\(\int \frac {(a+b x^2)^{3/4} (c+d x^2)}{(e x)^{5/2}} \, dx\) [427]

Optimal result

Integrand size = 26, antiderivative size = 173 \[ \int \frac {\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )}{(e x)^{5/2}} \, dx=\frac {(4 b c+3 a d) \sqrt {e x} \left (a+b x^2\right )^{3/4}}{6 a e^3}-\frac {2 c \left (a+b x^2\right )^{7/4}}{3 a e (e x)^{3/2}}+\frac {(4 b c+3 a d) \arctan \left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{4 \sqrt [4]{b} e^{5/2}}+\frac {(4 b c+3 a d) \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{4 \sqrt [4]{b} e^{5/2}} \] Output:

1/6*(3*a*d+4*b*c)*(e*x)^(1/2)*(b*x^2+a)^(3/4)/a/e^3-2/3*c*(b*x^2+a)^(7/4)/ 
a/e/(e*x)^(3/2)+1/4*(3*a*d+4*b*c)*arctan(b^(1/4)*(e*x)^(1/2)/e^(1/2)/(b*x^ 
2+a)^(1/4))/b^(1/4)/e^(5/2)+1/4*(3*a*d+4*b*c)*arctanh(b^(1/4)*(e*x)^(1/2)/ 
e^(1/2)/(b*x^2+a)^(1/4))/b^(1/4)/e^(5/2)
 

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.72 \[ \int \frac {\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )}{(e x)^{5/2}} \, dx=\frac {x \left (2 \sqrt [4]{b} \left (a+b x^2\right )^{3/4} \left (-4 c+3 d x^2\right )+3 (4 b c+3 a d) x^{3/2} \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a+b x^2}}\right )+3 (4 b c+3 a d) x^{3/2} \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a+b x^2}}\right )\right )}{12 \sqrt [4]{b} (e x)^{5/2}} \] Input:

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

Output:

(x*(2*b^(1/4)*(a + b*x^2)^(3/4)*(-4*c + 3*d*x^2) + 3*(4*b*c + 3*a*d)*x^(3/ 
2)*ArcTan[(b^(1/4)*Sqrt[x])/(a + b*x^2)^(1/4)] + 3*(4*b*c + 3*a*d)*x^(3/2) 
*ArcTanh[(b^(1/4)*Sqrt[x])/(a + b*x^2)^(1/4)]))/(12*b^(1/4)*(e*x)^(5/2))
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.99, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {359, 248, 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.

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

\(\Big \downarrow \) 359

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

\(\Big \downarrow \) 248

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 770

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

\(\Big \downarrow \) 756

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 221

\(\displaystyle \frac {(3 a d+4 b c) \left (\frac {3 a \left (\frac {\sqrt {e} \arctan \left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{2 \sqrt [4]{b}}+\frac {\sqrt {e} \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{2 \sqrt [4]{b}}\right )}{2 e}+\frac {\sqrt {e x} \left (a+b x^2\right )^{3/4}}{2 e}\right )}{3 a e^2}-\frac {2 c \left (a+b x^2\right )^{7/4}}{3 a e (e x)^{3/2}}\)

Input:

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

Output:

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

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 248

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

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

Input:

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

Output:

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 758, normalized size of antiderivative = 4.38 \[ \int \frac {\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )}{(e x)^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

