Integrand size = 30, antiderivative size = 403 \[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^{3/2}}{x} \, dx=\frac {4 a^2 f x \sqrt {a+b x^4}}{15 \sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {1}{16} a \left (8 c+3 e x^2\right ) \sqrt {a+b x^4}+\frac {2}{105} a x \left (15 d+7 f x^2\right ) \sqrt {a+b x^4}+\frac {1}{24} \left (4 c+3 e x^2\right ) \left (a+b x^4\right )^{3/2}+\frac {1}{63} x \left (9 d+7 f x^2\right ) \left (a+b x^4\right )^{3/2}+\frac {3 a^2 e \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{16 \sqrt {b}}-\frac {1}{2} a^{3/2} c \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )-\frac {4 a^{9/4} f \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 b^{3/4} \sqrt {a+b x^4}}+\frac {2 a^{7/4} \left (15 \sqrt {b} d+7 \sqrt {a} f\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{105 b^{3/4} \sqrt {a+b x^4}} \] Output:
4/15*a^2*f*x*(b*x^4+a)^(1/2)/b^(1/2)/(a^(1/2)+b^(1/2)*x^2)+1/16*a*(3*e*x^2 +8*c)*(b*x^4+a)^(1/2)+2/105*a*x*(7*f*x^2+15*d)*(b*x^4+a)^(1/2)+1/24*(3*e*x ^2+4*c)*(b*x^4+a)^(3/2)+1/63*x*(7*f*x^2+9*d)*(b*x^4+a)^(3/2)+3/16*a^2*e*ar ctanh(b^(1/2)*x^2/(b*x^4+a)^(1/2))/b^(1/2)-1/2*a^(3/2)*c*arctanh((b*x^4+a) ^(1/2)/a^(1/2))-4/15*a^(9/4)*f*(a^(1/2)+b^(1/2)*x^2)*((b*x^4+a)/(a^(1/2)+b ^(1/2)*x^2)^2)^(1/2)*EllipticE(sin(2*arctan(b^(1/4)*x/a^(1/4))),1/2*2^(1/2 ))/b^(3/4)/(b*x^4+a)^(1/2)+2/105*a^(7/4)*(15*b^(1/2)*d+7*a^(1/2)*f)*(a^(1/ 2)+b^(1/2)*x^2)*((b*x^4+a)/(a^(1/2)+b^(1/2)*x^2)^2)^(1/2)*InverseJacobiAM( 2*arctan(b^(1/4)*x/a^(1/4)),1/2*2^(1/2))/b^(3/4)/(b*x^4+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.52 (sec) , antiderivative size = 224, normalized size of antiderivative = 0.56 \[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^{3/2}}{x} \, dx=\frac {1}{16} e \sqrt {a+b x^4} \left (5 a x^2+2 b x^6+\frac {3 a^{3/2} \text {arcsinh}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{\sqrt {b} \sqrt {1+\frac {b x^4}{a}}}\right )+\frac {1}{6} c \left (\sqrt {a+b x^4} \left (4 a+b x^4\right )-3 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )\right )+\frac {a d x \sqrt {a+b x^4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{4},\frac {5}{4},-\frac {b x^4}{a}\right )}{\sqrt {1+\frac {b x^4}{a}}}+\frac {a f x^3 \sqrt {a+b x^4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {3}{4},\frac {7}{4},-\frac {b x^4}{a}\right )}{3 \sqrt {1+\frac {b x^4}{a}}} \] Input:
Integrate[((c + d*x + e*x^2 + f*x^3)*(a + b*x^4)^(3/2))/x,x]
Output:
(e*Sqrt[a + b*x^4]*(5*a*x^2 + 2*b*x^6 + (3*a^(3/2)*ArcSinh[(Sqrt[b]*x^2)/S qrt[a]])/(Sqrt[b]*Sqrt[1 + (b*x^4)/a])))/16 + (c*(Sqrt[a + b*x^4]*(4*a + b *x^4) - 3*a^(3/2)*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a]]))/6 + (a*d*x*Sqrt[a + b *x^4]*Hypergeometric2F1[-3/2, 1/4, 5/4, -((b*x^4)/a)])/Sqrt[1 + (b*x^4)/a] + (a*f*x^3*Sqrt[a + b*x^4]*Hypergeometric2F1[-3/2, 3/4, 7/4, -((b*x^4)/a) ])/(3*Sqrt[1 + (b*x^4)/a])
Time = 1.05 (sec) , antiderivative size = 403, 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.
