\(\int \frac {(a+c x^4)^2}{(d+e x^2)^{7/2}} \, dx\) [378]

Optimal result

Integrand size = 21, antiderivative size = 185 \[ \int \frac {\left (a+c x^4\right )^2}{\left (d+e x^2\right )^{7/2}} \, dx=\frac {\left (c d^2+a e^2\right )^2 x}{5 d e^4 \left (d+e x^2\right )^{5/2}}-\frac {4 \left (4 c d^2-a e^2\right ) \left (c d^2+a e^2\right ) x}{15 d^2 e^4 \left (d+e x^2\right )^{3/2}}+\frac {2 \left (29 c^2 d^4+3 a c d^2 e^2+4 a^2 e^4\right ) x}{15 d^3 e^4 \sqrt {d+e x^2}}+\frac {c^2 x \sqrt {d+e x^2}}{2 e^4}-\frac {7 c^2 d \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{2 e^{9/2}} \] Output:

1/5*(a*e^2+c*d^2)^2*x/d/e^4/(e*x^2+d)^(5/2)-4/15*(-a*e^2+4*c*d^2)*(a*e^2+c 
*d^2)*x/d^2/e^4/(e*x^2+d)^(3/2)+2/15*(4*a^2*e^4+3*a*c*d^2*e^2+29*c^2*d^4)* 
x/d^3/e^4/(e*x^2+d)^(1/2)+1/2*c^2*x*(e*x^2+d)^(1/2)/e^4-7/2*c^2*d*arctanh( 
e^(1/2)*x/(e*x^2+d)^(1/2))/e^(9/2)
 

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.75 \[ \int \frac {\left (a+c x^4\right )^2}{\left (d+e x^2\right )^{7/2}} \, dx=\frac {x \left (12 a c d^2 e^4 x^4+2 a^2 e^4 \left (15 d^2+20 d e x^2+8 e^2 x^4\right )+c^2 d^3 \left (105 d^3+245 d^2 e x^2+161 d e^2 x^4+15 e^3 x^6\right )\right )}{30 d^3 e^4 \left (d+e x^2\right )^{5/2}}+\frac {7 c^2 d \log \left (-\sqrt {e} x+\sqrt {d+e x^2}\right )}{2 e^{9/2}} \] Input:

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

Output:

(x*(12*a*c*d^2*e^4*x^4 + 2*a^2*e^4*(15*d^2 + 20*d*e*x^2 + 8*e^2*x^4) + c^2 
*d^3*(105*d^3 + 245*d^2*e*x^2 + 161*d*e^2*x^4 + 15*e^3*x^6)))/(30*d^3*e^4* 
(d + e*x^2)^(5/2)) + (7*c^2*d*Log[-(Sqrt[e]*x) + Sqrt[d + e*x^2]])/(2*e^(9 
/2))
 

Rubi [A] (verified)

Time = 0.96 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.06, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {1472, 25, 2345, 25, 1471, 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.

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

\(\Big \downarrow \) 1472

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2345

\(\displaystyle \frac {\frac {4 x \left (a^2-\frac {3 a c d^2}{e^2}-\frac {4 c^2 d^4}{e^4}\right )}{3 d \left (d+e x^2\right )^{3/2}}-\frac {\int -\frac {\frac {13 c^2 d^4}{e^4}-\frac {30 c^2 x^2 d^3}{e^3}+\frac {15 c^2 x^4 d^2}{e^2}+\frac {6 a c d^2}{e^2}+8 a^2}{\left (e x^2+d\right )^{3/2}}dx}{3 d}}{5 d}+\frac {x \left (a e^2+c d^2\right )^2}{5 d e^4 \left (d+e x^2\right )^{5/2}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {\frac {13 c^2 d^4}{e^4}-\frac {30 c^2 x^2 d^3}{e^3}+\frac {15 c^2 x^4 d^2}{e^2}+\frac {6 a c d^2}{e^2}+8 a^2}{\left (e x^2+d\right )^{3/2}}dx}{3 d}+\frac {4 x \left (a^2-\frac {3 a c d^2}{e^2}-\frac {4 c^2 d^4}{e^4}\right )}{3 d \left (d+e x^2\right )^{3/2}}}{5 d}+\frac {x \left (a e^2+c d^2\right )^2}{5 d e^4 \left (d+e x^2\right )^{5/2}}\)

