\(\int \frac {(e x)^{13/2} (c+d x^2)}{(a+b x^2)^{9/4}} \, dx\) [1132]

Optimal result

Integrand size = 26, antiderivative size = 230 \[ \int \frac {(e x)^{13/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\frac {2 (b c-a d) (e x)^{15/2}}{5 a b e \left (a+b x^2\right )^{5/4}}-\frac {77 a (2 b c-3 a d) e^5 (e x)^{3/2}}{60 b^4 \sqrt [4]{a+b x^2}}+\frac {11 (2 b c-3 a d) e^3 (e x)^{7/2}}{30 b^3 \sqrt [4]{a+b x^2}}-\frac {(2 b c-3 a d) e (e x)^{11/2}}{5 a b^2 \sqrt [4]{a+b x^2}}-\frac {77 a^{3/2} (2 b c-3 a d) e^6 \sqrt [4]{1+\frac {a}{b x^2}} \sqrt {e x} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{20 b^{9/2} \sqrt [4]{a+b x^2}} \]

[Out]

2/5*(-a*d+b*c)*(e*x)^(15/2)/a/b/e/(b*x^2+a)^(5/4)-77/60*a*(-3*a*d+2*b*c)*e^5*(e*x)^(3/2)/b^4/(b*x^2+a)^(1/4)+1
1/30*(-3*a*d+2*b*c)*e^3*(e*x)^(7/2)/b^3/(b*x^2+a)^(1/4)-1/5*(-3*a*d+2*b*c)*e*(e*x)^(11/2)/a/b^2/(b*x^2+a)^(1/4
)-77/20*a^(3/2)*(-3*a*d+2*b*c)*e^6*(1+a/b/x^2)^(1/4)*(cos(1/2*arccot(x*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arcc
ot(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^(9/2)/(b*x^2+a)^(1/
4)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {468, 291, 290, 342, 202} \[ \int \frac {(e x)^{13/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=-\frac {77 a^{3/2} e^6 \sqrt {e x} \sqrt [4]{\frac {a}{b x^2}+1} (2 b c-3 a d) E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{20 b^{9/2} \sqrt [4]{a+b x^2}}-\frac {77 a e^5 (e x)^{3/2} (2 b c-3 a d)}{60 b^4 \sqrt [4]{a+b x^2}}+\frac {11 e^3 (e x)^{7/2} (2 b c-3 a d)}{30 b^3 \sqrt [4]{a+b x^2}}-\frac {e (e x)^{11/2} (2 b c-3 a d)}{5 a b^2 \sqrt [4]{a+b x^2}}+\frac {2 (e x)^{15/2} (b c-a d)}{5 a b e \left (a+b x^2\right )^{5/4}} \]

[In]

Int[((e*x)^(13/2)*(c + d*x^2))/(a + b*x^2)^(9/4),x]

[Out]

(2*(b*c - a*d)*(e*x)^(15/2))/(5*a*b*e*(a + b*x^2)^(5/4)) - (77*a*(2*b*c - 3*a*d)*e^5*(e*x)^(3/2))/(60*b^4*(a +
b*x^2)^(1/4)) + (11*(2*b*c - 3*a*d)*e^3*(e*x)^(7/2))/(30*b^3*(a + b*x^2)^(1/4)) - ((2*b*c - 3*a*d)*e*(e*x)^(1
1/2))/(5*a*b^2*(a + b*x^2)^(1/4)) - (77*a^(3/2)*(2*b*c - 3*a*d)*e^6*(1 + a/(b*x^2))^(1/4)*Sqrt[e*x]*EllipticE[
ArcCot[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(20*b^(9/2)*(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)^{15/2}}{5 a b e \left (a+b x^2\right )^{5/4}}+\frac {\left (2 \left (-5 b c+\frac {15 a d}{2}\right )\right ) \int \frac {(e x)^{13/2}}{\left (a+b x^2\right )^{5/4}} \, dx}{5 a b} \\ & = \frac {2 (b c-a d) (e x)^{15/2}}{5 a b e \left (a+b x^2\right )^{5/4}}-\frac {(2 b c-3 a d) e (e x)^{11/2}}{5 a b^2 \sqrt [4]{a+b x^2}}+\frac {\left (11 (2 b c-3 a d) e^2\right ) \int \frac {(e x)^{9/2}}{\left (a+b x^2\right )^{5/4}} \, dx}{10 b^2} \\ & = \frac {2 (b c-a d) (e x)^{15/2}}{5 a b e \left (a+b x^2\right )^{5/4}}+\frac {11 (2 b c-3 a d) e^3 (e x)^{7/2}}{30 b^3 \sqrt [4]{a+b x^2}}-\frac {(2 b c-3 a d) e (e x)^{11/2}}{5 a b^2 \sqrt [4]{a+b x^2}}-\frac {\left (77 a (2 b c-3 a d) e^4\right ) \int \frac {(e x)^{5/2}}{\left (a+b x^2\right )^{5/4}} \, dx}{60 b^3} \\ & = \frac {2 (b c-a d) (e x)^{15/2}}{5 a b e \left (a+b x^2\right )^{5/4}}-\frac {77 a (2 b c-3 a d) e^5 (e x)^{3/2}}{60 b^4 \sqrt [4]{a+b x^2}}+\frac {11 (2 b c-3 a d) e^3 (e x)^{7/2}}{30 b^3 \sqrt [4]{a+b x^2}}-\frac {(2 b c-3 a d) e (e x)^{11/2}}{5 a b^2 \sqrt [4]{a+b x^2}}+\frac {\left (77 a^2 (2 b c-3 a d) e^6\right ) \int \frac {\sqrt {e x}}{\left (a+b x^2\right )^{5/4}} \, dx}{40 b^4} \\ & = \frac {2 (b c-a d) (e x)^{15/2}}{5 a b e \left (a+b x^2\right )^{5/4}}-\frac {77 a (2 b c-3 a d) e^5 (e x)^{3/2}}{60 b^4 \sqrt [4]{a+b x^2}}+\frac {11 (2 b c-3 a d) e^3 (e x)^{7/2}}{30 b^3 \sqrt [4]{a+b x^2}}-\frac {(2 b c-3 a d) e (e x)^{11/2}}{5 a b^2 \sqrt [4]{a+b x^2}}+\frac {\left (77 a^2 (2 b c-3 a d) e^6 \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^5 \sqrt [4]{a+b x^2}} \\ & = \frac {2 (b c-a d) (e x)^{15/2}}{5 a b e \left (a+b x^2\right )^{5/4}}-\frac {77 a (2 b c-3 a d) e^5 (e x)^{3/2}}{60 b^4 \sqrt [4]{a+b x^2}}+\frac {11 (2 b c-3 a d) e^3 (e x)^{7/2}}{30 b^3 \sqrt [4]{a+b x^2}}-\frac {(2 b c-3 a d) e (e x)^{11/2}}{5 a b^2 \sqrt [4]{a+b x^2}}-\frac {\left (77 a^2 (2 b c-3 a d) e^6 \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^5 \sqrt [4]{a+b x^2}} \\ & = \frac {2 (b c-a d) (e x)^{15/2}}{5 a b e \left (a+b x^2\right )^{5/4}}-\frac {77 a (2 b c-3 a d) e^5 (e x)^{3/2}}{60 b^4 \sqrt [4]{a+b x^2}}+\frac {11 (2 b c-3 a d) e^3 (e x)^{7/2}}{30 b^3 \sqrt [4]{a+b x^2}}-\frac {(2 b c-3 a d) e (e x)^{11/2}}{5 a b^2 \sqrt [4]{a+b x^2}}-\frac {77 a^{3/2} (2 b c-3 a d) e^6 \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^{9/2} \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 = 10.15 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.61 \[ \int \frac {(e x)^{13/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\frac {e^5 (e x)^{3/2} \left (1155 a^3 d-110 a^2 b \left (7 c-3 d x^2\right )+8 b^3 x^4 \left (5 c+3 d x^2\right )-20 a b^2 x^2 \left (11 c+3 d x^2\right )-385 a (-2 b c+3 a d) \left (a+b x^2\right ) \sqrt [4]{1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {9}{4},\frac {7}{4},-\frac {b x^2}{a}\right )\right )}{120 b^4 \left (a+b x^2\right )^{5/4}} \]

