### 3.731 $$\int (d+e x)^m (a+c x^2)^p \, dx$$

Optimal. Leaf size=152 $\frac{\left (a+c x^2\right )^p (d+e x)^{m+1} \left (1-\frac{d+e x}{d-\frac{\sqrt{-a} e}{\sqrt{c}}}\right )^{-p} \left (1-\frac{d+e x}{\frac{\sqrt{-a} e}{\sqrt{c}}+d}\right )^{-p} F_1\left (m+1;-p,-p;m+2;\frac{d+e x}{d-\frac{\sqrt{-a} e}{\sqrt{c}}},\frac{d+e x}{d+\frac{\sqrt{-a} e}{\sqrt{c}}}\right )}{e (m+1)}$

[Out]

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

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Rubi [A]  time = 0.064112, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 17, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.118, Rules used = {760, 133} $\frac{\left (a+c x^2\right )^p (d+e x)^{m+1} \left (1-\frac{d+e x}{d-\frac{\sqrt{-a} e}{\sqrt{c}}}\right )^{-p} \left (1-\frac{d+e x}{\frac{\sqrt{-a} e}{\sqrt{c}}+d}\right )^{-p} F_1\left (m+1;-p,-p;m+2;\frac{d+e x}{d-\frac{\sqrt{-a} e}{\sqrt{c}}},\frac{d+e x}{d+\frac{\sqrt{-a} e}{\sqrt{c}}}\right )}{e (m+1)}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)^m*(a + c*x^2)^p,x]

[Out]

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

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

Mathematica [A]  time = 0.136545, size = 157, normalized size = 1.03 $\frac{\left (a+c x^2\right )^p (d+e x)^{m+1} \left (\frac{e \left (\sqrt{-\frac{a}{c}}-x\right )}{e \sqrt{-\frac{a}{c}}+d}\right )^{-p} \left (\frac{e \left (\sqrt{-\frac{a}{c}}+x\right )}{e \sqrt{-\frac{a}{c}}-d}\right )^{-p} F_1\left (m+1;-p,-p;m+2;\frac{d+e x}{d-\sqrt{-\frac{a}{c}} e},\frac{d+e x}{d+\sqrt{-\frac{a}{c}} e}\right )}{e (m+1)}$

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [F]  time = 0.599, size = 0, normalized size = 0. \begin{align*} \int \left ( ex+d \right ) ^{m} \left ( c{x}^{2}+a \right ) ^{p}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (c x^{2} + a\right )}^{p}{\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)^p,x, algorithm="maxima")

[Out]

integrate((c*x^2 + a)^p*(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 )}^{p}{\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)^p,x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (c x^{2} + a\right )}^{p}{\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)^p,x, algorithm="giac")

[Out]

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