\(\int \frac {(d+e x)^m}{(b x+c x^2)^{3/2}} \, dx\) [449]

Optimal result

Integrand size = 21, antiderivative size = 105 \[ \int \frac {(d+e x)^m}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {\left (-\frac {e x}{d}\right )^{3/2} (d+e x)^{1+m} \left (1-\frac {c (d+e x)}{c d-b e}\right )^{3/2} \operatorname {AppellF1}\left (1+m,\frac {3}{2},\frac {3}{2},2+m,\frac {d+e x}{d},\frac {c (d+e x)}{c d-b e}\right )}{e (1+m) \left (b x+c x^2\right )^{3/2}} \]

[Out]

(-e*x/d)^(3/2)*(e*x+d)^(1+m)*(1-c*(e*x+d)/(-b*e+c*d))^(3/2)*AppellF1(1+m,3/2,3/2,2+m,(e*x+d)/d,c*(e*x+d)/(-b*e
+c*d))/e/(1+m)/(c*x^2+b*x)^(3/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {773, 138} \[ \int \frac {(d+e x)^m}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {\left (-\frac {e x}{d}\right )^{3/2} (d+e x)^{m+1} \left (1-\frac {c (d+e x)}{c d-b e}\right )^{3/2} \operatorname {AppellF1}\left (m+1,\frac {3}{2},\frac {3}{2},m+2,\frac {d+e x}{d},\frac {c (d+e x)}{c d-b e}\right )}{e (m+1) \left (b x+c x^2\right )^{3/2}} \]

[In]

Int[(d + e*x)^m/(b*x + c*x^2)^(3/2),x]

[Out]

((-((e*x)/d))^(3/2)*(d + e*x)^(1 + m)*(1 - (c*(d + e*x))/(c*d - b*e))^(3/2)*AppellF1[1 + m, 3/2, 3/2, 2 + m, (
d + e*x)/d, (c*(d + e*x))/(c*d - b*e)])/(e*(1 + m)*(b*x + c*x^2)^(3/2))

Rule 138

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[c^n*e^p*((b*x)^(m +
1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rule 773

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c,
2]}, Dist[(a + b*x + c*x^2)^p/(e*(1 - (d + e*x)/(d - e*((b - q)/(2*c))))^p*(1 - (d + e*x)/(d - e*((b + q)/(2*
c))))^p), Subst[Int[x^m*Simp[1 - x/(d - e*((b - q)/(2*c))), x]^p*Simp[1 - x/(d - e*((b + q)/(2*c))), x]^p, x],
x, d + e*x], x]] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] &
& NeQ[2*c*d - b*e, 0] &&  !IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (\left (1-\frac {d+e x}{d}\right )^{3/2} \left (1-\frac {d+e x}{d-\frac {b e}{c}}\right )^{3/2}\right ) \text {Subst}\left (\int \frac {x^m}{\left (1-\frac {x}{d}\right )^{3/2} \left (1-\frac {c x}{c d-b e}\right )^{3/2}} \, dx,x,d+e x\right )}{e \left (b x+c x^2\right )^{3/2}} \\ & = \frac {\left (-\frac {e x}{d}\right )^{3/2} (d+e x)^{1+m} \left (1-\frac {c (d+e x)}{c d-b e}\right )^{3/2} F_1\left (1+m;\frac {3}{2},\frac {3}{2};2+m;\frac {d+e x}{d},\frac {c (d+e x)}{c d-b e}\right )}{e (1+m) \left (b x+c x^2\right )^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.75 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.31 \[ \int \frac {(d+e x)^m}{\left (b x+c x^2\right )^{3/2}} \, dx=-\frac {2 \sqrt {x (b+c x)} (d+e x)^m \left (1+\frac {e x}{d}\right )^{-m} \left (b \operatorname {AppellF1}\left (-\frac {1}{2},-\frac {1}{2},-m,\frac {1}{2},-\frac {c x}{b},-\frac {e x}{d}\right )+c x \left (\operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-m,\frac {3}{2},-\frac {c x}{b},-\frac {e x}{d}\right )+\operatorname {AppellF1}\left (\frac {1}{2},\frac {3}{2},-m,\frac {3}{2},-\frac {c x}{b},-\frac {e x}{d}\right )\right )\right )}{b^3 x \sqrt {1+\frac {c x}{b}}} \]

[In]

Integrate[(d + e*x)^m/(b*x + c*x^2)^(3/2),x]

[Out]

(-2*Sqrt[x*(b + c*x)]*(d + e*x)^m*(b*AppellF1[-1/2, -1/2, -m, 1/2, -((c*x)/b), -((e*x)/d)] + c*x*(AppellF1[1/2
, 1/2, -m, 3/2, -((c*x)/b), -((e*x)/d)] + AppellF1[1/2, 3/2, -m, 3/2, -((c*x)/b), -((e*x)/d)])))/(b^3*x*Sqrt[1
+ (c*x)/b]*(1 + (e*x)/d)^m)

Maple [F]

\[\int \frac {\left (e x +d \right )^{m}}{\left (c \,x^{2}+b x \right )^{\frac {3}{2}}}d x\]

[In]

int((e*x+d)^m/(c*x^2+b*x)^(3/2),x)

[Out]

int((e*x+d)^m/(c*x^2+b*x)^(3/2),x)

Fricas [F]

\[ \int \frac {(d+e x)^m}{\left (b x+c x^2\right )^{3/2}} \, dx=\int { \frac {{\left (e x + d\right )}^{m}}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate((e*x+d)^m/(c*x^2+b*x)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*x^2 + b*x)*(e*x + d)^m/(c^2*x^4 + 2*b*c*x^3 + b^2*x^2), x)

Sympy [F]

\[ \int \frac {(d+e x)^m}{\left (b x+c x^2\right )^{3/2}} \, dx=\int \frac {\left (d + e x\right )^{m}}{\left (x \left (b + c x\right )\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate((e*x+d)**m/(c*x**2+b*x)**(3/2),x)

[Out]

Integral((d + e*x)**m/(x*(b + c*x))**(3/2), x)

Maxima [F]

\[ \int \frac {(d+e x)^m}{\left (b x+c x^2\right )^{3/2}} \, dx=\int { \frac {{\left (e x + d\right )}^{m}}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate((e*x+d)^m/(c*x^2+b*x)^(3/2),x, algorithm="maxima")

[Out]

integrate((e*x + d)^m/(c*x^2 + b*x)^(3/2), x)

Giac [F]

\[ \int \frac {(d+e x)^m}{\left (b x+c x^2\right )^{3/2}} \, dx=\int { \frac {{\left (e x + d\right )}^{m}}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate((e*x+d)^m/(c*x^2+b*x)^(3/2),x, algorithm="giac")

[Out]

integrate((e*x + d)^m/(c*x^2 + b*x)^(3/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(d+e x)^m}{\left (b x+c x^2\right )^{3/2}} \, dx=\int \frac {{\left (d+e\,x\right )}^m}{{\left (c\,x^2+b\,x\right )}^{3/2}} \,d x \]

[In]

int((d + e*x)^m/(b*x + c*x^2)^(3/2),x)

[Out]

int((d + e*x)^m/(b*x + c*x^2)^(3/2), x)