3.449 $$\int \frac{(d+e x)^m}{(b x+c x^2)^{3/2}} \, dx$$

Optimal. Leaf size=105 $\frac{\left (-\frac{e x}{d}\right )^{3/2} (d+e x)^{m+1} \left (1-\frac{c (d+e x)}{c d-b e}\right )^{3/2} F_1\left (m+1;\frac{3}{2},\frac{3}{2};m+2;\frac{d+e x}{d},\frac{c (d+e x)}{c d-b e}\right )}{e (m+1) \left (b x+c x^2\right )^{3/2}}$

[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))

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Rubi [A]  time = 0.0480589, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 21, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.095, Rules used = {759, 133} $\frac{\left (-\frac{e x}{d}\right )^{3/2} (d+e x)^{m+1} \left (1-\frac{c (d+e x)}{c d-b e}\right )^{3/2} F_1\left (m+1;\frac{3}{2},\frac{3}{2};m+2;\frac{d+e x}{d},\frac{c (d+e x)}{c d-b e}\right )}{e (m+1) \left (b x+c x^2\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[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 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]

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])

Rubi steps

\begin{align*} \int \frac{(d+e x)^m}{\left (b x+c x^2\right )^{3/2}} \, dx &=\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 ) \operatorname{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]  time = 0.133985, size = 138, normalized size = 1.31 $-\frac{2 \sqrt{x (b+c x)} (d+e x)^m \left (\frac{e x}{d}+1\right )^{-m} \left (b F_1\left (-\frac{1}{2};-\frac{1}{2},-m;\frac{1}{2};-\frac{c x}{b},-\frac{e x}{d}\right )+c x \left (F_1\left (\frac{1}{2};\frac{1}{2},-m;\frac{3}{2};-\frac{c x}{b},-\frac{e x}{d}\right )+F_1\left (\frac{1}{2};\frac{3}{2},-m;\frac{3}{2};-\frac{c x}{b},-\frac{e x}{d}\right )\right )\right )}{b^3 x \sqrt{\frac{c x}{b}+1}}$

Warning: Unable to verify antiderivative.

[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)

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Maple [F]  time = 0.613, size = 0, normalized size = 0. \begin{align*} \int{ \left ( ex+d \right ) ^{m} \left ( c{x}^{2}+bx \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x + d\right )}^{m}}{{\left (c x^{2} + b x\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{c x^{2} + b x}{\left (e x + d\right )}^{m}}{c^{2} x^{4} + 2 \, b c x^{3} + b^{2} x^{2}}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d + e x\right )^{m}}{\left (x \left (b + c x\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x + d\right )}^{m}}{{\left (c x^{2} + b x\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)