| \(\displaystyle \int \frac {\left (a+b x^4\right )^{3/2} \left (c+d x+e x^2+f x^3\right )}{x} \, dx\) |
| \(\Big \downarrow \) 2372 |
| \(\displaystyle \int \left (\frac {\left (a+b x^4\right )^{3/2} \left (c+e x^2\right )}{x}+\left (a+b x^4\right )^{3/2} \left (d+f x^2\right )\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 a^{7/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (7 \sqrt {a} f+15 \sqrt {b} d\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{105 b^{3/4} \sqrt {a+b x^4}}-\frac {4 a^{9/4} f \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 b^{3/4} \sqrt {a+b x^4}}-\frac {1}{2} a^{3/2} c \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )+\frac {3 a^2 e \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{16 \sqrt {b}}+\frac {4 a^2 f x \sqrt {a+b x^4}}{15 \sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {1}{16} a \sqrt {a+b x^4} \left (8 c+3 e x^2\right )+\frac {1}{24} \left (a+b x^4\right )^{3/2} \left (4 c+3 e x^2\right )+\frac {2}{105} a x \sqrt {a+b x^4} \left (15 d+7 f x^2\right )+\frac {1}{63} x \left (a+b x^4\right )^{3/2} \left (9 d+7 f x^2\right )\) |
Input:
Int[((c + d*x + e*x^2 + f*x^3)*(a + b*x^4)^(3/2))/x,x]
Output:
(4*a^2*f*x*Sqrt[a + b*x^4])/(15*Sqrt[b]*(Sqrt[a] + Sqrt[b]*x^2)) + (a*(8*c + 3*e*x^2)*Sqrt[a + b*x^4])/16 + (2*a*x*(15*d + 7*f*x^2)*Sqrt[a + b*x^4]) /105 + ((4*c + 3*e*x^2)*(a + b*x^4)^(3/2))/24 + (x*(9*d + 7*f*x^2)*(a + b* x^4)^(3/2))/63 + (3*a^2*e*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(16*Sqrt [b]) - (a^(3/2)*c*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a]])/2 - (4*a^(9/4)*f*(Sqrt [a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticE[2 *ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(15*b^(3/4)*Sqrt[a + b*x^4]) + (2*a^(7 /4)*(15*Sqrt[b]*d + 7*Sqrt[a]*f)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/ (Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticF[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/ (105*b^(3/4)*Sqrt[a + b*x^4])
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 = 0.85 (sec) , antiderivative size = 346, normalized size of antiderivative = 0.86
| method | result | size |
| elliptic | \(\frac {f b \,x^{7} \sqrt {b \,x^{4}+a}}{9}+\frac {b e \,x^{6} \sqrt {b \,x^{4}+a}}{8}+\frac {b d \,x^{5} \sqrt {b \,x^{4}+a}}{7}+\frac {c b \,x^{4} \sqrt {b \,x^{4}+a}}{6}+\frac {11 a f \,x^{3} \sqrt {b \,x^{4}+a}}{45}+\frac {5 a e \,x^{2} \sqrt {b \,x^{4}+a}}{16}+\frac {3 a d x \sqrt {b \,x^{4}+a}}{7}+\frac {2 a c \sqrt {b \,x^{4}+a}}{3}+\frac {4 a^{2} d \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{7 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {3 a^{2} e \ln \left (2 \sqrt {b}\, x^{2}+2 \sqrt {b \,x^{4}+a}\right )}{16 \sqrt {b}}+\frac {4 i f \,a^{\frac {5}{2}} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{15 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}-\frac {a^{\frac {3}{2}} c \,\operatorname {arctanh}\left (\frac {\sqrt {a}}{\sqrt {b \,x^{4}+a}}\right )}{2}\) | \(346\) |
