\(\int \frac {(d+e x)^m}{(a+b x+c x^2)^{5/2}} \, dx\) [2560]

Optimal result

Integrand size = 22, antiderivative size = 189 \[ \int \frac {(d+e x)^m}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\frac {(d+e x)^{1+m} \left (1-\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )^{5/2} \left (1-\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )^{5/2} \operatorname {AppellF1}\left (1+m,\frac {5}{2},\frac {5}{2},2+m,\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{e (1+m) \left (a+b x+c x^2\right )^{5/2}} \]

[Out]

(e*x+d)^(1+m)*AppellF1(1+m,5/2,5/2,2+m,2*c*(e*x+d)/(2*c*d-e*(b-(-4*a*c+b^2)^(1/2))),2*c*(e*x+d)/(2*c*d-e*(b+(-
4*a*c+b^2)^(1/2))))*(1-2*c*(e*x+d)/(2*c*d-e*(b-(-4*a*c+b^2)^(1/2))))^(5/2)*(1-2*c*(e*x+d)/(2*c*d-e*(b+(-4*a*c+
b^2)^(1/2))))^(5/2)/e/(1+m)/(c*x^2+b*x+a)^(5/2)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {773, 138} \[ \int \frac {(d+e x)^m}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\frac {(d+e x)^{m+1} \left (1-\frac {2 c (d+e x)}{2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}\right )^{5/2} \left (1-\frac {2 c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}\right )^{5/2} \operatorname {AppellF1}\left (m+1,\frac {5}{2},\frac {5}{2},m+2,\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{e (m+1) \left (a+b x+c x^2\right )^{5/2}} \]

[In]

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

[Out]

((d + e*x)^(1 + m)*(1 - (2*c*(d + e*x))/(2*c*d - (b - Sqrt[b^2 - 4*a*c])*e))^(5/2)*(1 - (2*c*(d + e*x))/(2*c*d
- (b + Sqrt[b^2 - 4*a*c])*e))^(5/2)*AppellF1[1 + m, 5/2, 5/2, 2 + m, (2*c*(d + e*x))/(2*c*d - (b - Sqrt[b^2 -
4*a*c])*e), (2*c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)])/(e*(1 + m)*(a + b*x + c*x^2)^(5/2))

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-\frac {\left (b-\sqrt {b^2-4 a c}\right ) e}{2 c}}\right )^{5/2} \left (1-\frac {d+e x}{d-\frac {\left (b+\sqrt {b^2-4 a c}\right ) e}{2 c}}\right )^{5/2}\right ) \text {Subst}\left (\int \frac {x^m}{\left (1-\frac {2 c x}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )^{5/2} \left (1-\frac {2 c x}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )^{5/2}} \, dx,x,d+e x\right )}{e \left (a+b x+c x^2\right )^{5/2}} \\ & = \frac {(d+e x)^{1+m} \left (1-\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )^{5/2} \left (1-\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )^{5/2} F_1\left (1+m;\frac {5}{2},\frac {5}{2};2+m;\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{e (1+m) \left (a+b x+c x^2\right )^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.51 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.19 \[ \int \frac {(d+e x)^m}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\frac {e^3 \sqrt {\frac {e \left (-b+\sqrt {b^2-4 a c}-2 c x\right )}{2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}} \sqrt {\frac {e \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e}} (d+e x)^{1+m} \operatorname {AppellF1}\left (1+m,\frac {5}{2},\frac {5}{2},2+m,\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}\right )}{\left (c d^2+e (-b d+a e)\right )^2 (1+m) \sqrt {a+x (b+c x)}} \]

[In]

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

[Out]

(e^3*Sqrt[(e*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x))/(2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e)]*Sqrt[(e*(b + Sqrt[b^2 - 4
*a*c] + 2*c*x))/(-2*c*d + (b + Sqrt[b^2 - 4*a*c])*e)]*(d + e*x)^(1 + m)*AppellF1[1 + m, 5/2, 5/2, 2 + m, (2*c*
(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e), (2*c*(d + e*x))/(2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e)])/((c*d^2
+ e*(-(b*d) + a*e))^2*(1 + m)*Sqrt[a + x*(b + c*x)])

Maple [F]

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

[In]

int((e*x+d)^m/(c*x^2+b*x+a)^(5/2),x)

[Out]

int((e*x+d)^m/(c*x^2+b*x+a)^(5/2),x)

Fricas [F]

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

[In]

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

[Out]

integral(sqrt(c*x^2 + b*x + a)*(e*x + d)^m/(c^3*x^6 + 3*b*c^2*x^5 + 3*(b^2*c + a*c^2)*x^4 + 3*a^2*b*x + (b^3 +
6*a*b*c)*x^3 + a^3 + 3*(a*b^2 + a^2*c)*x^2), x)

Sympy [F]

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

[In]

integrate((e*x+d)**m/(c*x**2+b*x+a)**(5/2),x)

[Out]

Integral((d + e*x)**m/(a + b*x + c*x**2)**(5/2), x)

Maxima [F]

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

[In]

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

[Out]

integrate((e*x + d)^m/(c*x^2 + b*x + a)^(5/2), x)

Giac [F]

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

[In]

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

[Out]

integrate((e*x + d)^m/(c*x^2 + b*x + a)^(5/2), x)

Mupad [F(-1)]

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

[In]

int((d + e*x)^m/(a + b*x + c*x^2)^(5/2),x)

[Out]

int((d + e*x)^m/(a + b*x + c*x^2)^(5/2), x)