\(\int \frac {c+d x^2}{(e x)^{5/2} (a+b x^2)^{7/4}} \, dx\) [469]

Optimal result

Integrand size = 26, antiderivative size = 144 \[ \int \frac {c+d x^2}{(e x)^{5/2} \left (a+b x^2\right )^{7/4}} \, dx=-\frac {2 c}{3 a e (e x)^{3/2} \left (a+b x^2\right )^{3/4}}-\frac {2 (2 b c-a d) \sqrt {e x}}{3 a^2 e^3 \left (a+b x^2\right )^{3/4}}+\frac {4 \sqrt {b} (2 b c-a d) \left (1+\frac {a}{b x^2}\right )^{3/4} (e x)^{3/2} \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{3 a^{5/2} e^4 \left (a+b x^2\right )^{3/4}} \] Output:

-2/3*c/a/e/(e*x)^(3/2)/(b*x^2+a)^(3/4)-2/3*(-a*d+2*b*c)*(e*x)^(1/2)/a^2/e^ 
3/(b*x^2+a)^(3/4)+4/3*b^(1/2)*(-a*d+2*b*c)*(1+a/b/x^2)^(3/4)*(e*x)^(3/2)*I 
nverseJacobiAM(1/2*arccot(b^(1/2)*x/a^(1/2)),2^(1/2))/a^(5/2)/e^4/(b*x^2+a 
)^(3/4)
 

Mathematica [C] (verified)

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

Time = 10.04 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.63 \[ \int \frac {c+d x^2}{(e x)^{5/2} \left (a+b x^2\right )^{7/4}} \, dx=\frac {x \left (-2 a c-4 b c x^2+2 a d x^2+4 (-2 b c+a d) x^2 \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3}{4},\frac {5}{4},-\frac {b x^2}{a}\right )\right )}{3 a^2 (e x)^{5/2} \left (a+b x^2\right )^{3/4}} \] Input:

Integrate[(c + d*x^2)/((e*x)^(5/2)*(a + b*x^2)^(7/4)),x]
 

Output:

(x*(-2*a*c - 4*b*c*x^2 + 2*a*d*x^2 + 4*(-2*b*c + a*d)*x^2*(1 + (b*x^2)/a)^ 
(3/4)*Hypergeometric2F1[1/4, 3/4, 5/4, -((b*x^2)/a)]))/(3*a^2*(e*x)^(5/2)* 
(a + b*x^2)^(3/4))
 

Rubi [A] (warning: unable to verify)

Time = 0.32 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {359, 253, 266, 768, 858, 807, 229}

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.

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

\(\Big \downarrow \) 359

\(\displaystyle -\frac {(2 b c-a d) \int \frac {1}{\sqrt {e x} \left (b x^2+a\right )^{7/4}}dx}{a e^2}-\frac {2 c}{3 a e (e x)^{3/2} \left (a+b x^2\right )^{3/4}}\)

\(\Big \downarrow \) 253

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

\(\Big \downarrow \) 266

\(\displaystyle -\frac {(2 b c-a d) \left (\frac {4 \int \frac {1}{\left (b x^2+a\right )^{3/4}}d\sqrt {e x}}{3 a e}+\frac {2 \sqrt {e x}}{3 a e \left (a+b x^2\right )^{3/4}}\right )}{a e^2}-\frac {2 c}{3 a e (e x)^{3/2} \left (a+b x^2\right )^{3/4}}\)

\(\Big \downarrow \) 768

\(\displaystyle -\frac {(2 b c-a d) \left (\frac {4 (e x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{3/4} (e x)^{3/2}}d\sqrt {e x}}{3 a e \left (a+b x^2\right )^{3/4}}+\frac {2 \sqrt {e x}}{3 a e \left (a+b x^2\right )^{3/4}}\right )}{a e^2}-\frac {2 c}{3 a e (e x)^{3/2} \left (a+b x^2\right )^{3/4}}\)

\(\Big \downarrow \) 858

\(\displaystyle -\frac {(2 b c-a d) \left (\frac {2 \sqrt {e x}}{3 a e \left (a+b x^2\right )^{3/4}}-\frac {4 (e x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} \int \frac {1}{\sqrt {e x} \left (\frac {a x^2 e^4}{b}+1\right )^{3/4}}d\frac {1}{\sqrt {e x}}}{3 a e \left (a+b x^2\right )^{3/4}}\right )}{a e^2}-\frac {2 c}{3 a e (e x)^{3/2} \left (a+b x^2\right )^{3/4}}\)

\(\Big \downarrow \) 807

\(\displaystyle -\frac {(2 b c-a d) \left (\frac {2 \sqrt {e x}}{3 a e \left (a+b x^2\right )^{3/4}}-\frac {2 (e x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} \int \frac {1}{\left (\frac {a x e^3}{b}+1\right )^{3/4}}d(e x)}{3 a e \left (a+b x^2\right )^{3/4}}\right )}{a e^2}-\frac {2 c}{3 a e (e x)^{3/2} \left (a+b x^2\right )^{3/4}}\)

\(\Big \downarrow \) 229

\(\displaystyle -\frac {(2 b c-a d) \left (\frac {2 \sqrt {e x}}{3 a e \left (a+b x^2\right )^{3/4}}-\frac {4 \sqrt {b} (e x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {a} e^2 x}{\sqrt {b}}\right ),2\right )}{3 a^{3/2} e^2 \left (a+b x^2\right )^{3/4}}\right )}{a e^2}-\frac {2 c}{3 a e (e x)^{3/2} \left (a+b x^2\right )^{3/4}}\)

