3.493 \(\int (e x)^m (A+B x) (a+c x^2)^{3/2} \, dx\)

Optimal. Leaf size=141 \[ \frac {a A \sqrt {a+c x^2} (e x)^{m+1} \, _2F_1\left (-\frac {3}{2},\frac {m+1}{2};\frac {m+3}{2};-\frac {c x^2}{a}\right )}{e (m+1) \sqrt {\frac {c x^2}{a}+1}}+\frac {a B \sqrt {a+c x^2} (e x)^{m+2} \, _2F_1\left (-\frac {3}{2},\frac {m+2}{2};\frac {m+4}{2};-\frac {c x^2}{a}\right )}{e^2 (m+2) \sqrt {\frac {c x^2}{a}+1}} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 141, 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} \[ \frac {a A \sqrt {a+c x^2} (e x)^{m+1} \, _2F_1\left (-\frac {3}{2},\frac {m+1}{2};\frac {m+3}{2};-\frac {c x^2}{a}\right )}{e (m+1) \sqrt {\frac {c x^2}{a}+1}}+\frac {a B \sqrt {a+c x^2} (e x)^{m+2} \, _2F_1\left (-\frac {3}{2},\frac {m+2}{2};\frac {m+4}{2};-\frac {c x^2}{a}\right )}{e^2 (m+2) \sqrt {\frac {c x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.05, size = 109, normalized size = 0.77 \[ \frac {a x \sqrt {a+c x^2} (e x)^m \left (A (m+2) \, _2F_1\left (-\frac {3}{2},\frac {m+1}{2};\frac {m+3}{2};-\frac {c x^2}{a}\right )+B (m+1) x \, _2F_1\left (-\frac {3}{2},\frac {m}{2}+1;\frac {m}{2}+2;-\frac {c x^2}{a}\right )\right )}{(m+1) (m+2) \sqrt {\frac {c x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.93, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B c x^{3} + A c x^{2} + B a x + A a\right )} \sqrt {c x^{2} + a} \left (e x\right )^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (c x^{2} + a\right )}^{\frac {3}{2}} {\left (B x + A\right )} \left (e x\right )^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.15, size = 0, normalized size = 0.00 \[ \int \left (B x +A \right ) \left (c \,x^{2}+a \right )^{\frac {3}{2}} \left (e x \right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (c x^{2} + a\right )}^{\frac {3}{2}} {\left (B x + A\right )} \left (e x\right )^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (e\,x\right )}^m\,{\left (c\,x^2+a\right )}^{3/2}\,\left (A+B\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 9.55, size = 238, normalized size = 1.69 \[ \frac {A a^{\frac {3}{2}} e^{m} x x^{m} \Gamma \left (\frac {m}{2} + \frac {1}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {m}{2} + \frac {1}{2} \\ \frac {m}{2} + \frac {3}{2} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {A \sqrt {a} c e^{m} x^{3} x^{m} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {m}{2} + \frac {3}{2} \\ \frac {m}{2} + \frac {5}{2} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {B a^{\frac {3}{2}} e^{m} x^{2} x^{m} \Gamma \left (\frac {m}{2} + 1\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {m}{2} + 1 \\ \frac {m}{2} + 2 \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {m}{2} + 2\right )} + \frac {B \sqrt {a} c e^{m} x^{4} x^{m} \Gamma \left (\frac {m}{2} + 2\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {m}{2} + 2 \\ \frac {m}{2} + 3 \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {m}{2} + 3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*a**(3/2)*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*ga
mma(m/2 + 3/2)) + A*sqrt(a)*c*e**m*x**3*x**m*gamma(m/2 + 3/2)*hyper((-1/2, m/2 + 3/2), (m/2 + 5/2,), c*x**2*ex
p_polar(I*pi)/a)/(2*gamma(m/2 + 5/2)) + B*a**(3/2)*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*gamma(m/2 + 2)) + B*sqrt(a)*c*e**m*x**4*x**m*gamma(m/2 + 2)*hyper((-1/2, m/
2 + 2), (m/2 + 3,), c*x**2*exp_polar(I*pi)/a)/(2*gamma(m/2 + 3))

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