\(\int \frac {x^4}{(c+d x)^2 (a+b x^2)^{5/2}} \, dx\) [1288]

Optimal result

Integrand size = 22, antiderivative size = 228 \[ \int \frac {x^4}{(c+d x)^2 \left (a+b x^2\right )^{5/2}} \, dx=\frac {a \left (2 a c d+\left (b c^2-a d^2\right ) x\right )}{3 b \left (b c^2+a d^2\right )^2 \left (a+b x^2\right )^{3/2}}-\frac {12 a b c^3 d+\left (4 b^2 c^4-9 a b c^2 d^2-a^2 d^4\right ) x}{3 b \left (b c^2+a d^2\right )^3 \sqrt {a+b x^2}}-\frac {c^4 d \sqrt {a+b x^2}}{\left (b c^2+a d^2\right )^3 (c+d x)}-\frac {c^3 \left (b c^2-4 a d^2\right ) \text {arctanh}\left (\frac {a d-b c x}{\sqrt {b c^2+a d^2} \sqrt {a+b x^2}}\right )}{\left (b c^2+a d^2\right )^{7/2}} \] Output:

1/3*a*(2*a*c*d+(-a*d^2+b*c^2)*x)/b/(a*d^2+b*c^2)^2/(b*x^2+a)^(3/2)-1/3*(12 
*a*b*c^3*d+(-a^2*d^4-9*a*b*c^2*d^2+4*b^2*c^4)*x)/b/(a*d^2+b*c^2)^3/(b*x^2+ 
a)^(1/2)-c^4*d*(b*x^2+a)^(1/2)/(a*d^2+b*c^2)^3/(d*x+c)-c^3*(-4*a*d^2+b*c^2 
)*arctanh((-b*c*x+a*d)/(a*d^2+b*c^2)^(1/2)/(b*x^2+a)^(1/2))/(a*d^2+b*c^2)^ 
(7/2)
 

Mathematica [A] (verified)

Time = 10.37 (sec) , antiderivative size = 224, normalized size of antiderivative = 0.98 \[ \int \frac {x^4}{(c+d x)^2 \left (a+b x^2\right )^{5/2}} \, dx=\frac {2 a^3 c d^3 (c+d x)-b^3 c^4 x^3 (4 c+7 d x)-3 a b^2 c^2 x \left (c^3+7 c^2 d x+c d^2 x^2-3 d^3 x^3\right )+a^2 b d \left (-13 c^4-c^3 d x+9 c^2 d^2 x^2+c d^3 x^3+d^4 x^4\right )}{3 b \left (b c^2+a d^2\right )^3 (c+d x) \left (a+b x^2\right )^{3/2}}-\frac {c^3 \left (b c^2-4 a d^2\right ) \text {arctanh}\left (\frac {a d-b c x}{\sqrt {b c^2+a d^2} \sqrt {a+b x^2}}\right )}{\left (b c^2+a d^2\right )^{7/2}} \] Input:

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

Output:

(2*a^3*c*d^3*(c + d*x) - b^3*c^4*x^3*(4*c + 7*d*x) - 3*a*b^2*c^2*x*(c^3 + 
7*c^2*d*x + c*d^2*x^2 - 3*d^3*x^3) + a^2*b*d*(-13*c^4 - c^3*d*x + 9*c^2*d^ 
2*x^2 + c*d^3*x^3 + d^4*x^4))/(3*b*(b*c^2 + a*d^2)^3*(c + d*x)*(a + b*x^2) 
^(3/2)) - (c^3*(b*c^2 - 4*a*d^2)*ArcTanh[(a*d - b*c*x)/(Sqrt[b*c^2 + a*d^2 
]*Sqrt[a + b*x^2])])/(b*c^2 + a*d^2)^(7/2)
 

Rubi [A] (verified)

Time = 1.47 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {601, 2178, 27, 679, 488, 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.

