\(\int \frac {(c+d x+e x^2+f x^3) \sqrt {a+b x^4}}{x^7} \, dx\) [51]

Optimal result

Integrand size = 30, antiderivative size = 352 \[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x^7} \, dx=-\frac {e \sqrt {a+b x^4}}{4 x^4}+\frac {2 f \sqrt {a+b x^4}}{3 x^3}-\frac {2 b d \sqrt {a+b x^4}}{5 \sqrt {a} x \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {\left (d+5 f x^2\right ) \sqrt {a+b x^4}}{5 x^5}-\frac {c \left (a+b x^4\right )^{3/2}}{6 a x^6}-\frac {b e \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{4 \sqrt {a}}-\frac {2 b^{5/4} d \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 a^{3/4} \sqrt {a+b x^4}}+\frac {b^{3/4} \left (3 \sqrt {b} d+5 \sqrt {a} f\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{15 a^{3/4} \sqrt {a+b x^4}} \] Output:

-1/4*e*(b*x^4+a)^(1/2)/x^4+2/3*f*(b*x^4+a)^(1/2)/x^3-2/5*b*d*(b*x^4+a)^(1/ 
2)/a^(1/2)/x/(a^(1/2)+b^(1/2)*x^2)-1/5*(5*f*x^2+d)*(b*x^4+a)^(1/2)/x^5-1/6 
*c*(b*x^4+a)^(3/2)/a/x^6-1/4*b*e*arctanh((b*x^4+a)^(1/2)/a^(1/2))/a^(1/2)- 
2/5*b^(5/4)*d*(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))/a^(3/4)/(b*x^4 
+a)^(1/2)+1/15*b^(3/4)*(3*b^(1/2)*d+5*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))/a^(3/4)/(b*x^4+a)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.17 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.41 \[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x^7} \, dx=-\frac {\sqrt {a+b x^4} \left (12 a d x \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},-\frac {1}{2},-\frac {1}{4},-\frac {b x^4}{a}\right )+5 \left (\sqrt {1+\frac {b x^4}{a}} \left (2 a c+3 a e x^2+2 b c x^4\right )+3 b e x^6 \text {arctanh}\left (\sqrt {1+\frac {b x^4}{a}}\right )+4 a f x^3 \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},-\frac {1}{2},\frac {1}{4},-\frac {b x^4}{a}\right )\right )\right )}{60 a x^6 \sqrt {1+\frac {b x^4}{a}}} \] Input:

Integrate[((c + d*x + e*x^2 + f*x^3)*Sqrt[a + b*x^4])/x^7,x]
 

Output:

-1/60*(Sqrt[a + b*x^4]*(12*a*d*x*Hypergeometric2F1[-5/4, -1/2, -1/4, -((b* 
x^4)/a)] + 5*(Sqrt[1 + (b*x^4)/a]*(2*a*c + 3*a*e*x^2 + 2*b*c*x^4) + 3*b*e* 
x^6*ArcTanh[Sqrt[1 + (b*x^4)/a]] + 4*a*f*x^3*Hypergeometric2F1[-3/4, -1/2, 
 1/4, -((b*x^4)/a)])))/(a*x^6*Sqrt[1 + (b*x^4)/a])
 

Rubi [A] (verified)

Time = 1.04 (sec) , antiderivative size = 345, normalized size of antiderivative = 0.98, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2364, 27, 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.

\(\displaystyle \int \frac {\sqrt {a+b x^4} \left (c+d x+e x^2+f x^3\right )}{x^7} \, dx\)

\(\Big \downarrow \) 2364

\(\displaystyle -2 b \int -\frac {20 f x^3+15 e x^2+12 d x+10 c}{60 x^3 \sqrt {b x^4+a}}dx-\frac {1}{60} \sqrt {a+b x^4} \left (\frac {10 c}{x^6}+\frac {12 d}{x^5}+\frac {15 e}{x^4}+\frac {20 f}{x^3}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{30} b \int \frac {20 f x^3+15 e x^2+12 d x+10 c}{x^3 \sqrt {b x^4+a}}dx-\frac {1}{60} \sqrt {a+b x^4} \left (\frac {10 c}{x^6}+\frac {12 d}{x^5}+\frac {15 e}{x^4}+\frac {20 f}{x^3}\right )\)

