\(\int \frac {x (c+d x)^{3/2}}{(a-b x^2)^3} \, dx\) [700]

Optimal result

Integrand size = 21, antiderivative size = 231 \[ \int \frac {x (c+d x)^{3/2}}{\left (a-b x^2\right )^3} \, dx=\frac {(c+d x)^{3/2}}{4 b \left (a-b x^2\right )^2}-\frac {3 d x \sqrt {c+d x}}{16 a b \left (a-b x^2\right )}+\frac {3 d \left (2 \sqrt {b} c-\sqrt {a} d\right ) \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {c+d x}}{\sqrt {\sqrt {b} c-\sqrt {a} d}}\right )}{32 a^{3/2} b^{7/4} \sqrt {\sqrt {b} c-\sqrt {a} d}}-\frac {3 d \left (2 \sqrt {b} c+\sqrt {a} d\right ) \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {c+d x}}{\sqrt {\sqrt {b} c+\sqrt {a} d}}\right )}{32 a^{3/2} b^{7/4} \sqrt {\sqrt {b} c+\sqrt {a} d}} \] Output:

1/4*(d*x+c)^(3/2)/b/(-b*x^2+a)^2-3/16*d*x*(d*x+c)^(1/2)/a/b/(-b*x^2+a)+3/3 
2*d*(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))/a^(3/2)/b^(7/4)/(b^(1/2)*c-a^(1/2)*d)^(1/2)-3/32*d*(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))/a 
^(3/2)/b^(7/4)/(b^(1/2)*c+a^(1/2)*d)^(1/2)
 

Mathematica [A] (verified)

Time = 1.54 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.05 \[ \int \frac {x (c+d x)^{3/2}}{\left (a-b x^2\right )^3} \, dx=\frac {\frac {2 \sqrt {a} \sqrt {b} \sqrt {c+d x} \left (4 a c+a d x+3 b d x^3\right )}{\left (a-b x^2\right )^2}-\frac {3 d \left (2 \sqrt {b} c+\sqrt {a} d\right ) \arctan \left (\frac {\sqrt {-b c-\sqrt {a} \sqrt {b} d} \sqrt {c+d x}}{\sqrt {b} c+\sqrt {a} d}\right )}{\sqrt {-b c-\sqrt {a} \sqrt {b} d}}-\frac {3 d \left (-2 \sqrt {b} c+\sqrt {a} d\right ) \arctan \left (\frac {\sqrt {-b c+\sqrt {a} \sqrt {b} d} \sqrt {c+d x}}{\sqrt {b} c-\sqrt {a} d}\right )}{\sqrt {-b c+\sqrt {a} \sqrt {b} d}}}{32 a^{3/2} b^{3/2}} \] Input:

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

Output:

((2*Sqrt[a]*Sqrt[b]*Sqrt[c + d*x]*(4*a*c + a*d*x + 3*b*d*x^3))/(a - b*x^2) 
^2 - (3*d*(2*Sqrt[b]*c + Sqrt[a]*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]*Sqrt[b]* 
d] - (3*d*(-2*Sqrt[b]*c + Sqrt[a]*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]*Sqrt[b] 
*d])/(32*a^(3/2)*b^(3/2))
 

Rubi [A] (verified)

Time = 0.82 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.33, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {561, 25, 27, 1598, 27, 1439, 27, 1480, 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.

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

\(\Big \downarrow \) 561

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1598

\(\displaystyle -\frac {2 \left (\frac {d^2 \int \frac {6 a (c+d x)}{\left (-\frac {b c^2}{d^2}+\frac {2 b (c+d x) c}{d^2}-\frac {b (c+d x)^2}{d^2}+a\right )^2}d\sqrt {c+d x}}{16 a b}-\frac {d^2 (c+d x)^{3/2}}{8 b \left (a-\frac {b c^2}{d^2}+\frac {2 b c (c+d x)}{d^2}-\frac {b (c+d x)^2}{d^2}\right )^2}\right )}{d^2}\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1439

