Integrand size = 30, antiderivative size = 406 \[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^{3/2}}{x^3} \, dx=\frac {12 a \sqrt {b} d x \sqrt {a+b x^4}}{5 \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {1}{4} \left (2 a e+3 b c x^2\right ) \sqrt {a+b x^4}+\frac {2}{35} x \left (5 a f+21 b d x^2\right ) \sqrt {a+b x^4}-\frac {\left (3 c-e x^2\right ) \left (a+b x^4\right )^{3/2}}{6 x^2}-\frac {\left (7 d-f x^2\right ) \left (a+b x^4\right )^{3/2}}{7 x}+\frac {3}{4} a \sqrt {b} c \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )-\frac {1}{2} a^{3/2} e \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )-\frac {12 a^{5/4} \sqrt [4]{b} d \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 \sqrt {a+b x^4}}+\frac {2 a^{5/4} \left (21 \sqrt {b} d+5 \sqrt {a} f\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{35 \sqrt [4]{b} \sqrt {a+b x^4}} \]
-1/6*(-e*x^2+3*c)*(b*x^4+a)^(3/2)/x^2-1/7*(-f*x^2+7*d)*(b*x^4+a)^(3/2)/x-1 /2*a^(3/2)*e*arctanh((b*x^4+a)^(1/2)/a^(1/2))+3/4*a*c*arctanh(x^2*b^(1/2)/ (b*x^4+a)^(1/2))*b^(1/2)+1/4*(3*b*c*x^2+2*a*e)*(b*x^4+a)^(1/2)+2/35*x*(21* b*d*x^2+5*a*f)*(b*x^4+a)^(1/2)+12/5*a*d*x*b^(1/2)*(b*x^4+a)^(1/2)/(a^(1/2) +x^2*b^(1/2))-12/5*a^(5/4)*b^(1/4)*d*(cos(2*arctan(b^(1/4)*x/a^(1/4)))^2)^ (1/2)/cos(2*arctan(b^(1/4)*x/a^(1/4)))*EllipticE(sin(2*arctan(b^(1/4)*x/a^ (1/4))),1/2*2^(1/2))*(a^(1/2)+x^2*b^(1/2))*((b*x^4+a)/(a^(1/2)+x^2*b^(1/2) )^2)^(1/2)/(b*x^4+a)^(1/2)+2/35*a^(5/4)*(cos(2*arctan(b^(1/4)*x/a^(1/4)))^ 2)^(1/2)/cos(2*arctan(b^(1/4)*x/a^(1/4)))*EllipticF(sin(2*arctan(b^(1/4)*x /a^(1/4))),1/2*2^(1/2))*(5*f*a^(1/2)+21*d*b^(1/2))*(a^(1/2)+x^2*b^(1/2))*( (b*x^4+a)/(a^(1/2)+x^2*b^(1/2))^2)^(1/2)/b^(1/4)/(b*x^4+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.34 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.48 \[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^{3/2}}{x^3} \, dx=\frac {-3 a c \sqrt {a+b x^4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {1}{2},\frac {1}{2},-\frac {b x^4}{a}\right )+x \left (e x \sqrt {1+\frac {b x^4}{a}} \left (\sqrt {a+b x^4} \left (4 a+b x^4\right )-3 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )\right )-6 a d \sqrt {a+b x^4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {1}{4},\frac {3}{4},-\frac {b x^4}{a}\right )+6 a f x^2 \sqrt {a+b x^4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{4},\frac {5}{4},-\frac {b x^4}{a}\right )\right )}{6 x^2 \sqrt {1+\frac {b x^4}{a}}} \]
(-3*a*c*Sqrt[a + b*x^4]*Hypergeometric2F1[-3/2, -1/2, 1/2, -((b*x^4)/a)] + x*(e*x*Sqrt[1 + (b*x^4)/a]*(Sqrt[a + b*x^4]*(4*a + b*x^4) - 3*a^(3/2)*Arc Tanh[Sqrt[a + b*x^4]/Sqrt[a]]) - 6*a*d*Sqrt[a + b*x^4]*Hypergeometric2F1[- 3/2, -1/4, 3/4, -((b*x^4)/a)] + 6*a*f*x^2*Sqrt[a + b*x^4]*Hypergeometric2F 1[-3/2, 1/4, 5/4, -((b*x^4)/a)]))/(6*x^2*Sqrt[1 + (b*x^4)/a])
Time = 0.62 (sec) , antiderivative size = 406, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2372, 2009}
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.