\(\Big \downarrow \) 2372

\(\displaystyle \frac {1}{30} b \int \left (\frac {15 e x^2+10 c}{x^3 \sqrt {b x^4+a}}+\frac {20 f x^2+12 d}{x^2 \sqrt {b x^4+a}}\right )dx-\frac {1}{60} \sqrt {a+b x^4} \left (\frac {10 c}{x^6}+\frac {12 d}{x^5}+\frac {15 e}{x^4}+\frac {20 f}{x^3}\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{30} b \left (\frac {2 \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (5 \sqrt {a} f+3 \sqrt {b} d\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{a^{3/4} \sqrt [4]{b} \sqrt {a+b x^4}}-\frac {12 \sqrt [4]{b} d \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{a^{3/4} \sqrt {a+b x^4}}-\frac {15 e \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{2 \sqrt {a}}-\frac {5 c \sqrt {a+b x^4}}{a x^2}-\frac {12 d \sqrt {a+b x^4}}{a x}+\frac {12 \sqrt {b} d x \sqrt {a+b x^4}}{a \left (\sqrt {a}+\sqrt {b} x^2\right )}\right )-\frac {1}{60} \sqrt {a+b x^4} \left (\frac {10 c}{x^6}+\frac {12 d}{x^5}+\frac {15 e}{x^4}+\frac {20 f}{x^3}\right )\)

Input:

Int[((c + d*x + e*x^2 + f*x^3)*Sqrt[a + b*x^4])/x^7,x]
 

Output:

-1/60*(((10*c)/x^6 + (12*d)/x^5 + (15*e)/x^4 + (20*f)/x^3)*Sqrt[a + b*x^4] 
) + (b*((-5*c*Sqrt[a + b*x^4])/(a*x^2) - (12*d*Sqrt[a + b*x^4])/(a*x) + (1 
2*Sqrt[b]*d*x*Sqrt[a + b*x^4])/(a*(Sqrt[a] + Sqrt[b]*x^2)) - (15*e*ArcTanh 
[Sqrt[a + b*x^4]/Sqrt[a]])/(2*Sqrt[a]) - (12*b^(1/4)*d*(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])/(a^(3/4)*Sqrt[a + b*x^4]) + (2*(3*Sqrt[b]*d + 5*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])/(a^(3/4)*b^(1/4)*Sqrt[a + 
b*x^4])))/30
 

Defintions of rubi rules used

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 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2364

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Module[{u 
 = IntHide[x^m*Pq, x]}, Simp[u*(a + b*x^n)^p, x] - Simp[b*n*p   Int[x^(m + 
n)*(a + b*x^n)^(p - 1)*ExpandToSum[u/x^(m + 1), x], x], x]] /; FreeQ[{a, b} 
, x] && PolyQ[Pq, x] && IGtQ[n, 0] && GtQ[p, 0] && LtQ[m + Expon[Pq, x] + 1 
, 0]
 

rule 2372

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)]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 2.16 (sec) , antiderivative size = 262, normalized size of antiderivative = 0.74

method	result	size
risch	\(-\frac {\sqrt {b \,x^{4}+a}\, \left (24 x^{5} b d +10 b c \,x^{4}+20 a f \,x^{3}+15 a e \,x^{2}+12 a d x +10 a c \right )}{60 x^{6} a}+\frac {b \left (\frac {20 a f \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {15 \sqrt {a}\, e \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{4}+a}}{x^{2}}\right )}{2}+\frac {12 i \sqrt {b}\, d \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )}{30 a}\)	\(262\)
elliptic	\(-\frac {c \sqrt {b \,x^{4}+a}}{6 x^{6}}-\frac {d \sqrt {b \,x^{4}+a}}{5 x^{5}}-\frac {e \sqrt {b \,x^{4}+a}}{4 x^{4}}-\frac {f \sqrt {b \,x^{4}+a}}{3 x^{3}}-\frac {b c \sqrt {b \,x^{4}+a}}{6 a \,x^{2}}-\frac {2 d b \sqrt {b \,x^{4}+a}}{5 a x}+\frac {2 f b \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {2 i d \,b^{\frac {3}{2}} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{5 \sqrt {a}\, \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {b e \,\operatorname {arctanh}\left (\frac {\sqrt {a}}{\sqrt {b \,x^{4}+a}}\right )}{4 \sqrt {a}}\)	\(290\)
default	\(-\frac {c \left (b \,x^{4}+a \right )^{\frac {3}{2}}}{6 a \,x^{6}}+d \left (-\frac {\sqrt {b \,x^{4}+a}}{5 x^{5}}-\frac {2 b \sqrt {b \,x^{4}+a}}{5 a x}+\frac {2 i b^{\frac {3}{2}} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{5 \sqrt {a}\, \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+e \left (-\frac {\left (b \,x^{4}+a \right )^{\frac {3}{2}}}{4 a \,x^{4}}-\frac {b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{4}+a}}{x^{2}}\right )}{4 \sqrt {a}}+\frac {b \sqrt {b \,x^{4}+a}}{4 a}\right )+f \left (-\frac {\sqrt {b \,x^{4}+a}}{3 x^{3}}+\frac {2 b \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )\)	\(303\)