\(\Big \downarrow \) 1471

\(\displaystyle \frac {\frac {\frac {2 x \left (4 a^2+\frac {3 a c d^2}{e^2}+\frac {29 c^2 d^4}{e^4}\right )}{d \sqrt {d+e x^2}}-\frac {\int \frac {15 c^2 d^3 \left (3 d-e x^2\right )}{e^4 \sqrt {e x^2+d}}dx}{d}}{3 d}+\frac {4 x \left (a^2-\frac {3 a c d^2}{e^2}-\frac {4 c^2 d^4}{e^4}\right )}{3 d \left (d+e x^2\right )^{3/2}}}{5 d}+\frac {x \left (a e^2+c d^2\right )^2}{5 d e^4 \left (d+e x^2\right )^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {2 x \left (4 a^2+\frac {3 a c d^2}{e^2}+\frac {29 c^2 d^4}{e^4}\right )}{d \sqrt {d+e x^2}}-\frac {15 c^2 d^2 \int \frac {3 d-e x^2}{\sqrt {e x^2+d}}dx}{e^4}}{3 d}+\frac {4 x \left (a^2-\frac {3 a c d^2}{e^2}-\frac {4 c^2 d^4}{e^4}\right )}{3 d \left (d+e x^2\right )^{3/2}}}{5 d}+\frac {x \left (a e^2+c d^2\right )^2}{5 d e^4 \left (d+e x^2\right )^{5/2}}\)

\(\Big \downarrow \) 299

\(\displaystyle \frac {\frac {\frac {2 x \left (4 a^2+\frac {3 a c d^2}{e^2}+\frac {29 c^2 d^4}{e^4}\right )}{d \sqrt {d+e x^2}}-\frac {15 c^2 d^2 \left (\frac {7}{2} d \int \frac {1}{\sqrt {e x^2+d}}dx-\frac {1}{2} x \sqrt {d+e x^2}\right )}{e^4}}{3 d}+\frac {4 x \left (a^2-\frac {3 a c d^2}{e^2}-\frac {4 c^2 d^4}{e^4}\right )}{3 d \left (d+e x^2\right )^{3/2}}}{5 d}+\frac {x \left (a e^2+c d^2\right )^2}{5 d e^4 \left (d+e x^2\right )^{5/2}}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {\frac {2 x \left (4 a^2+\frac {3 a c d^2}{e^2}+\frac {29 c^2 d^4}{e^4}\right )}{d \sqrt {d+e x^2}}-\frac {15 c^2 d^2 \left (\frac {7}{2} d \int \frac {1}{1-\frac {e x^2}{e x^2+d}}d\frac {x}{\sqrt {e x^2+d}}-\frac {1}{2} x \sqrt {d+e x^2}\right )}{e^4}}{3 d}+\frac {4 x \left (a^2-\frac {3 a c d^2}{e^2}-\frac {4 c^2 d^4}{e^4}\right )}{3 d \left (d+e x^2\right )^{3/2}}}{5 d}+\frac {x \left (a e^2+c d^2\right )^2}{5 d e^4 \left (d+e x^2\right )^{5/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\frac {2 x \left (4 a^2+\frac {3 a c d^2}{e^2}+\frac {29 c^2 d^4}{e^4}\right )}{d \sqrt {d+e x^2}}-\frac {15 c^2 d^2 \left (\frac {7 d \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{2 \sqrt {e}}-\frac {1}{2} x \sqrt {d+e x^2}\right )}{e^4}}{3 d}+\frac {4 x \left (a^2-\frac {3 a c d^2}{e^2}-\frac {4 c^2 d^4}{e^4}\right )}{3 d \left (d+e x^2\right )^{3/2}}}{5 d}+\frac {x \left (a e^2+c d^2\right )^2}{5 d e^4 \left (d+e x^2\right )^{5/2}}\)

