\(\int \frac {x^4}{(c+d x)^{3/2} (a-b x^2)^2} \, dx\) [671]

Optimal result

Integrand size = 23, antiderivative size = 263 \[ \int \frac {x^4}{(c+d x)^{3/2} \left (a-b x^2\right )^2} \, dx=-\frac {2 c^4}{d \left (b c^2-a d^2\right )^2 \sqrt {c+d x}}+\frac {a \sqrt {c+d x} \left (b c^2 x-a d (2 c-d x)\right )}{2 b \left (b c^2-a d^2\right )^2 \left (a-b x^2\right )}+\frac {3 \sqrt {a} \left (2 \sqrt {b} c-\sqrt {a} d\right ) \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {c+d x}}{\sqrt {\sqrt {b} c-\sqrt {a} d}}\right )}{4 b^{7/4} \left (\sqrt {b} c-\sqrt {a} d\right )^{5/2}}-\frac {3 \sqrt {a} \left (2 \sqrt {b} c+\sqrt {a} d\right ) \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {c+d x}}{\sqrt {\sqrt {b} c+\sqrt {a} d}}\right )}{4 b^{7/4} \left (\sqrt {b} c+\sqrt {a} d\right )^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 1.77 (sec) , antiderivative size = 325, normalized size of antiderivative = 1.24 \[ \int \frac {x^4}{(c+d x)^{3/2} \left (a-b x^2\right )^2} \, dx=\frac {-\frac {2 b \left (4 b^2 c^4 x^2+a^2 d^2 \left (-2 c^2-c d x+d^2 x^2\right )+a b c^2 \left (-4 c^2+c d x+d^2 x^2\right )\right )}{d \left (b c^2-a d^2\right )^2 \sqrt {c+d x} \left (-a+b x^2\right )}+\frac {3 \left (2 \sqrt {a} \sqrt {b} c+a d\right ) \sqrt {-b c-\sqrt {a} \sqrt {b} d} \arctan \left (\frac {\sqrt {-b c-\sqrt {a} \sqrt {b} d} \sqrt {c+d x}}{\sqrt {b} c+\sqrt {a} d}\right )}{\left (\sqrt {b} c+\sqrt {a} d\right )^3}-\frac {3 \left (2 \sqrt {a} \sqrt {b} c-a d\right ) \sqrt {-b c+\sqrt {a} \sqrt {b} d} \arctan \left (\frac {\sqrt {-b c+\sqrt {a} \sqrt {b} d} \sqrt {c+d x}}{\sqrt {b} c-\sqrt {a} d}\right )}{\left (\sqrt {b} c-\sqrt {a} d\right )^3}}{4 b^2} \] Input:

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

Output:

((-2*b*(4*b^2*c^4*x^2 + a^2*d^2*(-2*c^2 - c*d*x + d^2*x^2) + a*b*c^2*(-4*c 
^2 + c*d*x + d^2*x^2)))/(d*(b*c^2 - a*d^2)^2*Sqrt[c + d*x]*(-a + b*x^2)) + 
 (3*(2*Sqrt[a]*Sqrt[b]*c + a*d)*Sqrt[-(b*c) - Sqrt[a]*Sqrt[b]*d]*ArcTan[(S 
qrt[-(b*c) - Sqrt[a]*Sqrt[b]*d]*Sqrt[c + d*x])/(Sqrt[b]*c + Sqrt[a]*d)])/( 
Sqrt[b]*c + Sqrt[a]*d)^3 - (3*(2*Sqrt[a]*Sqrt[b]*c - a*d)*Sqrt[-(b*c) + Sq 
rt[a]*Sqrt[b]*d]*ArcTan[(Sqrt[-(b*c) + Sqrt[a]*Sqrt[b]*d]*Sqrt[c + d*x])/( 
Sqrt[b]*c - Sqrt[a]*d)])/(Sqrt[b]*c - Sqrt[a]*d)^3)/(4*b^2)
 

Rubi [A] (verified)

Time = 1.98 (sec) , antiderivative size = 401, normalized size of antiderivative = 1.52, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {561, 27, 1673, 27, 2195, 2009}

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.

