Integrand size = 26, antiderivative size = 196 \[ \int \frac {\left (d+e x^3\right )^{3/2}}{a+b x^3+c x^6} \, dx=-\frac {2 c d x \sqrt {d+e x^3} \operatorname {AppellF1}\left (\frac {1}{3},-\frac {3}{2},1,\frac {4}{3},-\frac {e x^3}{d},-\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) \sqrt {1+\frac {e x^3}{d}}}-\frac {2 c d x \sqrt {d+e x^3} \operatorname {AppellF1}\left (\frac {1}{3},-\frac {3}{2},1,\frac {4}{3},-\frac {e x^3}{d},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c+b \sqrt {b^2-4 a c}\right ) \sqrt {1+\frac {e x^3}{d}}} \] Output:
-2*c*d*x*(e*x^3+d)^(1/2)*AppellF1(1/3,1,-3/2,4/3,-2*c*x^3/(b-(-4*a*c+b^2)^ (1/2)),-e*x^3/d)/(b^2-4*a*c-b*(-4*a*c+b^2)^(1/2))/(1+e*x^3/d)^(1/2)-2*c*d* x*(e*x^3+d)^(1/2)*AppellF1(1/3,1,-3/2,4/3,-2*c*x^3/(b+(-4*a*c+b^2)^(1/2)), -e*x^3/d)/(b*(-4*a*c+b^2)^(1/2)-4*a*c+b^2)/(1+e*x^3/d)^(1/2)
\[ \int \frac {\left (d+e x^3\right )^{3/2}}{a+b x^3+c x^6} \, dx=\int \frac {\left (d+e x^3\right )^{3/2}}{a+b x^3+c x^6} \, dx \] Input:
Integrate[(d + e*x^3)^(3/2)/(a + b*x^3 + c*x^6),x]
Output:
Integrate[(d + e*x^3)^(3/2)/(a + b*x^3 + c*x^6), x]
Time = 0.38 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.05, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1758, 937, 936}
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.
\(\displaystyle \int \frac {\left (d+e x^3\right )^{3/2}}{a+b x^3+c x^6} \, dx\) |
\(\Big \downarrow \) 1758 |
\(\displaystyle \frac {2 c \int \frac {\left (e x^3+d\right )^{3/2}}{2 c x^3+b-\sqrt {b^2-4 a c}}dx}{\sqrt {b^2-4 a c}}-\frac {2 c \int \frac {\left (e x^3+d\right )^{3/2}}{2 c x^3+b+\sqrt {b^2-4 a c}}dx}{\sqrt {b^2-4 a c}}\) |
\(\Big \downarrow \) 937 |
\(\displaystyle \frac {2 c d \sqrt {d+e x^3} \int \frac {\left (\frac {e x^3}{d}+1\right )^{3/2}}{2 c x^3+b-\sqrt {b^2-4 a c}}dx}{\sqrt {b^2-4 a c} \sqrt {\frac {e x^3}{d}+1}}-\frac {2 c d \sqrt {d+e x^3} \int \frac {\left (\frac {e x^3}{d}+1\right )^{3/2}}{2 c x^3+b+\sqrt {b^2-4 a c}}dx}{\sqrt {b^2-4 a c} \sqrt {\frac {e x^3}{d}+1}}\) |
\(\Big \downarrow \) 936 |
\(\displaystyle \frac {2 c d x \sqrt {d+e x^3} \operatorname {AppellF1}\left (\frac {1}{3},1,-\frac {3}{2},\frac {4}{3},-\frac {2 c x^3}{b-\sqrt {b^2-4 a c}},-\frac {e x^3}{d}\right )}{\sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}\right ) \sqrt {\frac {e x^3}{d}+1}}-\frac {2 c d x \sqrt {d+e x^3} \operatorname {AppellF1}\left (\frac {1}{3},1,-\frac {3}{2},\frac {4}{3},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}},-\frac {e x^3}{d}\right )}{\sqrt {b^2-4 a c} \left (\sqrt {b^2-4 a c}+b\right ) \sqrt {\frac {e x^3}{d}+1}}\) |