Input:

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

Output:

((c*d^2 + a*e^2)^2*x)/(5*d*e^4*(d + e*x^2)^(5/2)) + ((4*(a^2 - (4*c^2*d^4) 
/e^4 - (3*a*c*d^2)/e^2)*x)/(3*d*(d + e*x^2)^(3/2)) + ((2*(4*a^2 + (29*c^2* 
d^4)/e^4 + (3*a*c*d^2)/e^2)*x)/(d*Sqrt[d + e*x^2]) - (15*c^2*d^2*(-1/2*(x* 
Sqrt[d + e*x^2]) + (7*d*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(2*Sqrt[e])) 
)/e^4)/(3*d))/(5*d)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 1471

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), 
x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x^2 + c*x^4)^p, d + e*x^2 
, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + 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, b, c, d, e}, 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 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 2345

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot 
ient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 
 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b 
*f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   In 
t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] 
/; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
 

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.75

method	result	size
pseudoelliptic	\(\frac {-\frac {7 \left (e \,x^{2}+d \right )^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {e \,x^{2}+d}}{x \sqrt {e}}\right ) c^{2} d^{4}}{2}+x \left (a \,d^{2} \left (\frac {2 c \,x^{4}}{5}+a \right ) e^{\frac {9}{2}}+\frac {4 e^{\frac {11}{2}} a^{2} d \,x^{2}}{3}+\frac {8 e^{\frac {13}{2}} a^{2} x^{4}}{15}+\frac {161 d^{3} \left (\frac {15 x^{6} e^{\frac {7}{2}}}{161}+e^{\frac {5}{2}} d \,x^{4}+\frac {35 e^{\frac {3}{2}} d^{2} x^{2}}{23}+\frac {15 \sqrt {e}\, d^{3}}{23}\right ) c^{2}}{30}\right )}{e^{\frac {9}{2}} \left (e \,x^{2}+d \right )^{\frac {5}{2}} d^{3}}\)	\(138\)
default	\(a^{2} \left (\frac {x}{5 d \left (e \,x^{2}+d \right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d \left (e \,x^{2}+d \right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{2} \sqrt {e \,x^{2}+d}}}{d}\right )+c^{2} \left (\frac {x^{7}}{2 e \left (e \,x^{2}+d \right )^{\frac {5}{2}}}-\frac {7 d \left (-\frac {x^{5}}{5 e \left (e \,x^{2}+d \right )^{\frac {5}{2}}}+\frac {-\frac {x^{3}}{3 e \left (e \,x^{2}+d \right )^{\frac {3}{2}}}+\frac {-\frac {x}{e \sqrt {e \,x^{2}+d}}+\frac {\ln \left (x \sqrt {e}+\sqrt {e \,x^{2}+d}\right )}{e^{\frac {3}{2}}}}{e}}{e}\right )}{2 e}\right )+2 a c \left (-\frac {x^{3}}{2 e \left (e \,x^{2}+d \right )^{\frac {5}{2}}}+\frac {3 d \left (-\frac {x}{4 e \left (e \,x^{2}+d \right )^{\frac {5}{2}}}+\frac {d \left (\frac {x}{5 d \left (e \,x^{2}+d \right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d \left (e \,x^{2}+d \right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{2} \sqrt {e \,x^{2}+d}}}{d}\right )}{4 e}\right )}{2 e}\right )\)	\(268\)
risch	\(\text {Expression too large to display}\)	\(1238\)

Input:

int((c*x^4+a)^2/(e*x^2+d)^(7/2),x,method=_RETURNVERBOSE)
 

Output:

1/e^(9/2)*(-7/2*(e*x^2+d)^(5/2)*arctanh((e*x^2+d)^(1/2)/x/e^(1/2))*c^2*d^4 
+x*(a*d^2*(2/5*c*x^4+a)*e^(9/2)+4/3*e^(11/2)*a^2*d*x^2+8/15*e^(13/2)*a^2*x 
^4+161/30*d^3*(15/161*x^6*e^(7/2)+e^(5/2)*d*x^4+35/23*e^(3/2)*d^2*x^2+15/2 
3*e^(1/2)*d^3)*c^2))/(e*x^2+d)^(5/2)/d^3
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 459, normalized size of antiderivative = 2.48 \[ \int \frac {\left (a+c x^4\right )^2}{\left (d+e x^2\right )^{7/2}} \, dx=\left [\frac {105 \, {\left (c^{2} d^{4} e^{3} x^{6} + 3 \, c^{2} d^{5} e^{2} x^{4} + 3 \, c^{2} d^{6} e x^{2} + c^{2} d^{7}\right )} \sqrt {e} \log \left (-2 \, e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {e} x - d\right ) + 2 \, {\left (15 \, c^{2} d^{3} e^{4} x^{7} + {\left (161 \, c^{2} d^{4} e^{3} + 12 \, a c d^{2} e^{5} + 16 \, a^{2} e^{7}\right )} x^{5} + 5 \, {\left (49 \, c^{2} d^{5} e^{2} + 8 \, a^{2} d e^{6}\right )} x^{3} + 15 \, {\left (7 \, c^{2} d^{6} e + 2 \, a^{2} d^{2} e^{5}\right )} x\right )} \sqrt {e x^{2} + d}}{60 \, {\left (d^{3} e^{8} x^{6} + 3 \, d^{4} e^{7} x^{4} + 3 \, d^{5} e^{6} x^{2} + d^{6} e^{5}\right )}}, \frac {105 \, {\left (c^{2} d^{4} e^{3} x^{6} + 3 \, c^{2} d^{5} e^{2} x^{4} + 3 \, c^{2} d^{6} e x^{2} + c^{2} d^{7}\right )} \sqrt {-e} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right ) + {\left (15 \, c^{2} d^{3} e^{4} x^{7} + {\left (161 \, c^{2} d^{4} e^{3} + 12 \, a c d^{2} e^{5} + 16 \, a^{2} e^{7}\right )} x^{5} + 5 \, {\left (49 \, c^{2} d^{5} e^{2} + 8 \, a^{2} d e^{6}\right )} x^{3} + 15 \, {\left (7 \, c^{2} d^{6} e + 2 \, a^{2} d^{2} e^{5}\right )} x\right )} \sqrt {e x^{2} + d}}{30 \, {\left (d^{3} e^{8} x^{6} + 3 \, d^{4} e^{7} x^{4} + 3 \, d^{5} e^{6} x^{2} + d^{6} e^{5}\right )}}\right ] \] Input:

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

Output:

[1/60*(105*(c^2*d^4*e^3*x^6 + 3*c^2*d^5*e^2*x^4 + 3*c^2*d^6*e*x^2 + c^2*d^ 
7)*sqrt(e)*log(-2*e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(e)*x - d) + 2*(15*c^2*d^3 
*e^4*x^7 + (161*c^2*d^4*e^3 + 12*a*c*d^2*e^5 + 16*a^2*e^7)*x^5 + 5*(49*c^2 
*d^5*e^2 + 8*a^2*d*e^6)*x^3 + 15*(7*c^2*d^6*e + 2*a^2*d^2*e^5)*x)*sqrt(e*x 
^2 + d))/(d^3*e^8*x^6 + 3*d^4*e^7*x^4 + 3*d^5*e^6*x^2 + d^6*e^5), 1/30*(10 
5*(c^2*d^4*e^3*x^6 + 3*c^2*d^5*e^2*x^4 + 3*c^2*d^6*e*x^2 + c^2*d^7)*sqrt(- 
e)*arctan(sqrt(-e)*x/sqrt(e*x^2 + d)) + (15*c^2*d^3*e^4*x^7 + (161*c^2*d^4 
*e^3 + 12*a*c*d^2*e^5 + 16*a^2*e^7)*x^5 + 5*(49*c^2*d^5*e^2 + 8*a^2*d*e^6) 
*x^3 + 15*(7*c^2*d^6*e + 2*a^2*d^2*e^5)*x)*sqrt(e*x^2 + d))/(d^3*e^8*x^6 + 
 3*d^4*e^7*x^4 + 3*d^5*e^6*x^2 + d^6*e^5)]
 

