\(\int \frac {(e x)^{7/2} (c+d x^2)}{(a+b x^2)^{3/4}} \, dx\) [444]

Optimal result

Integrand size = 26, antiderivative size = 180 \[ \int \frac {(e x)^{7/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{3/4}} \, dx=-\frac {a (10 b c-9 a d) e^3 \sqrt {e x} \sqrt [4]{a+b x^2}}{12 b^3}+\frac {(10 b c-9 a d) e (e x)^{5/2} \sqrt [4]{a+b x^2}}{30 b^2}+\frac {d (e x)^{9/2} \sqrt [4]{a+b x^2}}{5 b e}-\frac {a^{3/2} (10 b c-9 a d) e^2 \left (1+\frac {a}{b x^2}\right )^{3/4} (e x)^{3/2} \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{12 b^{5/2} \left (a+b x^2\right )^{3/4}} \] Output:

-1/12*a*(-9*a*d+10*b*c)*e^3*(e*x)^(1/2)*(b*x^2+a)^(1/4)/b^3+1/30*(-9*a*d+1 
0*b*c)*e*(e*x)^(5/2)*(b*x^2+a)^(1/4)/b^2+1/5*d*(e*x)^(9/2)*(b*x^2+a)^(1/4) 
/b/e-1/12*a^(3/2)*(-9*a*d+10*b*c)*e^2*(1+a/b/x^2)^(3/4)*(e*x)^(3/2)*Invers 
eJacobiAM(1/2*arccot(b^(1/2)*x/a^(1/2)),2^(1/2))/b^(5/2)/(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 = 10.10 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.68 \[ \int \frac {(e x)^{7/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{3/4}} \, dx=\frac {e^3 \sqrt {e x} \left (\left (a+b x^2\right ) \left (45 a^2 d+4 b^2 x^2 \left (5 c+3 d x^2\right )-2 a b \left (25 c+9 d x^2\right )\right )+5 a^2 (10 b c-9 a d) \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3}{4},\frac {5}{4},-\frac {b x^2}{a}\right )\right )}{60 b^3 \left (a+b x^2\right )^{3/4}} \] Input:

Integrate[((e*x)^(7/2)*(c + d*x^2))/(a + b*x^2)^(3/4),x]
 

Output:

(e^3*Sqrt[e*x]*((a + b*x^2)*(45*a^2*d + 4*b^2*x^2*(5*c + 3*d*x^2) - 2*a*b* 
(25*c + 9*d*x^2)) + 5*a^2*(10*b*c - 9*a*d)*(1 + (b*x^2)/a)^(3/4)*Hypergeom 
etric2F1[1/4, 3/4, 5/4, -((b*x^2)/a)]))/(60*b^3*(a + b*x^2)^(3/4))
 

Rubi [A] (warning: unable to verify)

Time = 0.35 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.96, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {363, 262, 262, 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.

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

\(\Big \downarrow \) 363

\(\displaystyle \frac {(10 b c-9 a d) \int \frac {(e x)^{7/2}}{\left (b x^2+a\right )^{3/4}}dx}{10 b}+\frac {d (e x)^{9/2} \sqrt [4]{a+b x^2}}{5 b e}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {(10 b c-9 a d) \left (\frac {e (e x)^{5/2} \sqrt [4]{a+b x^2}}{3 b}-\frac {5 a e^2 \int \frac {(e x)^{3/2}}{\left (b x^2+a\right )^{3/4}}dx}{6 b}\right )}{10 b}+\frac {d (e x)^{9/2} \sqrt [4]{a+b x^2}}{5 b e}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {(10 b c-9 a d) \left (\frac {e (e x)^{5/2} \sqrt [4]{a+b x^2}}{3 b}-\frac {5 a e^2 \left (\frac {e \sqrt {e x} \sqrt [4]{a+b x^2}}{b}-\frac {a e^2 \int \frac {1}{\sqrt {e x} \left (b x^2+a\right )^{3/4}}dx}{2 b}\right )}{6 b}\right )}{10 b}+\frac {d (e x)^{9/2} \sqrt [4]{a+b x^2}}{5 b e}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {(10 b c-9 a d) \left (\frac {e (e x)^{5/2} \sqrt [4]{a+b x^2}}{3 b}-\frac {5 a e^2 \left (\frac {e \sqrt {e x} \sqrt [4]{a+b x^2}}{b}-\frac {a e \int \frac {1}{\left (b x^2+a\right )^{3/4}}d\sqrt {e x}}{b}\right )}{6 b}\right )}{10 b}+\frac {d (e x)^{9/2} \sqrt [4]{a+b x^2}}{5 b e}\)