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

\(\Big \downarrow \) 561

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1673

\(\displaystyle \frac {2 \left (\frac {d^4 \int -\frac {2 \left (\frac {4 a b c^4}{d^2}-\frac {a \left (8 b^2 c^4-15 a b d^2 c^2+3 a^2 d^4\right ) (c+d x) c}{d^2 \left (b c^2-a d^2\right )}+\frac {a \left (4 b^2 c^4-9 a b d^2 c^2+3 a^2 d^4\right ) (c+d x)^2}{d^2 \left (b c^2-a d^2\right )}\right )}{(c+d x) \left (-\frac {b c^2}{d^2}+\frac {2 b (c+d x) c}{d^2}-\frac {b (c+d x)^2}{d^2}+a\right )}d\sqrt {c+d x}}{8 a b \left (b c^2-a d^2\right )}-\frac {a d^4 \sqrt {c+d x} \left (\frac {c \left (3 a d^2+b c^2\right )}{b c^2-a d^2}-\frac {(c+d x) \left (a d^2+b c^2\right )}{b c^2-a d^2}\right )}{4 b \left (b c^2-a d^2\right ) \left (a-\frac {b c^2}{d^2}+\frac {2 b c (c+d x)}{d^2}-\frac {b (c+d x)^2}{d^2}\right )}\right )}{d^5}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 \left (-\frac {d^4 \int \frac {\frac {4 a b c^4}{d^2}-\frac {a \left (8 b^2 c^4-15 a b d^2 c^2+3 a^2 d^4\right ) (c+d x) c}{d^2 \left (b c^2-a d^2\right )}+\frac {a \left (4 b^2 c^4-9 a b d^2 c^2+3 a^2 d^4\right ) (c+d x)^2}{d^2 \left (b c^2-a d^2\right )}}{(c+d x) \left (-\frac {b c^2}{d^2}+\frac {2 b (c+d x) c}{d^2}-\frac {b (c+d x)^2}{d^2}+a\right )}d\sqrt {c+d x}}{4 a b \left (b c^2-a d^2\right )}-\frac {a d^4 \sqrt {c+d x} \left (\frac {c \left (3 a d^2+b c^2\right )}{b c^2-a d^2}-\frac {(c+d x) \left (a d^2+b c^2\right )}{b c^2-a d^2}\right )}{4 b \left (b c^2-a d^2\right ) \left (a-\frac {b c^2}{d^2}+\frac {2 b c (c+d x)}{d^2}-\frac {b (c+d x)^2}{d^2}\right )}\right )}{d^5}\)

\(\Big \downarrow \) 2195

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 \left (-\frac {d^4 \left (-\frac {3 a^{3/2} d \left (2 \sqrt {b} c-\sqrt {a} d\right ) \left (\sqrt {a} d+\sqrt {b} c\right ) \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {c+d x}}{\sqrt {\sqrt {b} c-\sqrt {a} d}}\right )}{2 b^{3/4} \left (\sqrt {b} c-\sqrt {a} d\right )^{3/2}}+\frac {3 a^{3/2} d \left (\sqrt {b} c-\sqrt {a} d\right ) \left (\sqrt {a} d+2 \sqrt {b} c\right ) \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {c+d x}}{\sqrt {\sqrt {a} d+\sqrt {b} c}}\right )}{2 b^{3/4} \left (\sqrt {a} d+\sqrt {b} c\right )^{3/2}}+\frac {4 a b c^4}{\sqrt {c+d x} \left (b c^2-a d^2\right )}\right )}{4 a b \left (b c^2-a d^2\right )}-\frac {a d^4 \sqrt {c+d x} \left (\frac {c \left (3 a d^2+b c^2\right )}{b c^2-a d^2}-\frac {(c+d x) \left (a d^2+b c^2\right )}{b c^2-a d^2}\right )}{4 b \left (b c^2-a d^2\right ) \left (a-\frac {b c^2}{d^2}+\frac {2 b c (c+d x)}{d^2}-\frac {b (c+d x)^2}{d^2}\right )}\right )}{d^5}\)