Input:

int((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2)/x^7,x,method=_RETURNVERBOSE)
 

Output:

-1/60*(b*x^4+a)^(1/2)*(24*b*d*x^5+10*b*c*x^4+20*a*f*x^3+15*a*e*x^2+12*a*d* 
x+10*a*c)/x^6/a+1/30*b/a*(20*a*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)-15/2*a^(1/2)*e*ln((2*a+2*a^(1/2)*(b*x^4+a) 
^(1/2))/x^2)+12*I*b^(1/2)*d*a^(1/2)/(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)*(Ellip 
ticF(x*(I/a^(1/2)*b^(1/2))^(1/2),I)-EllipticE(x*(I/a^(1/2)*b^(1/2))^(1/2), 
I)))
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.47 \[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x^7} \, dx=-\frac {48 \, \sqrt {a} b d x^{6} \left (-\frac {b}{a}\right )^{\frac {3}{4}} E(\arcsin \left (x \left (-\frac {b}{a}\right )^{\frac {1}{4}}\right )\,|\,-1) - 15 \, \sqrt {a} b e x^{6} \log \left (-\frac {b x^{4} - 2 \, \sqrt {b x^{4} + a} \sqrt {a} + 2 \, a}{x^{4}}\right ) - 16 \, {\left (3 \, b d - 5 \, a f\right )} \sqrt {a} x^{6} \left (-\frac {b}{a}\right )^{\frac {3}{4}} F(\arcsin \left (x \left (-\frac {b}{a}\right )^{\frac {1}{4}}\right )\,|\,-1) + 2 \, {\left (24 \, b d x^{5} + 10 \, b c x^{4} + 20 \, a f x^{3} + 15 \, a e x^{2} + 12 \, a d x + 10 \, a c\right )} \sqrt {b x^{4} + a}}{120 \, a x^{6}} \] Input:

integrate((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2)/x^7,x, algorithm="fricas")
 

Output:

-1/120*(48*sqrt(a)*b*d*x^6*(-b/a)^(3/4)*elliptic_e(arcsin(x*(-b/a)^(1/4)), 
 -1) - 15*sqrt(a)*b*e*x^6*log(-(b*x^4 - 2*sqrt(b*x^4 + a)*sqrt(a) + 2*a)/x 
^4) - 16*(3*b*d - 5*a*f)*sqrt(a)*x^6*(-b/a)^(3/4)*elliptic_f(arcsin(x*(-b/ 
a)^(1/4)), -1) + 2*(24*b*d*x^5 + 10*b*c*x^4 + 20*a*f*x^3 + 15*a*e*x^2 + 12 
*a*d*x + 10*a*c)*sqrt(b*x^4 + a))/(a*x^6)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.58 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.54 \[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x^7} \, dx=\frac {\sqrt {a} d \Gamma \left (- \frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, - \frac {1}{2} \\ - \frac {1}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{5} \Gamma \left (- \frac {1}{4}\right )} + \frac {\sqrt {a} f \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, - \frac {1}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{3} \Gamma \left (\frac {1}{4}\right )} - \frac {\sqrt {b} c \sqrt {\frac {a}{b x^{4}} + 1}}{6 x^{4}} - \frac {\sqrt {b} e \sqrt {\frac {a}{b x^{4}} + 1}}{4 x^{2}} - \frac {b^{\frac {3}{2}} c \sqrt {\frac {a}{b x^{4}} + 1}}{6 a} - \frac {b e \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{2}} \right )}}{4 \sqrt {a}} \] Input:

integrate((f*x**3+e*x**2+d*x+c)*(b*x**4+a)**(1/2)/x**7,x)
 