| \(\displaystyle \int \frac {\left (a+b x^4\right )^{3/2} \left (c+d x+e x^2+f x^3\right )}{x^3} \, dx\) |
| \(\Big \downarrow \) 2372 |
| \(\displaystyle \int \left (\frac {\left (a+b x^4\right )^{3/2} \left (c+e x^2\right )}{x^3}+\frac {\left (a+b x^4\right )^{3/2} \left (d+f x^2\right )}{x^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 a^{5/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (5 \sqrt {a} f+21 \sqrt {b} d\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{35 \sqrt [4]{b} \sqrt {a+b x^4}}-\frac {12 a^{5/4} \sqrt [4]{b} d \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 \sqrt {a+b x^4}}-\frac {1}{2} a^{3/2} e \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )+\frac {3}{4} a \sqrt {b} c \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )-\frac {\left (a+b x^4\right )^{3/2} \left (3 c-e x^2\right )}{6 x^2}+\frac {1}{4} \sqrt {a+b x^4} \left (2 a e+3 b c x^2\right )-\frac {\left (a+b x^4\right )^{3/2} \left (7 d-f x^2\right )}{7 x}+\frac {2}{35} x \sqrt {a+b x^4} \left (5 a f+21 b d x^2\right )+\frac {12 a \sqrt {b} d x \sqrt {a+b x^4}}{5 \left (\sqrt {a}+\sqrt {b} x^2\right )}\) |
(12*a*Sqrt[b]*d*x*Sqrt[a + b*x^4])/(5*(Sqrt[a] + Sqrt[b]*x^2)) + ((2*a*e + 3*b*c*x^2)*Sqrt[a + b*x^4])/4 + (2*x*(5*a*f + 21*b*d*x^2)*Sqrt[a + b*x^4] )/35 - ((3*c - e*x^2)*(a + b*x^4)^(3/2))/(6*x^2) - ((7*d - f*x^2)*(a + b*x ^4)^(3/2))/(7*x) + (3*a*Sqrt[b]*c*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/ 4 - (a^(3/2)*e*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a]])/2 - (12*a^(5/4)*b^(1/4)*d *(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*Ellip ticE[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(5*Sqrt[a + b*x^4]) + (2*a^(5/4) *(21*Sqrt[b]*d + 5*Sqrt[a]*f)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sq rt[a] + Sqrt[b]*x^2)^2]*EllipticF[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(35 *b^(1/4)*Sqrt[a + b*x^4])
3.6.17.3.1 Defintions of rubi rules used
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Mo dule[{q = Expon[Pq, x], j, k}, Int[Sum[((c*x)^(m + j)/c^j)*Sum[Coeff[Pq, x, j + k*(n/2)]*x^(k*(n/2)), {k, 0, 2*((q - j)/n) + 1}]*(a + b*x^n)^p, {j, 0, n/2 - 1}], x]] /; FreeQ[{a, b, c, m, p}, x] && PolyQ[Pq, x] && IGtQ[n/2, 0 ] && !PolyQ[Pq, x^(n/2)]
Result contains complex when optimal does not.