Input:

Int[(c + d*x^2)/((e*x)^(5/2)*(a + b*x^2)^(7/4)),x]
 

Output:

(-2*c)/(3*a*e*(e*x)^(3/2)*(a + b*x^2)^(3/4)) - ((2*b*c - a*d)*((2*Sqrt[e*x 
])/(3*a*e*(a + b*x^2)^(3/4)) - (4*Sqrt[b]*(1 + a/(b*x^2))^(3/4)*(e*x)^(3/2 
)*EllipticF[ArcTan[(Sqrt[a]*e^2*x)/Sqrt[b]]/2, 2])/(3*a^(3/2)*e^2*(a + b*x 
^2)^(3/4))))/(a*e^2)
 

Defintions of rubi rules used

rule 229

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]) 
)*EllipticF[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a 
, 0] && PosQ[b/a]
 

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 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[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 768

Int[((a_) + (b_.)*(x_)^4)^(-3/4), x_Symbol] :> Simp[x^3*((1 + a/(b*x^4))^(3 
/4)/(a + b*x^4)^(3/4))   Int[1/(x^3*(1 + a/(b*x^4))^(3/4)), x], x] /; FreeQ 
[{a, b}, x]
 

rule 807

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

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]
 

Maple [F]

\[\int \frac {x^{2} d +c}{\left (e x \right )^{\frac {5}{2}} \left (b \,x^{2}+a \right )^{\frac {7}{4}}}d x\]

Input:

int((d*x^2+c)/(e*x)^(5/2)/(b*x^2+a)^(7/4),x)
 

Output:

int((d*x^2+c)/(e*x)^(5/2)/(b*x^2+a)^(7/4),x)
 

Fricas [F]

\[ \int \frac {c+d x^2}{(e x)^{5/2} \left (a+b x^2\right )^{7/4}} \, dx=\int { \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {7}{4}} \left (e x\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate((d*x^2+c)/(e*x)^(5/2)/(b*x^2+a)^(7/4),x, algorithm="fricas")
 

Output:

integral((b*x^2 + a)^(1/4)*(d*x^2 + c)*sqrt(e*x)/(b^2*e^3*x^7 + 2*a*b*e^3* 
x^5 + a^2*e^3*x^3), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 50.27 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.67 \[ \int \frac {c+d x^2}{(e x)^{5/2} \left (a+b x^2\right )^{7/4}} \, dx=\frac {c \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {7}{4} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {7}{4}} e^{\frac {5}{2}} x^{\frac {3}{2}} \Gamma \left (\frac {1}{4}\right )} + \frac {d \sqrt {x} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {7}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {7}{4}} e^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right )} \] Input:

integrate((d*x**2+c)/(e*x)**(5/2)/(b*x**2+a)**(7/4),x)
 

Output:

c*gamma(-3/4)*hyper((-3/4, 7/4), (1/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**( 
7/4)*e**(5/2)*x**(3/2)*gamma(1/4)) + d*sqrt(x)*gamma(1/4)*hyper((1/4, 7/4) 
, (5/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(7/4)*e**(5/2)*gamma(5/4))
 

Maxima [F]

\[ \int \frac {c+d x^2}{(e x)^{5/2} \left (a+b x^2\right )^{7/4}} \, dx=\int { \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {7}{4}} \left (e x\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate((d*x^2+c)/(e*x)^(5/2)/(b*x^2+a)^(7/4),x, algorithm="maxima")
 

Output:

integrate((d*x^2 + c)/((b*x^2 + a)^(7/4)*(e*x)^(5/2)), x)
 

Giac [F]

\[ \int \frac {c+d x^2}{(e x)^{5/2} \left (a+b x^2\right )^{7/4}} \, dx=\int { \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {7}{4}} \left (e x\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate((d*x^2+c)/(e*x)^(5/2)/(b*x^2+a)^(7/4),x, algorithm="giac")
 

Output:

integrate((d*x^2 + c)/((b*x^2 + a)^(7/4)*(e*x)^(5/2)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {c+d x^2}{(e x)^{5/2} \left (a+b x^2\right )^{7/4}} \, dx=\int \frac {d\,x^2+c}{{\left (e\,x\right )}^{5/2}\,{\left (b\,x^2+a\right )}^{7/4}} \,d x \] Input:

int((c + d*x^2)/((e*x)^(5/2)*(a + b*x^2)^(7/4)),x)
 

Output:

int((c + d*x^2)/((e*x)^(5/2)*(a + b*x^2)^(7/4)), x)
 

Reduce [B] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.38 \[ \int \frac {c+d x^2}{(e x)^{5/2} \left (a+b x^2\right )^{7/4}} \, dx=\frac {2 \sqrt {e}\, \left (3 a d \,x^{2}-4 b c \,x^{2}-a c \right )}{3 \left (b \,x^{2}+a \right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {b \,x^{2}+a}\, a^{2} e^{3} x} \] Input:

int((d*x^2+c)/(e*x)^(5/2)/(b*x^2+a)^(7/4),x)
 

Output:

(2*sqrt(e)*(a + b*x**2)**(3/4)*( - a*c + 3*a*d*x**2 - 4*b*c*x**2))/(3*sqrt 
(x)*sqrt(a + b*x**2)*a**2*e**3*x*(a + b*x**2))