\(\int \frac {(e x)^{9/2} (c+d x^2)}{(a+b x^2)^{9/4}} \, dx\) [1133]

Optimal result

Integrand size = 26, antiderivative size = 192 \[ \int \frac {(e x)^{9/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\frac {2 (b c-a d) (e x)^{11/2}}{5 a b e \left (a+b x^2\right )^{5/4}}+\frac {7 (6 b c-11 a d) e^3 (e x)^{3/2}}{30 b^3 \sqrt [4]{a+b x^2}}-\frac {(6 b c-11 a d) e (e x)^{7/2}}{15 a b^2 \sqrt [4]{a+b x^2}}+\frac {7 \sqrt {a} (6 b c-11 a d) e^4 \sqrt [4]{1+\frac {a}{b x^2}} \sqrt {e x} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{10 b^{7/2} \sqrt [4]{a+b x^2}} \]

[Out]

2/5*(-a*d+b*c)*(e*x)^(11/2)/a/b/e/(b*x^2+a)^(5/4)+7/30*(-11*a*d+6*b*c)*e^3*(e*x)^(3/2)/b^3/(b*x^2+a)^(1/4)-1/1
5*(-11*a*d+6*b*c)*e*(e*x)^(7/2)/a/b^2/(b*x^2+a)^(1/4)+7/10*(-11*a*d+6*b*c)*e^4*(1+a/b/x^2)^(1/4)*(cos(1/2*arcc
ot(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))*a^(1/2)*(e*x)^(1/2)/b^(7/2)/(b*x^2+a)^(1/4)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {468, 291, 290, 342, 202} \[ \int \frac {(e x)^{9/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\frac {7 \sqrt {a} e^4 \sqrt {e x} \sqrt [4]{\frac {a}{b x^2}+1} (6 b c-11 a d) E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{10 b^{7/2} \sqrt [4]{a+b x^2}}+\frac {7 e^3 (e x)^{3/2} (6 b c-11 a d)}{30 b^3 \sqrt [4]{a+b x^2}}-\frac {e (e x)^{7/2} (6 b c-11 a d)}{15 a b^2 \sqrt [4]{a+b x^2}}+\frac {2 (e x)^{11/2} (b c-a d)}{5 a b e \left (a+b x^2\right )^{5/4}} \]

[In]

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

[Out]

(2*(b*c - a*d)*(e*x)^(11/2))/(5*a*b*e*(a + b*x^2)^(5/4)) + (7*(6*b*c - 11*a*d)*e^3*(e*x)^(3/2))/(30*b^3*(a + b
*x^2)^(1/4)) - ((6*b*c - 11*a*d)*e*(e*x)^(7/2))/(15*a*b^2*(a + b*x^2)^(1/4)) + (7*Sqrt[a]*(6*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])/(10*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 468

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(-(b*c - a*d
))*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*b*e*n*(p + 1))), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a
*b*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0]
&& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0]
&& LeQ[-1, m, (-n)*(p + 1)]))

