\(\int \frac {(e x)^{3/2} (c+d x^2)}{(a+b x^2)^{9/4}} \, dx\) [472]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [F]
Fricas [C] (verification not implemented)
Sympy [C] (verification not implemented)
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 26, antiderivative size = 149 \[ \int \frac {(e x)^{3/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\frac {2 (b c-a d) (e x)^{5/2}}{5 a b e \left (a+b x^2\right )^{5/4}}-\frac {2 d e \sqrt {e x}}{b^2 \sqrt [4]{a+b x^2}}+\frac {d e^{3/2} \arctan \left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{b^{9/4}}+\frac {d e^{3/2} \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{b^{9/4}} \] Output:

2/5*(-a*d+b*c)*(e*x)^(5/2)/a/b/e/(b*x^2+a)^(5/4)-2*d*e*(e*x)^(1/2)/b^2/(b* 
x^2+a)^(1/4)+d*e^(3/2)*arctan(b^(1/4)*(e*x)^(1/2)/e^(1/2)/(b*x^2+a)^(1/4)) 
/b^(9/4)+d*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 = 1.32 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.83 \[ \int \frac {(e x)^{3/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\frac {e \sqrt {e x} \left (\frac {2 \sqrt [4]{b} \left (-5 a^2 d+b^2 c x^2-6 a b d x^2\right )}{a \left (a+b x^2\right )^{5/4}}+\frac {5 d \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a+b x^2}}\right )}{\sqrt {x}}+\frac {5 d \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a+b x^2}}\right )}{\sqrt {x}}\right )}{5 b^{9/4}} \] Input:

Integrate[((e*x)^(3/2)*(c + d*x^2))/(a + b*x^2)^(9/4),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {357, 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.

\(\displaystyle \int \frac {(e x)^{3/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx\)

\(\Big \downarrow \) 357

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

\(\Big \downarrow \) 252

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 770

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

\(\Big \downarrow \) 756

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 221

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

Input:

Int[((e*x)^(3/2)*(c + d*x^2))/(a + b*x^2)^(9/4),x]
 

Output:

(2*(b*c - a*d)*(e*x)^(5/2))/(5*a*b*e*(a + b*x^2)^(5/4)) + (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))/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 357
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_ 
Symbol] :> Simp[(b*c - a*d)*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*b*e*(m + 
1))), x] + Simp[d/b   Int[(e*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, 
b, c, d, e}, x] && NeQ[b*c - a*d, 0] && EqQ[m + 2*p + 3, 0] && LtQ[p, -1]
 

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]

\[\int \frac {\left (e x \right )^{\frac {3}{2}} \left (x^{2} d +c \right )}{\left (b \,x^{2}+a \right )^{\frac {9}{4}}}d x\]

Input:

int((e*x)^(3/2)*(d*x^2+c)/(b*x^2+a)^(9/4),x)
 

Output:

int((e*x)^(3/2)*(d*x^2+c)/(b*x^2+a)^(9/4),x)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 468, normalized size of antiderivative = 3.14 \[ \int \frac {(e x)^{3/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=-\frac {4 \, {\left (5 \, a^{2} d e - {\left (b^{2} c - 6 \, a b d\right )} e x^{2}\right )} {\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {e x} - 5 \, {\left (a b^{4} x^{4} + 2 \, a^{2} b^{3} x^{2} + a^{3} b^{2}\right )} \left (\frac {d^{4} e^{6}}{b^{9}}\right )^{\frac {1}{4}} \log \left (\frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {e x} d e + {\left (b^{3} x^{2} + a b^{2}\right )} \left (\frac {d^{4} e^{6}}{b^{9}}\right )^{\frac {1}{4}}}{b x^{2} + a}\right ) + 5 \, {\left (a b^{4} x^{4} + 2 \, a^{2} b^{3} x^{2} + a^{3} b^{2}\right )} \left (\frac {d^{4} e^{6}}{b^{9}}\right )^{\frac {1}{4}} \log \left (\frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {e x} d e - {\left (b^{3} x^{2} + a b^{2}\right )} \left (\frac {d^{4} e^{6}}{b^{9}}\right )^{\frac {1}{4}}}{b x^{2} + a}\right ) + 5 \, {\left (i \, a b^{4} x^{4} + 2 i \, a^{2} b^{3} x^{2} + i \, a^{3} b^{2}\right )} \left (\frac {d^{4} e^{6}}{b^{9}}\right )^{\frac {1}{4}} \log \left (\frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {e x} d e - {\left (i \, b^{3} x^{2} + i \, a b^{2}\right )} \left (\frac {d^{4} e^{6}}{b^{9}}\right )^{\frac {1}{4}}}{b x^{2} + a}\right ) + 5 \, {\left (-i \, a b^{4} x^{4} - 2 i \, a^{2} b^{3} x^{2} - i \, a^{3} b^{2}\right )} \left (\frac {d^{4} e^{6}}{b^{9}}\right )^{\frac {1}{4}} \log \left (\frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {e x} d e - {\left (-i \, b^{3} x^{2} - i \, a b^{2}\right )} \left (\frac {d^{4} e^{6}}{b^{9}}\right )^{\frac {1}{4}}}{b x^{2} + a}\right )}{10 \, {\left (a b^{4} x^{4} + 2 \, a^{2} b^{3} x^{2} + a^{3} b^{2}\right )}} \] Input:

integrate((e*x)^(3/2)*(d*x^2+c)/(b*x^2+a)^(9/4),x, algorithm="fricas")
 

