Integrand size = 35, antiderivative size = 640 \[ \int \frac {\sqrt {a+b x^3} \left (c+d x+e x^2+f x^3+g x^4\right )}{x^3} \, dx=\frac {3 c \sqrt {a+b x^3}}{2 x^2}-\frac {3 d \sqrt {a+b x^3}}{x}+\frac {3 (7 b d+2 a g) \sqrt {a+b x^3}}{7 b^{2/3} \left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )}-\frac {2 \sqrt {a+b x^3} \left (105 c x-105 d x^2-35 e x^3-21 f x^4-15 g x^5\right )}{105 x^3}-\frac {2}{3} \sqrt {a} e \text {arctanh}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )-\frac {3 \sqrt [4]{3} \sqrt {2-\sqrt {3}} \sqrt [3]{a} (7 b d+2 a g) \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} E\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}\right )|-7-4 \sqrt {3}\right )}{14 b^{2/3} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}}+\frac {3^{3/4} \sqrt {2+\sqrt {3}} \left (7 \sqrt [3]{b} (5 b c+4 a f)-10 \left (1-\sqrt {3}\right ) \sqrt [3]{a} (7 b d+2 a g)\right ) \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}\right ),-7-4 \sqrt {3}\right )}{70 b^{2/3} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}} \]
-2/3*e*arctanh((b*x^3+a)^(1/2)/a^(1/2))*a^(1/2)+3/2*c*(b*x^3+a)^(1/2)/x^2- 3*d*(b*x^3+a)^(1/2)/x-2/105*(-15*g*x^5-21*f*x^4-35*e*x^3-105*d*x^2+105*c*x )*(b*x^3+a)^(1/2)/x^3+3/7*(2*a*g+7*b*d)*(b*x^3+a)^(1/2)/b^(2/3)/(b^(1/3)*x +a^(1/3)*(1+3^(1/2)))-3/14*3^(1/4)*a^(1/3)*(2*a*g+7*b*d)*(a^(1/3)+b^(1/3)* x)*EllipticE((b^(1/3)*x+a^(1/3)*(1-3^(1/2)))/(b^(1/3)*x+a^(1/3)*(1+3^(1/2) )),I*3^(1/2)+2*I)*(1/2*6^(1/2)-1/2*2^(1/2))*((a^(2/3)-a^(1/3)*b^(1/3)*x+b^ (2/3)*x^2)/(b^(1/3)*x+a^(1/3)*(1+3^(1/2)))^2)^(1/2)/b^(2/3)/(b*x^3+a)^(1/2 )/(a^(1/3)*(a^(1/3)+b^(1/3)*x)/(b^(1/3)*x+a^(1/3)*(1+3^(1/2)))^2)^(1/2)+1/ 70*3^(3/4)*(a^(1/3)+b^(1/3)*x)*EllipticF((b^(1/3)*x+a^(1/3)*(1-3^(1/2)))/( b^(1/3)*x+a^(1/3)*(1+3^(1/2))),I*3^(1/2)+2*I)*(7*b^(1/3)*(4*a*f+5*b*c)-10* a^(1/3)*(2*a*g+7*b*d)*(1-3^(1/2)))*(1/2*6^(1/2)+1/2*2^(1/2))*((a^(2/3)-a^( 1/3)*b^(1/3)*x+b^(2/3)*x^2)/(b^(1/3)*x+a^(1/3)*(1+3^(1/2)))^2)^(1/2)/b^(2/ 3)/(b*x^3+a)^(1/2)/(a^(1/3)*(a^(1/3)+b^(1/3)*x)/(b^(1/3)*x+a^(1/3)*(1+3^(1 /2)))^2)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 9.78 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.34 \[ \int \frac {\sqrt {a+b x^3} \left (c+d x+e x^2+f x^3+g x^4\right )}{x^3} \, dx=\frac {-3 c \sqrt {a+b x^3} \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},-\frac {1}{2},\frac {1}{3},-\frac {b x^3}{a}\right )+x \left (-6 d \sqrt {a+b x^3} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {1}{3},\frac {2}{3},-\frac {b x^3}{a}\right )+x \left (4 e \sqrt {1+\frac {b x^3}{a}} \left (\sqrt {a+b x^3}-\sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )\right )+6 f x \sqrt {a+b x^3} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{3},\frac {4}{3},-\frac {b x^3}{a}\right )+3 g x^2 \sqrt {a+b x^3} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {2}{3},\frac {5}{3},-\frac {b x^3}{a}\right )\right )\right )}{6 x^2 \sqrt {1+\frac {b x^3}{a}}} \]