Input:
Int[(d + e*x^3)^(3/2)/(a + b*x^3 + c*x^6),x]
Output:
(2*c*d*x*Sqrt[d + e*x^3]*AppellF1[1/3, 1, -3/2, 4/3, (-2*c*x^3)/(b - Sqrt[ b^2 - 4*a*c]), -((e*x^3)/d)])/(Sqrt[b^2 - 4*a*c]*(b - Sqrt[b^2 - 4*a*c])*S qrt[1 + (e*x^3)/d]) - (2*c*d*x*Sqrt[d + e*x^3]*AppellF1[1/3, 1, -3/2, 4/3, (-2*c*x^3)/(b + Sqrt[b^2 - 4*a*c]), -((e*x^3)/d)])/(Sqrt[b^2 - 4*a*c]*(b + Sqrt[b^2 - 4*a*c])*Sqrt[1 + (e*x^3)/d])
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p, -q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c) ], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p]) Int[(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, p, q }, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] && !(IntegerQ[p] || GtQ[a, 0])
Int[((d_) + (e_.)*(x_)^(n_))^(q_)/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_ )), x_Symbol] :> With[{r = Rt[b^2 - 4*a*c, 2]}, Simp[2*(c/r) Int[(d + e*x ^n)^q/(b - r + 2*c*x^n), x], x] - Simp[2*(c/r) Int[(d + e*x^n)^q/(b + r + 2*c*x^n), x], x]] /; FreeQ[{a, b, c, d, e, n, q}, x] && EqQ[n2, 2*n] && Ne Q[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && !IntegerQ[q]
Result contains higher order function than in optimal. Order 9 vs. order 6.
Time = 0.47 (sec) , antiderivative size = 1091, normalized size of antiderivative = 5.57
method | result | size |
default | \(\text {Expression too large to display}\) | \(1091\) |
elliptic | \(\text {Expression too large to display}\) | \(1091\) |
Input:
int((e*x^3+d)^(3/2)/(c*x^6+b*x^3+a),x,method=_RETURNVERBOSE)
Output:
-2/3*I*e/c*3^(1/2)*(-d*e^2)^(1/3)*(I*(x+1/2/e*(-d*e^2)^(1/3)-1/2*I*3^(1/2) /e*(-d*e^2)^(1/3))*3^(1/2)*e/(-d*e^2)^(1/3))^(1/2)*((x-1/e*(-d*e^2)^(1/3)) /(-3/2/e*(-d*e^2)^(1/3)+1/2*I*3^(1/2)/e*(-d*e^2)^(1/3)))^(1/2)*(-I*(x+1/2/ e*(-d*e^2)^(1/3)+1/2*I*3^(1/2)/e*(-d*e^2)^(1/3))*3^(1/2)*e/(-d*e^2)^(1/3)) ^(1/2)/(e*x^3+d)^(1/2)*EllipticF(1/3*3^(1/2)*(I*(x+1/2/e*(-d*e^2)^(1/3)-1/ 2*I*3^(1/2)/e*(-d*e^2)^(1/3))*3^(1/2)*e/(-d*e^2)^(1/3))^(1/2),(I*3^(1/2)/e *(-d*e^2)^(1/3)/(-3/2/e*(-d*e^2)^(1/3)+1/2*I*3^(1/2)/e*(-d*e^2)^(1/3)))^(1 /2))+1/3*I/c/e^2*2^(1/2)*sum((_alpha^3*b*e^2-2*_alpha^3*c*d*e+a*e^2-c*d^2) /_alpha^2/(2*_alpha^3*c+b)/(a*e^2-b*d*e+c*d^2)*(-d*e^2)^(1/3)*(1/2*I*e*(2* x+1/e*(-I*3^(1/2)*(-d*e^2)^(1/3)+(-d*e^2)^(1/3)))/(-d*e^2)^(1/3))^(1/2)*(e *(x-1/e*(-d*e^2)^(1/3))/(-3*(-d*e^2)^(1/3)+I*3^(1/2)*(-d*e^2)^(1/3)))^(1/2 )*(-1/2*I*e*(2*x+1/e*(I*3^(1/2)*(-d*e^2)^(1/3)+(-d*e^2)^(1/3)))/(-d*e^2)^( 1/3))^(1/2)/(e*x^3+d)^(1/2)*(2*e^2*(_alpha^5*c*e+_alpha^2*b*e-_alpha^2*c*d )+I*(-d*e^2)^(1/3)*3^(1/2)*_alpha^4*c*e^2-I*(-d*e^2)^(2/3)*3^(1/2)*_alpha^ 3*c*e-(-d*e^2)^(1/3)*_alpha^4*c*e^2-(-d*e^2)^(2/3)*_alpha^3*c*e+I*(-d*e^2) ^(1/3)*3^(1/2)*_alpha*b*e^2-I*(-d*e^2)^(1/3)*3^(1/2)*_alpha*c*d*e-I*(-d*e^ 2)^(2/3)*3^(1/2)*b*e+I*(-d*e^2)^(2/3)*3^(1/2)*c*d-(-d*e^2)^(1/3)*_alpha*b* e^2+(-d*e^2)^(1/3)*_alpha*c*d*e-(-d*e^2)^(2/3)*b*e+(-d*e^2)^(2/3)*c*d)*Ell ipticPi(1/3*3^(1/2)*(I*(x+1/2/e*(-d*e^2)^(1/3)-1/2*I*3^(1/2)/e*(-d*e^2)^(1 /3))*3^(1/2)*e/(-d*e^2)^(1/3))^(1/2),1/2/e*(2*I*(-d*e^2)^(1/3)*3^(1/2)*...
Timed out. \[ \int \frac {\left (d+e x^3\right )^{3/2}}{a+b x^3+c x^6} \, dx=\text {Timed out} \] Input:
integrate((e*x^3+d)^(3/2)/(c*x^6+b*x^3+a),x, algorithm="fricas")
Output:
Timed out
Timed out. \[ \int \frac {\left (d+e x^3\right )^{3/2}}{a+b x^3+c x^6} \, dx=\text {Timed out} \] Input:
integrate((e*x**3+d)**(3/2)/(c*x**6+b*x**3+a),x)
Output:
Timed out
\[ \int \frac {\left (d+e x^3\right )^{3/2}}{a+b x^3+c x^6} \, dx=\int { \frac {{\left (e x^{3} + d\right )}^{\frac {3}{2}}}{c x^{6} + b x^{3} + a} \,d x } \] Input:
integrate((e*x^3+d)^(3/2)/(c*x^6+b*x^3+a),x, algorithm="maxima")
Output:
integrate((e*x^3 + d)^(3/2)/(c*x^6 + b*x^3 + a), x)
\[ \int \frac {\left (d+e x^3\right )^{3/2}}{a+b x^3+c x^6} \, dx=\int { \frac {{\left (e x^{3} + d\right )}^{\frac {3}{2}}}{c x^{6} + b x^{3} + a} \,d x } \] Input:
integrate((e*x^3+d)^(3/2)/(c*x^6+b*x^3+a),x, algorithm="giac")
Output:
integrate((e*x^3 + d)^(3/2)/(c*x^6 + b*x^3 + a), x)
Timed out. \[ \int \frac {\left (d+e x^3\right )^{3/2}}{a+b x^3+c x^6} \, dx=\text {Hanged} \] Input:
int((d + e*x^3)^(3/2)/(a + b*x^3 + c*x^6),x)
Output:
\text{Hanged}
\[ \int \frac {\left (d+e x^3\right )^{3/2}}{a+b x^3+c x^6} \, dx=\left (\int \frac {\sqrt {e \,x^{3}+d}}{c \,x^{6}+b \,x^{3}+a}d x \right ) d +\left (\int \frac {\sqrt {e \,x^{3}+d}\, x^{3}}{c \,x^{6}+b \,x^{3}+a}d x \right ) e \] Input:
int((e*x^3+d)^(3/2)/(c*x^6+b*x^3+a),x)
Output:
int(sqrt(d + e*x**3)/(a + b*x**3 + c*x**6),x)*d + int((sqrt(d + e*x**3)*x* *3)/(a + b*x**3 + c*x**6),x)*e