Output:

sqrt(a)*d*gamma(-5/4)*hyper((-5/4, -1/2), (-1/4,), b*x**4*exp_polar(I*pi)/ 
a)/(4*x**5*gamma(-1/4)) + sqrt(a)*f*gamma(-3/4)*hyper((-3/4, -1/2), (1/4,) 
, b*x**4*exp_polar(I*pi)/a)/(4*x**3*gamma(1/4)) - sqrt(b)*c*sqrt(a/(b*x**4 
) + 1)/(6*x**4) - sqrt(b)*e*sqrt(a/(b*x**4) + 1)/(4*x**2) - b**(3/2)*c*sqr 
t(a/(b*x**4) + 1)/(6*a) - b*e*asinh(sqrt(a)/(sqrt(b)*x**2))/(4*sqrt(a))
 

Maxima [F]

\[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x^7} \, dx=\int { \frac {\sqrt {b x^{4} + a} {\left (f x^{3} + e x^{2} + d x + c\right )}}{x^{7}} \,d x } \] Input:

integrate((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2)/x^7,x, algorithm="maxima")
 

Output:

-1/6*(b*x^4 + a)^(3/2)*c/(a*x^6) + integrate(sqrt(b*x^4 + a)*(f*x^2 + e*x 
+ d)/x^6, x)
 

Giac [F]

\[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x^7} \, dx=\int { \frac {\sqrt {b x^{4} + a} {\left (f x^{3} + e x^{2} + d x + c\right )}}{x^{7}} \,d x } \] Input:

integrate((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2)/x^7,x, algorithm="giac")
 

Output:

integrate(sqrt(b*x^4 + a)*(f*x^3 + e*x^2 + d*x + c)/x^7, x)
 

Mupad [F(-1)]

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

int(((a + b*x^4)^(1/2)*(c + d*x + e*x^2 + f*x^3))/x^7,x)
 

Output:

int(((a + b*x^4)^(1/2)*(c + d*x + e*x^2 + f*x^3))/x^7, x)
 

Reduce [F]

\[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x^7} \, dx=\frac {-4 \sqrt {b \,x^{4}+a}\, a c -8 \sqrt {b \,x^{4}+a}\, a d x -6 \sqrt {b \,x^{4}+a}\, a e \,x^{2}-24 \sqrt {b \,x^{4}+a}\, a f \,x^{3}-4 \sqrt {b \,x^{4}+a}\, b c \,x^{4}+3 \sqrt {a}\, \mathrm {log}\left (\sqrt {b \,x^{4}+a}-\sqrt {a}\right ) b e \,x^{6}-3 \sqrt {a}\, \mathrm {log}\left (\sqrt {b \,x^{4}+a}+\sqrt {a}\right ) b e \,x^{6}-16 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b \,x^{10}+a \,x^{6}}d x \right ) a^{2} d \,x^{6}-48 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b \,x^{8}+a \,x^{4}}d x \right ) a^{2} f \,x^{6}}{24 a \,x^{6}} \] Input:

int((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2)/x^7,x)
 

Output:

( - 4*sqrt(a + b*x**4)*a*c - 8*sqrt(a + b*x**4)*a*d*x - 6*sqrt(a + b*x**4) 
*a*e*x**2 - 24*sqrt(a + b*x**4)*a*f*x**3 - 4*sqrt(a + b*x**4)*b*c*x**4 + 3 
*sqrt(a)*log(sqrt(a + b*x**4) - sqrt(a))*b*e*x**6 - 3*sqrt(a)*log(sqrt(a + 
 b*x**4) + sqrt(a))*b*e*x**6 - 16*int(sqrt(a + b*x**4)/(a*x**6 + b*x**10), 
x)*a**2*d*x**6 - 48*int(sqrt(a + b*x**4)/(a*x**4 + b*x**8),x)*a**2*f*x**6) 
/(24*a*x**6)