### 3.727 $$\int (d+e x)^m (a+c x^2)^{3/2} \, dx$$

Optimal. Leaf size=154 $\frac{\left (a+c x^2\right )^{3/2} (d+e x)^{m+1} F_1\left (m+1;-\frac{3}{2},-\frac{3}{2};m+2;\frac{d+e x}{d-\frac{\sqrt{-a} e}{\sqrt{c}}},\frac{d+e x}{d+\frac{\sqrt{-a} e}{\sqrt{c}}}\right )}{e (m+1) \left (1-\frac{d+e x}{d-\frac{\sqrt{-a} e}{\sqrt{c}}}\right )^{3/2} \left (1-\frac{d+e x}{\frac{\sqrt{-a} e}{\sqrt{c}}+d}\right )^{3/2}}$

[Out]

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

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Rubi [A]  time = 0.135235, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 19, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.105, Rules used = {760, 133} $\frac{\left (a+c x^2\right )^{3/2} (d+e x)^{m+1} F_1\left (m+1;-\frac{3}{2},-\frac{3}{2};m+2;\frac{d+e x}{d-\frac{\sqrt{-a} e}{\sqrt{c}}},\frac{d+e x}{d+\frac{\sqrt{-a} e}{\sqrt{c}}}\right )}{e (m+1) \left (1-\frac{d+e x}{d-\frac{\sqrt{-a} e}{\sqrt{c}}}\right )^{3/2} \left (1-\frac{d+e x}{\frac{\sqrt{-a} e}{\sqrt{c}}+d}\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)^m*(a + c*x^2)^(3/2),x]

[Out]

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

Rule 760

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[(a + c*x^
2)^p/(e*(1 - (d + e*x)/(d + (e*q)/c))^p*(1 - (d + e*x)/(d - (e*q)/c))^p), Subst[Int[x^m*Simp[1 - x/(d + (e*q)/
c), x]^p*Simp[1 - x/(d - (e*q)/c), x]^p, x], x, d + e*x], x]] /; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a
*e^2, 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 (d+e x)^m \left (a+c x^2\right )^{3/2} \, dx &=\frac{\left (a+c x^2\right )^{3/2} \operatorname{Subst}\left (\int x^m \left (1-\frac{x}{d-\frac{\sqrt{-a} e}{\sqrt{c}}}\right )^{3/2} \left (1-\frac{x}{d+\frac{\sqrt{-a} e}{\sqrt{c}}}\right )^{3/2} \, dx,x,d+e x\right )}{e \left (1-\frac{d+e x}{d-\frac{\sqrt{-a} e}{\sqrt{c}}}\right )^{3/2} \left (1-\frac{d+e x}{d+\frac{\sqrt{-a} e}{\sqrt{c}}}\right )^{3/2}}\\ &=\frac{(d+e x)^{1+m} \left (a+c x^2\right )^{3/2} F_1\left (1+m;-\frac{3}{2},-\frac{3}{2};2+m;\frac{d+e x}{d-\frac{\sqrt{-a} e}{\sqrt{c}}},\frac{d+e x}{d+\frac{\sqrt{-a} e}{\sqrt{c}}}\right )}{e (1+m) \left (1-\frac{d+e x}{d-\frac{\sqrt{-a} e}{\sqrt{c}}}\right )^{3/2} \left (1-\frac{d+e x}{d+\frac{\sqrt{-a} e}{\sqrt{c}}}\right )^{3/2}}\\ \end{align*}

Mathematica [F]  time = 0.0809725, size = 0, normalized size = 0. $\int (d+e x)^m \left (a+c x^2\right )^{3/2} \, dx$

Veriﬁcation is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.558, size = 0, normalized size = 0. \begin{align*} \int \left ( ex+d \right ) ^{m} \left ( c{x}^{2}+a \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+a)^(3/2),x)

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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