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1480

\(\displaystyle -\frac {2 \left (\frac {3 d^2 \left (\frac {\frac {1}{2} \left (1-\frac {2 \sqrt {b} c}{\sqrt {a} d}\right ) \int \frac {1}{\frac {\sqrt {b} \left (\sqrt {b} c-\sqrt {a} d\right )}{d^2}-\frac {b (c+d x)}{d^2}}d\sqrt {c+d x}+\frac {1}{2} \left (\frac {2 \sqrt {b} c}{\sqrt {a} d}+1\right ) \int \frac {1}{\frac {\sqrt {b} \left (\sqrt {b} c+\sqrt {a} d\right )}{d^2}-\frac {b (c+d x)}{d^2}}d\sqrt {c+d x}}{4 a}+\frac {d x \sqrt {c+d x}}{4 a \left (a-\frac {b c^2}{d^2}+\frac {2 b c (c+d x)}{d^2}-\frac {b (c+d x)^2}{d^2}\right )}\right )}{8 b}-\frac {d^2 (c+d x)^{3/2}}{8 b \left (a-\frac {b c^2}{d^2}+\frac {2 b c (c+d x)}{d^2}-\frac {b (c+d x)^2}{d^2}\right )^2}\right )}{d^2}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {2 \left (\frac {3 d^2 \left (\frac {\frac {d^2 \left (1-\frac {2 \sqrt {b} c}{\sqrt {a} d}\right ) \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {c+d x}}{\sqrt {\sqrt {b} c-\sqrt {a} d}}\right )}{2 b^{3/4} \sqrt {\sqrt {b} c-\sqrt {a} d}}+\frac {d^2 \left (\frac {2 \sqrt {b} c}{\sqrt {a} d}+1\right ) \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {c+d x}}{\sqrt {\sqrt {a} d+\sqrt {b} c}}\right )}{2 b^{3/4} \sqrt {\sqrt {a} d+\sqrt {b} c}}}{4 a}+\frac {d x \sqrt {c+d x}}{4 a \left (a-\frac {b c^2}{d^2}+\frac {2 b c (c+d x)}{d^2}-\frac {b (c+d x)^2}{d^2}\right )}\right )}{8 b}-\frac {d^2 (c+d x)^{3/2}}{8 b \left (a-\frac {b c^2}{d^2}+\frac {2 b c (c+d x)}{d^2}-\frac {b (c+d x)^2}{d^2}\right )^2}\right )}{d^2}\)

Input:

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

Output:

(-2*(-1/8*(d^2*(c + d*x)^(3/2))/(b*(a - (b*c^2)/d^2 + (2*b*c*(c + d*x))/d^ 
2 - (b*(c + d*x)^2)/d^2)^2) + (3*d^2*((d*x*Sqrt[c + d*x])/(4*a*(a - (b*c^2 
)/d^2 + (2*b*c*(c + d*x))/d^2 - (b*(c + d*x)^2)/d^2)) + (((1 - (2*Sqrt[b]* 
c)/(Sqrt[a]*d))*d^2*ArcTanh[(b^(1/4)*Sqrt[c + d*x])/Sqrt[Sqrt[b]*c - Sqrt[ 
a]*d]])/(2*b^(3/4)*Sqrt[Sqrt[b]*c - Sqrt[a]*d]) + ((1 + (2*Sqrt[b]*c)/(Sqr 
t[a]*d))*d^2*ArcTanh[(b^(1/4)*Sqrt[c + d*x])/Sqrt[Sqrt[b]*c + Sqrt[a]*d]]) 
/(2*b^(3/4)*Sqrt[Sqrt[b]*c + Sqrt[a]*d]))/(4*a)))/(8*b)))/d^2
 

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 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 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 1439