(-3*c*Sqrt[a + b*x^3]*Hypergeometric2F1[-2/3, -1/2, 1/3, -((b*x^3)/a)] + x *(-6*d*Sqrt[a + b*x^3]*Hypergeometric2F1[-1/2, -1/3, 2/3, -((b*x^3)/a)] + x*(4*e*Sqrt[1 + (b*x^3)/a]*(Sqrt[a + b*x^3] - Sqrt[a]*ArcTanh[Sqrt[a + b*x ^3]/Sqrt[a]]) + 6*f*x*Sqrt[a + b*x^3]*Hypergeometric2F1[-1/2, 1/3, 4/3, -( (b*x^3)/a)] + 3*g*x^2*Sqrt[a + b*x^3]*Hypergeometric2F1[-1/2, 2/3, 5/3, -( (b*x^3)/a)])))/(6*x^2*Sqrt[1 + (b*x^3)/a])
Time = 1.15 (sec) , antiderivative size = 657, normalized size of antiderivative = 1.03, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.343, Rules used = {2365, 27, 2374, 2374, 27, 2371, 798, 73, 221, 2417, 759, 2416}
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.
\(\displaystyle \int \frac {\sqrt {a+b x^3} \left (c+d x+e x^2+f x^3+g x^4\right )}{x^3} \, dx\) |
\(\Big \downarrow \) 2365 |
\(\displaystyle \frac {3}{2} a \int -\frac {2 \left (-15 g x^4-21 f x^3-35 e x^2-105 d x+105 c\right )}{105 x^3 \sqrt {b x^3+a}}dx-\frac {2 \sqrt {a+b x^3} \left (105 c x-105 d x^2-35 e x^3-21 f x^4-15 g x^5\right )}{105 x^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {1}{35} a \int \frac {-15 g x^4-21 f x^3-35 e x^2-105 d x+105 c}{x^3 \sqrt {b x^3+a}}dx-\frac {2 \sqrt {a+b x^3} \left (105 c x-105 d x^2-35 e x^3-21 f x^4-15 g x^5\right )}{105 x^3}\) |
\(\Big \downarrow \) 2374 |
\(\displaystyle -\frac {1}{35} a \left (-\frac {\int \frac {60 a g x^3+21 (5 b c+4 a f) x^2+140 a e x+420 a d}{x^2 \sqrt {b x^3+a}}dx}{4 a}-\frac {105 c \sqrt {a+b x^3}}{2 a x^2}\right )-\frac {2 \sqrt {a+b x^3} \left (105 c x-105 d x^2-35 e x^3-21 f x^4-15 g x^5\right )}{105 x^3}\) |
\(\Big \downarrow \) 2374 |
\(\displaystyle -\frac {1}{35} a \left (-\frac {-\frac {\int -\frac {2 \left (140 e a^2+30 (7 b d+2 a g) x^2 a+21 (5 b c+4 a f) x a\right )}{x \sqrt {b x^3+a}}dx}{2 a}-\frac {420 d \sqrt {a+b x^3}}{x}}{4 a}-\frac {105 c \sqrt {a+b x^3}}{2 a x^2}\right )-\frac {2 \sqrt {a+b x^3} \left (105 c x-105 d x^2-35 e x^3-21 f x^4-15 g x^5\right )}{105 x^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {1}{35} a \left (-\frac {\frac {\int \frac {140 e a^2+30 (7 b d+2 a g) x^2 a+21 (5 b c+4 a f) x a}{x \sqrt {b x^3+a}}dx}{a}-\frac {420 d \sqrt {a+b x^3}}{x}}{4 a}-\frac {105 c \sqrt {a+b x^3}}{2 a x^2}\right )-\frac {2 \sqrt {a+b x^3} \left (105 c x-105 d x^2-35 e x^3-21 f x^4-15 g x^5\right )}{105 x^3}\) |
\(\Big \downarrow \) 2371 |
