Integrand size = 26, antiderivative size = 100 \[ \int \frac {\sqrt [3]{a+b x} \sqrt {c+d x}}{e+f x} \, dx=\frac {3 (a+b x)^{4/3} \sqrt {c+d x} \operatorname {AppellF1}\left (\frac {4}{3},-\frac {1}{2},1,\frac {7}{3},-\frac {d (a+b x)}{b c-a d},-\frac {f (a+b x)}{b e-a f}\right )}{4 (b e-a f) \sqrt {\frac {b (c+d x)}{b c-a d}}} \] Output:
3/4*(b*x+a)^(4/3)*(d*x+c)^(1/2)*AppellF1(4/3,-1/2,1,7/3,-d*(b*x+a)/(-a*d+b *c),-f*(b*x+a)/(-a*f+b*e))/(-a*f+b*e)/(b*(d*x+c)/(-a*d+b*c))^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(201\) vs. \(2(100)=200\).
Time = 21.38 (sec) , antiderivative size = 201, normalized size of antiderivative = 2.01 \[ \int \frac {\sqrt [3]{a+b x} \sqrt {c+d x}}{e+f x} \, dx=\frac {6 \sqrt {c+d x} \left (7 f (a+b x)+\frac {\left (\frac {d (a+b x)}{b (c+d x)}\right )^{2/3} \left (7 (5 b d e-3 b c f-2 a d f) \operatorname {AppellF1}\left (\frac {1}{6},\frac {2}{3},1,\frac {7}{6},\frac {b c-a d}{b c+b d x},\frac {-d e+c f}{f (c+d x)}\right )+\frac {3 (b c-a d) (-d e+c f) \operatorname {AppellF1}\left (\frac {7}{6},\frac {2}{3},1,\frac {13}{6},\frac {b c-a d}{b c+b d x},\frac {-d e+c f}{f (c+d x)}\right )}{c+d x}\right )}{d}\right )}{35 f^2 (a+b x)^{2/3}} \] Input:
Integrate[((a + b*x)^(1/3)*Sqrt[c + d*x])/(e + f*x),x]
Output:
(6*Sqrt[c + d*x]*(7*f*(a + b*x) + (((d*(a + b*x))/(b*(c + d*x)))^(2/3)*(7* (5*b*d*e - 3*b*c*f - 2*a*d*f)*AppellF1[1/6, 2/3, 1, 7/6, (b*c - a*d)/(b*c + b*d*x), (-(d*e) + c*f)/(f*(c + d*x))] + (3*(b*c - a*d)*(-(d*e) + c*f)*Ap pellF1[7/6, 2/3, 1, 13/6, (b*c - a*d)/(b*c + b*d*x), (-(d*e) + c*f)/(f*(c + d*x))])/(c + d*x)))/d))/(35*f^2*(a + b*x)^(2/3))
Time = 0.30 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.16, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {149, 27, 1013, 1012}
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.
\(\displaystyle \int \frac {\sqrt [3]{a+b x} \sqrt {c+d x}}{e+f x} \, dx\) |
\(\Big \downarrow \) 149 |
\(\displaystyle \frac {3 \int \frac {b (a+b x) \sqrt {c-\frac {a d}{b}+\frac {d (a+b x)}{b}}}{b \left (e-\frac {a f}{b}\right )+f (a+b x)}d\sqrt [3]{a+b x}}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 3 \int \frac {(a+b x) \sqrt {c-\frac {a d}{b}+\frac {d (a+b x)}{b}}}{b e-a f+f (a+b x)}d\sqrt [3]{a+b x}\) |
\(\Big \downarrow \) 1013 |
\(\displaystyle \frac {3 \sqrt {\frac {d (a+b x)}{b}-\frac {a d}{b}+c} \int \frac {(a+b x) \sqrt {\frac {d (a+b x)}{b c-a d}+1}}{b e-a f+f (a+b x)}d\sqrt [3]{a+b x}}{\sqrt {\frac {d (a+b x)}{b c-a d}+1}}\) |
\(\Big \downarrow \) 1012 |
\(\displaystyle \frac {3 (a+b x)^{4/3} \sqrt {\frac {d (a+b x)}{b}-\frac {a d}{b}+c} \operatorname {AppellF1}\left (\frac {4}{3},-\frac {1}{2},1,\frac {7}{3},-\frac {d (a+b x)}{b c-a d},-\frac {f (a+b x)}{b e-a f}\right )}{4 (b e-a f) \sqrt {\frac {d (a+b x)}{b c-a d}+1}}\) |
