\(\int \frac {(b x)^{5/2} (c+d x)^n}{e+f x} \, dx\) [950]

Optimal result

Integrand size = 22, antiderivative size = 61 \[ \int \frac {(b x)^{5/2} (c+d x)^n}{e+f x} \, dx=\frac {2 (b x)^{7/2} (c+d x)^n \left (1+\frac {d x}{c}\right )^{-n} \operatorname {AppellF1}\left (\frac {7}{2},-n,1,\frac {9}{2},-\frac {d x}{c},-\frac {f x}{e}\right )}{7 b e} \]

[Out]

2/7*(b*x)^(7/2)*(d*x+c)^n*AppellF1(7/2,-n,1,9/2,-d*x/c,-f*x/e)/b/e/((1+d*x/c)^n)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {129, 525, 524} \[ \int \frac {(b x)^{5/2} (c+d x)^n}{e+f x} \, dx=\frac {2 (b x)^{7/2} (c+d x)^n \left (\frac {d x}{c}+1\right )^{-n} \operatorname {AppellF1}\left (\frac {7}{2},-n,1,\frac {9}{2},-\frac {d x}{c},-\frac {f x}{e}\right )}{7 b e} \]

[In]

Int[((b*x)^(5/2)*(c + d*x)^n)/(e + f*x),x]

[Out]

(2*(b*x)^(7/2)*(c + d*x)^n*AppellF1[7/2, -n, 1, 9/2, -((d*x)/c), -((f*x)/e)])/(7*b*e*(1 + (d*x)/c)^n)

Rule 129

Int[((e_.)*(x_))^(p_)*((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> With[{k = Denominator[p]
}, Dist[k/e, Subst[Int[x^(k*(p + 1) - 1)*(a + b*(x^k/e))^m*(c + d*(x^k/e))^n, x], x, (e*x)^(1/k)], x]] /; Free
Q[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && FractionQ[p] && IntegerQ[m]

Rule 524

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] /; Free
Q[{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])

Rule 525

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPar
t[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])

Rubi steps \begin{align*} \text {integral}& = \frac {2 \text {Subst}\left (\int \frac {x^6 \left (c+\frac {d x^2}{b}\right )^n}{e+\frac {f x^2}{b}} \, dx,x,\sqrt {b x}\right )}{b} \\ & = \frac {\left (2 (c+d x)^n \left (1+\frac {d x}{c}\right )^{-n}\right ) \text {Subst}\left (\int \frac {x^6 \left (1+\frac {d x^2}{b c}\right )^n}{e+\frac {f x^2}{b}} \, dx,x,\sqrt {b x}\right )}{b} \\ & = \frac {2 (b x)^{7/2} (c+d x)^n \left (1+\frac {d x}{c}\right )^{-n} F_1\left (\frac {7}{2};-n,1;\frac {9}{2};-\frac {d x}{c},-\frac {f x}{e}\right )}{7 b e} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(134\) vs. \(2(61)=122\).

Time = 0.66 (sec) , antiderivative size = 134, normalized size of antiderivative = 2.20 \[ \int \frac {(b x)^{5/2} (c+d x)^n}{e+f x} \, dx=\frac {2 b^2 \sqrt {b x} (c+d x)^n \left (1+\frac {d x}{c}\right )^{-n} \left (-15 e^2 \operatorname {AppellF1}\left (\frac {1}{2},-n,1,\frac {3}{2},-\frac {d x}{c},-\frac {f x}{e}\right )+15 e^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-n,\frac {3}{2},-\frac {d x}{c}\right )+f x \left (-5 e \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-n,\frac {5}{2},-\frac {d x}{c}\right )+3 f x \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-n,\frac {7}{2},-\frac {d x}{c}\right )\right )\right )}{15 f^3} \]

[In]

Integrate[((b*x)^(5/2)*(c + d*x)^n)/(e + f*x),x]

[Out]

(2*b^2*Sqrt[b*x]*(c + d*x)^n*(-15*e^2*AppellF1[1/2, -n, 1, 3/2, -((d*x)/c), -((f*x)/e)] + 15*e^2*Hypergeometri
c2F1[1/2, -n, 3/2, -((d*x)/c)] + f*x*(-5*e*Hypergeometric2F1[3/2, -n, 5/2, -((d*x)/c)] + 3*f*x*Hypergeometric2
F1[5/2, -n, 7/2, -((d*x)/c)])))/(15*f^3*(1 + (d*x)/c)^n)

Maple [F]

\[\int \frac {\left (b x \right )^{\frac {5}{2}} \left (d x +c \right )^{n}}{f x +e}d x\]

[In]

int((b*x)^(5/2)*(d*x+c)^n/(f*x+e),x)

[Out]

int((b*x)^(5/2)*(d*x+c)^n/(f*x+e),x)

Fricas [F]

\[ \int \frac {(b x)^{5/2} (c+d x)^n}{e+f x} \, dx=\int { \frac {\left (b x\right )^{\frac {5}{2}} {\left (d x + c\right )}^{n}}{f x + e} \,d x } \]

[In]

integrate((b*x)^(5/2)*(d*x+c)^n/(f*x+e),x, algorithm="fricas")

[Out]

integral(sqrt(b*x)*(d*x + c)^n*b^2*x^2/(f*x + e), x)

Sympy [F(-1)]

Timed out. \[ \int \frac {(b x)^{5/2} (c+d x)^n}{e+f x} \, dx=\text {Timed out} \]

[In]

integrate((b*x)**(5/2)*(d*x+c)**n/(f*x+e),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(b x)^{5/2} (c+d x)^n}{e+f x} \, dx=\int { \frac {\left (b x\right )^{\frac {5}{2}} {\left (d x + c\right )}^{n}}{f x + e} \,d x } \]

[In]

integrate((b*x)^(5/2)*(d*x+c)^n/(f*x+e),x, algorithm="maxima")

[Out]

integrate((b*x)^(5/2)*(d*x + c)^n/(f*x + e), x)

Giac [F]

\[ \int \frac {(b x)^{5/2} (c+d x)^n}{e+f x} \, dx=\int { \frac {\left (b x\right )^{\frac {5}{2}} {\left (d x + c\right )}^{n}}{f x + e} \,d x } \]

[In]

integrate((b*x)^(5/2)*(d*x+c)^n/(f*x+e),x, algorithm="giac")

[Out]

integrate((b*x)^(5/2)*(d*x + c)^n/(f*x + e), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(b x)^{5/2} (c+d x)^n}{e+f x} \, dx=\int \frac {{\left (b\,x\right )}^{5/2}\,{\left (c+d\,x\right )}^n}{e+f\,x} \,d x \]

[In]

int(((b*x)^(5/2)*(c + d*x)^n)/(e + f*x),x)

[Out]

int(((b*x)^(5/2)*(c + d*x)^n)/(e + f*x), x)