\(\int \frac {(e x)^{3/2} (c+d x^2)}{(a+b x^2)^{5/4}} \, dx\) [450]

Optimal result

Integrand size = 26, antiderivative size = 171 \[ \int \frac {(e x)^{3/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=-\frac {(4 b c-5 a d) e \sqrt {e x}}{2 b^2 \sqrt [4]{a+b x^2}}+\frac {d (e x)^{5/2}}{2 b e \sqrt [4]{a+b x^2}}+\frac {(4 b c-5 a d) e^{3/2} \arctan \left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{4 b^{9/4}}+\frac {(4 b c-5 a d) e^{3/2} \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{4 b^{9/4}} \] Output:

-1/2*(-5*a*d+4*b*c)*e*(e*x)^(1/2)/b^2/(b*x^2+a)^(1/4)+1/2*d*(e*x)^(5/2)/b/ 
e/(b*x^2+a)^(1/4)+1/4*(-5*a*d+4*b*c)*e^(3/2)*arctan(b^(1/4)*(e*x)^(1/2)/e^ 
(1/2)/(b*x^2+a)^(1/4))/b^(9/4)+1/4*(-5*a*d+4*b*c)*e^(3/2)*arctanh(b^(1/4)* 
(e*x)^(1/2)/e^(1/2)/(b*x^2+a)^(1/4))/b^(9/4)
 

Mathematica [A] (verified)

Time = 1.02 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.87 \[ \int \frac {(e x)^{3/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\frac {(e x)^{3/2} \left (2 \sqrt [4]{b} \sqrt {x} \left (-4 b c+5 a d+b d x^2\right )+(4 b c-5 a d) \sqrt [4]{a+b x^2} \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a+b x^2}}\right )+(4 b c-5 a d) \sqrt [4]{a+b x^2} \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a+b x^2}}\right )\right )}{4 b^{9/4} x^{3/2} \sqrt [4]{a+b x^2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.97, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {363, 252, 266, 770, 756, 218, 221}

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.

\(\displaystyle \int \frac {(e x)^{3/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx\)

\(\Big \downarrow \) 363

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

\(\Big \downarrow \) 252

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 770

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

\(\Big \downarrow \) 756

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 221

\(\displaystyle \frac {(4 b c-5 a d) \left (\frac {2 e \left (\frac {\sqrt {e} \arctan \left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{2 \sqrt [4]{b}}+\frac {\sqrt {e} \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{2 \sqrt [4]{b}}\right )}{b}-\frac {2 e \sqrt {e x}}{b \sqrt [4]{a+b x^2}}\right )}{4 b}+\frac {d (e x)^{5/2}}{2 b e \sqrt [4]{a+b x^2}}\)

Input:

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

Output:

(d*(e*x)^(5/2))/(2*b*e*(a + b*x^2)^(1/4)) + ((4*b*c - 5*a*d)*((-2*e*Sqrt[e 
*x])/(b*(a + b*x^2)^(1/4)) + (2*e*((Sqrt[e]*ArcTan[(b^(1/4)*Sqrt[e*x])/(Sq 
rt[e]*(a + b*x^2)^(1/4))])/(2*b^(1/4)) + (Sqrt[e]*ArcTanh[(b^(1/4)*Sqrt[e* 
x])/(Sqrt[e]*(a + b*x^2)^(1/4))])/(2*b^(1/4))))/b))/(4*b)
 

Defintions of rubi rules used

rule 218

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

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 252

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x 
)^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* 
(p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c 
}, x] && LtQ[p, -1] && GtQ[m, 1] &&  !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi 
alQ[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 756

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 
]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a)   Int[1/(r - s*x^2), x], x] 
 + Simp[r/(2*a)   Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ[a 
/b, 0]
 

rule 770

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + 1/n)   Subst[In 
t[1/(1 - b*x^n)^(p + 1/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, 
 b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p + 1 
/n]
 

Maple [F]

\[\int \frac {\left (e x \right )^{\frac {3}{2}} \left (x^{2} d +c \right )}{\left (b \,x^{2}+a \right )^{\frac {5}{4}}}d x\]

Input:

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

Output:

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 810, normalized size of antiderivative = 4.74 \[ \int \frac {(e x)^{3/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx =\text {Too large to display} \] Input:

integrate((e*x)^(3/2)*(d*x^2+c)/(b*x^2+a)^(5/4),x, algorithm="fricas")
 

Output:

