3.12.10 \(\int \frac {(e x)^{9/2} (c+d x^2)}{(a+b x^2)^{5/4}} \, dx\) [1110]

Optimal. Leaf size=180 \[ -\frac {7 a (10 b c-11 a d) e^3 (e x)^{3/2}}{60 b^3 \sqrt [4]{a+b x^2}}+\frac {(10 b c-11 a d) e (e x)^{7/2}}{30 b^2 \sqrt [4]{a+b x^2}}+\frac {d (e x)^{11/2}}{5 b e \sqrt [4]{a+b x^2}}-\frac {7 a^{3/2} (10 b c-11 a d) e^4 \sqrt [4]{1+\frac {a}{b x^2}} \sqrt {e x} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{20 b^{7/2} \sqrt [4]{a+b x^2}} \]

[Out]

-7/60*a*(-11*a*d+10*b*c)*e^3*(e*x)^(3/2)/b^3/(b*x^2+a)^(1/4)+1/30*(-11*a*d+10*b*c)*e*(e*x)^(7/2)/b^2/(b*x^2+a)
^(1/4)+1/5*d*(e*x)^(11/2)/b/e/(b*x^2+a)^(1/4)-7/20*a^(3/2)*(-11*a*d+10*b*c)*e^4*(1+a/b/x^2)^(1/4)*(cos(1/2*arc
cot(x*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arccot(x*b^(1/2)/a^(1/2)))*EllipticE(sin(1/2*arccot(x*b^(1/2)/a^(1/2)
)),2^(1/2))*(e*x)^(1/2)/b^(7/2)/(b*x^2+a)^(1/4)

________________________________________________________________________________________

Rubi [A]
time = 0.07, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {470, 291, 290, 342, 202} \begin {gather*} -\frac {7 a^{3/2} e^4 \sqrt {e x} \sqrt [4]{\frac {a}{b x^2}+1} (10 b c-11 a d) E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{20 b^{7/2} \sqrt [4]{a+b x^2}}-\frac {7 a e^3 (e x)^{3/2} (10 b c-11 a d)}{60 b^3 \sqrt [4]{a+b x^2}}+\frac {e (e x)^{7/2} (10 b c-11 a d)}{30 b^2 \sqrt [4]{a+b x^2}}+\frac {d (e x)^{11/2}}{5 b e \sqrt [4]{a+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((e*x)^(9/2)*(c + d*x^2))/(a + b*x^2)^(5/4),x]

[Out]

(-7*a*(10*b*c - 11*a*d)*e^3*(e*x)^(3/2))/(60*b^3*(a + b*x^2)^(1/4)) + ((10*b*c - 11*a*d)*e*(e*x)^(7/2))/(30*b^
2*(a + b*x^2)^(1/4)) + (d*(e*x)^(11/2))/(5*b*e*(a + b*x^2)^(1/4)) - (7*a^(3/2)*(10*b*c - 11*a*d)*e^4*(1 + a/(b
*x^2))^(1/4)*Sqrt[e*x]*EllipticE[ArcCot[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(20*b^(7/2)*(a + b*x^2)^(1/4))

Rule 202

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]))*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]
*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 290

Int[Sqrt[(c_.)*(x_)]/((a_) + (b_.)*(x_)^2)^(5/4), x_Symbol] :> Dist[Sqrt[c*x]*((1 + a/(b*x^2))^(1/4)/(b*(a + b
*x^2)^(1/4))), Int[1/(x^2*(1 + a/(b*x^2))^(5/4)), x], x] /; FreeQ[{a, b, c}, x] && PosQ[b/a]

Rule 291

Int[((c_.)*(x_))^(m_)/((a_) + (b_.)*(x_)^2)^(5/4), x_Symbol] :> Simp[2*c*((c*x)^(m - 1)/(b*(2*m - 3)*(a + b*x^
2)^(1/4))), x] - Dist[2*a*c^2*((m - 1)/(b*(2*m - 3))), Int[(c*x)^(m - 2)/(a + b*x^2)^(5/4), x], x] /; FreeQ[{a
, b, c}, x] && PosQ[b/a] && IntegerQ[2*m] && GtQ[m, 3/2]

