3.12.38 \(\int \frac {c+d x^2}{(e x)^{11/2} (a+b x^2)^{9/4}} \, dx\) [1138]

3.12.38.1 Optimal result

Integrand size = 26, antiderivative size = 219 \[ \int \frac {c+d x^2}{(e x)^{11/2} \left (a+b x^2\right )^{9/4}} \, dx=-\frac {2 c}{9 a e (e x)^{9/2} \left (a+b x^2\right )^{5/4}}-\frac {2 (14 b c-9 a d)}{45 a^2 e^3 (e x)^{5/2} \left (a+b x^2\right )^{5/4}}+\frac {4 (14 b c-9 a d)}{45 a^3 e^3 (e x)^{5/2} \sqrt [4]{a+b x^2}}-\frac {8 b (14 b c-9 a d)}{15 a^4 e^5 \sqrt {e x} \sqrt [4]{a+b x^2}}+\frac {16 b^{3/2} (14 b c-9 a d) \sqrt [4]{1+\frac {a}{b x^2}} \sqrt {e x} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{15 a^{9/2} e^6 \sqrt [4]{a+b x^2}} \]

output

-2/9*c/a/e/(e*x)^(9/2)/(b*x^2+a)^(5/4)-2/45*(-9*a*d+14*b*c)/a^2/e^3/(e*x)^ 
(5/2)/(b*x^2+a)^(5/4)+4/45*(-9*a*d+14*b*c)/a^3/e^3/(e*x)^(5/2)/(b*x^2+a)^( 
1/4)-8/15*b*(-9*a*d+14*b*c)/a^4/e^5/(b*x^2+a)^(1/4)/(e*x)^(1/2)+16/15*b^(3 
/2)*(-9*a*d+14*b*c)*(1+a/b/x^2)^(1/4)*(cos(1/2*arccot(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)/a^(9/2)/e^6/(b*x^2+a)^(1/4)
 

3.12.38.2 Mathematica [C] (verified)

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

Time = 10.07 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.40 \[ \int \frac {c+d x^2}{(e x)^{11/2} \left (a+b x^2\right )^{9/4}} \, dx=\frac {2 x \left (-5 a^2 c-(-14 b c+9 a d) x^2 \left (a+b x^2\right ) \sqrt [4]{1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {9}{4},-\frac {1}{4},-\frac {b x^2}{a}\right )\right )}{45 a^3 (e x)^{11/2} \left (a+b x^2\right )^{5/4}} \]

input

Integrate[(c + d*x^2)/((e*x)^(11/2)*(a + b*x^2)^(9/4)),x]
 

output

(2*x*(-5*a^2*c - (-14*b*c + 9*a*d)*x^2*(a + b*x^2)*(1 + (b*x^2)/a)^(1/4)*H 
ypergeometric2F1[-5/4, 9/4, -1/4, -((b*x^2)/a)]))/(45*a^3*(e*x)^(11/2)*(a 
+ b*x^2)^(5/4))
 

3.12.38.3 Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {359, 253, 251, 251, 249, 858, 212}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {c+d x^2}{(e x)^{11/2} \left (a+b x^2\right )^{9/4}} \, dx\)

\(\Big \downarrow \) 359

\(\displaystyle -\frac {(14 b c-9 a d) \int \frac {1}{(e x)^{7/2} \left (b x^2+a\right )^{9/4}}dx}{9 a e^2}-\frac {2 c}{9 a e (e x)^{9/2} \left (a+b x^2\right )^{5/4}}\)

\(\Big \downarrow \) 253

\(\displaystyle -\frac {(14 b c-9 a d) \left (\frac {2 \int \frac {1}{(e x)^{7/2} \left (b x^2+a\right )^{5/4}}dx}{a}+\frac {2}{5 a e (e x)^{5/2} \left (a+b x^2\right )^{5/4}}\right )}{9 a e^2}-\frac {2 c}{9 a e (e x)^{9/2} \left (a+b x^2\right )^{5/4}}\)