Rubi steps \begin{align*} \text {integral}& = \frac {2 (b c-a d) (e x)^{11/2}}{5 a b e \left (a+b x^2\right )^{5/4}}+\frac {\left (2 \left (-3 b c+\frac {11 a d}{2}\right )\right ) \int \frac {(e x)^{9/2}}{\left (a+b x^2\right )^{5/4}} \, dx}{5 a b} \\ & = \frac {2 (b c-a d) (e x)^{11/2}}{5 a b e \left (a+b x^2\right )^{5/4}}-\frac {(6 b c-11 a d) e (e x)^{7/2}}{15 a b^2 \sqrt [4]{a+b x^2}}+\frac {\left (7 (6 b c-11 a d) e^2\right ) \int \frac {(e x)^{5/2}}{\left (a+b x^2\right )^{5/4}} \, dx}{30 b^2} \\ & = \frac {2 (b c-a d) (e x)^{11/2}}{5 a b e \left (a+b x^2\right )^{5/4}}+\frac {7 (6 b c-11 a d) e^3 (e x)^{3/2}}{30 b^3 \sqrt [4]{a+b x^2}}-\frac {(6 b c-11 a d) e (e x)^{7/2}}{15 a b^2 \sqrt [4]{a+b x^2}}-\frac {\left (7 a (6 b c-11 a d) e^4\right ) \int \frac {\sqrt {e x}}{\left (a+b x^2\right )^{5/4}} \, dx}{20 b^3} \\ & = \frac {2 (b c-a d) (e x)^{11/2}}{5 a b e \left (a+b x^2\right )^{5/4}}+\frac {7 (6 b c-11 a d) e^3 (e x)^{3/2}}{30 b^3 \sqrt [4]{a+b x^2}}-\frac {(6 b c-11 a d) e (e x)^{7/2}}{15 a b^2 \sqrt [4]{a+b x^2}}-\frac {\left (7 a (6 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}{20 b^4 \sqrt [4]{a+b x^2}} \\ & = \frac {2 (b c-a d) (e x)^{11/2}}{5 a b e \left (a+b x^2\right )^{5/4}}+\frac {7 (6 b c-11 a d) e^3 (e x)^{3/2}}{30 b^3 \sqrt [4]{a+b x^2}}-\frac {(6 b c-11 a d) e (e x)^{7/2}}{15 a b^2 \sqrt [4]{a+b x^2}}+\frac {\left (7 a (6 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 )}{20 b^4 \sqrt [4]{a+b x^2}} \\ & = \frac {2 (b c-a d) (e x)^{11/2}}{5 a b e \left (a+b x^2\right )^{5/4}}+\frac {7 (6 b c-11 a d) e^3 (e x)^{3/2}}{30 b^3 \sqrt [4]{a+b x^2}}-\frac {(6 b c-11 a d) e (e x)^{7/2}}{15 a b^2 \sqrt [4]{a+b x^2}}+\frac {7 \sqrt {a} (6 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 )}{10 b^{7/2} \sqrt [4]{a+b x^2}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.10 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.60 \[ \int \frac {(e x)^{9/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\frac {e^3 (e x)^{3/2} \left (-77 a^2 d+a b \left (42 c-22 d x^2\right )+4 b^2 x^2 \left (3 c+d x^2\right )+7 (-6 b c+11 a d) \left (a+b x^2\right ) \sqrt [4]{1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {9}{4},\frac {7}{4},-\frac {b x^2}{a}\right )\right )}{12 b^3 \left (a+b x^2\right )^{5/4}} \]

[In]

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

[Out]

(e^3*(e*x)^(3/2)*(-77*a^2*d + a*b*(42*c - 22*d*x^2) + 4*b^2*x^2*(3*c + d*x^2) + 7*(-6*b*c + 11*a*d)*(a + b*x^2
)*(1 + (b*x^2)/a)^(1/4)*Hypergeometric2F1[3/4, 9/4, 7/4, -((b*x^2)/a)]))/(12*b^3*(a + b*x^2)^(5/4))

Maple [F]

\[\int \frac {\left (e x \right )^{\frac {9}{2}} \left (d \,x^{2}+c \right )}{\left (b \,x^{2}+a \right )^{\frac {9}{4}}}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {(e x)^{9/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} \left (e x\right )^{\frac {9}{2}}}{{\left (b x^{2} + a\right )}^{\frac {9}{4}}} \,d x } \]

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {(e x)^{9/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {(e x)^{9/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} \left (e x\right )^{\frac {9}{2}}}{{\left (b x^{2} + a\right )}^{\frac {9}{4}}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {(e x)^{9/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} \left (e x\right )^{\frac {9}{2}}}{{\left (b x^{2} + a\right )}^{\frac {9}{4}}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(e x)^{9/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\int \frac {{\left (e\,x\right )}^{9/2}\,\left (d\,x^2+c\right )}{{\left (b\,x^2+a\right )}^{9/4}} \,d x \]

[In]

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

[Out]

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