Input:
Int[((a + b*x)^(1/3)*Sqrt[c + d*x])/(e + f*x),x]
Output:
(3*(a + b*x)^(4/3)*Sqrt[c - (a*d)/b + (d*(a + b*x))/b]*AppellF1[4/3, -1/2, 1, 7/3, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f))])/(4*( b*e - a*f)*Sqrt[1 + (d*(a + b*x))/(b*c - a*d)])
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_.)*((e_.) + (f_.)*(x_) )^(p_.), x_] :> With[{k = Denominator[m]}, Simp[k/b Subst[Int[x^(k*(m + 1 ) - 1)*(c - a*(d/b) + d*(x^k/b))^n*(e - a*(f/b) + f*(x^k/b))^p, x], x, (a + b*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && FractionQ[m] && IntegerQ[2*n] && IntegerQ[p]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ ))^(q_), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[((e_.)*(x_))^(m_.)*((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[(e*x)^m*(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] & & NeQ[m, n - 1] && !(IntegerQ[p] || GtQ[a, 0])
\[\int \frac {\left (b x +a \right )^{\frac {1}{3}} \sqrt {x d +c}}{f x +e}d x\]
Input:
int((b*x+a)^(1/3)*(d*x+c)^(1/2)/(f*x+e),x)
Output:
int((b*x+a)^(1/3)*(d*x+c)^(1/2)/(f*x+e),x)
Timed out. \[ \int \frac {\sqrt [3]{a+b x} \sqrt {c+d x}}{e+f x} \, dx=\text {Timed out} \] Input:
integrate((b*x+a)^(1/3)*(d*x+c)^(1/2)/(f*x+e),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {\sqrt [3]{a+b x} \sqrt {c+d x}}{e+f x} \, dx=\int \frac {\sqrt [3]{a + b x} \sqrt {c + d x}}{e + f x}\, dx \] Input:
integrate((b*x+a)**(1/3)*(d*x+c)**(1/2)/(f*x+e),x)
Output:
Integral((a + b*x)**(1/3)*sqrt(c + d*x)/(e + f*x), x)
\[ \int \frac {\sqrt [3]{a+b x} \sqrt {c+d x}}{e+f x} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {1}{3}} \sqrt {d x + c}}{f x + e} \,d x } \] Input:
integrate((b*x+a)^(1/3)*(d*x+c)^(1/2)/(f*x+e),x, algorithm="maxima")
Output:
integrate((b*x + a)^(1/3)*sqrt(d*x + c)/(f*x + e), x)
\[ \int \frac {\sqrt [3]{a+b x} \sqrt {c+d x}}{e+f x} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {1}{3}} \sqrt {d x + c}}{f x + e} \,d x } \] Input:
integrate((b*x+a)^(1/3)*(d*x+c)^(1/2)/(f*x+e),x, algorithm="giac")
Output:
integrate((b*x + a)^(1/3)*sqrt(d*x + c)/(f*x + e), x)
Timed out. \[ \int \frac {\sqrt [3]{a+b x} \sqrt {c+d x}}{e+f x} \, dx=\int \frac {{\left (a+b\,x\right )}^{1/3}\,\sqrt {c+d\,x}}{e+f\,x} \,d x \] Input:
int(((a + b*x)^(1/3)*(c + d*x)^(1/2))/(e + f*x),x)
Output:
int(((a + b*x)^(1/3)*(c + d*x)^(1/2))/(e + f*x), x)
\[ \int \frac {\sqrt [3]{a+b x} \sqrt {c+d x}}{e+f x} \, dx=\int \frac {\left (b x +a \right )^{\frac {1}{3}} \sqrt {d x +c}}{f x +e}d x \] Input:
int((b*x+a)^(1/3)*(d*x+c)^(1/2)/(f*x+e),x)
Output:
int((b*x+a)^(1/3)*(d*x+c)^(1/2)/(f*x+e),x)