3.8.98 \(\int \frac {(a+b x^2)^{3/2} (A+B x^2)}{(e x)^{5/2}} \, dx\) [798]

Optimal. Leaf size=210 \[ \frac {4 (7 A b+3 a B) \sqrt {e x} \sqrt {a+b x^2}}{21 e^3}+\frac {2 (7 A b+3 a B) \sqrt {e x} \left (a+b x^2\right )^{3/2}}{21 a e^3}-\frac {2 A \left (a+b x^2\right )^{5/2}}{3 a e (e x)^{3/2}}+\frac {4 a^{3/4} (7 A b+3 a B) \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{21 \sqrt [4]{b} e^{5/2} \sqrt {a+b x^2}} \]

[Out]

-2/3*A*(b*x^2+a)^(5/2)/a/e/(e*x)^(3/2)+2/21*(7*A*b+3*B*a)*(b*x^2+a)^(3/2)*(e*x)^(1/2)/a/e^3+4/21*(7*A*b+3*B*a)
*(e*x)^(1/2)*(b*x^2+a)^(1/2)/e^3+4/21*a^(3/4)*(7*A*b+3*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^(1/4)/e^(5/2)/(b*x^2+a
)^(1/2)

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Rubi [A]
time = 0.10, antiderivative size = 210, 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 = {464, 285, 335, 226} \begin {gather*} \frac {4 a^{3/4} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (3 a B+7 A b) F\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{21 \sqrt [4]{b} e^{5/2} \sqrt {a+b x^2}}+\frac {2 \sqrt {e x} \left (a+b x^2\right )^{3/2} (3 a B+7 A b)}{21 a e^3}+\frac {4 \sqrt {e x} \sqrt {a+b x^2} (3 a B+7 A b)}{21 e^3}-\frac {2 A \left (a+b x^2\right )^{5/2}}{3 a e (e x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)^(3/2)*(A + B*x^2))/(e*x)^(5/2),x]

[Out]

(4*(7*A*b + 3*a*B)*Sqrt[e*x]*Sqrt[a + b*x^2])/(21*e^3) + (2*(7*A*b + 3*a*B)*Sqrt[e*x]*(a + b*x^2)^(3/2))/(21*a
*e^3) - (2*A*(a + b*x^2)^(5/2))/(3*a*e*(e*x)^(3/2)) + (4*a^(3/4)*(7*A*b + 3*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])/(21*b^(1/4
)*e^(5/2)*Sqrt[a + b*x^2])

Rule 226

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)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 285

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 335

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 464

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

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{(e x)^{5/2}} \, dx &=-\frac {2 A \left (a+b x^2\right )^{5/2}}{3 a e (e x)^{3/2}}+\frac {(7 A b+3 a B) \int \frac {\left (a+b x^2\right )^{3/2}}{\sqrt {e x}} \, dx}{3 a e^2}\\ &=\frac {2 (7 A b+3 a B) \sqrt {e x} \left (a+b x^2\right )^{3/2}}{21 a e^3}-\frac {2 A \left (a+b x^2\right )^{5/2}}{3 a e (e x)^{3/2}}+\frac {(2 (7 A b+3 a B)) \int \frac {\sqrt {a+b x^2}}{\sqrt {e x}} \, dx}{7 e^2}\\ &=\frac {4 (7 A b+3 a B) \sqrt {e x} \sqrt {a+b x^2}}{21 e^3}+\frac {2 (7 A b+3 a B) \sqrt {e x} \left (a+b x^2\right )^{3/2}}{21 a e^3}-\frac {2 A \left (a+b x^2\right )^{5/2}}{3 a e (e x)^{3/2}}+\frac {(4 a (7 A b+3 a B)) \int \frac {1}{\sqrt {e x} \sqrt {a+b x^2}} \, dx}{21 e^2}\\ &=\frac {4 (7 A b+3 a B) \sqrt {e x} \sqrt {a+b x^2}}{21 e^3}+\frac {2 (7 A b+3 a B) \sqrt {e x} \left (a+b x^2\right )^{3/2}}{21 a e^3}-\frac {2 A \left (a+b x^2\right )^{5/2}}{3 a e (e x)^{3/2}}+\frac {(8 a (7 A b+3 a B)) \text {Subst}\left (\int \frac {1}{\sqrt {a+\frac {b x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{21 e^3}\\ &=\frac {4 (7 A b+3 a B) \sqrt {e x} \sqrt {a+b x^2}}{21 e^3}+\frac {2 (7 A b+3 a B) \sqrt {e x} \left (a+b x^2\right )^{3/2}}{21 a e^3}-\frac {2 A \left (a+b x^2\right )^{5/2}}{3 a e (e x)^{3/2}}+\frac {4 a^{3/4} (7 A b+3 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 )}{21 \sqrt [4]{b} e^{5/2} \sqrt {a+b x^2}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.06, size = 85, normalized size = 0.40 \begin {gather*} \frac {2 x \sqrt {a+b x^2} \left (-\frac {A \left (a+b x^2\right )^2}{a}+\frac {(7 A b+3 a B) x^2 \, _2F_1\left (-\frac {3}{2},\frac {1}{4};\frac {5}{4};-\frac {b x^2}{a}\right )}{\sqrt {1+\frac {b x^2}{a}}}\right )}{3 (e x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)^(3/2)*(A + B*x^2))/(e*x)^(5/2),x]

[Out]