\(\displaystyle -\frac {1}{35} a \left (-\frac {\frac {140 a^2 e \int \frac {1}{x \sqrt {b x^3+a}}dx+\int \frac {21 a (5 b c+4 a f)+30 a (7 b d+2 a g) x}{\sqrt {b x^3+a}}dx}{a}-\frac {420 d \sqrt {a+b x^3}}{x}}{4 a}-\frac {105 c \sqrt {a+b x^3}}{2 a x^2}\right )-\frac {2 \sqrt {a+b x^3} \left (105 c x-105 d x^2-35 e x^3-21 f x^4-15 g x^5\right )}{105 x^3}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle -\frac {1}{35} a \left (-\frac {\frac {\frac {140}{3} a^2 e \int \frac {1}{x^3 \sqrt {b x^3+a}}dx^3+\int \frac {21 a (5 b c+4 a f)+30 a (7 b d+2 a g) x}{\sqrt {b x^3+a}}dx}{a}-\frac {420 d \sqrt {a+b x^3}}{x}}{4 a}-\frac {105 c \sqrt {a+b x^3}}{2 a x^2}\right )-\frac {2 \sqrt {a+b x^3} \left (105 c x-105 d x^2-35 e x^3-21 f x^4-15 g x^5\right )}{105 x^3}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {1}{35} a \left (-\frac {\frac {\frac {280 a^2 e \int \frac {1}{\frac {x^6}{b}-\frac {a}{b}}d\sqrt {b x^3+a}}{3 b}+\int \frac {21 a (5 b c+4 a f)+30 a (7 b d+2 a g) x}{\sqrt {b x^3+a}}dx}{a}-\frac {420 d \sqrt {a+b x^3}}{x}}{4 a}-\frac {105 c \sqrt {a+b x^3}}{2 a x^2}\right )-\frac {2 \sqrt {a+b x^3} \left (105 c x-105 d x^2-35 e x^3-21 f x^4-15 g x^5\right )}{105 x^3}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {1}{35} a \left (-\frac {\frac {\int \frac {21 a (5 b c+4 a f)+30 a (7 b d+2 a g) x}{\sqrt {b x^3+a}}dx-\frac {280}{3} a^{3/2} e \text {arctanh}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{a}-\frac {420 d \sqrt {a+b x^3}}{x}}{4 a}-\frac {105 c \sqrt {a+b x^3}}{2 a x^2}\right )-\frac {2 \sqrt {a+b x^3} \left (105 c x-105 d x^2-35 e x^3-21 f x^4-15 g x^5\right )}{105 x^3}\) |
\(\Big \downarrow \) 2417 |
\(\displaystyle -\frac {1}{35} a \left (-\frac {\frac {3 a \left (7 (4 a f+5 b c)-\frac {10 \left (1-\sqrt {3}\right ) \sqrt [3]{a} (2 a g+7 b d)}{\sqrt [3]{b}}\right ) \int \frac {1}{\sqrt {b x^3+a}}dx+\frac {30 a (2 a g+7 b d) \int \frac {\sqrt [3]{b} x+\left (1-\sqrt {3}\right ) \sqrt [3]{a}}{\sqrt {b x^3+a}}dx}{\sqrt [3]{b}}-\frac {280}{3} a^{3/2} e \text {arctanh}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{a}-\frac {420 d \sqrt {a+b x^3}}{x}}{4 a}-\frac {105 c \sqrt {a+b x^3}}{2 a x^2}\right )-\frac {2 \sqrt {a+b x^3} \left (105 c x-105 d x^2-35 e x^3-21 f x^4-15 g x^5\right )}{105 x^3}\) |
\(\Big \downarrow \) 759 |
\(\displaystyle -\frac {1}{35} a \left (-\frac {\frac {\frac {30 a (2 a g+7 b d) \int \frac {\sqrt [3]{b} x+\left (1-\sqrt {3}\right ) \sqrt [3]{a}}{\sqrt {b x^3+a}}dx}{\sqrt [3]{b}}+\frac {2\ 3^{3/4} \sqrt {2+\sqrt {3}} a \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [3]{b} x+\left (1-\sqrt {3}\right ) \sqrt [3]{a}}{\sqrt [3]{b} x+\left (1+\sqrt {3}\right ) \sqrt [3]{a}}\right ),-7-4 \sqrt {3}\right ) \left (7 (4 a f+5 b c)-\frac {10 \left (1-\sqrt {3}\right ) \sqrt [3]{a} (2 a g+7 b d)}{\sqrt [3]{b}}\right )}{\sqrt [3]{b} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}}-\frac {280}{3} a^{3/2} e \text {arctanh}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{a}-\frac {420 d \sqrt {a+b x^3}}{x}}{4 a}-\frac {105 c \sqrt {a+b x^3}}{2 a x^2}\right )-\frac {2 \sqrt {a+b x^3} \left (105 c x-105 d x^2-35 e x^3-21 f x^4-15 g x^5\right )}{105 x^3}\) |