Int[((d_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] 
:> Simp[d*(d*x)^(m - 1)*(b + 2*c*x^2)*((a + b*x^2 + c*x^4)^(p + 1)/(2*(p + 
1)*(b^2 - 4*a*c))), x] - Simp[d^2/(2*(p + 1)*(b^2 - 4*a*c))   Int[(d*x)^(m 
- 2)*(b*(m - 1) + 2*c*(m + 4*p + 5)*x^2)*(a + b*x^2 + c*x^4)^(p + 1), x], x 
] /; FreeQ[{a, b, c, d}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && GtQ[m, 
1] && LeQ[m, 3] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])
 

rule 1480

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : 
> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q))   Int[1/( 
b/2 - q/2 + c*x^2), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q))   Int[1/(b/2 
+ q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] 
 && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^2 - 4*a*c]
 

rule 1598

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*( 
x_)^4)^(p_.), x_Symbol] :> Simp[f*(f*x)^(m - 1)*(a + b*x^2 + c*x^4)^(p + 1) 
*((b*d - 2*a*e - (b*e - 2*c*d)*x^2)/(2*(p + 1)*(b^2 - 4*a*c))), x] - Simp[f 
^2/(2*(p + 1)*(b^2 - 4*a*c))   Int[(f*x)^(m - 2)*(a + b*x^2 + c*x^4)^(p + 1 
)*Simp[(m - 1)*(b*d - 2*a*e) - (4*p + 4 + m + 1)*(b*e - 2*c*d)*x^2, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && 
 GtQ[m, 1] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])
 

Maple [A] (verified)

Time = 0.51 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.10

method	result	size
pseudoelliptic	\(\frac {-\frac {3 \left (-\frac {\sqrt {a b \,d^{2}}}{2}+b c \right ) d^{2} \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}\, \left (-b \,x^{2}+a \right )^{2} \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{4}+\sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}\, \left (-\frac {3 d^{2} \left (b c +\frac {\sqrt {a b \,d^{2}}}{2}\right ) \left (-b \,x^{2}+a \right )^{2} \operatorname {arctanh}\left (\frac {b \sqrt {d x +c}}{\sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{4}+\left (\frac {3 b d \,x^{3}}{4}+a \left (\frac {d x}{4}+c \right )\right ) \sqrt {d x +c}\, \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}\, \sqrt {a b \,d^{2}}\right )}{4 \sqrt {a b \,d^{2}}\, \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}\, \sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}\, a b \left (-b \,x^{2}+a \right )^{2}}\)	\(255\)
derivativedivides	\(-2 d^{4} \left (\frac {-\frac {3 \left (d x +c \right )^{\frac {7}{2}}}{32 a \,d^{2}}+\frac {9 c \left (d x +c \right )^{\frac {5}{2}}}{32 a \,d^{2}}-\frac {\left (a \,d^{2}+9 b \,c^{2}\right ) \left (d x +c \right )^{\frac {3}{2}}}{32 a b \,d^{2}}-\frac {3 \left (a \,d^{2}-b \,c^{2}\right ) c \sqrt {d x +c}}{32 a b \,d^{2}}}{\left (-b \left (d x +c \right )^{2}+2 b c \left (d x +c \right )+a \,d^{2}-b \,c^{2}\right )^{2}}+\frac {-\frac {3 \left (-2 b c -\sqrt {a b \,d^{2}}\right ) \operatorname {arctanh}\left (\frac {b \sqrt {d x +c}}{\sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{64 b \sqrt {a b \,d^{2}}\, \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}+\frac {3 \left (2 b c -\sqrt {a b \,d^{2}}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{64 b \sqrt {a b \,d^{2}}\, \sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}}{a \,d^{2}}\right )\)	\(285\)
default	\(2 d^{4} \left (-\frac {-\frac {3 \left (d x +c \right )^{\frac {7}{2}}}{32 a \,d^{2}}+\frac {9 c \left (d x +c \right )^{\frac {5}{2}}}{32 a \,d^{2}}-\frac {\left (a \,d^{2}+9 b \,c^{2}\right ) \left (d x +c \right )^{\frac {3}{2}}}{32 a b \,d^{2}}-\frac {3 \left (a \,d^{2}-b \,c^{2}\right ) c \sqrt {d x +c}}{32 a b \,d^{2}}}{\left (-b \left (d x +c \right )^{2}+2 b c \left (d x +c \right )+a \,d^{2}-b \,c^{2}\right )^{2}}-\frac {3 \left (-\frac {\left (-2 b c -\sqrt {a b \,d^{2}}\right ) \operatorname {arctanh}\left (\frac {b \sqrt {d x +c}}{\sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{2 b \sqrt {a b \,d^{2}}\, \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}+\frac {\left (2 b c -\sqrt {a b \,d^{2}}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{2 b \sqrt {a b \,d^{2}}\, \sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{32 a \,d^{2}}\right )\)	\(286\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1533 vs. \(2 (173) = 346\).