[In]

Integrate[((e*x)^(13/2)*(c + d*x^2))/(a + b*x^2)^(9/4),x]

[Out]

(e^5*(e*x)^(3/2)*(1155*a^3*d - 110*a^2*b*(7*c - 3*d*x^2) + 8*b^3*x^4*(5*c + 3*d*x^2) - 20*a*b^2*x^2*(11*c + 3*
d*x^2) - 385*a*(-2*b*c + 3*a*d)*(a + b*x^2)*(1 + (b*x^2)/a)^(1/4)*Hypergeometric2F1[3/4, 9/4, 7/4, -((b*x^2)/a
)]))/(120*b^4*(a + b*x^2)^(5/4))

Maple [F]

\[\int \frac {\left (e x \right )^{\frac {13}{2}} \left (d \,x^{2}+c \right )}{\left (b \,x^{2}+a \right )^{\frac {9}{4}}}d x\]

[In]

int((e*x)^(13/2)*(d*x^2+c)/(b*x^2+a)^(9/4),x)

[Out]

int((e*x)^(13/2)*(d*x^2+c)/(b*x^2+a)^(9/4),x)

Fricas [F]

\[ \int \frac {(e x)^{13/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} \left (e x\right )^{\frac {13}{2}}}{{\left (b x^{2} + a\right )}^{\frac {9}{4}}} \,d x } \]

[In]

integrate((e*x)^(13/2)*(d*x^2+c)/(b*x^2+a)^(9/4),x, algorithm="fricas")

[Out]

integral((d*e^6*x^8 + c*e^6*x^6)*(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. \[ \int \frac {(e x)^{13/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\text {Timed out} \]

[In]

integrate((e*x)**(13/2)*(d*x**2+c)/(b*x**2+a)**(9/4),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(e x)^{13/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} \left (e x\right )^{\frac {13}{2}}}{{\left (b x^{2} + a\right )}^{\frac {9}{4}}} \,d x } \]

[In]

integrate((e*x)^(13/2)*(d*x^2+c)/(b*x^2+a)^(9/4),x, algorithm="maxima")

[Out]

integrate((d*x^2 + c)*(e*x)^(13/2)/(b*x^2 + a)^(9/4), x)

Giac [F]

\[ \int \frac {(e x)^{13/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} \left (e x\right )^{\frac {13}{2}}}{{\left (b x^{2} + a\right )}^{\frac {9}{4}}} \,d x } \]

[In]

integrate((e*x)^(13/2)*(d*x^2+c)/(b*x^2+a)^(9/4),x, algorithm="giac")

[Out]

integrate((d*x^2 + c)*(e*x)^(13/2)/(b*x^2 + a)^(9/4), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(e x)^{13/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\int \frac {{\left (e\,x\right )}^{13/2}\,\left (d\,x^2+c\right )}{{\left (b\,x^2+a\right )}^{9/4}} \,d x \]

[In]

int(((e*x)^(13/2)*(c + d*x^2))/(a + b*x^2)^(9/4),x)

[Out]

int(((e*x)^(13/2)*(c + d*x^2))/(a + b*x^2)^(9/4), x)