\(\int \frac {(a+c x^4)^2}{(d+e x^2)^{3/2}} \, dx\) [376]

Optimal result

Integrand size = 21, antiderivative size = 158 \[ \int \frac {\left (a+c x^4\right )^2}{\left (d+e x^2\right )^{3/2}} \, dx=\frac {\left (c d^2+a e^2\right )^2 x}{d e^4 \sqrt {d+e x^2}}+\frac {c \left (19 c d^2+16 a e^2\right ) x \sqrt {d+e x^2}}{16 e^4}-\frac {11 c^2 d x^3 \sqrt {d+e x^2}}{24 e^3}+\frac {c^2 x^5 \sqrt {d+e x^2}}{6 e^2}-\frac {c d \left (35 c d^2+48 a e^2\right ) \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{16 e^{9/2}} \] Output:

(a*e^2+c*d^2)^2*x/d/e^4/(e*x^2+d)^(1/2)+1/16*c*(16*a*e^2+19*c*d^2)*x*(e*x^ 
2+d)^(1/2)/e^4-11/24*c^2*d*x^3*(e*x^2+d)^(1/2)/e^3+1/6*c^2*x^5*(e*x^2+d)^( 
1/2)/e^2-1/16*c*d*(48*a*e^2+35*c*d^2)*arctanh(e^(1/2)*x/(e*x^2+d)^(1/2))/e 
^(9/2)
 

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.83 \[ \int \frac {\left (a+c x^4\right )^2}{\left (d+e x^2\right )^{3/2}} \, dx=\frac {x \left (48 a^2 e^4+48 a c d e^2 \left (3 d+e x^2\right )+c^2 d \left (105 d^3+35 d^2 e x^2-14 d e^2 x^4+8 e^3 x^6\right )\right )}{48 d e^4 \sqrt {d+e x^2}}+\frac {c d \left (35 c d^2+48 a e^2\right ) \log \left (-\sqrt {e} x+\sqrt {d+e x^2}\right )}{16 e^{9/2}} \] Input:

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

Output:

(x*(48*a^2*e^4 + 48*a*c*d*e^2*(3*d + e*x^2) + c^2*d*(105*d^3 + 35*d^2*e*x^ 
2 - 14*d*e^2*x^4 + 8*e^3*x^6)))/(48*d*e^4*Sqrt[d + e*x^2]) + (c*d*(35*c*d^ 
2 + 48*a*e^2)*Log[-(Sqrt[e]*x) + Sqrt[d + e*x^2]])/(16*e^(9/2))
 

Rubi [A] (verified)

Time = 0.86 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.14, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1472, 2346, 1473, 27, 299, 224, 219}

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.

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

\(\Big \downarrow \) 1472

\(\displaystyle \frac {x \left (a e^2+c d^2\right )^2}{d e^4 \sqrt {d+e x^2}}-\frac {\int \frac {-\frac {c^2 d x^6}{e}+\frac {c^2 d^2 x^4}{e^2}-\frac {c d \left (c d^2+2 a e^2\right ) x^2}{e^3}+\frac {c d^2 \left (c d^2+2 a e^2\right )}{e^4}}{\sqrt {e x^2+d}}dx}{d}\)

\(\Big \downarrow \) 2346

\(\displaystyle \frac {x \left (a e^2+c d^2\right )^2}{d e^4 \sqrt {d+e x^2}}-\frac {\frac {\int \frac {\frac {11 c^2 d^2 x^4}{e}-6 c d \left (\frac {c d^2}{e^2}+2 a\right ) x^2+\frac {6 c d^2 \left (c d^2+2 a e^2\right )}{e^3}}{\sqrt {e x^2+d}}dx}{6 e}-\frac {c^2 d x^5 \sqrt {d+e x^2}}{6 e^2}}{d}\)