1/24*(3*e^3*x^2*((256*b^4*c^4 + 768*a*b^3*c^3*d + 864*a^2*b^2*c^2*d^2 + 43 
2*a^3*b*c*d^3 + 81*a^4*d^4)/(b*e^10))^(1/4)*log(((b*x^2 + a)^(3/4)*(4*b*c 
+ 3*a*d)*sqrt(e*x) + (b*e^3*x^2 + a*e^3)*((256*b^4*c^4 + 768*a*b^3*c^3*d + 
 864*a^2*b^2*c^2*d^2 + 432*a^3*b*c*d^3 + 81*a^4*d^4)/(b*e^10))^(1/4))/(b*x 
^2 + a)) - 3*e^3*x^2*((256*b^4*c^4 + 768*a*b^3*c^3*d + 864*a^2*b^2*c^2*d^2 
 + 432*a^3*b*c*d^3 + 81*a^4*d^4)/(b*e^10))^(1/4)*log(((b*x^2 + a)^(3/4)*(4 
*b*c + 3*a*d)*sqrt(e*x) - (b*e^3*x^2 + a*e^3)*((256*b^4*c^4 + 768*a*b^3*c^ 
3*d + 864*a^2*b^2*c^2*d^2 + 432*a^3*b*c*d^3 + 81*a^4*d^4)/(b*e^10))^(1/4)) 
/(b*x^2 + a)) - 3*I*e^3*x^2*((256*b^4*c^4 + 768*a*b^3*c^3*d + 864*a^2*b^2* 
c^2*d^2 + 432*a^3*b*c*d^3 + 81*a^4*d^4)/(b*e^10))^(1/4)*log(((b*x^2 + a)^( 
3/4)*(4*b*c + 3*a*d)*sqrt(e*x) - (I*b*e^3*x^2 + I*a*e^3)*((256*b^4*c^4 + 7 
68*a*b^3*c^3*d + 864*a^2*b^2*c^2*d^2 + 432*a^3*b*c*d^3 + 81*a^4*d^4)/(b*e^ 
10))^(1/4))/(b*x^2 + a)) + 3*I*e^3*x^2*((256*b^4*c^4 + 768*a*b^3*c^3*d + 8 
64*a^2*b^2*c^2*d^2 + 432*a^3*b*c*d^3 + 81*a^4*d^4)/(b*e^10))^(1/4)*log(((b 
*x^2 + a)^(3/4)*(4*b*c + 3*a*d)*sqrt(e*x) - (-I*b*e^3*x^2 - I*a*e^3)*((256 
*b^4*c^4 + 768*a*b^3*c^3*d + 864*a^2*b^2*c^2*d^2 + 432*a^3*b*c*d^3 + 81*a^ 
4*d^4)/(b*e^10))^(1/4))/(b*x^2 + a)) + 4*(b*x^2 + a)^(3/4)*(3*d*x^2 - 4*c) 
*sqrt(e*x))/(e^3*x^2)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 8.93 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.58 \[ \int \frac {\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )}{(e x)^{5/2}} \, dx=\frac {a^{\frac {3}{4}} c \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, - \frac {3}{4} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac {5}{2}} x^{\frac {3}{2}} \Gamma \left (\frac {1}{4}\right )} + \frac {a^{\frac {3}{4}} d \sqrt {x} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right )} \] Input:

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )}{(e x)^{5/2}} \, dx=\frac {\sqrt {e}\, \left (-8 \left (b \,x^{2}+a \right )^{\frac {3}{4}} c +6 \left (b \,x^{2}+a \right )^{\frac {3}{4}} d \,x^{2}+9 \sqrt {x}\, \left (\int \frac {\sqrt {x}\, \left (b \,x^{2}+a \right )^{\frac {3}{4}}}{b \,x^{3}+a x}d x \right ) a d x +12 \sqrt {x}\, \left (\int \frac {\sqrt {x}\, \left (b \,x^{2}+a \right )^{\frac {3}{4}}}{b \,x^{3}+a x}d x \right ) b c x \right )}{12 \sqrt {x}\, e^{3} x} \] Input:

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

Output:

(sqrt(e)*( - 8*(a + b*x**2)**(3/4)*c + 6*(a + b*x**2)**(3/4)*d*x**2 + 9*sq 
rt(x)*int((sqrt(x)*(a + b*x**2)**(3/4))/(a*x + b*x**3),x)*a*d*x + 12*sqrt( 
x)*int((sqrt(x)*(a + b*x**2)**(3/4))/(a*x + b*x**3),x)*b*c*x))/(12*sqrt(x) 
*e**3*x)