\(\Big \downarrow \) 768

\(\displaystyle \frac {(10 b c-9 a d) \left (\frac {e (e x)^{5/2} \sqrt [4]{a+b x^2}}{3 b}-\frac {5 a e^2 \left (\frac {e \sqrt {e x} \sqrt [4]{a+b x^2}}{b}-\frac {a e (e x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{3/4} (e x)^{3/2}}d\sqrt {e x}}{b \left (a+b x^2\right )^{3/4}}\right )}{6 b}\right )}{10 b}+\frac {d (e x)^{9/2} \sqrt [4]{a+b x^2}}{5 b e}\)

\(\Big \downarrow \) 858

\(\displaystyle \frac {(10 b c-9 a d) \left (\frac {e (e x)^{5/2} \sqrt [4]{a+b x^2}}{3 b}-\frac {5 a e^2 \left (\frac {a e (e x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} \int \frac {1}{\sqrt {e x} \left (\frac {a x^2 e^4}{b}+1\right )^{3/4}}d\frac {1}{\sqrt {e x}}}{b \left (a+b x^2\right )^{3/4}}+\frac {e \sqrt {e x} \sqrt [4]{a+b x^2}}{b}\right )}{6 b}\right )}{10 b}+\frac {d (e x)^{9/2} \sqrt [4]{a+b x^2}}{5 b e}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {(10 b c-9 a d) \left (\frac {e (e x)^{5/2} \sqrt [4]{a+b x^2}}{3 b}-\frac {5 a e^2 \left (\frac {a e (e x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} \int \frac {1}{\left (\frac {a x e^3}{b}+1\right )^{3/4}}d(e x)}{2 b \left (a+b x^2\right )^{3/4}}+\frac {e \sqrt {e x} \sqrt [4]{a+b x^2}}{b}\right )}{6 b}\right )}{10 b}+\frac {d (e x)^{9/2} \sqrt [4]{a+b x^2}}{5 b e}\)

\(\Big \downarrow \) 229

\(\displaystyle \frac {(10 b c-9 a d) \left (\frac {e (e x)^{5/2} \sqrt [4]{a+b x^2}}{3 b}-\frac {5 a e^2 \left (\frac {\sqrt {a} (e x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {a} e^2 x}{\sqrt {b}}\right ),2\right )}{\sqrt {b} \left (a+b x^2\right )^{3/4}}+\frac {e \sqrt {e x} \sqrt [4]{a+b x^2}}{b}\right )}{6 b}\right )}{10 b}+\frac {d (e x)^{9/2} \sqrt [4]{a+b x^2}}{5 b e}\)

Input:

Int[((e*x)^(7/2)*(c + d*x^2))/(a + b*x^2)^(3/4),x]
 

Output:

(d*(e*x)^(9/2)*(a + b*x^2)^(1/4))/(5*b*e) + ((10*b*c - 9*a*d)*((e*(e*x)^(5 
/2)*(a + b*x^2)^(1/4))/(3*b) - (5*a*e^2*((e*Sqrt[e*x]*(a + b*x^2)^(1/4))/b 
 + (Sqrt[a]*(1 + a/(b*x^2))^(3/4)*(e*x)^(3/2)*EllipticF[ArcTan[(Sqrt[a]*e^ 
2*x)/Sqrt[b]]/2, 2])/(Sqrt[b]*(a + b*x^2)^(3/4))))/(6*b)))/(10*b)
 

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 262

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) 
^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ 
(b*(m + 2*p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b 
, c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && 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 363

