3.495 \(\int \frac {(e x)^m (A+B x)}{\sqrt {a+c x^2}} \, dx\)

Optimal. Leaf size=139 \[ \frac {A \sqrt {\frac {c x^2}{a}+1} (e x)^{m+1} \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};-\frac {c x^2}{a}\right )}{e (m+1) \sqrt {a+c x^2}}+\frac {B \sqrt {\frac {c x^2}{a}+1} (e x)^{m+2} \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};-\frac {c x^2}{a}\right )}{e^2 (m+2) \sqrt {a+c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 139, normalized size of antiderivative = 1.00, 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} \[ \frac {A \sqrt {\frac {c x^2}{a}+1} (e x)^{m+1} \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};-\frac {c x^2}{a}\right )}{e (m+1) \sqrt {a+c x^2}}+\frac {B \sqrt {\frac {c x^2}{a}+1} (e x)^{m+2} \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};-\frac {c x^2}{a}\right )}{e^2 (m+2) \sqrt {a+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 108, normalized size = 0.78 \[ \frac {x \sqrt {\frac {c x^2}{a}+1} (e x)^m \left (A (m+2) \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};-\frac {c x^2}{a}\right )+B (m+1) x \, _2F_1\left (\frac {1}{2},\frac {m}{2}+1;\frac {m}{2}+2;-\frac {c x^2}{a}\right )\right )}{(m+1) (m+2) \sqrt {a+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.96, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B x + A\right )} \left (e x\right )^{m}}{\sqrt {c x^{2} + a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*(B*x+A)/(c*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*(B*x+A)/(c*x^2+a)^(1/2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.66, size = 0, normalized size = 0.00 \[ \int \frac {\left (B x +A \right ) \left (e x \right )^{m}}{\sqrt {c \,x^{2}+a}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^m*(B*x+A)/(c*x^2+a)^(1/2),x)

[Out]

int((e*x)^m*(B*x+A)/(c*x^2+a)^(1/2),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*(B*x+A)/(c*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((e*x)^m*(A + B*x))/(a + c*x^2)^(1/2),x)

[Out]

int(((e*x)^m*(A + B*x))/(a + c*x^2)^(1/2), x)

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sympy [C]  time = 3.65, size = 112, normalized size = 0.81 \[ \frac {A e^{m} x x^{m} \Gamma \left (\frac {m}{2} + \frac {1}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {m}{2} + \frac {1}{2} \\ \frac {m}{2} + \frac {3}{2} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {B e^{m} x^{2} x^{m} \Gamma \left (\frac {m}{2} + 1\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {m}{2} + 1 \\ \frac {m}{2} + 2 \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} \Gamma \left (\frac {m}{2} + 2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**m*(B*x+A)/(c*x**2+a)**(1/2),x)

[Out]

A*e**m*x*x**m*gamma(m/2 + 1/2)*hyper((1/2, m/2 + 1/2), (m/2 + 3/2,), c*x**2*exp_polar(I*pi)/a)/(2*sqrt(a)*gamm
a(m/2 + 3/2)) + B*e**m*x**2*x**m*gamma(m/2 + 1)*hyper((1/2, m/2 + 1), (m/2 + 2,), c*x**2*exp_polar(I*pi)/a)/(2
*sqrt(a)*gamma(m/2 + 2))

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