\(\Big \downarrow \) 251

\(\displaystyle -\frac {(14 b c-9 a d) \left (\frac {2 \left (-\frac {6 b \int \frac {1}{(e x)^{3/2} \left (b x^2+a\right )^{5/4}}dx}{5 a e^2}-\frac {2}{5 a e (e x)^{5/2} \sqrt [4]{a+b x^2}}\right )}{a}+\frac {2}{5 a e (e x)^{5/2} \left (a+b x^2\right )^{5/4}}\right )}{9 a e^2}-\frac {2 c}{9 a e (e x)^{9/2} \left (a+b x^2\right )^{5/4}}\)

\(\Big \downarrow \) 251

\(\displaystyle -\frac {(14 b c-9 a d) \left (\frac {2 \left (-\frac {6 b \left (-\frac {2 b \int \frac {\sqrt {e x}}{\left (b x^2+a\right )^{5/4}}dx}{a e^2}-\frac {2}{a e \sqrt {e x} \sqrt [4]{a+b x^2}}\right )}{5 a e^2}-\frac {2}{5 a e (e x)^{5/2} \sqrt [4]{a+b x^2}}\right )}{a}+\frac {2}{5 a e (e x)^{5/2} \left (a+b x^2\right )^{5/4}}\right )}{9 a e^2}-\frac {2 c}{9 a e (e x)^{9/2} \left (a+b x^2\right )^{5/4}}\)

\(\Big \downarrow \) 249

\(\displaystyle -\frac {(14 b c-9 a d) \left (\frac {2 \left (-\frac {6 b \left (-\frac {2 \sqrt {e x} \sqrt [4]{\frac {a}{b x^2}+1} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{5/4} x^2}dx}{a e^2 \sqrt [4]{a+b x^2}}-\frac {2}{a e \sqrt {e x} \sqrt [4]{a+b x^2}}\right )}{5 a e^2}-\frac {2}{5 a e (e x)^{5/2} \sqrt [4]{a+b x^2}}\right )}{a}+\frac {2}{5 a e (e x)^{5/2} \left (a+b x^2\right )^{5/4}}\right )}{9 a e^2}-\frac {2 c}{9 a e (e x)^{9/2} \left (a+b x^2\right )^{5/4}}\)

\(\Big \downarrow \) 858

\(\displaystyle -\frac {(14 b c-9 a d) \left (\frac {2 \left (-\frac {6 b \left (\frac {2 \sqrt {e x} \sqrt [4]{\frac {a}{b x^2}+1} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{5/4}}d\frac {1}{x}}{a e^2 \sqrt [4]{a+b x^2}}-\frac {2}{a e \sqrt {e x} \sqrt [4]{a+b x^2}}\right )}{5 a e^2}-\frac {2}{5 a e (e x)^{5/2} \sqrt [4]{a+b x^2}}\right )}{a}+\frac {2}{5 a e (e x)^{5/2} \left (a+b x^2\right )^{5/4}}\right )}{9 a e^2}-\frac {2 c}{9 a e (e x)^{9/2} \left (a+b x^2\right )^{5/4}}\)

\(\Big \downarrow \) 212

\(\displaystyle -\frac {(14 b c-9 a d) \left (\frac {2 \left (-\frac {6 b \left (\frac {4 \sqrt {b} \sqrt {e x} \sqrt [4]{\frac {a}{b x^2}+1} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x}\right )\right |2\right )}{a^{3/2} e^2 \sqrt [4]{a+b x^2}}-\frac {2}{a e \sqrt {e x} \sqrt [4]{a+b x^2}}\right )}{5 a e^2}-\frac {2}{5 a e (e x)^{5/2} \sqrt [4]{a+b x^2}}\right )}{a}+\frac {2}{5 a e (e x)^{5/2} \left (a+b x^2\right )^{5/4}}\right )}{9 a e^2}-\frac {2 c}{9 a e (e x)^{9/2} \left (a+b x^2\right )^{5/4}}\)

input

Int[(c + d*x^2)/((e*x)^(11/2)*(a + b*x^2)^(9/4)),x]
 

output

(-2*c)/(9*a*e*(e*x)^(9/2)*(a + b*x^2)^(5/4)) - ((14*b*c - 9*a*d)*(2/(5*a*e 
*(e*x)^(5/2)*(a + b*x^2)^(5/4)) + (2*(-2/(5*a*e*(e*x)^(5/2)*(a + b*x^2)^(1 
/4)) - (6*b*(-2/(a*e*Sqrt[e*x]*(a + b*x^2)^(1/4)) + (4*Sqrt[b]*(1 + a/(b*x 
^2))^(1/4)*Sqrt[e*x]*EllipticE[ArcTan[Sqrt[a]/(Sqrt[b]*x)]/2, 2])/(a^(3/2) 
*e^2*(a + b*x^2)^(1/4))))/(5*a*e^2)))/a))/(9*a*e^2)
 