(2*x*Sqrt[a + b*x^2]*(-((A*(a + b*x^2)^2)/a) + ((7*A*b + 3*a*B)*x^2*Hypergeometric2F1[-3/2, 1/4, 5/4, -((b*x^2
)/a)])/Sqrt[1 + (b*x^2)/a]))/(3*(e*x)^(5/2))

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Maple [A]
time = 0.12, size = 255, normalized size = 1.21

method	result	size
risch	\(-\frac {2 \sqrt {b \,x^{2}+a}\, \left (-3 b B \,x^{4}-7 A b \,x^{2}-9 B a \,x^{2}+7 A a \right )}{21 x \,e^{2} \sqrt {e x}}+\frac {4 a \left (7 A b +3 B a \right ) \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {\left (b \,x^{2}+a \right ) e x}}{21 b \sqrt {b e \,x^{3}+a e x}\, e^{2} \sqrt {e x}\, \sqrt {b \,x^{2}+a}}\)	\(199\)
default	\(\frac {\frac {4 A \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-a b}\, a b x}{3}+\frac {4 B \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-a b}\, a^{2} x}{7}+\frac {2 B \,b^{3} x^{6}}{7}+\frac {2 A \,b^{3} x^{4}}{3}+\frac {8 B a \,b^{2} x^{4}}{7}+\frac {6 B \,a^{2} b \,x^{2}}{7}-\frac {2 A \,a^{2} b}{3}}{\sqrt {b \,x^{2}+a}\, x b \,e^{2} \sqrt {e x}}\)	\(255\)
elliptic	\(\frac {\sqrt {\left (b \,x^{2}+a \right ) e x}\, \left (-\frac {2 a A \sqrt {b e \,x^{3}+a e x}}{3 e^{3} x^{2}}+\frac {2 B b \,x^{2} \sqrt {b e \,x^{3}+a e x}}{7 e^{3}}+\frac {2 \left (\frac {b \left (A b +2 B a \right )}{e^{2}}-\frac {5 B b a}{7 e^{2}}\right ) \sqrt {b e \,x^{3}+a e x}}{3 b e}+\frac {\left (\frac {a \left (2 A b +B a \right )}{e^{2}}-\frac {b a A}{3 e^{2}}-\frac {\left (\frac {b \left (A b +2 B a \right )}{e^{2}}-\frac {5 B b a}{7 e^{2}}\right ) a}{3 b}\right ) \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b \sqrt {b e \,x^{3}+a e x}}\right )}{\sqrt {e x}\, \sqrt {b \,x^{2}+a}}\)	\(277\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(3/2)*(B*x^2+A)/(e*x)^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/21/(b*x^2+a)^(1/2)/x*(14*A*((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*2^(1/2)*((-b*x+(-a*b)^(1/2))/(-a*b)^(1/2)
)^(1/2)*(-x*b/(-a*b)^(1/2))^(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*b*x+6*B*((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*2^(1/2)*((-b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*(-x*b/(-a*b
)^(1/2))^(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*x+3*B*b^3*x^6+7
*A*b^3*x^4+12*B*a*b^2*x^4+9*B*a^2*b*x^2-7*A*a^2*b)/b/e^2/(e*x)^(1/2)

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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((b*x^2+a)^(3/2)*(B*x^2+A)/(e*x)^(5/2),x, algorithm="maxima")

[Out]

e^(-5/2)*integrate((B*x^2 + A)*(b*x^2 + a)^(3/2)/x^(5/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.26, size = 84, normalized size = 0.40 \begin {gather*} \frac {2 \, {\left (4 \, {\left (3 \, B a^{2} + 7 \, A a b\right )} \sqrt {b} x^{2} {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right ) + {\left (3 \, B b^{2} x^{4} - 7 \, A a b + {\left (9 \, B a b + 7 \, A b^{2}\right )} x^{2}\right )} \sqrt {b x^{2} + a} \sqrt {x}\right )} e^{\left (-\frac {5}{2}\right )}}{21 \, b x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)*(B*x^2+A)/(e*x)^(5/2),x, algorithm="fricas")

[Out]

2/21*(4*(3*B*a^2 + 7*A*a*b)*sqrt(b)*x^2*weierstrassPInverse(-4*a/b, 0, x) + (3*B*b^2*x^4 - 7*A*a*b + (9*B*a*b
+ 7*A*b^2)*x^2)*sqrt(b*x^2 + a)*sqrt(x))*e^(-5/2)/(b*x^2)

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Sympy [C] Result contains complex when optimal does not.
time = 6.96, size = 202, normalized size = 0.96 \begin {gather*} \frac {A a^{\frac {3}{2}} \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, - \frac {1}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac {5}{2}} x^{\frac {3}{2}} \Gamma \left (\frac {1}{4}\right )} + \frac {A \sqrt {a} b \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 e^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right )} + \frac {B 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 e^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right )} + \frac {B \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 e^{\frac {5}{2}} \Gamma \left (\frac {9}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(3/2)*(B*x**2+A)/(e*x)**(5/2),x)

[Out]

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

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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((b*x^2+a)^(3/2)*(B*x^2+A)/(e*x)^(5/2),x, algorithm="giac")

[Out]

integrate((B*x^2 + A)*(b*x^2 + a)^(3/2)*e^(-5/2)/x^(5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (B\,x^2+A\right )\,{\left (b\,x^2+a\right )}^{3/2}}{{\left (e\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^2)*(a + b*x^2)^(3/2))/(e*x)^(5/2),x)

[Out]

int(((A + B*x^2)*(a + b*x^2)^(3/2))/(e*x)^(5/2), x)

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