\(\Big \downarrow \) 2416 |
\(\displaystyle -\frac {1}{35} a \left (-\frac {\frac {\frac {2\ 3^{3/4} \sqrt {2+\sqrt {3}} a \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [3]{b} x+\left (1-\sqrt {3}\right ) \sqrt [3]{a}}{\sqrt [3]{b} x+\left (1+\sqrt {3}\right ) \sqrt [3]{a}}\right ),-7-4 \sqrt {3}\right ) \left (7 (4 a f+5 b c)-\frac {10 \left (1-\sqrt {3}\right ) \sqrt [3]{a} (2 a g+7 b d)}{\sqrt [3]{b}}\right )}{\sqrt [3]{b} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}}+\frac {30 a (2 a g+7 b d) \left (\frac {2 \sqrt {a+b x^3}}{\sqrt [3]{b} \left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )}-\frac {\sqrt [4]{3} \sqrt {2-\sqrt {3}} \sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} E\left (\arcsin \left (\frac {\sqrt [3]{b} x+\left (1-\sqrt {3}\right ) \sqrt [3]{a}}{\sqrt [3]{b} x+\left (1+\sqrt {3}\right ) \sqrt [3]{a}}\right )|-7-4 \sqrt {3}\right )}{\sqrt [3]{b} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}}\right )}{\sqrt [3]{b}}-\frac {280}{3} a^{3/2} e \text {arctanh}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{a}-\frac {420 d \sqrt {a+b x^3}}{x}}{4 a}-\frac {105 c \sqrt {a+b x^3}}{2 a x^2}\right )-\frac {2 \sqrt {a+b x^3} \left (105 c x-105 d x^2-35 e x^3-21 f x^4-15 g x^5\right )}{105 x^3}\) |
(-2*Sqrt[a + b*x^3]*(105*c*x - 105*d*x^2 - 35*e*x^3 - 21*f*x^4 - 15*g*x^5) )/(105*x^3) - (a*((-105*c*Sqrt[a + b*x^3])/(2*a*x^2) - ((-420*d*Sqrt[a + b *x^3])/x + ((-280*a^(3/2)*e*ArcTanh[Sqrt[a + b*x^3]/Sqrt[a]])/3 + (30*a*(7 *b*d + 2*a*g)*((2*Sqrt[a + b*x^3])/(b^(1/3)*((1 + Sqrt[3])*a^(1/3) + b^(1/ 3)*x)) - (3^(1/4)*Sqrt[2 - Sqrt[3]]*a^(1/3)*(a^(1/3) + b^(1/3)*x)*Sqrt[(a^ (2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2)/((1 + Sqrt[3])*a^(1/3) + b^(1/3)* x)^2]*EllipticE[ArcSin[((1 - Sqrt[3])*a^(1/3) + b^(1/3)*x)/((1 + Sqrt[3])* a^(1/3) + b^(1/3)*x)], -7 - 4*Sqrt[3]])/(b^(1/3)*Sqrt[(a^(1/3)*(a^(1/3) + b^(1/3)*x))/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)^2]*Sqrt[a + b*x^3])))/b^(1 /3) + (2*3^(3/4)*Sqrt[2 + Sqrt[3]]*a*(7*(5*b*c + 4*a*f) - (10*(1 - Sqrt[3] )*a^(1/3)*(7*b*d + 2*a*g))/b^(1/3))*(a^(1/3) + b^(1/3)*x)*Sqrt[(a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2)/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)^2]*El lipticF[ArcSin[((1 - Sqrt[3])*a^(1/3) + b^(1/3)*x)/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)], -7 - 4*Sqrt[3]])/(b^(1/3)*Sqrt[(a^(1/3)*(a^(1/3) + b^(1/3)* x))/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)^2]*Sqrt[a + b*x^3]))/a)/(4*a)))/35
