^2 + b)^(3/2)*A*b^3*c^6*sgn(x) + 15*sqrt(c*x^2 + b)*A*b^4*c^6*sg
n(x))/(b^3*c^5*x^10))/c

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (B\,x^2+A\right )\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{x^{14}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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