3.785 \(\int (e x)^{3/2} \sqrt {a+b x^2} (A+B x^2) \, dx\)

Optimal. Leaf size=212 \[ -\frac {2 a^{7/4} e^{3/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (11 A b-5 a B) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{231 b^{9/4} \sqrt {a+b x^2}}+\frac {4 a e \sqrt {e x} \sqrt {a+b x^2} (11 A b-5 a B)}{231 b^2}+\frac {2 (e x)^{5/2} \sqrt {a+b x^2} (11 A b-5 a B)}{77 b e}+\frac {2 B (e x)^{5/2} \left (a+b x^2\right )^{3/2}}{11 b e} \]

[Out]

2/11*B*(e*x)^(5/2)*(b*x^2+a)^(3/2)/b/e+2/77*(11*A*b-5*B*a)*(e*x)^(5/2)*(b*x^2+a)^(1/2)/b/e+4/231*a*(11*A*b-5*B
*a)*e*(e*x)^(1/2)*(b*x^2+a)^(1/2)/b^2-2/231*a^(7/4)*(11*A*b-5*B*a)*e^(3/2)*(cos(2*arctan(b^(1/4)*(e*x)^(1/2)/a
^(1/4)/e^(1/2)))^2)^(1/2)/cos(2*arctan(b^(1/4)*(e*x)^(1/2)/a^(1/4)/e^(1/2)))*EllipticF(sin(2*arctan(b^(1/4)*(e
*x)^(1/2)/a^(1/4)/e^(1/2))),1/2*2^(1/2))*(a^(1/2)+x*b^(1/2))*((b*x^2+a)/(a^(1/2)+x*b^(1/2))^2)^(1/2)/b^(9/4)/(
b*x^2+a)^(1/2)

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Rubi [A]  time = 0.16, antiderivative size = 212, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {459, 279, 321, 329, 220} \[ -\frac {2 a^{7/4} e^{3/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (11 A b-5 a B) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{231 b^{9/4} \sqrt {a+b x^2}}+\frac {4 a e \sqrt {e x} \sqrt {a+b x^2} (11 A b-5 a B)}{231 b^2}+\frac {2 (e x)^{5/2} \sqrt {a+b x^2} (11 A b-5 a B)}{77 b e}+\frac {2 B (e x)^{5/2} \left (a+b x^2\right )^{3/2}}{11 b e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a*(11*A*b - 5*a*B)*e*Sqrt[e*x]*Sqrt[a + b*x^2])/(231*b^2) + (2*(11*A*b - 5*a*B)*(e*x)^(5/2)*Sqrt[a + b*x^2]
)/(77*b*e) + (2*B*(e*x)^(5/2)*(a + b*x^2)^(3/2))/(11*b*e) - (2*a^(7/4)*(11*A*b - 5*a*B)*e^(3/2)*(Sqrt[a] + Sqr
t[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticF[2*ArcTan[(b^(1/4)*Sqrt[e*x])/(a^(1/4)*Sqrt[e])], 1
/2])/(231*b^(9/4)*Sqrt[a + b*x^2])

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 279

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

Rule 321

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

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 459

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(b*e*(m + n*(p + 1) + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]