Rule 342

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rule 470

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 {(e x)^{9/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx &=\frac {d (e x)^{11/2}}{5 b e \sqrt [4]{a+b x^2}}-\frac {\left (-5 b c+\frac {11 a d}{2}\right ) \int \frac {(e x)^{9/2}}{\left (a+b x^2\right )^{5/4}} \, dx}{5 b}\\ &=\frac {(10 b c-11 a d) e (e x)^{7/2}}{30 b^2 \sqrt [4]{a+b x^2}}+\frac {d (e x)^{11/2}}{5 b e \sqrt [4]{a+b x^2}}-\frac {\left (7 a (10 b c-11 a d) e^2\right ) \int \frac {(e x)^{5/2}}{\left (a+b x^2\right )^{5/4}} \, dx}{60 b^2}\\ &=-\frac {7 a (10 b c-11 a d) e^3 (e x)^{3/2}}{60 b^3 \sqrt [4]{a+b x^2}}+\frac {(10 b c-11 a d) e (e x)^{7/2}}{30 b^2 \sqrt [4]{a+b x^2}}+\frac {d (e x)^{11/2}}{5 b e \sqrt [4]{a+b x^2}}+\frac {\left (7 a^2 (10 b c-11 a d) e^4\right ) \int \frac {\sqrt {e x}}{\left (a+b x^2\right )^{5/4}} \, dx}{40 b^3}\\ &=-\frac {7 a (10 b c-11 a d) e^3 (e x)^{3/2}}{60 b^3 \sqrt [4]{a+b x^2}}+\frac {(10 b c-11 a d) e (e x)^{7/2}}{30 b^2 \sqrt [4]{a+b x^2}}+\frac {d (e x)^{11/2}}{5 b e \sqrt [4]{a+b x^2}}+\frac {\left (7 a^2 (10 b c-11 a d) e^4 \sqrt [4]{1+\frac {a}{b x^2}} \sqrt {e x}\right ) \int \frac {1}{\left (1+\frac {a}{b x^2}\right )^{5/4} x^2} \, dx}{40 b^4 \sqrt [4]{a+b x^2}}\\ &=-\frac {7 a (10 b c-11 a d) e^3 (e x)^{3/2}}{60 b^3 \sqrt [4]{a+b x^2}}+\frac {(10 b c-11 a d) e (e x)^{7/2}}{30 b^2 \sqrt [4]{a+b x^2}}+\frac {d (e x)^{11/2}}{5 b e \sqrt [4]{a+b x^2}}-\frac {\left (7 a^2 (10 b c-11 a d) e^4 \sqrt [4]{1+\frac {a}{b x^2}} \sqrt {e x}\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {a x^2}{b}\right )^{5/4}} \, dx,x,\frac {1}{x}\right )}{40 b^4 \sqrt [4]{a+b x^2}}\\ &=-\frac {7 a (10 b c-11 a d) e^3 (e x)^{3/2}}{60 b^3 \sqrt [4]{a+b x^2}}+\frac {(10 b c-11 a d) e (e x)^{7/2}}{30 b^2 \sqrt [4]{a+b x^2}}+\frac {d (e x)^{11/2}}{5 b e \sqrt [4]{a+b x^2}}-\frac {7 a^{3/2} (10 b c-11 a d) e^4 \sqrt [4]{1+\frac {a}{b x^2}} \sqrt {e x} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{20 b^{7/2} \sqrt [4]{a+b x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.11, size = 112, normalized size = 0.62 \begin {gather*} \frac {e^3 (e x)^{3/2} \left (77 a^2 d+4 b^2 x^2 \left (5 c+3 d x^2\right )-2 a b \left (35 c+11 d x^2\right )+7 a (10 b c-11 a d) \sqrt [4]{1+\frac {b x^2}{a}} \, _2F_1\left (\frac {3}{4},\frac {5}{4};\frac {7}{4};-\frac {b x^2}{a}\right )\right )}{60 b^3 \sqrt [4]{a+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((e*x)^(9/2)*(c + d*x^2))/(a + b*x^2)^(5/4),x]

[Out]

(e^3*(e*x)^(3/2)*(77*a^2*d + 4*b^2*x^2*(5*c + 3*d*x^2) - 2*a*b*(35*c + 11*d*x^2) + 7*a*(10*b*c - 11*a*d)*(1 +
(b*x^2)/a)^(1/4)*Hypergeometric2F1[3/4, 5/4, 7/4, -((b*x^2)/a)]))/(60*b^3*(a + b*x^2)^(1/4))

________________________________________________________________________________________

Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (e x \right )^{\frac {9}{2}} \left (d \,x^{2}+c \right )}{\left (b \,x^{2}+a \right )^{\frac {5}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^(9/2)*(d*x^2+c)/(b*x^2+a)^(5/4),x)

[Out]

int((e*x)^(9/2)*(d*x^2+c)/(b*x^2+a)^(5/4),x)

________________________________________________________________________________________

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)^(9/2)*(d*x^2+c)/(b*x^2+a)^(5/4),x, algorithm="maxima")

[Out]

e^(9/2)*integrate((d*x^2 + c)*x^(9/2)/(b*x^2 + a)^(5/4), x)

________________________________________________________________________________________

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)^(9/2)*(d*x^2+c)/(b*x^2+a)^(5/4),x, algorithm="fricas")

[Out]

integral((d*x^6 + c*x^4)*(b*x^2 + a)^(3/4)*sqrt(x)*e^(9/2)/(b^2*x^4 + 2*a*b*x^2 + a^2), x)

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**(9/2)*(d*x**2+c)/(b*x**2+a)**(5/4),x)

[Out]

Timed out

________________________________________________________________________________________

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)^(9/2)*(d*x^2+c)/(b*x^2+a)^(5/4),x, algorithm="giac")

[Out]

integrate((d*x^2 + c)*x^(9/2)*e^(9/2)/(b*x^2 + a)^(5/4), x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (e\,x\right )}^{9/2}\,\left (d\,x^2+c\right )}{{\left (b\,x^2+a\right )}^{5/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((e*x)^(9/2)*(c + d*x^2))/(a + b*x^2)^(5/4),x)

[Out]

int(((e*x)^(9/2)*(c + d*x^2))/(a + b*x^2)^(5/4), x)

________________________________________________________________________________________