3.5.51.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt[2 + Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s *x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[s* ((s + r*x)/((1 + Sqrt[3])*s + r*x)^2)]))*EllipticF[ArcSin[((1 - Sqrt[3])*s + r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]], x]] /; FreeQ[{a, b}, x] & & PosQ[a]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n Subst [Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> M odule[{q = Expon[Pq, x], i}, Simp[(c*x)^m*(a + b*x^n)^p*Sum[Coeff[Pq, x, i] *(x^(i + 1)/(m + n*p + i + 1)), {i, 0, q}], x] + Simp[a*n*p Int[(c*x)^m*( a + b*x^n)^(p - 1)*Sum[Coeff[Pq, x, i]*(x^i/(m + n*p + i + 1)), {i, 0, q}], x], x]] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[(n - 1)/2, 0] && GtQ[p, 0]
Int[(Pq_)/((x_)*Sqrt[(a_) + (b_.)*(x_)^(n_)]), x_Symbol] :> Simp[Coeff[Pq, x, 0] Int[1/(x*Sqrt[a + b*x^n]), x], x] + Int[ExpandToSum[(Pq - Coeff[Pq, x, 0])/x, x]/Sqrt[a + b*x^n], x] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IG tQ[n, 0] && NeQ[Coeff[Pq, x, 0], 0]
Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Wit h[{Pq0 = Coeff[Pq, x, 0]}, Simp[Pq0*(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c *(m + 1))), x] + Simp[1/(2*a*c*(m + 1)) Int[(c*x)^(m + 1)*ExpandToSum[2*a *(m + 1)*((Pq - Pq0)/x) - 2*b*Pq0*(m + n*(p + 1) + 1)*x^(n - 1), x]*(a + b* x^n)^p, x], x] /; NeQ[Pq0, 0]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[m, -1] && LeQ[n - 1, Expon[Pq, x]]
Int[((c_) + (d_.)*(x_))/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = N umer[Simplify[(1 - Sqrt[3])*(d/c)]], s = Denom[Simplify[(1 - Sqrt[3])*(d/c) ]]}, Simp[2*d*s^3*(Sqrt[a + b*x^3]/(a*r^2*((1 + Sqrt[3])*s + r*x))), x] - S imp[3^(1/4)*Sqrt[2 - Sqrt[3]]*d*s*(s + r*x)*(Sqrt[(s^2 - r*s*x + r^2*x^2)/( (1 + Sqrt[3])*s + r*x)^2]/(r^2*Sqrt[a + b*x^3]*Sqrt[s*((s + r*x)/((1 + Sqrt [3])*s + r*x)^2)]))*EllipticE[ArcSin[((1 - Sqrt[3])*s + r*x)/((1 + Sqrt[3]) *s + r*x)], -7 - 4*Sqrt[3]], x]] /; FreeQ[{a, b, c, d}, x] && PosQ[a] && Eq Q[b*c^3 - 2*(5 - 3*Sqrt[3])*a*d^3, 0]
Int[((c_) + (d_.)*(x_))/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = N umer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(c*r - (1 - Sqrt[3])*d*s)/r Int[1/Sqrt[a + b*x^3], x], x] + Simp[d/r Int[((1 - Sqrt[3])*s + r*x)/Sq rt[a + b*x^3], x], x]] /; FreeQ[{a, b, c, d}, x] && PosQ[a] && NeQ[b*c^3 - 2*(5 - 3*Sqrt[3])*a*d^3, 0]
Time = 1.93 (sec) , antiderivative size = 826, normalized size of antiderivative = 1.29
method | result | size |
elliptic | \(\text {Expression too large to display}\) | \(826\) |
default | \(\text {Expression too large to display}\) | \(1529\) |
risch | \(\text {Expression too large to display}\) | \(2244\) |