Time = 3.31 (sec) , antiderivative size = 344, normalized size of antiderivative = 0.85
| method | result | size |
| elliptic | \(-\frac {a c \sqrt {b \,x^{4}+a}}{2 x^{2}}-\frac {a d \sqrt {b \,x^{4}+a}}{x}+\frac {b f \,x^{5} \sqrt {b \,x^{4}+a}}{7}+\frac {b e \,x^{4} \sqrt {b \,x^{4}+a}}{6}+\frac {b d \,x^{3} \sqrt {b \,x^{4}+a}}{5}+\frac {b c \,x^{2} \sqrt {b \,x^{4}+a}}{4}+\frac {3 a f x \sqrt {b \,x^{4}+a}}{7}+\frac {2 a e \sqrt {b \,x^{4}+a}}{3}+\frac {4 a^{2} f \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{7 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {3 a \sqrt {b}\, c \ln \left (2 x^{2} \sqrt {b}+2 \sqrt {b \,x^{4}+a}\right )}{4}+\frac {12 i a^{\frac {3}{2}} \sqrt {b}\, d \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{5 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {a^{\frac {3}{2}} e \,\operatorname {arctanh}\left (\frac {\sqrt {a}}{\sqrt {b \,x^{4}+a}}\right )}{2}\) | \(344\) |
| default | \(f \left (\frac {b \,x^{5} \sqrt {b \,x^{4}+a}}{7}+\frac {3 a x \sqrt {b \,x^{4}+a}}{7}+\frac {4 a^{2} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{7 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+e \left (\frac {b \,x^{4} \sqrt {b \,x^{4}+a}}{6}+\frac {2 a \sqrt {b \,x^{4}+a}}{3}-\frac {a^{\frac {3}{2}} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{4}+a}}{x^{2}}\right )}{2}\right )+d \left (-\frac {a \sqrt {b \,x^{4}+a}}{x}+\frac {\sqrt {b \,x^{4}+a}\, b \,x^{3}}{5}+\frac {12 i a^{\frac {3}{2}} \sqrt {b}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{5 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+c \left (\frac {b \,x^{2} \sqrt {b \,x^{4}+a}}{4}+\frac {3 a \sqrt {b}\, \ln \left (x^{2} \sqrt {b}+\sqrt {b \,x^{4}+a}\right )}{4}-\frac {a \sqrt {b \,x^{4}+a}}{2 x^{2}}\right )\) | \(350\) |
| risch | \(-\frac {a \sqrt {b \,x^{4}+a}\, \left (2 d x +c \right )}{2 x^{2}}+\frac {4 a^{2} f \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{7 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {b f \,x^{5} \sqrt {b \,x^{4}+a}}{7}+\frac {3 a f x \sqrt {b \,x^{4}+a}}{7}-\frac {e \sqrt {b \,x^{4}+a}\, \left (-b \,x^{4}+2 a \right )}{6}+\frac {b d \,x^{3} \sqrt {b \,x^{4}+a}}{5}-\frac {3 i \sqrt {b}\, d \,a^{\frac {3}{2}} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{5 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {3 i a^{\frac {3}{2}} \sqrt {b}\, d \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {b c \,x^{2} \sqrt {b \,x^{4}+a}}{4}+\frac {3 a \sqrt {b}\, c \ln \left (x^{2} \sqrt {b}+\sqrt {b \,x^{4}+a}\right )}{4}-\frac {a^{\frac {3}{2}} e \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{4}+a}}{x^{2}}\right )}{2}+a e \sqrt {b \,x^{4}+a}+\frac {3 i \sqrt {b}\, d \,a^{\frac {3}{2}} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, E\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{5 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\) | \(500\) |
-1/2*a*c*(b*x^4+a)^(1/2)/x^2-a*d*(b*x^4+a)^(1/2)/x+1/7*b*f*x^5*(b*x^4+a)^( 1/2)+1/6*b*e*x^4*(b*x^4+a)^(1/2)+1/5*b*d*x^3*(b*x^4+a)^(1/2)+1/4*b*c*x^2*( b*x^4+a)^(1/2)+3/7*a*f*x*(b*x^4+a)^(1/2)+2/3*a*e*(b*x^4+a)^(1/2)+4/7*a^2*f /(I/a^(1/2)*b^(1/2))^(1/2)*(1-I/a^(1/2)*b^(1/2)*x^2)^(1/2)*(1+I/a^(1/2)*b^ (1/2)*x^2)^(1/2)/(b*x^4+a)^(1/2)*EllipticF(x*(I/a^(1/2)*b^(1/2))^(1/2),I)+ 3/4*a*b^(1/2)*c*ln(2*x^2*b^(1/2)+2*(b*x^4+a)^(1/2))+12/5*I*a^(3/2)*b^(1/2) *d/(I/a^(1/2)*b^(1/2))^(1/2)*(1-I/a^(1/2)*b^(1/2)*x^2)^(1/2)*(1+I/a^(1/2)* b^(1/2)*x^2)^(1/2)/(b*x^4+a)^(1/2)*(EllipticF(x*(I/a^(1/2)*b^(1/2))^(1/2), I)-EllipticE(x*(I/a^(1/2)*b^(1/2))^(1/2),I))-1/2*a^(3/2)*e*arctanh(a^(1/2) /(b*x^4+a)^(1/2))