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

\(\Big \downarrow \) 601

\(\displaystyle \frac {a \left (x \left (b c^2-a d^2\right )+2 a c d\right )}{3 b \left (a+b x^2\right )^{3/2} \left (a d^2+b c^2\right )^2}-\frac {\int \frac {-\frac {2 c d \left (2 b c^2+a d^2\right ) x a^2}{b \left (b c^2+a d^2\right )^2}+\frac {c^2 \left (b c^2-a d^2\right ) a^2}{b \left (b c^2+a d^2\right )^2}-\frac {\left (3 b^2 c^4+8 a b d^2 c^2+a^2 d^4\right ) x^2 a}{b \left (b c^2+a d^2\right )^2}}{(c+d x)^2 \left (b x^2+a\right )^{3/2}}dx}{3 a}\)

\(\Big \downarrow \) 2178

\(\displaystyle \frac {a \left (x \left (b c^2-a d^2\right )+2 a c d\right )}{3 b \left (a+b x^2\right )^{3/2} \left (a d^2+b c^2\right )^2}-\frac {\frac {a \left (x \left (-a^2 d^4-9 a b c^2 d^2+4 b^2 c^4\right )+12 a b c^3 d\right )}{b \sqrt {a+b x^2} \left (a d^2+b c^2\right )^3}-\frac {\int \frac {3 a^2 b c^3 \left (c \left (b c^2-3 a d^2\right )-4 a d^3 x\right )}{\left (b c^2+a d^2\right )^3 (c+d x)^2 \sqrt {b x^2+a}}dx}{a b}}{3 a}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {a \left (x \left (b c^2-a d^2\right )+2 a c d\right )}{3 b \left (a+b x^2\right )^{3/2} \left (a d^2+b c^2\right )^2}-\frac {\frac {a \left (x \left (-a^2 d^4-9 a b c^2 d^2+4 b^2 c^4\right )+12 a b c^3 d\right )}{b \sqrt {a+b x^2} \left (a d^2+b c^2\right )^3}-\frac {3 a c^3 \int \frac {c \left (b c^2-3 a d^2\right )-4 a d^3 x}{(c+d x)^2 \sqrt {b x^2+a}}dx}{\left (a d^2+b c^2\right )^3}}{3 a}\)

\(\Big \downarrow \) 679

\(\displaystyle \frac {a \left (x \left (b c^2-a d^2\right )+2 a c d\right )}{3 b \left (a+b x^2\right )^{3/2} \left (a d^2+b c^2\right )^2}-\frac {\frac {a \left (x \left (-a^2 d^4-9 a b c^2 d^2+4 b^2 c^4\right )+12 a b c^3 d\right )}{b \sqrt {a+b x^2} \left (a d^2+b c^2\right )^3}-\frac {3 a c^3 \left (\left (b c^2-4 a d^2\right ) \int \frac {1}{(c+d x) \sqrt {b x^2+a}}dx-\frac {c d \sqrt {a+b x^2}}{c+d x}\right )}{\left (a d^2+b c^2\right )^3}}{3 a}\)

\(\Big \downarrow \) 488

\(\displaystyle \frac {a \left (x \left (b c^2-a d^2\right )+2 a c d\right )}{3 b \left (a+b x^2\right )^{3/2} \left (a d^2+b c^2\right )^2}-\frac {\frac {a \left (x \left (-a^2 d^4-9 a b c^2 d^2+4 b^2 c^4\right )+12 a b c^3 d\right )}{b \sqrt {a+b x^2} \left (a d^2+b c^2\right )^3}-\frac {3 a c^3 \left (-\left (b c^2-4 a d^2\right ) \int \frac {1}{b c^2+a d^2-\frac {(a d-b c x)^2}{b x^2+a}}d\frac {a d-b c x}{\sqrt {b x^2+a}}-\frac {c d \sqrt {a+b x^2}}{c+d x}\right )}{\left (a d^2+b c^2\right )^3}}{3 a}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {a \left (x \left (b c^2-a d^2\right )+2 a c d\right )}{3 b \left (a+b x^2\right )^{3/2} \left (a d^2+b c^2\right )^2}-\frac {\frac {a \left (x \left (-a^2 d^4-9 a b c^2 d^2+4 b^2 c^4\right )+12 a b c^3 d\right )}{b \sqrt {a+b x^2} \left (a d^2+b c^2\right )^3}-\frac {3 a c^3 \left (-\frac {\left (b c^2-4 a d^2\right ) \text {arctanh}\left (\frac {a d-b c x}{\sqrt {a+b x^2} \sqrt {a d^2+b c^2}}\right )}{\sqrt {a d^2+b c^2}}-\frac {c d \sqrt {a+b x^2}}{c+d x}\right )}{\left (a d^2+b c^2\right )^3}}{3 a}\)

Input:

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

Output:

(a*(2*a*c*d + (b*c^2 - a*d^2)*x))/(3*b*(b*c^2 + a*d^2)^2*(a + b*x^2)^(3/2) 
) - ((a*(12*a*b*c^3*d + (4*b^2*c^4 - 9*a*b*c^2*d^2 - a^2*d^4)*x))/(b*(b*c^ 
2 + a*d^2)^3*Sqrt[a + b*x^2]) - (3*a*c^3*(-((c*d*Sqrt[a + b*x^2])/(c + d*x 
)) - ((b*c^2 - 4*a*d^2)*ArcTanh[(a*d - b*c*x)/(Sqrt[b*c^2 + a*d^2]*Sqrt[a 
+ b*x^2])])/Sqrt[b*c^2 + a*d^2]))/(b*c^2 + a*d^2)^3)/(3*a)
 

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 488

Int[1/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> -Subst[ 
Int[1/(b*c^2 + a*d^2 - x^2), x], x, (a*d - b*c*x)/Sqrt[a + b*x^2]] /; FreeQ 
[{a, b, c, d}, x]
 

rule 601

Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Coe 
ff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Pol 
ynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(a*f - b*e*x) 
*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   Int[(c 
+ d*x)^n*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Qx)/(c + d*x)^n + (e* 
(2*p + 3))/(c + d*x)^n, x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 1] 
 && LtQ[p, -1] && ILtQ[n, 0] && NeQ[b*c^2 + a*d^2, 0]
 

rule 679

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p 
_.), x_Symbol] :> Simp[(-(e*f - d*g))*(d + e*x)^(m + 1)*((a + c*x^2)^(p + 1 
)/(2*(p + 1)*(c*d^2 + a*e^2))), x] + Simp[(c*d*f + a*e*g)/(c*d^2 + a*e^2) 
 Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, 
 p}, x] && EqQ[Simplify[m + 2*p + 3], 0]
 

rule 2178

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : 
> With[{Qx = PolynomialQuotient[(d + e*x)^m*Pq, a + b*x^2, x], R = Coeff[Po 
lynomialRemainder[(d + e*x)^m*Pq, a + b*x^2, x], x, 0], S = Coeff[Polynomia 
lRemainder[(d + e*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[(a*S - b*R*x)*((a + 
b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*b*(p + 1))   Int[(d + e*x 
)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*b*(p + 1)*Qx)/(d + e*x)^m + (b*R*( 
2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, b, d, e}, x] && PolyQ[Pq, x 
] && NeQ[b*d^2 + a*e^2, 0] && LtQ[p, -1] && ILtQ[m, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1620\) vs. \(2(212)=424\).

Time = 0.39 (sec) , antiderivative size = 1621, normalized size of antiderivative = 7.11

method	result	size
default	\(\text {Expression too large to display}\)	\(1621\)

Input:

int(x^4/(d*x+c)^2/(b*x^2+a)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/d^2*(-1/2*x/b/(b*x^2+a)^(3/2)+1/2*a/b*(1/3*x/a/(b*x^2+a)^(3/2)+2/3/a^2/( 
b*x^2+a)^(1/2)*x))+c^4/d^6*(-1/(a*d^2+b*c^2)*d^2/(x+c/d)/(b*(x+c/d)^2-2*b* 
c/d*(x+c/d)+(a*d^2+b*c^2)/d^2)^(3/2)+5*b*c*d/(a*d^2+b*c^2)*(1/3/(a*d^2+b*c 
^2)*d^2/(b*(x+c/d)^2-2*b*c/d*(x+c/d)+(a*d^2+b*c^2)/d^2)^(3/2)+b*c*d/(a*d^2 
+b*c^2)*(2/3*(2*b*(x+c/d)-2*b*c/d)/(4*b*(a*d^2+b*c^2)/d^2-4*b^2*c^2/d^2)/( 
b*(x+c/d)^2-2*b*c/d*(x+c/d)+(a*d^2+b*c^2)/d^2)^(3/2)+16/3*b/(4*b*(a*d^2+b* 
c^2)/d^2-4*b^2*c^2/d^2)^2*(2*b*(x+c/d)-2*b*c/d)/(b*(x+c/d)^2-2*b*c/d*(x+c/ 
d)+(a*d^2+b*c^2)/d^2)^(1/2))+1/(a*d^2+b*c^2)*d^2*(1/(a*d^2+b*c^2)*d^2/(b*( 
x+c/d)^2-2*b*c/d*(x+c/d)+(a*d^2+b*c^2)/d^2)^(1/2)+2*b*c*d/(a*d^2+b*c^2)*(2 
*b*(x+c/d)-2*b*c/d)/(4*b*(a*d^2+b*c^2)/d^2-4*b^2*c^2/d^2)/(b*(x+c/d)^2-2*b 
*c/d*(x+c/d)+(a*d^2+b*c^2)/d^2)^(1/2)-1/(a*d^2+b*c^2)*d^2/((a*d^2+b*c^2)/d 
^2)^(1/2)*ln((2*(a*d^2+b*c^2)/d^2-2*b*c/d*(x+c/d)+2*((a*d^2+b*c^2)/d^2)^(1 
/2)*(b*(x+c/d)^2-2*b*c/d*(x+c/d)+(a*d^2+b*c^2)/d^2)^(1/2))/(x+c/d))))-4*b/ 
(a*d^2+b*c^2)*d^2*(2/3*(2*b*(x+c/d)-2*b*c/d)/(4*b*(a*d^2+b*c^2)/d^2-4*b^2* 
c^2/d^2)/(b*(x+c/d)^2-2*b*c/d*(x+c/d)+(a*d^2+b*c^2)/d^2)^(3/2)+16/3*b/(4*b 
*(a*d^2+b*c^2)/d^2-4*b^2*c^2/d^2)^2*(2*b*(x+c/d)-2*b*c/d)/(b*(x+c/d)^2-2*b 
*c/d*(x+c/d)+(a*d^2+b*c^2)/d^2)^(1/2)))+3*c^2/d^4*(1/3*x/a/(b*x^2+a)^(3/2) 
+2/3/a^2/(b*x^2+a)^(1/2)*x)+2/3*c/d^3/b/(b*x^2+a)^(3/2)-4/d^5*c^3*(1/3/(a* 
d^2+b*c^2)*d^2/(b*(x+c/d)^2-2*b*c/d*(x+c/d)+(a*d^2+b*c^2)/d^2)^(3/2)+b*c*d 
/(a*d^2+b*c^2)*(2/3*(2*b*(x+c/d)-2*b*c/d)/(4*b*(a*d^2+b*c^2)/d^2-4*b^2*...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 865 vs. \(2 (213) = 426\).

Time = 0.37 (sec) , antiderivative size = 1757, normalized size of antiderivative = 7.71 \[ \int \frac {x^4}{(c+d x)^2 \left (a+b x^2\right )^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

[-1/6*(3*(a^2*b^2*c^6 - 4*a^3*b*c^4*d^2 + (b^4*c^5*d - 4*a*b^3*c^3*d^3)*x^ 
5 + (b^4*c^6 - 4*a*b^3*c^4*d^2)*x^4 + 2*(a*b^3*c^5*d - 4*a^2*b^2*c^3*d^3)* 
x^3 + 2*(a*b^3*c^6 - 4*a^2*b^2*c^4*d^2)*x^2 + (a^2*b^2*c^5*d - 4*a^3*b*c^3 
*d^3)*x)*sqrt(b*c^2 + a*d^2)*log((2*a*b*c*d*x - a*b*c^2 - 2*a^2*d^2 - (2*b 
^2*c^2 + a*b*d^2)*x^2 + 2*sqrt(b*c^2 + a*d^2)*(b*c*x - a*d)*sqrt(b*x^2 + a 
))/(d^2*x^2 + 2*c*d*x + c^2)) + 2*(13*a^2*b^2*c^6*d + 11*a^3*b*c^4*d^3 - 2 
*a^4*c^2*d^5 + (7*b^4*c^6*d - 2*a*b^3*c^4*d^3 - 10*a^2*b^2*c^2*d^5 - a^3*b 
*d^7)*x^4 + (4*b^4*c^7 + 7*a*b^3*c^5*d^2 + 2*a^2*b^2*c^3*d^4 - a^3*b*c*d^6 
)*x^3 + 3*(7*a*b^3*c^6*d + 4*a^2*b^2*c^4*d^3 - 3*a^3*b*c^2*d^5)*x^2 + (3*a 
*b^3*c^7 + 4*a^2*b^2*c^5*d^2 - a^3*b*c^3*d^4 - 2*a^4*c*d^6)*x)*sqrt(b*x^2 
+ a))/(a^2*b^5*c^9 + 4*a^3*b^4*c^7*d^2 + 6*a^4*b^3*c^5*d^4 + 4*a^5*b^2*c^3 
*d^6 + a^6*b*c*d^8 + (b^7*c^8*d + 4*a*b^6*c^6*d^3 + 6*a^2*b^5*c^4*d^5 + 4* 
a^3*b^4*c^2*d^7 + a^4*b^3*d^9)*x^5 + (b^7*c^9 + 4*a*b^6*c^7*d^2 + 6*a^2*b^ 
5*c^5*d^4 + 4*a^3*b^4*c^3*d^6 + a^4*b^3*c*d^8)*x^4 + 2*(a*b^6*c^8*d + 4*a^ 
2*b^5*c^6*d^3 + 6*a^3*b^4*c^4*d^5 + 4*a^4*b^3*c^2*d^7 + a^5*b^2*d^9)*x^3 + 
 2*(a*b^6*c^9 + 4*a^2*b^5*c^7*d^2 + 6*a^3*b^4*c^5*d^4 + 4*a^4*b^3*c^3*d^6 
+ a^5*b^2*c*d^8)*x^2 + (a^2*b^5*c^8*d + 4*a^3*b^4*c^6*d^3 + 6*a^4*b^3*c^4* 
d^5 + 4*a^5*b^2*c^2*d^7 + a^6*b*d^9)*x), -1/3*(3*(a^2*b^2*c^6 - 4*a^3*b*c^ 
4*d^2 + (b^4*c^5*d - 4*a*b^3*c^3*d^3)*x^5 + (b^4*c^6 - 4*a*b^3*c^4*d^2)*x^ 
4 + 2*(a*b^3*c^5*d - 4*a^2*b^2*c^3*d^3)*x^3 + 2*(a*b^3*c^6 - 4*a^2*b^2*...
 

