3.497 \(\int \frac{(e x)^m (A+B x)}{(a+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=145 \[ \frac{A \sqrt{\frac{c x^2}{a}+1} (e x)^{m+1} \, _2F_1\left (\frac{5}{2},\frac{m+1}{2};\frac{m+3}{2};-\frac{c x^2}{a}\right )}{a^2 e (m+1) \sqrt{a+c x^2}}+\frac{B \sqrt{\frac{c x^2}{a}+1} (e x)^{m+2} \, _2F_1\left (\frac{5}{2},\frac{m+2}{2};\frac{m+4}{2};-\frac{c x^2}{a}\right )}{a^2 e^2 (m+2) \sqrt{a+c x^2}} \]

[Out]

(A*(e*x)^(1 + m)*Sqrt[1 + (c*x^2)/a]*Hypergeometric2F1[5/2, (1 + m)/2, (3 + m)/2, -((c*x^2)/a)])/(a^2*e*(1 + m
)*Sqrt[a + c*x^2]) + (B*(e*x)^(2 + m)*Sqrt[1 + (c*x^2)/a]*Hypergeometric2F1[5/2, (2 + m)/2, (4 + m)/2, -((c*x^
2)/a)])/(a^2*e^2*(2 + m)*Sqrt[a + c*x^2])

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Rubi [A]  time = 0.0674913, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {808, 365, 364} \[ \frac{A \sqrt{\frac{c x^2}{a}+1} (e x)^{m+1} \, _2F_1\left (\frac{5}{2},\frac{m+1}{2};\frac{m+3}{2};-\frac{c x^2}{a}\right )}{a^2 e (m+1) \sqrt{a+c x^2}}+\frac{B \sqrt{\frac{c x^2}{a}+1} (e x)^{m+2} \, _2F_1\left (\frac{5}{2},\frac{m+2}{2};\frac{m+4}{2};-\frac{c x^2}{a}\right )}{a^2 e^2 (m+2) \sqrt{a+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(A*(e*x)^(1 + m)*Sqrt[1 + (c*x^2)/a]*Hypergeometric2F1[5/2, (1 + m)/2, (3 + m)/2, -((c*x^2)/a)])/(a^2*e*(1 + m
)*Sqrt[a + c*x^2]) + (B*(e*x)^(2 + m)*Sqrt[1 + (c*x^2)/a]*Hypergeometric2F1[5/2, (2 + m)/2, (4 + m)/2, -((c*x^
2)/a)])/(a^2*e^2*(2 + m)*Sqrt[a + c*x^2])

Rule 808

Int[((e_.)*(x_))^(m_)*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[f, Int[(e*x)^m*(a + c*
x^2)^p, x], x] + Dist[g/e, Int[(e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, e, f, g, p}, x] &&  !Ration
alQ[m] &&  !IGtQ[p, 0]

Rule 365

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])
/(1 + (b*x^n)/a)^FracPart[p], Int[(c*x)^m*(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
 (ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \frac{(e x)^m (A+B x)}{\left (a+c x^2\right )^{5/2}} \, dx &=A \int \frac{(e x)^m}{\left (a+c x^2\right )^{5/2}} \, dx+\frac{B \int \frac{(e x)^{1+m}}{\left (a+c x^2\right )^{5/2}} \, dx}{e}\\ &=\frac{\left (A \sqrt{1+\frac{c x^2}{a}}\right ) \int \frac{(e x)^m}{\left (1+\frac{c x^2}{a}\right )^{5/2}} \, dx}{a^2 \sqrt{a+c x^2}}+\frac{\left (B \sqrt{1+\frac{c x^2}{a}}\right ) \int \frac{(e x)^{1+m}}{\left (1+\frac{c x^2}{a}\right )^{5/2}} \, dx}{a^2 e \sqrt{a+c x^2}}\\ &=\frac{A (e x)^{1+m} \sqrt{1+\frac{c x^2}{a}} \, _2F_1\left (\frac{5}{2},\frac{1+m}{2};\frac{3+m}{2};-\frac{c x^2}{a}\right )}{a^2 e (1+m) \sqrt{a+c x^2}}+\frac{B (e x)^{2+m} \sqrt{1+\frac{c x^2}{a}} \, _2F_1\left (\frac{5}{2},\frac{2+m}{2};\frac{4+m}{2};-\frac{c x^2}{a}\right )}{a^2 e^2 (2+m) \sqrt{a+c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0489288, size = 111, normalized size = 0.77 \[ \frac{x \sqrt{\frac{c x^2}{a}+1} (e x)^m \left (A (m+2) \, _2F_1\left (\frac{5}{2},\frac{m+1}{2};\frac{m+3}{2};-\frac{c x^2}{a}\right )+B (m+1) x \, _2F_1\left (\frac{5}{2},\frac{m}{2}+1;\frac{m}{2}+2;-\frac{c x^2}{a}\right )\right )}{a^2 (m+1) (m+2) \sqrt{a+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(e*x)^m*Sqrt[1 + (c*x^2)/a]*(B*(1 + m)*x*Hypergeometric2F1[5/2, 1 + m/2, 2 + m/2, -((c*x^2)/a)] + A*(2 + m)
*Hypergeometric2F1[5/2, (1 + m)/2, (3 + m)/2, -((c*x^2)/a)]))/(a^2*(1 + m)*(2 + m)*Sqrt[a + c*x^2])

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Maple [F]  time = 0.018, size = 0, normalized size = 0. \begin{align*} \int{ \left ( ex \right ) ^{m} \left ( Bx+A \right ) \left ( c{x}^{2}+a \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^m*(B*x+A)/(c*x^2+a)^(5/2),x)

[Out]

int((e*x)^m*(B*x+A)/(c*x^2+a)^(5/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x + A\right )} \left (e x\right )^{m}}{{\left (c x^{2} + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*x + A)*(e*x)^m/(c*x^2 + a)^(5/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{c x^{2} + a}{\left (B x + A\right )} \left (e x\right )^{m}}{c^{3} x^{6} + 3 \, a c^{2} x^{4} + 3 \, a^{2} c x^{2} + a^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x + A\right )} \left (e x\right )^{m}}{{\left (c x^{2} + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*x + A)*(e*x)^m/(c*x^2 + a)^(5/2), x)