Time = 0.12 (sec) , antiderivative size = 1533, normalized size of antiderivative = 6.64 \[ \int \frac {x (c+d x)^{3/2}}{\left (a-b x^2\right )^3} \, dx=\text {Too large to display} \] Input:

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

Output:

1/64*(3*(a*b^3*x^4 - 2*a^2*b^2*x^2 + a^3*b)*sqrt((4*b*c^3*d^2 - 3*a*c*d^4 
+ (a^3*b^4*c^2 - a^4*b^3*d^2)*sqrt(d^10/(a^3*b^9*c^4 - 2*a^4*b^8*c^2*d^2 + 
 a^5*b^7*d^4)))/(a^3*b^4*c^2 - a^4*b^3*d^2))*log(-27*(4*b*c^2*d^6 - a*d^8) 
*sqrt(d*x + c) + 27*(a^2*b^2*c*d^6 - (2*a^3*b^7*c^4 - 3*a^4*b^6*c^2*d^2 + 
a^5*b^5*d^4)*sqrt(d^10/(a^3*b^9*c^4 - 2*a^4*b^8*c^2*d^2 + a^5*b^7*d^4)))*s 
qrt((4*b*c^3*d^2 - 3*a*c*d^4 + (a^3*b^4*c^2 - a^4*b^3*d^2)*sqrt(d^10/(a^3* 
b^9*c^4 - 2*a^4*b^8*c^2*d^2 + a^5*b^7*d^4)))/(a^3*b^4*c^2 - a^4*b^3*d^2))) 
 - 3*(a*b^3*x^4 - 2*a^2*b^2*x^2 + a^3*b)*sqrt((4*b*c^3*d^2 - 3*a*c*d^4 + ( 
a^3*b^4*c^2 - a^4*b^3*d^2)*sqrt(d^10/(a^3*b^9*c^4 - 2*a^4*b^8*c^2*d^2 + a^ 
5*b^7*d^4)))/(a^3*b^4*c^2 - a^4*b^3*d^2))*log(-27*(4*b*c^2*d^6 - a*d^8)*sq 
rt(d*x + c) - 27*(a^2*b^2*c*d^6 - (2*a^3*b^7*c^4 - 3*a^4*b^6*c^2*d^2 + a^5 
*b^5*d^4)*sqrt(d^10/(a^3*b^9*c^4 - 2*a^4*b^8*c^2*d^2 + a^5*b^7*d^4)))*sqrt 
((4*b*c^3*d^2 - 3*a*c*d^4 + (a^3*b^4*c^2 - a^4*b^3*d^2)*sqrt(d^10/(a^3*b^9 
*c^4 - 2*a^4*b^8*c^2*d^2 + a^5*b^7*d^4)))/(a^3*b^4*c^2 - a^4*b^3*d^2))) + 
3*(a*b^3*x^4 - 2*a^2*b^2*x^2 + a^3*b)*sqrt((4*b*c^3*d^2 - 3*a*c*d^4 - (a^3 
*b^4*c^2 - a^4*b^3*d^2)*sqrt(d^10/(a^3*b^9*c^4 - 2*a^4*b^8*c^2*d^2 + a^5*b 
^7*d^4)))/(a^3*b^4*c^2 - a^4*b^3*d^2))*log(-27*(4*b*c^2*d^6 - a*d^8)*sqrt( 
d*x + c) + 27*(a^2*b^2*c*d^6 + (2*a^3*b^7*c^4 - 3*a^4*b^6*c^2*d^2 + a^5*b^ 
5*d^4)*sqrt(d^10/(a^3*b^9*c^4 - 2*a^4*b^8*c^2*d^2 + a^5*b^7*d^4)))*sqrt((4 
*b*c^3*d^2 - 3*a*c*d^4 - (a^3*b^4*c^2 - a^4*b^3*d^2)*sqrt(d^10/(a^3*b^9...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 438 vs. \(2 (173) = 346\).