-1/2*c*(b*x^3+a)^(1/2)/x^2-d*(b*x^3+a)^(1/2)/x+2/7*g*x^2*(b*x^3+a)^(1/2)+2 /5*f*x*(b*x^3+a)^(1/2)+2/3*e*(b*x^3+a)^(1/2)-2/3*I*(3/5*a*f+3/4*b*c)*3^(1/ 2)/b*(-a*b^2)^(1/3)*(I*(x+1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1 /3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2)*((x-1/b*(-a*b^2)^(1/3))/(-3/2/b*(-a*b ^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2)*(-I*(x+1/2/b*(-a*b^2)^(1/ 3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2)/(b*x^3+ a)^(1/2)*EllipticF(1/3*3^(1/2)*(I*(x+1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b* (-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2),(I*3^(1/2)/b*(-a*b^2)^(1/3 )/(-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2))-2/3*I*(3/ 7*a*g+3/2*b*d)*3^(1/2)/b*(-a*b^2)^(1/3)*(I*(x+1/2/b*(-a*b^2)^(1/3)-1/2*I*3 ^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2)*((x-1/b*(-a*b^2)^ (1/3))/(-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2)*(-I*( x+1/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^ (1/3))^(1/2)/(b*x^3+a)^(1/2)*((-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b ^2)^(1/3))*EllipticE(1/3*3^(1/2)*(I*(x+1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/ b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2),(I*3^(1/2)/b*(-a*b^2)^(1 /3)/(-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2))+1/b*(-a *b^2)^(1/3)*EllipticF(1/3*3^(1/2)*(I*(x+1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2) /b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2),(I*3^(1/2)/b*(-a*b^2)^( 1/3)/(-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2)))-2/...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.24 (sec) , antiderivative size = 320, normalized size of antiderivative = 0.50 \[ \int \frac {\sqrt {a+b x^3} \left (c+d x+e x^2+f x^3+g x^4\right )}{x^3} \, dx=\left [\frac {35 \, \sqrt {a} b e x^{2} \log \left (-\frac {b^{2} x^{6} + 8 \, a b x^{3} - 4 \, {\left (b x^{3} + 2 \, a\right )} \sqrt {b x^{3} + a} \sqrt {a} + 8 \, a^{2}}{x^{6}}\right ) + 63 \, {\left (5 \, b c + 4 \, a f\right )} \sqrt {b} x^{2} {\rm weierstrassPInverse}\left (0, -\frac {4 \, a}{b}, x\right ) - 90 \, {\left (7 \, b d + 2 \, a g\right )} \sqrt {b} x^{2} {\rm weierstrassZeta}\left (0, -\frac {4 \, a}{b}, {\rm weierstrassPInverse}\left (0, -\frac {4 \, a}{b}, x\right )\right ) + {\left (60 \, b g x^{4} + 84 \, b f x^{3} + 140 \, b e x^{2} - 210 \, b d x - 105 \, b c\right )} \sqrt {b x^{3} + a}}{210 \, b x^{2}}, \frac {70 \, \sqrt {-a} b e x^{2} \arctan \left (\frac {2 \, \sqrt {b x^{3} + a} \sqrt {-a}}{b x^{3} + 2 \, a}\right ) + 63 \, {\left (5 \, b c + 4 \, a f\right )} \sqrt {b} x^{2} {\rm weierstrassPInverse}\left (0, -\frac {4 \, a}{b}, x\right ) - 90 \, {\left (7 \, b d + 2 \, a g\right )} \sqrt {b} x^{2} {\rm weierstrassZeta}\left (0, -\frac {4 \, a}{b}, {\rm weierstrassPInverse}\left (0, -\frac {4 \, a}{b}, x\right )\right ) + {\left (60 \, b g x^{4} + 84 \, b f x^{3} + 140 \, b e x^{2} - 210 \, b d x - 105 \, b c\right )} \sqrt {b x^{3} + a}}{210 \, b x^{2}}\right ] \]