Sympy [F]

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

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

Output:

Integral(x**4/((a + b*x**2)**(5/2)*(c + d*x)**2), x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 920 vs. \(2 (213) = 426\).

Time = 0.15 (sec) , antiderivative size = 920, normalized size of antiderivative = 4.04 \[ \int \frac {x^4}{(c+d x)^2 \left (a+b x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

5*b^2*c^6*x/(sqrt(b*x^2 + a)*a*b^3*c^6*d^2 + 3*sqrt(b*x^2 + a)*a^2*b^2*c^4 
*d^4 + 3*sqrt(b*x^2 + a)*a^3*b*c^2*d^6 + sqrt(b*x^2 + a)*a^4*d^8) + 5/3*b^ 
2*c^6*x/((b*x^2 + a)^(3/2)*a*b^2*c^4*d^4 + 2*(b*x^2 + a)^(3/2)*a^2*b*c^2*d 
^6 + (b*x^2 + a)^(3/2)*a^3*d^8) + 10/3*b^2*c^6*x/(sqrt(b*x^2 + a)*a^2*b^2* 
c^4*d^4 + 2*sqrt(b*x^2 + a)*a^3*b*c^2*d^6 + sqrt(b*x^2 + a)*a^4*d^8) + 5*b 
*c^5/(sqrt(b*x^2 + a)*b^3*c^6*d + 3*sqrt(b*x^2 + a)*a*b^2*c^4*d^3 + 3*sqrt 
(b*x^2 + a)*a^2*b*c^2*d^5 + sqrt(b*x^2 + a)*a^3*d^7) + 5/3*b*c^5/((b*x^2 + 
 a)^(3/2)*b^2*c^4*d^3 + 2*(b*x^2 + a)^(3/2)*a*b*c^2*d^5 + (b*x^2 + a)^(3/2 
)*a^2*d^7) - 4*b*c^4*x/(sqrt(b*x^2 + a)*a*b^2*c^4*d^2 + 2*sqrt(b*x^2 + a)* 
a^2*b*c^2*d^4 + sqrt(b*x^2 + a)*a^3*d^6) - 8/3*b*c^4*x/((b*x^2 + a)^(3/2)* 
a*b*c^2*d^4 + (b*x^2 + a)^(3/2)*a^2*d^6) - 16/3*b*c^4*x/(sqrt(b*x^2 + a)*a 
^2*b*c^2*d^4 + sqrt(b*x^2 + a)*a^3*d^6) - c^4/((b*x^2 + a)^(3/2)*b*c^2*d^4 
*x + (b*x^2 + a)^(3/2)*a*d^6*x + (b*x^2 + a)^(3/2)*b*c^3*d^3 + (b*x^2 + a) 
^(3/2)*a*c*d^5) - 4*c^3/(sqrt(b*x^2 + a)*b^2*c^4*d + 2*sqrt(b*x^2 + a)*a*b 
*c^2*d^3 + sqrt(b*x^2 + a)*a^2*d^5) - 4/3*c^3/((b*x^2 + a)^(3/2)*b*c^2*d^3 
 + (b*x^2 + a)^(3/2)*a*d^5) + 2*c^2*x/(sqrt(b*x^2 + a)*a^2*d^4) + c^2*x/(( 
b*x^2 + a)^(3/2)*a*d^4) - 1/3*x/((b*x^2 + a)^(3/2)*b*d^2) + 1/3*x/(sqrt(b* 
x^2 + a)*a*b*d^2) + 5*b*c^5*arcsinh(b*c*x/(sqrt(a*b)*abs(d*x + c)) - a*d/( 
sqrt(a*b)*abs(d*x + c)))/((a + b*c^2/d^2)^(7/2)*d^7) - 4*c^3*arcsinh(b*c*x 
/(sqrt(a*b)*abs(d*x + c)) - a*d/(sqrt(a*b)*abs(d*x + c)))/((a + b*c^2/d...
 