\[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^{3/2}}{x^3} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{\frac {3}{2}} {\left (f x^{3} + e x^{2} + d x + c\right )}}{x^{3}} \,d x } \]
integral((b*f*x^7 + b*e*x^6 + b*d*x^5 + b*c*x^4 + a*f*x^3 + a*e*x^2 + a*d* x + a*c)*sqrt(b*x^4 + a)/x^3, x)
Time = 4.90 (sec) , antiderivative size = 377, normalized size of antiderivative = 0.93 \[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^{3/2}}{x^3} \, dx=- \frac {a^{\frac {3}{2}} c}{2 x^{2} \sqrt {1 + \frac {b x^{4}}{a}}} + \frac {a^{\frac {3}{2}} d \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 x \Gamma \left (\frac {3}{4}\right )} - \frac {a^{\frac {3}{2}} e \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{2}} \right )}}{2} + \frac {a^{\frac {3}{2}} f x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} + \frac {\sqrt {a} b c x^{2} \sqrt {1 + \frac {b x^{4}}{a}}}{4} - \frac {\sqrt {a} b c x^{2}}{2 \sqrt {1 + \frac {b x^{4}}{a}}} + \frac {\sqrt {a} b d x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} + \frac {\sqrt {a} b f x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {9}{4}\right )} + \frac {a^{2} e}{2 \sqrt {b} x^{2} \sqrt {\frac {a}{b x^{4}} + 1}} + \frac {3 a \sqrt {b} c \operatorname {asinh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{4} + \frac {a \sqrt {b} e x^{2}}{2 \sqrt {\frac {a}{b x^{4}} + 1}} + b e \left (\begin {cases} \frac {\sqrt {a} x^{4}}{4} & \text {for}\: b = 0 \\\frac {\left (a + b x^{4}\right )^{\frac {3}{2}}}{6 b} & \text {otherwise} \end {cases}\right ) \]
-a**(3/2)*c/(2*x**2*sqrt(1 + b*x**4/a)) + a**(3/2)*d*gamma(-1/4)*hyper((-1 /2, -1/4), (3/4,), b*x**4*exp_polar(I*pi)/a)/(4*x*gamma(3/4)) - a**(3/2)*e *asinh(sqrt(a)/(sqrt(b)*x**2))/2 + a**(3/2)*f*x*gamma(1/4)*hyper((-1/2, 1/ 4), (5/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(5/4)) + sqrt(a)*b*c*x**2*sq rt(1 + b*x**4/a)/4 - sqrt(a)*b*c*x**2/(2*sqrt(1 + b*x**4/a)) + sqrt(a)*b*d *x**3*gamma(3/4)*hyper((-1/2, 3/4), (7/4,), b*x**4*exp_polar(I*pi)/a)/(4*g amma(7/4)) + sqrt(a)*b*f*x**5*gamma(5/4)*hyper((-1/2, 5/4), (9/4,), b*x**4 *exp_polar(I*pi)/a)/(4*gamma(9/4)) + a**2*e/(2*sqrt(b)*x**2*sqrt(a/(b*x**4 ) + 1)) + 3*a*sqrt(b)*c*asinh(sqrt(b)*x**2/sqrt(a))/4 + a*sqrt(b)*e*x**2/( 2*sqrt(a/(b*x**4) + 1)) + b*e*Piecewise((sqrt(a)*x**4/4, Eq(b, 0)), ((a + b*x**4)**(3/2)/(6*b), True))
\[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^{3/2}}{x^3} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{\frac {3}{2}} {\left (f x^{3} + e x^{2} + d x + c\right )}}{x^{3}} \,d x } \]
\[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^{3/2}}{x^3} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{\frac {3}{2}} {\left (f x^{3} + e x^{2} + d x + c\right )}}{x^{3}} \,d x } \]
Timed out. \[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^{3/2}}{x^3} \, dx=\int \frac {{\left (b\,x^4+a\right )}^{3/2}\,\left (f\,x^3+e\,x^2+d\,x+c\right )}{x^3} \,d x \]