3.2559 \(\int \frac {(d+e x)^m}{(a+b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=189 \[ \frac {(d+e x)^{m+1} \left (1-\frac {2 c (d+e x)}{2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}\right )^{3/2} \left (1-\frac {2 c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}\right )^{3/2} F_1\left (m+1;\frac {3}{2},\frac {3}{2};m+2;\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{e (m+1) \left (a+b x+c x^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {759, 133} \[ \frac {(d+e x)^{m+1} \left (1-\frac {2 c (d+e x)}{2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}\right )^{3/2} \left (1-\frac {2 c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}\right )^{3/2} F_1\left (m+1;\frac {3}{2},\frac {3}{2};m+2;\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{e (m+1) \left (a+b x+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 133

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

Rule 759

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*} \int \frac {(d+e x)^m}{\left (a+b x+c x^2\right )^{3/2}} \, dx &=\frac {\left (\left (1-\frac {d+e x}{d-\frac {\left (b-\sqrt {b^2-4 a c}\right ) e}{2 c}}\right )^{3/2} \left (1-\frac {d+e x}{d-\frac {\left (b+\sqrt {b^2-4 a c}\right ) e}{2 c}}\right )^{3/2}\right ) \operatorname {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 )^{3/2} \left (1-\frac {2 c x}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )^{3/2}} \, dx,x,d+e x\right )}{e \left (a+b x+c x^2\right )^{3/2}}\\ &=\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 )^{3/2} \left (1-\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )^{3/2} F_1\left (1+m;\frac {3}{2},\frac {3}{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 )^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.35, size = 267, normalized size = 1.41 \[ -\frac {e \left (\sqrt {b^2-4 a c}-b-2 c x\right ) \left (\sqrt {b^2-4 a c}+b+2 c x\right ) (d+e x)^{m+1} \sqrt {\frac {e \left (\sqrt {b^2-4 a c}-b-2 c x\right )}{e \left (\sqrt {b^2-4 a c}-b\right )+2 c d}} \sqrt {\frac {e \left (\sqrt {b^2-4 a c}+b+2 c x\right )}{e \left (\sqrt {b^2-4 a c}+b\right )-2 c d}} F_1\left (m+1;\frac {3}{2},\frac {3}{2};m+2;\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d+\left (\sqrt {b^2-4 a c}-b\right ) e}\right )}{4 c (m+1) (a+x (b+c x))^{3/2} \left (e (a e-b d)+c d^2\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 1.21, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{2} + b x + a} {\left (e x + d\right )}^{m}}{c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x + {\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x + d\right )}^{m}}{{\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 11.99, size = 0, normalized size = 0.00 \[ \int \frac {\left (e x +d \right )^{m}}{\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x + d\right )}^{m}}{{\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (d+e\,x\right )}^m}{{\left (c\,x^2+b\,x+a\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d + e x\right )^{m}}{\left (a + b x + c x^{2}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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