3.5.92 \(\int (e x)^m (A+B x) (a+c x^2)^{5/2} \, dx\) [492]

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

[Out]

a^2*A*(e*x)^(1+m)*hypergeom([-5/2, 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^2*B*(e*x)^(2+m)*hypergeom([-5/2, 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 = 0.05, antiderivative size = 145, 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 = {822, 372, 371} \begin {gather*} \frac {a^2 A \sqrt {a+c x^2} (e x)^{m+1} \, _2F_1\left (-\frac {5}{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^2 B \sqrt {a+c x^2} (e x)^{m+2} \, _2F_1\left (-\frac {5}{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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

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 822

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 )^{5/2} \, dx &=A \int (e x)^m \left (a+c x^2\right )^{5/2} \, dx+\frac {B \int (e x)^{1+m} \left (a+c x^2\right )^{5/2} \, dx}{e}\\ &=\frac {\left (a^2 A \sqrt {a+c x^2}\right ) \int (e x)^m \left (1+\frac {c x^2}{a}\right )^{5/2} \, dx}{\sqrt {1+\frac {c x^2}{a}}}+\frac {\left (a^2 B \sqrt {a+c x^2}\right ) \int (e x)^{1+m} \left (1+\frac {c x^2}{a}\right )^{5/2} \, dx}{e \sqrt {1+\frac {c x^2}{a}}}\\ &=\frac {a^2 A (e x)^{1+m} \sqrt {a+c x^2} \, _2F_1\left (-\frac {5}{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^2 B (e x)^{2+m} \sqrt {a+c x^2} \, _2F_1\left (-\frac {5}{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 = 0.43, size = 111, normalized size = 0.77 \begin {gather*} \frac {a^2 x (e x)^m \sqrt {a+c x^2} \left (B (1+m) x \, _2F_1\left (-\frac {5}{2},1+\frac {m}{2};2+\frac {m}{2};-\frac {c x^2}{a}\right )+A (2+m) \, _2F_1\left (-\frac {5}{2},\frac {1+m}{2};\frac {3+m}{2};-\frac {c x^2}{a}\right )\right )}{(1+m) (2+m) \sqrt {1+\frac {c x^2}{a}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*x*(e*x)^m*Sqrt[a + c*x^2]*(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)]))/((1 + m)*(2 + m)*Sqrt[1 + (c*x^2)/a])

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Maple [F]
time = 0.09, size = 0, normalized size = 0.00 \[\int \left (e x \right )^{m} \left (B x +A \right ) \left (c \,x^{2}+a \right )^{\frac {5}{2}}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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Sympy [C] Result contains complex when optimal does not.
time = 11.89, size = 360, normalized size = 2.48 \begin {gather*} \frac {A a^{\frac {5}{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 a^{\frac {3}{2}} 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 )}}{\Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {A \sqrt {a} c^{2} e^{m} x^{5} x^{m} \Gamma \left (\frac {m}{2} + \frac {5}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {m}{2} + \frac {5}{2} \\ \frac {m}{2} + \frac {7}{2} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {m}{2} + \frac {7}{2}\right )} + \frac {B a^{\frac {5}{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 a^{\frac {3}{2}} 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 )}}{\Gamma \left (\frac {m}{2} + 3\right )} + \frac {B \sqrt {a} c^{2} e^{m} x^{6} x^{m} \Gamma \left (\frac {m}{2} + 3\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {m}{2} + 3 \\ \frac {m}{2} + 4 \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {m}{2} + 4\right )} \end {gather*}

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]

A*a**(5/2)*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*a**(3/2)*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*e
xp_polar(I*pi)/a)/gamma(m/2 + 5/2) + A*sqrt(a)*c**2*e**m*x**5*x**m*gamma(m/2 + 5/2)*hyper((-1/2, m/2 + 5/2), (
m/2 + 7/2,), c*x**2*exp_polar(I*pi)/a)/(2*gamma(m/2 + 7/2)) + B*a**(5/2)*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*a**(3/2)*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)/gamma(m/2 + 3) + B*sqrt(a)*c**2*e**m*x**6
*x**m*gamma(m/2 + 3)*hyper((-1/2, m/2 + 3), (m/2 + 4,), c*x**2*exp_polar(I*pi)/a)/(2*gamma(m/2 + 4))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (e\,x\right )}^m\,{\left (c\,x^2+a\right )}^{5/2}\,\left (A+B\,x\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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