3.12.38.3.1 Defintions of rubi rules used

rule 212

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 249

Int[Sqrt[(c_.)*(x_)]/((a_) + (b_.)*(x_)^2)^(5/4), x_Symbol] :> Simp[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 251

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

rule 253

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(c*x 
)^(m + 1))*((a + b*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + Simp[(m + 2*p + 3)/( 
2*a*(p + 1))   Int[(c*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, m 
}, x] && LtQ[p, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
 

rule 359

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + 
Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1))   Int[(e*x)^(m + 2)* 
(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] 
&& LtQ[m, -1] &&  !ILtQ[p, -1]
 

rule 858

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] && Int 
egerQ[m]
 

3.12.38.4 Maple [F]

\[\int \frac {d \,x^{2}+c}{\left (e x \right )^{\frac {11}{2}} \left (b \,x^{2}+a \right )^{\frac {9}{4}}}d x\]

input

int((d*x^2+c)/(e*x)^(11/2)/(b*x^2+a)^(9/4),x)
 

output

int((d*x^2+c)/(e*x)^(11/2)/(b*x^2+a)^(9/4),x)
 

3.12.38.5 Fricas [F]

\[ \int \frac {c+d x^2}{(e x)^{11/2} \left (a+b x^2\right )^{9/4}} \, dx=\int { \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {9}{4}} \left (e x\right )^{\frac {11}{2}}} \,d x } \]

input

integrate((d*x^2+c)/(e*x)^(11/2)/(b*x^2+a)^(9/4),x, algorithm="fricas")
 

output

integral((b*x^2 + a)^(3/4)*(d*x^2 + c)*sqrt(e*x)/(b^3*e^6*x^12 + 3*a*b^2*e 
^6*x^10 + 3*a^2*b*e^6*x^8 + a^3*e^6*x^6), x)
 

3.12.38.6 Sympy [F(-1)]

Timed out. \[ \int \frac {c+d x^2}{(e x)^{11/2} \left (a+b x^2\right )^{9/4}} \, dx=\text {Timed out} \]

input

integrate((d*x**2+c)/(e*x)**(11/2)/(b*x**2+a)**(9/4),x)
 

output

Timed out
 

3.12.38.7 Maxima [F]

\[ \int \frac {c+d x^2}{(e x)^{11/2} \left (a+b x^2\right )^{9/4}} \, dx=\int { \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {9}{4}} \left (e x\right )^{\frac {11}{2}}} \,d x } \]

input

integrate((d*x^2+c)/(e*x)^(11/2)/(b*x^2+a)^(9/4),x, algorithm="maxima")
 

output

integrate((d*x^2 + c)/((b*x^2 + a)^(9/4)*(e*x)^(11/2)), x)
 

3.12.38.8 Giac [F]

\[ \int \frac {c+d x^2}{(e x)^{11/2} \left (a+b x^2\right )^{9/4}} \, dx=\int { \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {9}{4}} \left (e x\right )^{\frac {11}{2}}} \,d x } \]

input

integrate((d*x^2+c)/(e*x)^(11/2)/(b*x^2+a)^(9/4),x, algorithm="giac")
 

output

integrate((d*x^2 + c)/((b*x^2 + a)^(9/4)*(e*x)^(11/2)), x)
 

3.12.38.9 Mupad [F(-1)]

Timed out. \[ \int \frac {c+d x^2}{(e x)^{11/2} \left (a+b x^2\right )^{9/4}} \, dx=\int \frac {d\,x^2+c}{{\left (e\,x\right )}^{11/2}\,{\left (b\,x^2+a\right )}^{9/4}} \,d x \]

input

int((c + d*x^2)/((e*x)^(11/2)*(a + b*x^2)^(9/4)),x)
 

output

int((c + d*x^2)/((e*x)^(11/2)*(a + b*x^2)^(9/4)), x)