Time = 0.25 (sec) , antiderivative size = 438, normalized size of antiderivative = 1.90 \[ \int \frac {x (c+d x)^{3/2}}{\left (a-b x^2\right )^3} \, dx=-\frac {3 \, {\left (2 \, a b^{3} c^{2} d^{2} - a^{2} b^{2} d^{4} - \sqrt {a b} b c d^{2} {\left | a \right |} {\left | b \right |} {\left | d \right |}\right )} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-\frac {a b^{2} c + \sqrt {a^{2} b^{4} c^{2} - {\left (a b^{2} c^{2} - a^{2} b d^{2}\right )} a b^{2}}}{a b^{2}}}}\right )}{32 \, {\left (a^{2} b^{3} d - \sqrt {a b} a b^{3} c\right )} \sqrt {-b^{2} c - \sqrt {a b} b d} {\left | a \right |} {\left | d \right |}} - \frac {3 \, {\left (2 \, a b^{3} c^{2} d^{2} - a^{2} b^{2} d^{4} + \sqrt {a b} b c d^{2} {\left | a \right |} {\left | b \right |} {\left | d \right |}\right )} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-\frac {a b^{2} c - \sqrt {a^{2} b^{4} c^{2} - {\left (a b^{2} c^{2} - a^{2} b d^{2}\right )} a b^{2}}}{a b^{2}}}}\right )}{32 \, {\left (a^{2} b^{3} d + \sqrt {a b} a b^{3} c\right )} \sqrt {-b^{2} c + \sqrt {a b} b d} {\left | a \right |} {\left | d \right |}} + \frac {3 \, {\left (d x + c\right )}^{\frac {7}{2}} b d^{2} - 9 \, {\left (d x + c\right )}^{\frac {5}{2}} b c d^{2} + 9 \, {\left (d x + c\right )}^{\frac {3}{2}} b c^{2} d^{2} - 3 \, \sqrt {d x + c} b c^{3} d^{2} + {\left (d x + c\right )}^{\frac {3}{2}} a d^{4} + 3 \, \sqrt {d x + c} a c d^{4}}{16 \, {\left ({\left (d x + c\right )}^{2} b - 2 \, {\left (d x + c\right )} b c + b c^{2} - a d^{2}\right )}^{2} a b} \] Input:

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

Output:

-3/32*(2*a*b^3*c^2*d^2 - a^2*b^2*d^4 - sqrt(a*b)*b*c*d^2*abs(a)*abs(b)*abs 
(d))*arctan(sqrt(d*x + c)/sqrt(-(a*b^2*c + sqrt(a^2*b^4*c^2 - (a*b^2*c^2 - 
 a^2*b*d^2)*a*b^2))/(a*b^2)))/((a^2*b^3*d - sqrt(a*b)*a*b^3*c)*sqrt(-b^2*c 
 - sqrt(a*b)*b*d)*abs(a)*abs(d)) - 3/32*(2*a*b^3*c^2*d^2 - a^2*b^2*d^4 + s 
qrt(a*b)*b*c*d^2*abs(a)*abs(b)*abs(d))*arctan(sqrt(d*x + c)/sqrt(-(a*b^2*c 
 - sqrt(a^2*b^4*c^2 - (a*b^2*c^2 - a^2*b*d^2)*a*b^2))/(a*b^2)))/((a^2*b^3* 
d + sqrt(a*b)*a*b^3*c)*sqrt(-b^2*c + sqrt(a*b)*b*d)*abs(a)*abs(d)) + 1/16* 
(3*(d*x + c)^(7/2)*b*d^2 - 9*(d*x + c)^(5/2)*b*c*d^2 + 9*(d*x + c)^(3/2)*b 
*c^2*d^2 - 3*sqrt(d*x + c)*b*c^3*d^2 + (d*x + c)^(3/2)*a*d^4 + 3*sqrt(d*x 
+ c)*a*c*d^4)/(((d*x + c)^2*b - 2*(d*x + c)*b*c + b*c^2 - a*d^2)^2*a*b)
 