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 1875, normalized size of antiderivative = 8.22 \[ \int \frac {x^4}{(c+d x)^2 \left (a+b x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

(12*sqrt(a*d**2 + b*c**2)*log( - sqrt(a + b*x**2)*sqrt(a*d**2 + b*c**2) - 
a*d + b*c*x)*a**3*b*c**4*d**2 + 12*sqrt(a*d**2 + b*c**2)*log( - sqrt(a + b 
*x**2)*sqrt(a*d**2 + b*c**2) - a*d + b*c*x)*a**3*b*c**3*d**3*x - 3*sqrt(a* 
d**2 + b*c**2)*log( - sqrt(a + b*x**2)*sqrt(a*d**2 + b*c**2) - a*d + b*c*x 
)*a**2*b**2*c**6 - 3*sqrt(a*d**2 + b*c**2)*log( - sqrt(a + b*x**2)*sqrt(a* 
d**2 + b*c**2) - a*d + b*c*x)*a**2*b**2*c**5*d*x + 24*sqrt(a*d**2 + b*c**2 
)*log( - sqrt(a + b*x**2)*sqrt(a*d**2 + b*c**2) - a*d + b*c*x)*a**2*b**2*c 
**4*d**2*x**2 + 24*sqrt(a*d**2 + b*c**2)*log( - sqrt(a + b*x**2)*sqrt(a*d* 
*2 + b*c**2) - a*d + b*c*x)*a**2*b**2*c**3*d**3*x**3 - 6*sqrt(a*d**2 + b*c 
**2)*log( - sqrt(a + b*x**2)*sqrt(a*d**2 + b*c**2) - a*d + b*c*x)*a*b**3*c 
**6*x**2 - 6*sqrt(a*d**2 + b*c**2)*log( - sqrt(a + b*x**2)*sqrt(a*d**2 + b 
*c**2) - a*d + b*c*x)*a*b**3*c**5*d*x**3 + 12*sqrt(a*d**2 + b*c**2)*log( - 
 sqrt(a + b*x**2)*sqrt(a*d**2 + b*c**2) - a*d + b*c*x)*a*b**3*c**4*d**2*x* 
*4 + 12*sqrt(a*d**2 + b*c**2)*log( - sqrt(a + b*x**2)*sqrt(a*d**2 + b*c**2 
) - a*d + b*c*x)*a*b**3*c**3*d**3*x**5 - 3*sqrt(a*d**2 + b*c**2)*log( - sq 
rt(a + b*x**2)*sqrt(a*d**2 + b*c**2) - a*d + b*c*x)*b**4*c**6*x**4 - 3*sqr 
t(a*d**2 + b*c**2)*log( - sqrt(a + b*x**2)*sqrt(a*d**2 + b*c**2) - a*d + b 
*c*x)*b**4*c**5*d*x**5 - 12*sqrt(a*d**2 + b*c**2)*log(c + d*x)*a**3*b*c**4 
*d**2 - 12*sqrt(a*d**2 + b*c**2)*log(c + d*x)*a**3*b*c**3*d**3*x + 3*sqrt( 
a*d**2 + b*c**2)*log(c + d*x)*a**2*b**2*c**6 + 3*sqrt(a*d**2 + b*c**2)*...