Output:

-1/10*(4*(5*a^2*d*e - (b^2*c - 6*a*b*d)*e*x^2)*(b*x^2 + a)^(3/4)*sqrt(e*x) 
 - 5*(a*b^4*x^4 + 2*a^2*b^3*x^2 + a^3*b^2)*(d^4*e^6/b^9)^(1/4)*log(((b*x^2 
 + a)^(3/4)*sqrt(e*x)*d*e + (b^3*x^2 + a*b^2)*(d^4*e^6/b^9)^(1/4))/(b*x^2 
+ a)) + 5*(a*b^4*x^4 + 2*a^2*b^3*x^2 + a^3*b^2)*(d^4*e^6/b^9)^(1/4)*log((( 
b*x^2 + a)^(3/4)*sqrt(e*x)*d*e - (b^3*x^2 + a*b^2)*(d^4*e^6/b^9)^(1/4))/(b 
*x^2 + a)) + 5*(I*a*b^4*x^4 + 2*I*a^2*b^3*x^2 + I*a^3*b^2)*(d^4*e^6/b^9)^( 
1/4)*log(((b*x^2 + a)^(3/4)*sqrt(e*x)*d*e - (I*b^3*x^2 + I*a*b^2)*(d^4*e^6 
/b^9)^(1/4))/(b*x^2 + a)) + 5*(-I*a*b^4*x^4 - 2*I*a^2*b^3*x^2 - I*a^3*b^2) 
*(d^4*e^6/b^9)^(1/4)*log(((b*x^2 + a)^(3/4)*sqrt(e*x)*d*e - (-I*b^3*x^2 - 
I*a*b^2)*(d^4*e^6/b^9)^(1/4))/(b*x^2 + a)))/(a*b^4*x^4 + 2*a^2*b^3*x^2 + a 
^3*b^2)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 64.87 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.78 \[ \int \frac {(e x)^{3/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\frac {c e^{\frac {3}{2}} x^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right )}{2 a^{\frac {9}{4}} \sqrt [4]{1 + \frac {b x^{2}}{a}} \Gamma \left (\frac {9}{4}\right ) + 2 a^{\frac {5}{4}} b x^{2} \sqrt [4]{1 + \frac {b x^{2}}{a}} \Gamma \left (\frac {9}{4}\right )} + \frac {d e^{\frac {3}{2}} x^{\frac {9}{2}} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {9}{4}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {9}{4}} \Gamma \left (\frac {13}{4}\right )} \] Input:

integrate((e*x)**(3/2)*(d*x**2+c)/(b*x**2+a)**(9/4),x)
 

Output:

c*e**(3/2)*x**(5/2)*gamma(5/4)/(2*a**(9/4)*(1 + b*x**2/a)**(1/4)*gamma(9/4 
) + 2*a**(5/4)*b*x**2*(1 + b*x**2/a)**(1/4)*gamma(9/4)) + d*e**(3/2)*x**(9 
/2)*gamma(9/4)*hyper((9/4, 9/4), (13/4,), b*x**2*exp_polar(I*pi)/a)/(2*a** 
(9/4)*gamma(13/4))
 

Maxima [F]

\[ \int \frac {(e x)^{3/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} \left (e x\right )^{\frac {3}{2}}}{{\left (b x^{2} + a\right )}^{\frac {9}{4}}} \,d x } \] Input:

integrate((e*x)^(3/2)*(d*x^2+c)/(b*x^2+a)^(9/4),x, algorithm="maxima")
 

Output:

integrate((d*x^2 + c)*(e*x)^(3/2)/(b*x^2 + a)^(9/4), x)
                                                                                    
                                                                                    
 

Giac [F]

\[ \int \frac {(e x)^{3/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} \left (e x\right )^{\frac {3}{2}}}{{\left (b x^{2} + a\right )}^{\frac {9}{4}}} \,d x } \] Input:

integrate((e*x)^(3/2)*(d*x^2+c)/(b*x^2+a)^(9/4),x, algorithm="giac")
 

Output:

integrate((d*x^2 + c)*(e*x)^(3/2)/(b*x^2 + a)^(9/4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e x)^{3/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\int \frac {{\left (e\,x\right )}^{3/2}\,\left (d\,x^2+c\right )}{{\left (b\,x^2+a\right )}^{9/4}} \,d x \] Input:

int(((e*x)^(3/2)*(c + d*x^2))/(a + b*x^2)^(9/4),x)
 

Output:

int(((e*x)^(3/2)*(c + d*x^2))/(a + b*x^2)^(9/4), x)
 

Reduce [F]

\[ \int \frac {(e x)^{3/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\sqrt {e}\, e \left (\left (\int \frac {\sqrt {x}\, x^{3}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} a^{2}+2 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a b \,x^{2}+\left (b \,x^{2}+a \right )^{\frac {1}{4}} b^{2} x^{4}}d x \right ) d +\left (\int \frac {\sqrt {x}\, x}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} a^{2}+2 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a b \,x^{2}+\left (b \,x^{2}+a \right )^{\frac {1}{4}} b^{2} x^{4}}d x \right ) c \right ) \] Input:

int((e*x)^(3/2)*(d*x^2+c)/(b*x^2+a)^(9/4),x)
 

Output:

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