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

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]

\[\int \frac {\left (e x \right )^{\frac {7}{2}} \left (x^{2} d +c \right )}{\left (b \,x^{2}+a \right )^{\frac {3}{4}}}d x\]

Input:

int((e*x)^(7/2)*(d*x^2+c)/(b*x^2+a)^(3/4),x)
 

Output:

int((e*x)^(7/2)*(d*x^2+c)/(b*x^2+a)^(3/4),x)
 

Fricas [F]

\[ \int \frac {(e x)^{7/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{3/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} \left (e x\right )^{\frac {7}{2}}}{{\left (b x^{2} + a\right )}^{\frac {3}{4}}} \,d x } \] Input:

integrate((e*x)^(7/2)*(d*x^2+c)/(b*x^2+a)^(3/4),x, algorithm="fricas")
 

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 39.48 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.52 \[ \int \frac {(e x)^{7/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{3/4}} \, dx=\frac {c e^{\frac {7}{2}} x^{\frac {9}{2}} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{4}} \Gamma \left (\frac {13}{4}\right )} + \frac {d e^{\frac {7}{2}} x^{\frac {13}{2}} \Gamma \left (\frac {13}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {13}{4} \\ \frac {17}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{4}} \Gamma \left (\frac {17}{4}\right )} \] Input:

integrate((e*x)**(7/2)*(d*x**2+c)/(b*x**2+a)**(3/4),x)
 

Output:

c*e**(7/2)*x**(9/2)*gamma(9/4)*hyper((3/4, 9/4), (13/4,), b*x**2*exp_polar 
(I*pi)/a)/(2*a**(3/4)*gamma(13/4)) + d*e**(7/2)*x**(13/2)*gamma(13/4)*hype 
r((3/4, 13/4), (17/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(3/4)*gamma(17/4))
 

Maxima [F]

\[ \int \frac {(e x)^{7/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{3/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} \left (e x\right )^{\frac {7}{2}}}{{\left (b x^{2} + a\right )}^{\frac {3}{4}}} \,d x } \] Input:

integrate((e*x)^(7/2)*(d*x^2+c)/(b*x^2+a)^(3/4),x, algorithm="maxima")
 

Output:

integrate((d*x^2 + c)*(e*x)^(7/2)/(b*x^2 + a)^(3/4), x)
 

Giac [F]

\[ \int \frac {(e x)^{7/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{3/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} \left (e x\right )^{\frac {7}{2}}}{{\left (b x^{2} + a\right )}^{\frac {3}{4}}} \,d x } \] Input:

integrate((e*x)^(7/2)*(d*x^2+c)/(b*x^2+a)^(3/4),x, algorithm="giac")
 

Output:

integrate((d*x^2 + c)*(e*x)^(7/2)/(b*x^2 + a)^(3/4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e x)^{7/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{3/4}} \, dx=\int \frac {{\left (e\,x\right )}^{7/2}\,\left (d\,x^2+c\right )}{{\left (b\,x^2+a\right )}^{3/4}} \,d x \] Input:

int(((e*x)^(7/2)*(c + d*x^2))/(a + b*x^2)^(3/4),x)
 

Output:

int(((e*x)^(7/2)*(c + d*x^2))/(a + b*x^2)^(3/4), x)
 

Reduce [F]

\[ \int \frac {(e x)^{7/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{3/4}} \, dx=\sqrt {e}\, e^{3} \left (\left (\int \frac {\sqrt {x}\, x^{5}}{\left (b \,x^{2}+a \right )^{\frac {3}{4}}}d x \right ) d +\left (\int \frac {\sqrt {x}\, x^{3}}{\left (b \,x^{2}+a \right )^{\frac {3}{4}}}d x \right ) c \right ) \] Input:

int((e*x)^(7/2)*(d*x^2+c)/(b*x^2+a)^(3/4),x)
 

Output:

sqrt(e)*e**3*(int((sqrt(x)*x**5)/(a + b*x**2)**(3/4),x)*d + int((sqrt(x)*x 
**3)/(a + b*x**2)**(3/4),x)*c)