| default | \(d \left (\frac {b \,x^{5} \sqrt {b \,x^{4}+a}}{7}+\frac {3 a x \sqrt {b \,x^{4}+a}}{7}+\frac {4 a^{2} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{7 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+e \left (\frac {3 a^{2} \ln \left (\sqrt {b}\, x^{2}+\sqrt {b \,x^{4}+a}\right )}{16 \sqrt {b}}+\frac {b \,x^{6} \sqrt {b \,x^{4}+a}}{8}+\frac {5 a \,x^{2} \sqrt {b \,x^{4}+a}}{16}\right )+f \left (\frac {b \,x^{7} \sqrt {b \,x^{4}+a}}{9}+\frac {11 a \,x^{3} \sqrt {b \,x^{4}+a}}{45}+\frac {4 i a^{\frac {5}{2}} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{15 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}\right )+c \left (-\frac {a^{\frac {3}{2}} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{4}+a}}{x^{2}}\right )}{2}+\frac {b \,x^{4} \sqrt {b \,x^{4}+a}}{6}+\frac {2 a \sqrt {b \,x^{4}+a}}{3}\right )\) | \(352\) |
Input:
int((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(3/2)/x,x,method=_RETURNVERBOSE)
Output:
1/9*f*b*x^7*(b*x^4+a)^(1/2)+1/8*b*e*x^6*(b*x^4+a)^(1/2)+1/7*b*d*x^5*(b*x^4 +a)^(1/2)+1/6*c*b*x^4*(b*x^4+a)^(1/2)+11/45*a*f*x^3*(b*x^4+a)^(1/2)+5/16*a *e*x^2*(b*x^4+a)^(1/2)+3/7*a*d*x*(b*x^4+a)^(1/2)+2/3*a*c*(b*x^4+a)^(1/2)+4 /7*a^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)+3/16*a^2*e*ln(2*b^(1/2)*x^2+2*(b*x^4+a)^(1/2))/b^(1/2)+4/15*I*f*a^ (5/2)/(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)/b^(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)*c*arct anh(a^(1/2)/(b*x^4+a)^(1/2))
\[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^{3/2}}{x} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{\frac {3}{2}} {\left (f x^{3} + e x^{2} + d x + c\right )}}{x} \,d x } \] Input:
integrate((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(3/2)/x,x, algorithm="fricas")
Output:
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, x)
Time = 11.46 (sec) , antiderivative size = 405, normalized size of antiderivative = 1.00 \[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^{3/2}}{x} \, dx=- \frac {a^{\frac {3}{2}} c \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{2}} \right )}}{2} + \frac {a^{\frac {3}{2}} d x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} + \frac {a^{\frac {3}{2}} e x^{2} \sqrt {1 + \frac {b x^{4}}{a}}}{4} + \frac {a^{\frac {3}{2}} e x^{2}}{16 \sqrt {1 + \frac {b x^{4}}{a}}} + \frac {a^{\frac {3}{2}} f x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} + \frac {\sqrt {a} b d x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {9}{4}\right )} + \frac {3 \sqrt {a} b e x^{6}}{16 \sqrt {1 + \frac {b x^{4}}{a}}} + \frac {\sqrt {a} b f x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {11}{4}\right )} + \frac {a^{2} c}{2 \sqrt {b} x^{2} \sqrt {\frac {a}{b x^{4}} + 1}} + \frac {3 a^{2} e \operatorname {asinh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{16 \sqrt {b}} + \frac {a \sqrt {b} c x^{2}}{2 \sqrt {\frac {a}{b x^{4}} + 1}} + b c \left (\begin {cases} \frac {\sqrt {a} x^{4}}{4} & \text {for}\: b = 0 \\\frac {\left (a + b x^{4}\right )^{\frac {3}{2}}}{6 b} & \text {otherwise} \end {cases}\right ) + \frac {b^{2} e x^{10}}{8 \sqrt {a} \sqrt {1 + \frac {b x^{4}}{a}}} \] Input:
integrate((f*x**3+e*x**2+d*x+c)*(b*x**4+a)**(3/2)/x,x)
Output:
-a**(3/2)*c*asinh(sqrt(a)/(sqrt(b)*x**2))/2 + a**(3/2)*d*x*gamma(1/4)*hype r((-1/2, 1/4), (5/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(5/4)) + a**(3/2) *e*x**2*sqrt(1 + b*x**4/a)/4 + a**(3/2)*e*x**2/(16*sqrt(1 + b*x**4/a)) + a **(3/2)*f*x**3*gamma(3/4)*hyper((-1/2, 3/4), (7/4,), b*x**4*exp_polar(I*pi )/a)/(4*gamma(7/4)) + sqrt(a)*b*d*x**5*gamma(5/4)*hyper((-1/2, 5/4), (9/4, ), b*x**4*exp_polar(I*pi)/a)/(4*gamma(9/4)) + 3*sqrt(a)*b*e*x**6/(16*sqrt( 1 + b*x**4/a)) + sqrt(a)*b*f*x**7*gamma(7/4)*hyper((-1/2, 7/4), (11/4,), b *x**4*exp_polar(I*pi)/a)/(4*gamma(11/4)) + a**2*c/(2*sqrt(b)*x**2*sqrt(a/( b*x**4) + 1)) + 3*a**2*e*asinh(sqrt(b)*x**2/sqrt(a))/(16*sqrt(b)) + a*sqrt (b)*c*x**2/(2*sqrt(a/(b*x**4) + 1)) + b*c*Piecewise((sqrt(a)*x**4/4, Eq(b, 0)), ((a + b*x**4)**(3/2)/(6*b), True)) + b**2*e*x**10/(8*sqrt(a)*sqrt(1 + b*x**4/a))
\[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^{3/2}}{x} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{\frac {3}{2}} {\left (f x^{3} + e x^{2} + d x + c\right )}}{x} \,d x } \] Input:
integrate((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(3/2)/x,x, algorithm="maxima")
Output:
integrate((b*x^4 + a)^(3/2)*(f*x^3 + e*x^2 + d*x + c)/x, x)
\[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^{3/2}}{x} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{\frac {3}{2}} {\left (f x^{3} + e x^{2} + d x + c\right )}}{x} \,d x } \] Input:
integrate((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(3/2)/x,x, algorithm="giac")
Output:
integrate((b*x^4 + a)^(3/2)*(f*x^3 + e*x^2 + d*x + c)/x, x)
Timed out. \[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^{3/2}}{x} \, dx=\int \frac {{\left (b\,x^4+a\right )}^{3/2}\,\left (f\,x^3+e\,x^2+d\,x+c\right )}{x} \,d x \] Input:
int(((a + b*x^4)^(3/2)*(c + d*x + e*x^2 + f*x^3))/x,x)
Output:
int(((a + b*x^4)^(3/2)*(c + d*x + e*x^2 + f*x^3))/x, x)
\[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^{3/2}}{x} \, dx=\frac {6720 \sqrt {b \,x^{4}+a}\, a b c +4320 \sqrt {b \,x^{4}+a}\, a b d x +3150 \sqrt {b \,x^{4}+a}\, a b e \,x^{2}+2464 \sqrt {b \,x^{4}+a}\, a b f \,x^{3}+1680 \sqrt {b \,x^{4}+a}\, b^{2} c \,x^{4}+1440 \sqrt {b \,x^{4}+a}\, b^{2} d \,x^{5}+1260 \sqrt {b \,x^{4}+a}\, b^{2} e \,x^{6}+1120 \sqrt {b \,x^{4}+a}\, b^{2} f \,x^{7}+2520 \sqrt {a}\, \mathrm {log}\left (\sqrt {b \,x^{4}+a}-\sqrt {a}\right ) a b c -2520 \sqrt {a}\, \mathrm {log}\left (\sqrt {b \,x^{4}+a}+\sqrt {a}\right ) a b c -945 \sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{4}+a}-\sqrt {b}\, x^{2}\right ) a^{2} e +945 \sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{4}+a}+\sqrt {b}\, x^{2}\right ) a^{2} e +5760 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b \,x^{4}+a}d x \right ) a^{2} b d +2688 \left (\int \frac {\sqrt {b \,x^{4}+a}\, x^{2}}{b \,x^{4}+a}d x \right ) a^{2} b f}{10080 b} \] Input:
int((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(3/2)/x,x)
Output:
(6720*sqrt(a + b*x**4)*a*b*c + 4320*sqrt(a + b*x**4)*a*b*d*x + 3150*sqrt(a + b*x**4)*a*b*e*x**2 + 2464*sqrt(a + b*x**4)*a*b*f*x**3 + 1680*sqrt(a + b *x**4)*b**2*c*x**4 + 1440*sqrt(a + b*x**4)*b**2*d*x**5 + 1260*sqrt(a + b*x **4)*b**2*e*x**6 + 1120*sqrt(a + b*x**4)*b**2*f*x**7 + 2520*sqrt(a)*log(sq rt(a + b*x**4) - sqrt(a))*a*b*c - 2520*sqrt(a)*log(sqrt(a + b*x**4) + sqrt (a))*a*b*c - 945*sqrt(b)*log(sqrt(a + b*x**4) - sqrt(b)*x**2)*a**2*e + 945 *sqrt(b)*log(sqrt(a + b*x**4) + sqrt(b)*x**2)*a**2*e + 5760*int(sqrt(a + b *x**4)/(a + b*x**4),x)*a**2*b*d + 2688*int((sqrt(a + b*x**4)*x**2)/(a + b* x**4),x)*a**2*b*f)/(10080*b)