\(\Big \downarrow \) 1473

\(\displaystyle \frac {x \left (a e^2+c d^2\right )^2}{d e^4 \sqrt {d+e x^2}}-\frac {\frac {\frac {\int \frac {3 c d \left (8 d \left (\frac {c d^2}{e^2}+2 a\right )-\left (\frac {19 c d^2}{e}+16 a e\right ) x^2\right )}{\sqrt {e x^2+d}}dx}{4 e}+\frac {11 c^2 d^2 x^3 \sqrt {d+e x^2}}{4 e^2}}{6 e}-\frac {c^2 d x^5 \sqrt {d+e x^2}}{6 e^2}}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {x \left (a e^2+c d^2\right )^2}{d e^4 \sqrt {d+e x^2}}-\frac {\frac {\frac {3 c d \int \frac {8 d \left (\frac {c d^2}{e^2}+2 a\right )-\left (\frac {19 c d^2}{e}+16 a e\right ) x^2}{\sqrt {e x^2+d}}dx}{4 e}+\frac {11 c^2 d^2 x^3 \sqrt {d+e x^2}}{4 e^2}}{6 e}-\frac {c^2 d x^5 \sqrt {d+e x^2}}{6 e^2}}{d}\)

\(\Big \downarrow \) 299

\(\displaystyle \frac {x \left (a e^2+c d^2\right )^2}{d e^4 \sqrt {d+e x^2}}-\frac {\frac {\frac {3 c d \left (\frac {1}{2} d \left (48 a+\frac {35 c d^2}{e^2}\right ) \int \frac {1}{\sqrt {e x^2+d}}dx-\frac {1}{2} x \sqrt {d+e x^2} \left (16 a+\frac {19 c d^2}{e^2}\right )\right )}{4 e}+\frac {11 c^2 d^2 x^3 \sqrt {d+e x^2}}{4 e^2}}{6 e}-\frac {c^2 d x^5 \sqrt {d+e x^2}}{6 e^2}}{d}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {x \left (a e^2+c d^2\right )^2}{d e^4 \sqrt {d+e x^2}}-\frac {\frac {\frac {3 c d \left (\frac {1}{2} d \left (48 a+\frac {35 c d^2}{e^2}\right ) \int \frac {1}{1-\frac {e x^2}{e x^2+d}}d\frac {x}{\sqrt {e x^2+d}}-\frac {1}{2} x \sqrt {d+e x^2} \left (16 a+\frac {19 c d^2}{e^2}\right )\right )}{4 e}+\frac {11 c^2 d^2 x^3 \sqrt {d+e x^2}}{4 e^2}}{6 e}-\frac {c^2 d x^5 \sqrt {d+e x^2}}{6 e^2}}{d}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {x \left (a e^2+c d^2\right )^2}{d e^4 \sqrt {d+e x^2}}-\frac {\frac {\frac {3 c d \left (\frac {d \left (48 a+\frac {35 c d^2}{e^2}\right ) \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{2 \sqrt {e}}-\frac {1}{2} x \sqrt {d+e x^2} \left (16 a+\frac {19 c d^2}{e^2}\right )\right )}{4 e}+\frac {11 c^2 d^2 x^3 \sqrt {d+e x^2}}{4 e^2}}{6 e}-\frac {c^2 d x^5 \sqrt {d+e x^2}}{6 e^2}}{d}\)

Input:

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

Output:

((c*d^2 + a*e^2)^2*x)/(d*e^4*Sqrt[d + e*x^2]) - (-1/6*(c^2*d*x^5*Sqrt[d + 
e*x^2])/e^2 + ((11*c^2*d^2*x^3*Sqrt[d + e*x^2])/(4*e^2) + (3*c*d*(-1/2*((1 
6*a + (19*c*d^2)/e^2)*x*Sqrt[d + e*x^2]) + (d*(48*a + (35*c*d^2)/e^2)*ArcT 
anh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(2*Sqrt[e])))/(4*e))/(6*e))/d
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

rule 299

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

rule 1472

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Wi 
th[{Qx = PolynomialQuotient[(a + c*x^4)^p, d + e*x^2, x], R = Coeff[Polynom 
ialRemainder[(a + c*x^4)^p, d + e*x^2, x], x, 0]}, Simp[(-R)*x*((d + e*x^2) 
^(q + 1)/(2*d*(q + 1))), x] + Simp[1/(2*d*(q + 1))   Int[(d + e*x^2)^(q + 1 
)*ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, c, d, 
e}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]
 

rule 1473

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), 
x_Symbol] :> Simp[c^p*x^(4*p - 1)*((d + e*x^2)^(q + 1)/(e*(4*p + 2*q + 1))) 
, x] + Simp[1/(e*(4*p + 2*q + 1))   Int[(d + e*x^2)^q*ExpandToSum[e*(4*p + 
2*q + 1)*(a + b*x^2 + c*x^4)^p - d*c^p*(4*p - 1)*x^(4*p - 2) - e*c^p*(4*p + 
 2*q + 1)*x^(4*p), x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b^2 - 
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] &&  !LtQ[q, -1]
 

rule 2346

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], 
e = Coeff[Pq, x, Expon[Pq, x]]}, Simp[e*x^(q - 1)*((a + b*x^2)^(p + 1)/(b*( 
q + 2*p + 1))), x] + Simp[1/(b*(q + 2*p + 1))   Int[(a + b*x^2)^p*ExpandToS 
um[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], 
x], x]] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] &&  !LeQ[p, -1]
 