Input:

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

Output:

(2*(-1/4*(a*d^4*Sqrt[c + d*x]*((c*(b*c^2 + 3*a*d^2))/(b*c^2 - a*d^2) - ((b 
*c^2 + a*d^2)*(c + d*x))/(b*c^2 - a*d^2)))/(b*(b*c^2 - a*d^2)*(a - (b*c^2) 
/d^2 + (2*b*c*(c + d*x))/d^2 - (b*(c + d*x)^2)/d^2)) - (d^4*((4*a*b*c^4)/( 
(b*c^2 - a*d^2)*Sqrt[c + d*x]) - (3*a^(3/2)*d*(2*Sqrt[b]*c - Sqrt[a]*d)*(S 
qrt[b]*c + Sqrt[a]*d)*ArcTanh[(b^(1/4)*Sqrt[c + d*x])/Sqrt[Sqrt[b]*c - Sqr 
t[a]*d]])/(2*b^(3/4)*(Sqrt[b]*c - Sqrt[a]*d)^(3/2)) + (3*a^(3/2)*d*(Sqrt[b 
]*c - Sqrt[a]*d)*(2*Sqrt[b]*c + Sqrt[a]*d)*ArcTanh[(b^(1/4)*Sqrt[c + d*x]) 
/Sqrt[Sqrt[b]*c + Sqrt[a]*d]])/(2*b^(3/4)*(Sqrt[b]*c + Sqrt[a]*d)^(3/2)))) 
/(4*a*b*(b*c^2 - a*d^2))))/d^5
 

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 561

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> With[{k = Denominator[n]}, Simp[k/d   Subst[Int[x^(k*(n + 1) - 1)*(-c 
/d + x^k/d)^m*Simp[(b*c^2 + a*d^2)/d^2 - 2*b*c*(x^k/d^2) + b*(x^(2*k)/d^2), 
 x]^p, x], x, (c + d*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, m, p}, x] && Frac 
tionQ[n] && IntegerQ[p] && IntegerQ[m]
 

rule 1673

Int[(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^ 
4)^(p_), x_Symbol] :> With[{f = Coeff[PolynomialRemainder[x^m*(d + e*x^2)^q 
, a + b*x^2 + c*x^4, x], x, 0], g = Coeff[PolynomialRemainder[x^m*(d + e*x^ 
2)^q, a + b*x^2 + c*x^4, x], x, 2]}, Simp[x*(a + b*x^2 + c*x^4)^(p + 1)*((a 
*b*g - f*(b^2 - 2*a*c) - c*(b*f - 2*a*g)*x^2)/(2*a*(p + 1)*(b^2 - 4*a*c))), 
 x] + Simp[1/(2*a*(p + 1)*(b^2 - 4*a*c))   Int[x^m*(a + b*x^2 + c*x^4)^(p + 
 1)*Simp[ExpandToSum[(2*a*(p + 1)*(b^2 - 4*a*c)*PolynomialQuotient[x^m*(d + 
 e*x^2)^q, a + b*x^2 + c*x^4, x])/x^m + (b^2*f*(2*p + 3) - 2*a*c*f*(4*p + 5 
) - a*b*g)/x^m + c*(4*p + 7)*(b*f - 2*a*g)*x^(2 - m), x], x], x], x]] /; Fr 
eeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && IGtQ[q, 1] 
&& ILtQ[m/2, 0]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2195

Int[(Pq_)*((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_ 
Symbol] :> Int[ExpandIntegrand[(d*x)^m*Pq*(a + b*x^2 + c*x^4)^p, x], x] /; 
FreeQ[{a, b, c, d, m}, x] && PolyQ[Pq, x^2] && IGtQ[p, -2]
 

