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

Optimal. Leaf size=214 \[ \frac {4 a^{7/4} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (11 A b-a B) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{77 b^{5/4} \sqrt {e} \sqrt {a+b x^2}}+\frac {2 \sqrt {e x} \left (a+b x^2\right )^{3/2} (11 A b-a B)}{77 b e}+\frac {4 a \sqrt {e x} \sqrt {a+b x^2} (11 A b-a B)}{77 b e}+\frac {2 B \sqrt {e x} \left (a+b x^2\right )^{5/2}}{11 b e} \]

[Out]

2/77*(11*A*b-B*a)*(b*x^2+a)^(3/2)*(e*x)^(1/2)/b/e+2/11*B*(b*x^2+a)^(5/2)*(e*x)^(1/2)/b/e+4/77*a*(11*A*b-B*a)*(
e*x)^(1/2)*(b*x^2+a)^(1/2)/b/e+4/77*a^(7/4)*(11*A*b-B*a)*(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^(5/4)/e^(1/2)/(b*x^2+a)^(
1/2)

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Rubi [A]  time = 0.14, antiderivative size = 214, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {459, 279, 329, 220} \[ \frac {4 a^{7/4} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (11 A b-a B) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{77 b^{5/4} \sqrt {e} \sqrt {a+b x^2}}+\frac {2 \sqrt {e x} \left (a+b x^2\right )^{3/2} (11 A b-a B)}{77 b e}+\frac {4 a \sqrt {e x} \sqrt {a+b x^2} (11 A b-a B)}{77 b e}+\frac {2 B \sqrt {e x} \left (a+b x^2\right )^{5/2}}{11 b e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a*(11*A*b - a*B)*Sqrt[e*x]*Sqrt[a + b*x^2])/(77*b*e) + (2*(11*A*b - a*B)*Sqrt[e*x]*(a + b*x^2)^(3/2))/(77*b
*e) + (2*B*Sqrt[e*x]*(a + b*x^2)^(5/2))/(11*b*e) + (4*a^(7/4)*(11*A*b - a*B)*(Sqrt[a] + Sqrt[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])/(77*b^(5/4)*Sq
rt[e]*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 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 \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{\sqrt {e x}} \, dx &=\frac {2 B \sqrt {e x} \left (a+b x^2\right )^{5/2}}{11 b e}-\frac {\left (2 \left (-\frac {11 A b}{2}+\frac {a B}{2}\right )\right ) \int \frac {\left (a+b x^2\right )^{3/2}}{\sqrt {e x}} \, dx}{11 b}\\ &=\frac {2 (11 A b-a B) \sqrt {e x} \left (a+b x^2\right )^{3/2}}{77 b e}+\frac {2 B \sqrt {e x} \left (a+b x^2\right )^{5/2}}{11 b e}+\frac {(6 a (11 A b-a B)) \int \frac {\sqrt {a+b x^2}}{\sqrt {e x}} \, dx}{77 b}\\ &=\frac {4 a (11 A b-a B) \sqrt {e x} \sqrt {a+b x^2}}{77 b e}+\frac {2 (11 A b-a B) \sqrt {e x} \left (a+b x^2\right )^{3/2}}{77 b e}+\frac {2 B \sqrt {e x} \left (a+b x^2\right )^{5/2}}{11 b e}+\frac {\left (4 a^2 (11 A b-a B)\right ) \int \frac {1}{\sqrt {e x} \sqrt {a+b x^2}} \, dx}{77 b}\\ &=\frac {4 a (11 A b-a B) \sqrt {e x} \sqrt {a+b x^2}}{77 b e}+\frac {2 (11 A b-a B) \sqrt {e x} \left (a+b x^2\right )^{3/2}}{77 b e}+\frac {2 B \sqrt {e x} \left (a+b x^2\right )^{5/2}}{11 b e}+\frac {\left (8 a^2 (11 A b-a B)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+\frac {b x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{77 b e}\\ &=\frac {4 a (11 A b-a B) \sqrt {e x} \sqrt {a+b x^2}}{77 b e}+\frac {2 (11 A b-a B) \sqrt {e x} \left (a+b x^2\right )^{3/2}}{77 b e}+\frac {2 B \sqrt {e x} \left (a+b x^2\right )^{5/2}}{11 b e}+\frac {4 a^{7/4} (11 A b-a B) \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 )}{77 b^{5/4} \sqrt {e} \sqrt {a+b x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.08, size = 96, normalized size = 0.45 \[ \frac {2 x \sqrt {a+b x^2} \left (a (11 A b-a B) \, _2F_1\left (-\frac {3}{2},\frac {1}{4};\frac {5}{4};-\frac {b x^2}{a}\right )+B \sqrt {\frac {b x^2}{a}+1} \left (a+b x^2\right )^2\right )}{11 b \sqrt {e x} \sqrt {\frac {b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*Sqrt[a + b*x^2]*(B*(a + b*x^2)^2*Sqrt[1 + (b*x^2)/a] + a*(11*A*b - a*B)*Hypergeometric2F1[-3/2, 1/4, 5/4,
-((b*x^2)/a)]))/(11*b*Sqrt[e*x]*Sqrt[1 + (b*x^2)/a])

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fricas [F]  time = 0.65, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B b x^{4} + {\left (B a + A b\right )} x^{2} + A a\right )} \sqrt {b x^{2} + a} \sqrt {e x}}{e x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*b*x^4 + (B*a + A*b)*x^2 + A*a)*sqrt(b*x^2 + a)*sqrt(e*x)/(e*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 272, normalized size = 1.27 \[ \frac {\frac {2 B \,b^{4} x^{7}}{11}+\frac {2 A \,b^{4} x^{5}}{7}+\frac {40 B a \,b^{3} x^{5}}{77}+\frac {8 A a \,b^{3} x^{3}}{7}+\frac {34 B \,a^{2} b^{2} x^{3}}{77}+\frac {6 A \,a^{2} b^{2} x}{7}+\frac {8 B \,a^{3} b x}{77}+\frac {4 \sqrt {2}\, \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \sqrt {-a b}\, A \,a^{2} b \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{7}-\frac {4 \sqrt {2}\, \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \sqrt {-a b}\, B \,a^{3} \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{77}}{\sqrt {b \,x^{2}+a}\, \sqrt {e x}\, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/77/(b*x^2+a)^(1/2)*(7*B*b^4*x^7+22*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)*(-1/(-a*b)^(1/2)*b*x)^(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+11*A*b^4*x^5-2*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)*(-1/(-a*b)^(1/2)*b*x)^(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+20*B*a*b^3*x^5+44*A*a*b^3*x^3+17*B*a^2*b^2*x^3+33*A*a^2*b^2*x+4*B*a^3*b*x)/b^2/(e*x)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 10.41, size = 199, normalized size = 0.93 \[ \frac {A a^{\frac {3}{2}} \sqrt {x} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {e} \Gamma \left (\frac {5}{4}\right )} + \frac {A \sqrt {a} b x^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {e} \Gamma \left (\frac {9}{4}\right )} + \frac {B a^{\frac {3}{2}} x^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {e} \Gamma \left (\frac {9}{4}\right )} + \frac {B \sqrt {a} b x^{\frac {9}{2}} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {e} \Gamma \left (\frac {13}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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