1/8*(4*(b*d*e*x^2 - (4*b*c - 5*a*d)*e)*(b*x^2 + a)^(3/4)*sqrt(e*x) + (b^3* 
x^2 + a*b^2)*((256*b^4*c^4 - 1280*a*b^3*c^3*d + 2400*a^2*b^2*c^2*d^2 - 200 
0*a^3*b*c*d^3 + 625*a^4*d^4)*e^6/b^9)^(1/4)*log(-((b*x^2 + a)^(3/4)*(4*b*c 
 - 5*a*d)*sqrt(e*x)*e + (b^3*x^2 + a*b^2)*((256*b^4*c^4 - 1280*a*b^3*c^3*d 
 + 2400*a^2*b^2*c^2*d^2 - 2000*a^3*b*c*d^3 + 625*a^4*d^4)*e^6/b^9)^(1/4))/ 
(b*x^2 + a)) - (b^3*x^2 + a*b^2)*((256*b^4*c^4 - 1280*a*b^3*c^3*d + 2400*a 
^2*b^2*c^2*d^2 - 2000*a^3*b*c*d^3 + 625*a^4*d^4)*e^6/b^9)^(1/4)*log(-((b*x 
^2 + a)^(3/4)*(4*b*c - 5*a*d)*sqrt(e*x)*e - (b^3*x^2 + a*b^2)*((256*b^4*c^ 
4 - 1280*a*b^3*c^3*d + 2400*a^2*b^2*c^2*d^2 - 2000*a^3*b*c*d^3 + 625*a^4*d 
^4)*e^6/b^9)^(1/4))/(b*x^2 + a)) - (-I*b^3*x^2 - I*a*b^2)*((256*b^4*c^4 - 
1280*a*b^3*c^3*d + 2400*a^2*b^2*c^2*d^2 - 2000*a^3*b*c*d^3 + 625*a^4*d^4)* 
e^6/b^9)^(1/4)*log(-((b*x^2 + a)^(3/4)*(4*b*c - 5*a*d)*sqrt(e*x)*e + (I*b^ 
3*x^2 + I*a*b^2)*((256*b^4*c^4 - 1280*a*b^3*c^3*d + 2400*a^2*b^2*c^2*d^2 - 
 2000*a^3*b*c*d^3 + 625*a^4*d^4)*e^6/b^9)^(1/4))/(b*x^2 + a)) - (I*b^3*x^2 
 + I*a*b^2)*((256*b^4*c^4 - 1280*a*b^3*c^3*d + 2400*a^2*b^2*c^2*d^2 - 2000 
*a^3*b*c*d^3 + 625*a^4*d^4)*e^6/b^9)^(1/4)*log(-((b*x^2 + a)^(3/4)*(4*b*c 
- 5*a*d)*sqrt(e*x)*e + (-I*b^3*x^2 - I*a*b^2)*((256*b^4*c^4 - 1280*a*b^3*c 
^3*d + 2400*a^2*b^2*c^2*d^2 - 2000*a^3*b*c*d^3 + 625*a^4*d^4)*e^6/b^9)^(1/ 
4))/(b*x^2 + a)))/(b^3*x^2 + a*b^2)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 12.79 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.55 \[ \int \frac {(e x)^{3/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\frac {c e^{\frac {3}{2}} x^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {5}{4}} \Gamma \left (\frac {9}{4}\right )} + \frac {d e^{\frac {3}{2}} x^{\frac {9}{2}} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {5}{4}} \Gamma \left (\frac {13}{4}\right )} \] Input:

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

Output:

c*e**(3/2)*x**(5/2)*gamma(5/4)*hyper((5/4, 5/4), (9/4,), b*x**2*exp_polar( 
I*pi)/a)/(2*a**(5/4)*gamma(9/4)) + d*e**(3/2)*x**(9/2)*gamma(9/4)*hyper((5 
/4, 9/4), (13/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(5/4)*gamma(13/4))
 

Maxima [F]

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

integrate((e*x)^(3/2)*(d*x^2+c)/(b*x^2+a)^(5/4),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((e*x)^(3/2)*(d*x^2+c)/(b*x^2+a)^(5/4),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(e x)^{3/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\sqrt {e}\, e \left (\left (\int \frac {\sqrt {x}\, x^{3}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} a +\left (b \,x^{2}+a \right )^{\frac {1}{4}} b \,x^{2}}d x \right ) d +\left (\int \frac {\sqrt {x}\, x}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} a +\left (b \,x^{2}+a \right )^{\frac {1}{4}} b \,x^{2}}d x \right ) c \right ) \] Input:

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

Output:

sqrt(e)*e*(int((sqrt(x)*x**3)/((a + b*x**2)**(1/4)*a + (a + b*x**2)**(1/4) 
*b*x**2),x)*d + int((sqrt(x)*x)/((a + b*x**2)**(1/4)*a + (a + b*x**2)**(1/ 
4)*b*x**2),x)*c)