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 323, normalized size of antiderivative = 1.23

method	result	size
derivativedivides	\(\frac {-\frac {2 c^{4}}{\left (a \,d^{2}-b \,c^{2}\right )^{2} \sqrt {d x +c}}-\frac {2 a \,d^{2} \left (\frac {-\frac {\left (a \,d^{2}+b \,c^{2}\right ) \left (d x +c \right )^{\frac {3}{2}}}{4 b}+\frac {c \left (3 a \,d^{2}+b \,c^{2}\right ) \sqrt {d x +c}}{4 b}}{-b \left (d x +c \right )^{2}+2 b c \left (d x +c \right )+a \,d^{2}-b \,c^{2}}+\frac {3 \left (2 c^{3} b^{2}-\sqrt {a b \,d^{2}}\, a \,d^{2}+3 \sqrt {a b \,d^{2}}\, b \,c^{2}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{8 b \sqrt {a b \,d^{2}}\, \sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}-\frac {3 \left (-2 c^{3} b^{2}-\sqrt {a b \,d^{2}}\, a \,d^{2}+3 \sqrt {a b \,d^{2}}\, b \,c^{2}\right ) \operatorname {arctanh}\left (\frac {b \sqrt {d x +c}}{\sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{8 b \sqrt {a b \,d^{2}}\, \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{\left (a \,d^{2}-b \,c^{2}\right )^{2}}}{d}\)	\(323\)
default	\(\frac {-\frac {2 c^{4}}{\left (a \,d^{2}-b \,c^{2}\right )^{2} \sqrt {d x +c}}-\frac {2 a \,d^{2} \left (\frac {-\frac {\left (a \,d^{2}+b \,c^{2}\right ) \left (d x +c \right )^{\frac {3}{2}}}{4 b}+\frac {c \left (3 a \,d^{2}+b \,c^{2}\right ) \sqrt {d x +c}}{4 b}}{-b \left (d x +c \right )^{2}+2 b c \left (d x +c \right )+a \,d^{2}-b \,c^{2}}+\frac {3 \left (2 c^{3} b^{2}-\sqrt {a b \,d^{2}}\, a \,d^{2}+3 \sqrt {a b \,d^{2}}\, b \,c^{2}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{8 b \sqrt {a b \,d^{2}}\, \sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}-\frac {3 \left (-2 c^{3} b^{2}-\sqrt {a b \,d^{2}}\, a \,d^{2}+3 \sqrt {a b \,d^{2}}\, b \,c^{2}\right ) \operatorname {arctanh}\left (\frac {b \sqrt {d x +c}}{\sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{8 b \sqrt {a b \,d^{2}}\, \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{\left (a \,d^{2}-b \,c^{2}\right )^{2}}}{d}\)	\(323\)
pseudoelliptic	\(-\frac {-\frac {3 d^{2} \left (\left (a \,d^{2}-3 b \,c^{2}\right ) \sqrt {a b \,d^{2}}-2 c^{3} b^{2}\right ) a \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}\, \sqrt {d x +c}\, \left (-b \,x^{2}+a \right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{4}+\left (\frac {3 d^{2} \left (\left (a \,d^{2}-3 b \,c^{2}\right ) \sqrt {a b \,d^{2}}+2 c^{3} b^{2}\right ) a \sqrt {d x +c}\, \left (-b \,x^{2}+a \right ) \operatorname {arctanh}\left (\frac {b \sqrt {d x +c}}{\sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{4}+\left (-2 b^{2} c^{4} x^{2}+2 \left (-\frac {1}{4} d^{2} x^{2}-\frac {1}{4} c d x +c^{2}\right ) a \,c^{2} b +\left (d x +c \right ) d^{2} \left (-\frac {d x}{2}+c \right ) a^{2}\right ) \sqrt {a b \,d^{2}}\, \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}\right ) \sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}{\sqrt {a b \,d^{2}}\, \sqrt {d x +c}\, \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}\, \sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}\, d \left (-b \,x^{2}+a \right ) b \left (a \,d^{2}-b \,c^{2}\right )^{2}}\)	\(351\)

Input:

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

Output:

2/d*(-c^4/(a*d^2-b*c^2)^2/(d*x+c)^(1/2)-a*d^2/(a*d^2-b*c^2)^2*((-1/4*(a*d^ 
2+b*c^2)/b*(d*x+c)^(3/2)+1/4*c*(3*a*d^2+b*c^2)/b*(d*x+c)^(1/2))/(-b*(d*x+c 
)^2+2*b*c*(d*x+c)+a*d^2-b*c^2)+3/8*(2*c^3*b^2-(a*b*d^2)^(1/2)*a*d^2+3*(a*b 
*d^2)^(1/2)*b*c^2)/b/(a*b*d^2)^(1/2)/((-b*c+(a*b*d^2)^(1/2))*b)^(1/2)*arct 
an(b*(d*x+c)^(1/2)/((-b*c+(a*b*d^2)^(1/2))*b)^(1/2))-3/8*(-2*c^3*b^2-(a*b* 
d^2)^(1/2)*a*d^2+3*(a*b*d^2)^(1/2)*b*c^2)/b/(a*b*d^2)^(1/2)/((b*c+(a*b*d^2 
)^(1/2))*b)^(1/2)*arctanh(b*(d*x+c)^(1/2)/((b*c+(a*b*d^2)^(1/2))*b)^(1/2)) 
))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 6209 vs. \(2 (204) = 408\).

Time = 2.53 (sec) , antiderivative size = 6209, normalized size of antiderivative = 23.61 \[ \int \frac {x^4}{(c+d x)^{3/2} \left (a-b x^2\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1457 vs. \(2 (204) = 408\).

Time = 0.33 (sec) , antiderivative size = 1457, normalized size of antiderivative = 5.54 \[ \int \frac {x^4}{(c+d x)^{3/2} \left (a-b x^2\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

1/4*(3*((3*sqrt(a*b)*a*b*c^2*d^5 - sqrt(a*b)*a^2*d^7)*(b^3*c^4*d - 2*a*b^2 
*c^2*d^3 + a^2*b*d^5)^2*abs(b) - (5*a*b^5*c^7*d^5 - 11*a^2*b^4*c^5*d^7 + 7 
*a^3*b^3*c^3*d^9 - a^4*b^2*c*d^11)*abs(b^3*c^4*d - 2*a*b^2*c^2*d^3 + a^2*b 
*d^5)*abs(b) + 2*(sqrt(a*b)*b^8*c^12*d^5 - 4*sqrt(a*b)*a*b^7*c^10*d^7 + 6* 
sqrt(a*b)*a^2*b^6*c^8*d^9 - 4*sqrt(a*b)*a^3*b^5*c^6*d^11 + sqrt(a*b)*a^4*b 
^4*c^4*d^13)*abs(b))*arctan(sqrt(d*x + c)/sqrt(-(b^4*c^5 - 2*a*b^3*c^3*d^2 
 + a^2*b^2*c*d^4 + sqrt((b^4*c^5 - 2*a*b^3*c^3*d^2 + a^2*b^2*c*d^4)^2 - (b 
^4*c^6 - 3*a*b^3*c^4*d^2 + 3*a^2*b^2*c^2*d^4 - a^3*b*d^6)*(b^4*c^4 - 2*a*b 
^3*c^2*d^2 + a^2*b^2*d^4)))/(b^4*c^4 - 2*a*b^3*c^2*d^2 + a^2*b^2*d^4)))/(( 
b^8*c^9 - 4*a*b^7*c^7*d^2 + 6*a^2*b^6*c^5*d^4 - 4*a^3*b^5*c^3*d^6 + a^4*b^ 
4*c*d^8 - sqrt(a*b)*b^7*c^8*d + 4*sqrt(a*b)*a*b^6*c^6*d^3 - 6*sqrt(a*b)*a^ 
2*b^5*c^4*d^5 + 4*sqrt(a*b)*a^3*b^4*c^2*d^7 - sqrt(a*b)*a^4*b^3*d^9)*sqrt( 
-b^2*c - sqrt(a*b)*b*d)*abs(b^3*c^4*d - 2*a*b^2*c^2*d^3 + a^2*b*d^5)) - 3* 
((3*sqrt(a*b)*a*b*c^2*d^5 - sqrt(a*b)*a^2*d^7)*(b^3*c^4*d - 2*a*b^2*c^2*d^ 
3 + a^2*b*d^5)^2*abs(b) + (5*a*b^5*c^7*d^5 - 11*a^2*b^4*c^5*d^7 + 7*a^3*b^ 
3*c^3*d^9 - a^4*b^2*c*d^11)*abs(b^3*c^4*d - 2*a*b^2*c^2*d^3 + a^2*b*d^5)*a 
bs(b) + 2*(sqrt(a*b)*b^8*c^12*d^5 - 4*sqrt(a*b)*a*b^7*c^10*d^7 + 6*sqrt(a* 
b)*a^2*b^6*c^8*d^9 - 4*sqrt(a*b)*a^3*b^5*c^6*d^11 + sqrt(a*b)*a^4*b^4*c^4* 
d^13)*abs(b))*arctan(sqrt(d*x + c)/sqrt(-(b^4*c^5 - 2*a*b^3*c^3*d^2 + a^2* 
b^2*c*d^4 - sqrt((b^4*c^5 - 2*a*b^3*c^3*d^2 + a^2*b^2*c*d^4)^2 - (b^4*c...
 