Rubi steps

\begin {align*} \int (e x)^{3/2} \sqrt {a+b x^2} \left (A+B x^2\right ) \, dx &=\frac {2 B (e x)^{5/2} \left (a+b x^2\right )^{3/2}}{11 b e}-\frac {\left (2 \left (-\frac {11 A b}{2}+\frac {5 a B}{2}\right )\right ) \int (e x)^{3/2} \sqrt {a+b x^2} \, dx}{11 b}\\ &=\frac {2 (11 A b-5 a B) (e x)^{5/2} \sqrt {a+b x^2}}{77 b e}+\frac {2 B (e x)^{5/2} \left (a+b x^2\right )^{3/2}}{11 b e}+\frac {(2 a (11 A b-5 a B)) \int \frac {(e x)^{3/2}}{\sqrt {a+b x^2}} \, dx}{77 b}\\ &=\frac {4 a (11 A b-5 a B) e \sqrt {e x} \sqrt {a+b x^2}}{231 b^2}+\frac {2 (11 A b-5 a B) (e x)^{5/2} \sqrt {a+b x^2}}{77 b e}+\frac {2 B (e x)^{5/2} \left (a+b x^2\right )^{3/2}}{11 b e}-\frac {\left (2 a^2 (11 A b-5 a B) e^2\right ) \int \frac {1}{\sqrt {e x} \sqrt {a+b x^2}} \, dx}{231 b^2}\\ &=\frac {4 a (11 A b-5 a B) e \sqrt {e x} \sqrt {a+b x^2}}{231 b^2}+\frac {2 (11 A b-5 a B) (e x)^{5/2} \sqrt {a+b x^2}}{77 b e}+\frac {2 B (e x)^{5/2} \left (a+b x^2\right )^{3/2}}{11 b e}-\frac {\left (4 a^2 (11 A b-5 a B) e\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+\frac {b x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{231 b^2}\\ &=\frac {4 a (11 A b-5 a B) e \sqrt {e x} \sqrt {a+b x^2}}{231 b^2}+\frac {2 (11 A b-5 a B) (e x)^{5/2} \sqrt {a+b x^2}}{77 b e}+\frac {2 B (e x)^{5/2} \left (a+b x^2\right )^{3/2}}{11 b e}-\frac {2 a^{7/4} (11 A b-5 a B) e^{3/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{231 b^{9/4} \sqrt {a+b x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.16, size = 110, normalized size = 0.52 \[ \frac {2 e \sqrt {e x} \sqrt {a+b x^2} \left (a (5 a B-11 A b) \, _2F_1\left (-\frac {1}{2},\frac {1}{4};\frac {5}{4};-\frac {b x^2}{a}\right )-\left (a+b x^2\right ) \sqrt {\frac {b x^2}{a}+1} \left (5 a B-11 A b-7 b B x^2\right )\right )}{77 b^2 \sqrt {\frac {b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*e*Sqrt[e*x]*Sqrt[a + b*x^2]*(-((a + b*x^2)*Sqrt[1 + (b*x^2)/a]*(-11*A*b + 5*a*B - 7*b*B*x^2)) + a*(-11*A*b
+ 5*a*B)*Hypergeometric2F1[-1/2, 1/4, 5/4, -((b*x^2)/a)]))/(77*b^2*Sqrt[1 + (b*x^2)/a])

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fricas [F]  time = 0.74, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B e x^{3} + A e x\right )} \sqrt {b x^{2} + a} \sqrt {e x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(3/2)*(B*x^2+A)*(b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(3/2)*(B*x^2+A)*(b*x^2+a)^(1/2),x, algorithm="giac")

[Out]

integrate((B*x^2 + A)*sqrt(b*x^2 + a)*(e*x)^(3/2), x)

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maple [A]  time = 0.13, size = 276, normalized size = 1.30 \[ -\frac {2 \sqrt {e x}\, \left (-21 B \,b^{4} x^{7}-33 A \,b^{4} x^{5}-27 B a \,b^{3} x^{5}-55 A a \,b^{3} x^{3}+4 B \,a^{2} b^{2} x^{3}-22 A \,a^{2} b^{2} x +10 B \,a^{3} b x +11 \sqrt {2}\, \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \sqrt {-a b}\, A \,a^{2} b \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )-5 \sqrt {2}\, \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \sqrt {-a b}\, B \,a^{3} \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )\right ) e}{231 \sqrt {b \,x^{2}+a}\, b^{3} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^(3/2)*(B*x^2+A)*(b*x^2+a)^(1/2),x)

[Out]

-2/231*e/x*(e*x)^(1/2)/(b*x^2+a)^(1/2)*(-21*B*x^7*b^4+11*A*2^(1/2)*((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*((-
b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*(-x/(-a*b)^(1/2)*b)^(1/2)*EllipticF(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1
/2),1/2*2^(1/2))*(-a*b)^(1/2)*a^2*b-33*A*x^5*b^4-5*B*2^(1/2)*((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*((-b*x+(-
a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*(-x/(-a*b)^(1/2)*b)^(1/2)*EllipticF(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/
2*2^(1/2))*(-a*b)^(1/2)*a^3-27*B*x^5*a*b^3-55*A*x^3*a*b^3+4*B*x^3*a^2*b^2-22*A*x*a^2*b^2+10*B*x*a^3*b)/b^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(3/2)*(B*x^2+A)*(b*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)*sqrt(b*x^2 + a)*(e*x)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x^2)*(e*x)^(3/2)*(a + b*x^2)^(1/2),x)

[Out]

int((A + B*x^2)*(e*x)^(3/2)*(a + b*x^2)^(1/2), x)

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sympy [C]  time = 10.44, size = 97, normalized size = 0.46 \[ \frac {A \sqrt {a} e^{\frac {3}{2}} x^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {9}{4}\right )} + \frac {B \sqrt {a} e^{\frac {3}{2}} x^{\frac {9}{2}} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {13}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**(3/2)*(B*x**2+A)*(b*x**2+a)**(1/2),x)

[Out]

A*sqrt(a)*e**(3/2)*x**(5/2)*gamma(5/4)*hyper((-1/2, 5/4), (9/4,), b*x**2*exp_polar(I*pi)/a)/(2*gamma(9/4)) + B
*sqrt(a)*e**(3/2)*x**(9/2)*gamma(9/4)*hyper((-1/2, 9/4), (13/4,), b*x**2*exp_polar(I*pi)/a)/(2*gamma(13/4))

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