Sympy [F]

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

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

Output:

Integral((a + c*x**4)**2/(d + e*x**2)**(7/2), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (a+c x^4\right )^2}{\left (d+e x^2\right )^{7/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((c*x^4+a)^2/(e*x^2+d)^(7/2),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 = 0.29 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.86 \[ \int \frac {\left (a+c x^4\right )^2}{\left (d+e x^2\right )^{7/2}} \, dx=\frac {{\left ({\left ({\left (\frac {15 \, c^{2} x^{2}}{e} + \frac {161 \, c^{2} d^{4} e^{5} + 12 \, a c d^{2} e^{7} + 16 \, a^{2} e^{9}}{d^{3} e^{7}}\right )} x^{2} + \frac {5 \, {\left (49 \, c^{2} d^{5} e^{4} + 8 \, a^{2} d e^{8}\right )}}{d^{3} e^{7}}\right )} x^{2} + \frac {15 \, {\left (7 \, c^{2} d^{6} e^{3} + 2 \, a^{2} d^{2} e^{7}\right )}}{d^{3} e^{7}}\right )} x}{30 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}}} + \frac {7 \, c^{2} d \log \left ({\left | -\sqrt {e} x + \sqrt {e x^{2} + d} \right |}\right )}{2 \, e^{\frac {9}{2}}} \] Input:

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

Output:

1/30*(((15*c^2*x^2/e + (161*c^2*d^4*e^5 + 12*a*c*d^2*e^7 + 16*a^2*e^9)/(d^ 
3*e^7))*x^2 + 5*(49*c^2*d^5*e^4 + 8*a^2*d*e^8)/(d^3*e^7))*x^2 + 15*(7*c^2* 
d^6*e^3 + 2*a^2*d^2*e^7)/(d^3*e^7))*x/(e*x^2 + d)^(5/2) + 7/2*c^2*d*log(ab 
s(-sqrt(e)*x + sqrt(e*x^2 + d)))/e^(9/2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 505, normalized size of antiderivative = 2.73 \[ \int \frac {\left (a+c x^4\right )^2}{\left (d+e x^2\right )^{7/2}} \, dx=\frac {120 \sqrt {e \,x^{2}+d}\, a^{2} d^{2} e^{5} x +160 \sqrt {e \,x^{2}+d}\, a^{2} d \,e^{6} x^{3}+64 \sqrt {e \,x^{2}+d}\, a^{2} e^{7} x^{5}+48 \sqrt {e \,x^{2}+d}\, a c \,d^{2} e^{5} x^{5}+420 \sqrt {e \,x^{2}+d}\, c^{2} d^{6} e x +980 \sqrt {e \,x^{2}+d}\, c^{2} d^{5} e^{2} x^{3}+644 \sqrt {e \,x^{2}+d}\, c^{2} d^{4} e^{3} x^{5}+60 \sqrt {e \,x^{2}+d}\, c^{2} d^{3} e^{4} x^{7}-420 \sqrt {e}\, \mathrm {log}\left (\frac {\sqrt {e \,x^{2}+d}+\sqrt {e}\, x}{\sqrt {d}}\right ) c^{2} d^{7}-1260 \sqrt {e}\, \mathrm {log}\left (\frac {\sqrt {e \,x^{2}+d}+\sqrt {e}\, x}{\sqrt {d}}\right ) c^{2} d^{6} e \,x^{2}-1260 \sqrt {e}\, \mathrm {log}\left (\frac {\sqrt {e \,x^{2}+d}+\sqrt {e}\, x}{\sqrt {d}}\right ) c^{2} d^{5} e^{2} x^{4}-420 \sqrt {e}\, \mathrm {log}\left (\frac {\sqrt {e \,x^{2}+d}+\sqrt {e}\, x}{\sqrt {d}}\right ) c^{2} d^{4} e^{3} x^{6}-64 \sqrt {e}\, a^{2} d^{3} e^{4}-192 \sqrt {e}\, a^{2} d^{2} e^{5} x^{2}-192 \sqrt {e}\, a^{2} d \,e^{6} x^{4}-64 \sqrt {e}\, a^{2} e^{7} x^{6}+48 \sqrt {e}\, a c \,d^{5} e^{2}+144 \sqrt {e}\, a c \,d^{4} e^{3} x^{2}+144 \sqrt {e}\, a c \,d^{3} e^{4} x^{4}+48 \sqrt {e}\, a c \,d^{2} e^{5} x^{6}-203 \sqrt {e}\, c^{2} d^{7}-609 \sqrt {e}\, c^{2} d^{6} e \,x^{2}-609 \sqrt {e}\, c^{2} d^{5} e^{2} x^{4}-203 \sqrt {e}\, c^{2} d^{4} e^{3} x^{6}}{120 d^{3} e^{5} \left (e^{3} x^{6}+3 d \,e^{2} x^{4}+3 d^{2} e \,x^{2}+d^{3}\right )} \] Input:

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

Output:

(120*sqrt(d + e*x**2)*a**2*d**2*e**5*x + 160*sqrt(d + e*x**2)*a**2*d*e**6* 
x**3 + 64*sqrt(d + e*x**2)*a**2*e**7*x**5 + 48*sqrt(d + e*x**2)*a*c*d**2*e 
**5*x**5 + 420*sqrt(d + e*x**2)*c**2*d**6*e*x + 980*sqrt(d + e*x**2)*c**2* 
d**5*e**2*x**3 + 644*sqrt(d + e*x**2)*c**2*d**4*e**3*x**5 + 60*sqrt(d + e* 
x**2)*c**2*d**3*e**4*x**7 - 420*sqrt(e)*log((sqrt(d + e*x**2) + sqrt(e)*x) 
/sqrt(d))*c**2*d**7 - 1260*sqrt(e)*log((sqrt(d + e*x**2) + sqrt(e)*x)/sqrt 
(d))*c**2*d**6*e*x**2 - 1260*sqrt(e)*log((sqrt(d + e*x**2) + sqrt(e)*x)/sq 
rt(d))*c**2*d**5*e**2*x**4 - 420*sqrt(e)*log((sqrt(d + e*x**2) + sqrt(e)*x 
)/sqrt(d))*c**2*d**4*e**3*x**6 - 64*sqrt(e)*a**2*d**3*e**4 - 192*sqrt(e)*a 
**2*d**2*e**5*x**2 - 192*sqrt(e)*a**2*d*e**6*x**4 - 64*sqrt(e)*a**2*e**7*x 
**6 + 48*sqrt(e)*a*c*d**5*e**2 + 144*sqrt(e)*a*c*d**4*e**3*x**2 + 144*sqrt 
(e)*a*c*d**3*e**4*x**4 + 48*sqrt(e)*a*c*d**2*e**5*x**6 - 203*sqrt(e)*c**2* 
d**7 - 609*sqrt(e)*c**2*d**6*e*x**2 - 609*sqrt(e)*c**2*d**5*e**2*x**4 - 20 
3*sqrt(e)*c**2*d**4*e**3*x**6)/(120*d**3*e**5*(d**3 + 3*d**2*e*x**2 + 3*d* 
e**2*x**4 + e**3*x**6))