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.83

method	result	size
pseudoelliptic	\(\frac {-3 d^{2} \sqrt {e \,x^{2}+d}\, \left (a \,e^{2}+\frac {35 c \,d^{2}}{48}\right ) c \,\operatorname {arctanh}\left (\frac {\sqrt {e \,x^{2}+d}}{x \sqrt {e}}\right )+x \left (3 d^{2} \left (-\frac {7 c \,x^{4}}{72}+a \right ) c \,e^{\frac {5}{2}}+c d \,x^{2} \left (\frac {c \,x^{4}}{6}+a \right ) e^{\frac {7}{2}}+\frac {35 e^{\frac {3}{2}} c^{2} d^{3} x^{2}}{48}+\frac {35 \sqrt {e}\, c^{2} d^{4}}{16}+e^{\frac {9}{2}} a^{2}\right )}{e^{\frac {9}{2}} \sqrt {e \,x^{2}+d}\, d}\)	\(131\)
risch	\(\frac {c x \left (8 c \,x^{4} e^{2}-22 d e \,x^{2} c +48 a \,e^{2}+57 c \,d^{2}\right ) \sqrt {e \,x^{2}+d}}{48 e^{4}}+\frac {2 c d x a}{e^{2} \sqrt {e \,x^{2}+d}}+\frac {c^{2} d^{3} x}{e^{4} \sqrt {e \,x^{2}+d}}-\frac {3 c d \ln \left (x \sqrt {e}+\sqrt {e \,x^{2}+d}\right ) a}{e^{\frac {5}{2}}}-\frac {35 c^{2} d^{3} \ln \left (x \sqrt {e}+\sqrt {e \,x^{2}+d}\right )}{16 e^{\frac {9}{2}}}+\frac {a^{2} x}{d \sqrt {e \,x^{2}+d}}\)	\(154\)
default	\(\frac {a^{2} x}{d \sqrt {e \,x^{2}+d}}+c^{2} \left (\frac {x^{7}}{6 e \sqrt {e \,x^{2}+d}}-\frac {7 d \left (\frac {x^{5}}{4 e \sqrt {e \,x^{2}+d}}-\frac {5 d \left (\frac {x^{3}}{2 e \sqrt {e \,x^{2}+d}}-\frac {3 d \left (-\frac {x}{e \sqrt {e \,x^{2}+d}}+\frac {\ln \left (x \sqrt {e}+\sqrt {e \,x^{2}+d}\right )}{e^{\frac {3}{2}}}\right )}{2 e}\right )}{4 e}\right )}{6 e}\right )+2 a c \left (\frac {x^{3}}{2 e \sqrt {e \,x^{2}+d}}-\frac {3 d \left (-\frac {x}{e \sqrt {e \,x^{2}+d}}+\frac {\ln \left (x \sqrt {e}+\sqrt {e \,x^{2}+d}\right )}{e^{\frac {3}{2}}}\right )}{2 e}\right )\)	\(195\)

Input:

int((c*x^4+a)^2/(e*x^2+d)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/e^(9/2)/(e*x^2+d)^(1/2)*(-3*d^2*(e*x^2+d)^(1/2)*(a*e^2+35/48*c*d^2)*c*ar 
ctanh((e*x^2+d)^(1/2)/x/e^(1/2))+x*(3*d^2*(-7/72*c*x^4+a)*c*e^(5/2)+c*d*x^ 
2*(1/6*c*x^4+a)*e^(7/2)+35/48*e^(3/2)*c^2*d^3*x^2+35/16*e^(1/2)*c^2*d^4+e^ 
(9/2)*a^2))/d
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 369, normalized size of antiderivative = 2.34 \[ \int \frac {\left (a+c x^4\right )^2}{\left (d+e x^2\right )^{3/2}} \, dx=\left [\frac {3 \, {\left (35 \, c^{2} d^{5} + 48 \, a c d^{3} e^{2} + {\left (35 \, c^{2} d^{4} e + 48 \, a c d^{2} e^{3}\right )} x^{2}\right )} \sqrt {e} \log \left (-2 \, e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {e} x - d\right ) + 2 \, {\left (8 \, c^{2} d e^{4} x^{7} - 14 \, c^{2} d^{2} e^{3} x^{5} + {\left (35 \, c^{2} d^{3} e^{2} + 48 \, a c d e^{4}\right )} x^{3} + 3 \, {\left (35 \, c^{2} d^{4} e + 48 \, a c d^{2} e^{3} + 16 \, a^{2} e^{5}\right )} x\right )} \sqrt {e x^{2} + d}}{96 \, {\left (d e^{6} x^{2} + d^{2} e^{5}\right )}}, \frac {3 \, {\left (35 \, c^{2} d^{5} + 48 \, a c d^{3} e^{2} + {\left (35 \, c^{2} d^{4} e + 48 \, a c d^{2} e^{3}\right )} x^{2}\right )} \sqrt {-e} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right ) + {\left (8 \, c^{2} d e^{4} x^{7} - 14 \, c^{2} d^{2} e^{3} x^{5} + {\left (35 \, c^{2} d^{3} e^{2} + 48 \, a c d e^{4}\right )} x^{3} + 3 \, {\left (35 \, c^{2} d^{4} e + 48 \, a c d^{2} e^{3} + 16 \, a^{2} e^{5}\right )} x\right )} \sqrt {e x^{2} + d}}{48 \, {\left (d e^{6} x^{2} + d^{2} e^{5}\right )}}\right ] \] Input:

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

Output:

[1/96*(3*(35*c^2*d^5 + 48*a*c*d^3*e^2 + (35*c^2*d^4*e + 48*a*c*d^2*e^3)*x^ 
2)*sqrt(e)*log(-2*e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(e)*x - d) + 2*(8*c^2*d*e^ 
4*x^7 - 14*c^2*d^2*e^3*x^5 + (35*c^2*d^3*e^2 + 48*a*c*d*e^4)*x^3 + 3*(35*c 
^2*d^4*e + 48*a*c*d^2*e^3 + 16*a^2*e^5)*x)*sqrt(e*x^2 + d))/(d*e^6*x^2 + d 
^2*e^5), 1/48*(3*(35*c^2*d^5 + 48*a*c*d^3*e^2 + (35*c^2*d^4*e + 48*a*c*d^2 
*e^3)*x^2)*sqrt(-e)*arctan(sqrt(-e)*x/sqrt(e*x^2 + d)) + (8*c^2*d*e^4*x^7 
- 14*c^2*d^2*e^3*x^5 + (35*c^2*d^3*e^2 + 48*a*c*d*e^4)*x^3 + 3*(35*c^2*d^4 
*e + 48*a*c*d^2*e^3 + 16*a^2*e^5)*x)*sqrt(e*x^2 + d))/(d*e^6*x^2 + d^2*e^5 
)]
 

Sympy [F]

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

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

Output:

Integral((a + c*x**4)**2/(d + e*x**2)**(3/2), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (a+c x^4\right )^2}{\left (d+e x^2\right )^{3/2}} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(e>0)', see `assume?` for more de 
tails)Is e
 

Giac [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.95 \[ \int \frac {\left (a+c x^4\right )^2}{\left (d+e x^2\right )^{3/2}} \, dx=\frac {{\left ({\left (2 \, {\left (\frac {4 \, c^{2} x^{2}}{e} - \frac {7 \, c^{2} d}{e^{2}}\right )} x^{2} + \frac {35 \, c^{2} d^{3} e^{4} + 48 \, a c d e^{6}}{d e^{7}}\right )} x^{2} + \frac {3 \, {\left (35 \, c^{2} d^{4} e^{3} + 48 \, a c d^{2} e^{5} + 16 \, a^{2} e^{7}\right )}}{d e^{7}}\right )} x}{48 \, \sqrt {e x^{2} + d}} + \frac {{\left (35 \, c^{2} d^{3} + 48 \, a c d e^{2}\right )} \log \left ({\left | -\sqrt {e} x + \sqrt {e x^{2} + d} \right |}\right )}{16 \, e^{\frac {9}{2}}} \] Input:

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

Output:

1/48*((2*(4*c^2*x^2/e - 7*c^2*d/e^2)*x^2 + (35*c^2*d^3*e^4 + 48*a*c*d*e^6) 
/(d*e^7))*x^2 + 3*(35*c^2*d^4*e^3 + 48*a*c*d^2*e^5 + 16*a^2*e^7)/(d*e^7))* 
x/sqrt(e*x^2 + d) + 1/16*(35*c^2*d^3 + 48*a*c*d*e^2)*log(abs(-sqrt(e)*x + 
sqrt(e*x^2 + d)))/e^(9/2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 357, normalized size of antiderivative = 2.26 \[ \int \frac {\left (a+c x^4\right )^2}{\left (d+e x^2\right )^{3/2}} \, dx=\frac {384 \sqrt {e \,x^{2}+d}\, a^{2} e^{5} x +1152 \sqrt {e \,x^{2}+d}\, a c \,d^{2} e^{3} x +384 \sqrt {e \,x^{2}+d}\, a c d \,e^{4} x^{3}+840 \sqrt {e \,x^{2}+d}\, c^{2} d^{4} e x +280 \sqrt {e \,x^{2}+d}\, c^{2} d^{3} e^{2} x^{3}-112 \sqrt {e \,x^{2}+d}\, c^{2} d^{2} e^{3} x^{5}+64 \sqrt {e \,x^{2}+d}\, c^{2} d \,e^{4} x^{7}-1152 \sqrt {e}\, \mathrm {log}\left (\frac {\sqrt {e \,x^{2}+d}+\sqrt {e}\, x}{\sqrt {d}}\right ) a c \,d^{3} e^{2}-1152 \sqrt {e}\, \mathrm {log}\left (\frac {\sqrt {e \,x^{2}+d}+\sqrt {e}\, x}{\sqrt {d}}\right ) a c \,d^{2} e^{3} x^{2}-840 \sqrt {e}\, \mathrm {log}\left (\frac {\sqrt {e \,x^{2}+d}+\sqrt {e}\, x}{\sqrt {d}}\right ) c^{2} d^{5}-840 \sqrt {e}\, \mathrm {log}\left (\frac {\sqrt {e \,x^{2}+d}+\sqrt {e}\, x}{\sqrt {d}}\right ) c^{2} d^{4} e \,x^{2}+384 \sqrt {e}\, a^{2} d \,e^{4}+384 \sqrt {e}\, a^{2} e^{5} x^{2}+864 \sqrt {e}\, a c \,d^{3} e^{2}+864 \sqrt {e}\, a c \,d^{2} e^{3} x^{2}+525 \sqrt {e}\, c^{2} d^{5}+525 \sqrt {e}\, c^{2} d^{4} e \,x^{2}}{384 d \,e^{5} \left (e \,x^{2}+d \right )} \] Input:

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

Output:

(384*sqrt(d + e*x**2)*a**2*e**5*x + 1152*sqrt(d + e*x**2)*a*c*d**2*e**3*x 
+ 384*sqrt(d + e*x**2)*a*c*d*e**4*x**3 + 840*sqrt(d + e*x**2)*c**2*d**4*e* 
x + 280*sqrt(d + e*x**2)*c**2*d**3*e**2*x**3 - 112*sqrt(d + e*x**2)*c**2*d 
**2*e**3*x**5 + 64*sqrt(d + e*x**2)*c**2*d*e**4*x**7 - 1152*sqrt(e)*log((s 
qrt(d + e*x**2) + sqrt(e)*x)/sqrt(d))*a*c*d**3*e**2 - 1152*sqrt(e)*log((sq 
rt(d + e*x**2) + sqrt(e)*x)/sqrt(d))*a*c*d**2*e**3*x**2 - 840*sqrt(e)*log( 
(sqrt(d + e*x**2) + sqrt(e)*x)/sqrt(d))*c**2*d**5 - 840*sqrt(e)*log((sqrt( 
d + e*x**2) + sqrt(e)*x)/sqrt(d))*c**2*d**4*e*x**2 + 384*sqrt(e)*a**2*d*e* 
*4 + 384*sqrt(e)*a**2*e**5*x**2 + 864*sqrt(e)*a*c*d**3*e**2 + 864*sqrt(e)* 
a*c*d**2*e**3*x**2 + 525*sqrt(e)*c**2*d**5 + 525*sqrt(e)*c**2*d**4*e*x**2) 
/(384*d*e**5*(d + e*x**2))