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

Optimal. Leaf size=337 \[ \frac {3 \sqrt [4]{a} e^{5/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (5 A b-7 a B) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{10 b^{11/4} \sqrt {a+b x^2}}-\frac {3 \sqrt [4]{a} e^{5/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (5 A b-7 a B) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{5 b^{11/4} \sqrt {a+b x^2}}+\frac {3 e^2 \sqrt {e x} \sqrt {a+b x^2} (5 A b-7 a B)}{5 b^{5/2} \left (\sqrt {a}+\sqrt {b} x\right )}-\frac {e (e x)^{3/2} (5 A b-7 a B)}{5 b^2 \sqrt {a+b x^2}}+\frac {2 B (e x)^{7/2}}{5 b e \sqrt {a+b x^2}} \]

[Out]

-1/5*(5*A*b-7*B*a)*e*(e*x)^(3/2)/b^2/(b*x^2+a)^(1/2)+2/5*B*(e*x)^(7/2)/b/e/(b*x^2+a)^(1/2)+3/5*(5*A*b-7*B*a)*e
^2*(e*x)^(1/2)*(b*x^2+a)^(1/2)/b^(5/2)/(a^(1/2)+x*b^(1/2))-3/5*a^(1/4)*(5*A*b-7*B*a)*e^(5/2)*(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)))*EllipticE(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^(11/4)/(b*x^2+a)^(1/2)+3/10*a^(1/4)*(5*A*b-7*B*a)*e^(5/2)*(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^(11/4)/(b*x^2
+a)^(1/2)

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Rubi [A]  time = 0.25, antiderivative size = 337, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {459, 288, 329, 305, 220, 1196} \[ \frac {3 e^2 \sqrt {e x} \sqrt {a+b x^2} (5 A b-7 a B)}{5 b^{5/2} \left (\sqrt {a}+\sqrt {b} x\right )}+\frac {3 \sqrt [4]{a} e^{5/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (5 A b-7 a B) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{10 b^{11/4} \sqrt {a+b x^2}}-\frac {3 \sqrt [4]{a} e^{5/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (5 A b-7 a B) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{5 b^{11/4} \sqrt {a+b x^2}}-\frac {e (e x)^{3/2} (5 A b-7 a B)}{5 b^2 \sqrt {a+b x^2}}+\frac {2 B (e x)^{7/2}}{5 b e \sqrt {a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((5*A*b - 7*a*B)*e*(e*x)^(3/2))/(5*b^2*Sqrt[a + b*x^2]) + (2*B*(e*x)^(7/2))/(5*b*e*Sqrt[a + b*x^2]) + (3*(5*A
*b - 7*a*B)*e^2*Sqrt[e*x]*Sqrt[a + b*x^2])/(5*b^(5/2)*(Sqrt[a] + Sqrt[b]*x)) - (3*a^(1/4)*(5*A*b - 7*a*B)*e^(5
/2)*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticE[2*ArcTan[(b^(1/4)*Sqrt[e*x])/(a^
(1/4)*Sqrt[e])], 1/2])/(5*b^(11/4)*Sqrt[a + b*x^2]) + (3*a^(1/4)*(5*A*b - 7*a*B)*e^(5/2)*(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])/(1
0*b^(11/4)*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 288

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*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 305

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

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]

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

\begin {align*} \int \frac {(e x)^{5/2} \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx &=\frac {2 B (e x)^{7/2}}{5 b e \sqrt {a+b x^2}}-\frac {\left (2 \left (-\frac {5 A b}{2}+\frac {7 a B}{2}\right )\right ) \int \frac {(e x)^{5/2}}{\left (a+b x^2\right )^{3/2}} \, dx}{5 b}\\ &=-\frac {(5 A b-7 a B) e (e x)^{3/2}}{5 b^2 \sqrt {a+b x^2}}+\frac {2 B (e x)^{7/2}}{5 b e \sqrt {a+b x^2}}+\frac {\left (3 (5 A b-7 a B) e^2\right ) \int \frac {\sqrt {e x}}{\sqrt {a+b x^2}} \, dx}{10 b^2}\\ &=-\frac {(5 A b-7 a B) e (e x)^{3/2}}{5 b^2 \sqrt {a+b x^2}}+\frac {2 B (e x)^{7/2}}{5 b e \sqrt {a+b x^2}}+\frac {(3 (5 A b-7 a B) e) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {a+\frac {b x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{5 b^2}\\ &=-\frac {(5 A b-7 a B) e (e x)^{3/2}}{5 b^2 \sqrt {a+b x^2}}+\frac {2 B (e x)^{7/2}}{5 b e \sqrt {a+b x^2}}+\frac {\left (3 \sqrt {a} (5 A b-7 a B) e^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+\frac {b x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{5 b^{5/2}}-\frac {\left (3 \sqrt {a} (5 A b-7 a B) e^2\right ) \operatorname {Subst}\left (\int \frac {1-\frac {\sqrt {b} x^2}{\sqrt {a} e}}{\sqrt {a+\frac {b x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{5 b^{5/2}}\\ &=-\frac {(5 A b-7 a B) e (e x)^{3/2}}{5 b^2 \sqrt {a+b x^2}}+\frac {2 B (e x)^{7/2}}{5 b e \sqrt {a+b x^2}}+\frac {3 (5 A b-7 a B) e^2 \sqrt {e x} \sqrt {a+b x^2}}{5 b^{5/2} \left (\sqrt {a}+\sqrt {b} x\right )}-\frac {3 \sqrt [4]{a} (5 A b-7 a B) e^{5/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{5 b^{11/4} \sqrt {a+b x^2}}+\frac {3 \sqrt [4]{a} (5 A b-7 a B) e^{5/2} \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 )}{10 b^{11/4} \sqrt {a+b x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.12, size = 84, normalized size = 0.25 \[ \frac {2 e (e x)^{3/2} \left (\sqrt {\frac {b x^2}{a}+1} (7 a B-5 A b) \, _2F_1\left (\frac {3}{4},\frac {3}{2};\frac {7}{4};-\frac {b x^2}{a}\right )-7 a B+5 A b+b B x^2\right )}{5 b^2 \sqrt {a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B e^{2} x^{4} + A e^{2} x^{2}\right )} \sqrt {b x^{2} + a} \sqrt {e x}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B x^{2} + A\right )} \left (e x\right )^{\frac {5}{2}}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 391, normalized size = 1.16 \[ \frac {\sqrt {e x}\, \left (4 B \,b^{2} x^{4}-10 A \,b^{2} x^{2}+14 B a b \,x^{2}+30 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, A a b \EllipticE \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )-15 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, A a b \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )-42 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, B \,a^{2} \EllipticE \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )+21 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, B \,a^{2} \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )\right ) e^{2}}{10 \sqrt {b \,x^{2}+a}\, b^{3} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*e^2/x*(e*x)^(1/2)*(30*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)*(-1/(-a*b)^(1/2)*b*x)^(1/2)*EllipticE(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*a*b-15*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)*(-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-42*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)*(-1/(-a*b)^(1/2)*b*x)^(1/2)*EllipticE(((b*x+(-a*b)^(1/
2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*a^2+21*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)*(-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^2+4*B*b^2*x^4-10*A*b^2*x^2+14*B*a*b*x^2)/(b*x^2+a)^(1/2)/b^3

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B x^{2} + A\right )} \left (e x\right )^{\frac {5}{2}}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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