Mupad [B] (verification not implemented)

Time = 11.16 (sec) , antiderivative size = 9287, normalized size of antiderivative = 35.31 \[ \int \frac {x^4}{(c+d x)^{3/2} \left (a-b x^2\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

- ((2*c^4)/(a*d^2 - b*c^2) - ((c + d*x)^2*(a^2*d^4 + 4*b^2*c^4 + a*b*c^2*d 
^2))/(2*b*(a*d^2 - b*c^2)^2) + (c*(c + d*x)*(3*a^2*d^4 + 8*b^2*c^4 + a*b*c 
^2*d^2))/(2*b*(a*d^2 - b*c^2)^2))/((a*d^3 - b*c^2*d)*(c + d*x)^(1/2) - b*d 
*(c + d*x)^(5/2) + 2*b*c*d*(c + d*x)^(3/2)) - atan(((((9*(a^3*d^7*(a^3*b^7 
)^(1/2) + 4*a*b^7*c^7 + a^4*b^4*c*d^6 + 21*a^2*b^6*c^5*d^2 - 10*a^3*b^5*c^ 
3*d^4 + 16*b^3*c^6*d*(a^3*b^7)^(1/2) + 5*a*b^2*c^4*d^3*(a^3*b^7)^(1/2) - 6 
*a^2*b*c^2*d^5*(a^3*b^7)^(1/2)))/(64*(b^12*c^10 - a^5*b^7*d^10 - 5*a*b^11* 
c^8*d^2 + 10*a^2*b^10*c^6*d^4 - 10*a^3*b^9*c^4*d^6 + 5*a^4*b^8*c^2*d^8)))^ 
(1/2)*((c + d*x)^(1/2)*((9*(a^3*d^7*(a^3*b^7)^(1/2) + 4*a*b^7*c^7 + a^4*b^ 
4*c*d^6 + 21*a^2*b^6*c^5*d^2 - 10*a^3*b^5*c^3*d^4 + 16*b^3*c^6*d*(a^3*b^7) 
^(1/2) + 5*a*b^2*c^4*d^3*(a^3*b^7)^(1/2) - 6*a^2*b*c^2*d^5*(a^3*b^7)^(1/2) 
))/(64*(b^12*c^10 - a^5*b^7*d^10 - 5*a*b^11*c^8*d^2 + 10*a^2*b^10*c^6*d^4 
- 10*a^3*b^9*c^4*d^6 + 5*a^4*b^8*c^2*d^8)))^(1/2)*(2048*a*b^19*c^21*d^2 + 
2048*a^11*b^9*c*d^22 - 20480*a^2*b^18*c^19*d^4 + 92160*a^3*b^17*c^17*d^6 - 
 245760*a^4*b^16*c^15*d^8 + 430080*a^5*b^15*c^13*d^10 - 516096*a^6*b^14*c^ 
11*d^12 + 430080*a^7*b^13*c^9*d^14 - 245760*a^8*b^12*c^7*d^16 + 92160*a^9* 
b^11*c^5*d^18 - 20480*a^10*b^10*c^3*d^20) - 768*a^11*b^7*c*d^21 + 3840*a^2 
*b^16*c^19*d^3 - 31488*a^3*b^15*c^17*d^5 + 113664*a^4*b^14*c^15*d^7 - 2365 
44*a^5*b^13*c^13*d^9 + 311808*a^6*b^12*c^11*d^11 - 268800*a^7*b^11*c^9*d^1 
3 + 150528*a^8*b^10*c^7*d^15 - 52224*a^9*b^9*c^5*d^17 + 9984*a^10*b^8*c...
 