[1/210*(35*sqrt(a)*b*e*x^2*log(-(b^2*x^6 + 8*a*b*x^3 - 4*(b*x^3 + 2*a)*sqr t(b*x^3 + a)*sqrt(a) + 8*a^2)/x^6) + 63*(5*b*c + 4*a*f)*sqrt(b)*x^2*weiers trassPInverse(0, -4*a/b, x) - 90*(7*b*d + 2*a*g)*sqrt(b)*x^2*weierstrassZe ta(0, -4*a/b, weierstrassPInverse(0, -4*a/b, x)) + (60*b*g*x^4 + 84*b*f*x^ 3 + 140*b*e*x^2 - 210*b*d*x - 105*b*c)*sqrt(b*x^3 + a))/(b*x^2), 1/210*(70 *sqrt(-a)*b*e*x^2*arctan(2*sqrt(b*x^3 + a)*sqrt(-a)/(b*x^3 + 2*a)) + 63*(5 *b*c + 4*a*f)*sqrt(b)*x^2*weierstrassPInverse(0, -4*a/b, x) - 90*(7*b*d + 2*a*g)*sqrt(b)*x^2*weierstrassZeta(0, -4*a/b, weierstrassPInverse(0, -4*a/ b, x)) + (60*b*g*x^4 + 84*b*f*x^3 + 140*b*e*x^2 - 210*b*d*x - 105*b*c)*sqr t(b*x^3 + a))/(b*x^2)]
Time = 2.98 (sec) , antiderivative size = 255, normalized size of antiderivative = 0.40 \[ \int \frac {\sqrt {a+b x^3} \left (c+d x+e x^2+f x^3+g x^4\right )}{x^3} \, dx=\frac {\sqrt {a} c \Gamma \left (- \frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, - \frac {1}{2} \\ \frac {1}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 x^{2} \Gamma \left (\frac {1}{3}\right )} + \frac {\sqrt {a} d \Gamma \left (- \frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{3} \\ \frac {2}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 x \Gamma \left (\frac {2}{3}\right )} - \frac {2 \sqrt {a} e \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{\frac {3}{2}}} \right )}}{3} + \frac {\sqrt {a} f x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} + \frac {\sqrt {a} g x^{2} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {5}{3}\right )} + \frac {2 a e}{3 \sqrt {b} x^{\frac {3}{2}} \sqrt {\frac {a}{b x^{3}} + 1}} + \frac {2 \sqrt {b} e x^{\frac {3}{2}}}{3 \sqrt {\frac {a}{b x^{3}} + 1}} \]
sqrt(a)*c*gamma(-2/3)*hyper((-2/3, -1/2), (1/3,), b*x**3*exp_polar(I*pi)/a )/(3*x**2*gamma(1/3)) + sqrt(a)*d*gamma(-1/3)*hyper((-1/2, -1/3), (2/3,), b*x**3*exp_polar(I*pi)/a)/(3*x*gamma(2/3)) - 2*sqrt(a)*e*asinh(sqrt(a)/(sq rt(b)*x**(3/2)))/3 + sqrt(a)*f*x*gamma(1/3)*hyper((-1/2, 1/3), (4/3,), b*x **3*exp_polar(I*pi)/a)/(3*gamma(4/3)) + sqrt(a)*g*x**2*gamma(2/3)*hyper((- 1/2, 2/3), (5/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(5/3)) + 2*a*e/(3*sqr t(b)*x**(3/2)*sqrt(a/(b*x**3) + 1)) + 2*sqrt(b)*e*x**(3/2)/(3*sqrt(a/(b*x* *3) + 1))
\[ \int \frac {\sqrt {a+b x^3} \left (c+d x+e x^2+f x^3+g x^4\right )}{x^3} \, dx=\int { \frac {{\left (g x^{4} + f x^{3} + e x^{2} + d x + c\right )} \sqrt {b x^{3} + a}}{x^{3}} \,d x } \]
\[ \int \frac {\sqrt {a+b x^3} \left (c+d x+e x^2+f x^3+g x^4\right )}{x^3} \, dx=\int { \frac {{\left (g x^{4} + f x^{3} + e x^{2} + d x + c\right )} \sqrt {b x^{3} + a}}{x^{3}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {a+b x^3} \left (c+d x+e x^2+f x^3+g x^4\right )}{x^3} \, dx=\int \frac {\sqrt {b\,x^3+a}\,\left (g\,x^4+f\,x^3+e\,x^2+d\,x+c\right )}{x^3} \,d x \]