Reduce [B] (verification not implemented)

Time = 0.62 (sec) , antiderivative size = 1890, normalized size of antiderivative = 7.19 \[ \int \frac {x^4}{(c+d x)^{3/2} \left (a-b x^2\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(6*sqrt(a)*sqrt(c + d*x)*sqrt(sqrt(b)*sqrt(a)*d - b*c)*atan((sqrt(c + d*x) 
*b)/(sqrt(b)*sqrt(sqrt(b)*sqrt(a)*d - b*c)))*a**3*d**5 - 18*sqrt(a)*sqrt(c 
 + d*x)*sqrt(sqrt(b)*sqrt(a)*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt 
(sqrt(b)*sqrt(a)*d - b*c)))*a**2*b*c**2*d**3 - 6*sqrt(a)*sqrt(c + d*x)*sqr 
t(sqrt(b)*sqrt(a)*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(sqrt(b)*sq 
rt(a)*d - b*c)))*a**2*b*d**5*x**2 - 12*sqrt(a)*sqrt(c + d*x)*sqrt(sqrt(b)* 
sqrt(a)*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(sqrt(b)*sqrt(a)*d - 
b*c)))*a*b**2*c**4*d + 18*sqrt(a)*sqrt(c + d*x)*sqrt(sqrt(b)*sqrt(a)*d - b 
*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(sqrt(b)*sqrt(a)*d - b*c)))*a*b**2 
*c**2*d**3*x**2 + 12*sqrt(a)*sqrt(c + d*x)*sqrt(sqrt(b)*sqrt(a)*d - b*c)*a 
tan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(sqrt(b)*sqrt(a)*d - b*c)))*b**3*c**4*d 
*x**2 + 6*sqrt(b)*sqrt(c + d*x)*sqrt(sqrt(b)*sqrt(a)*d - b*c)*atan((sqrt(c 
 + d*x)*b)/(sqrt(b)*sqrt(sqrt(b)*sqrt(a)*d - b*c)))*a**3*c*d**4 - 30*sqrt( 
b)*sqrt(c + d*x)*sqrt(sqrt(b)*sqrt(a)*d - b*c)*atan((sqrt(c + d*x)*b)/(sqr 
t(b)*sqrt(sqrt(b)*sqrt(a)*d - b*c)))*a**2*b*c**3*d**2 - 6*sqrt(b)*sqrt(c + 
 d*x)*sqrt(sqrt(b)*sqrt(a)*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(s 
qrt(b)*sqrt(a)*d - b*c)))*a**2*b*c*d**4*x**2 + 30*sqrt(b)*sqrt(c + d*x)*sq 
rt(sqrt(b)*sqrt(a)*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(sqrt(b)*s 
qrt(a)*d - b*c)))*a*b**2*c**3*d**2*x**2 + 3*sqrt(a)*sqrt(c + d*x)*sqrt(sqr 
t(b)*sqrt(a)*d + b*c)*log( - sqrt(sqrt(b)*sqrt(a)*d + b*c) + sqrt(b)*sq...