Mupad [B] (verification not implemented)

Time = 9.90 (sec) , antiderivative size = 2552, normalized size of antiderivative = 11.05 \[ \int \frac {x (c+d x)^{3/2}}{\left (a-b x^2\right )^3} \, dx=\text {Too large to display} \] Input:

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

Output:

((3*d^2*(c + d*x)^(7/2))/(16*a) + ((a*d^4 + 9*b*c^2*d^2)*(c + d*x)^(3/2))/ 
(16*a*b) - (9*c*d^2*(c + d*x)^(5/2))/(16*a) - (3*(b*c^3*d^2 - a*c*d^4)*(c 
+ d*x)^(1/2))/(16*a*b))/(b^2*(c + d*x)^4 + a^2*d^4 + b^2*c^4 + (6*b^2*c^2 
- 2*a*b*d^2)*(c + d*x)^2 - (4*b^2*c^3 - 4*a*b*c*d^2)*(c + d*x) - 4*b^2*c*( 
c + d*x)^3 - 2*a*b*c^2*d^2) + atan((((3*b^2*c*d^4 - 64*a*b^4*c*d^2*(c + d* 
x)^(1/2)*((9*(d^5*(a^9*b^7)^(1/2) - 3*a^4*b^4*c*d^4 + 4*a^3*b^5*c^3*d^2))/ 
(4096*(a^6*b^8*c^2 - a^7*b^7*d^2)))^(1/2))*((9*(d^5*(a^9*b^7)^(1/2) - 3*a^ 
4*b^4*c*d^4 + 4*a^3*b^5*c^3*d^2))/(4096*(a^6*b^8*c^2 - a^7*b^7*d^2)))^(1/2 
) + ((9*a*d^6 + 36*b*c^2*d^4)*(c + d*x)^(1/2))/(64*a^2))*((9*(d^5*(a^9*b^7 
)^(1/2) - 3*a^4*b^4*c*d^4 + 4*a^3*b^5*c^3*d^2))/(4096*(a^6*b^8*c^2 - a^7*b 
^7*d^2)))^(1/2)*1i - ((3*b^2*c*d^4 + 64*a*b^4*c*d^2*(c + d*x)^(1/2)*((9*(d 
^5*(a^9*b^7)^(1/2) - 3*a^4*b^4*c*d^4 + 4*a^3*b^5*c^3*d^2))/(4096*(a^6*b^8* 
c^2 - a^7*b^7*d^2)))^(1/2))*((9*(d^5*(a^9*b^7)^(1/2) - 3*a^4*b^4*c*d^4 + 4 
*a^3*b^5*c^3*d^2))/(4096*(a^6*b^8*c^2 - a^7*b^7*d^2)))^(1/2) - ((9*a*d^6 + 
 36*b*c^2*d^4)*(c + d*x)^(1/2))/(64*a^2))*((9*(d^5*(a^9*b^7)^(1/2) - 3*a^4 
*b^4*c*d^4 + 4*a^3*b^5*c^3*d^2))/(4096*(a^6*b^8*c^2 - a^7*b^7*d^2)))^(1/2) 
*1i)/(((3*b^2*c*d^4 - 64*a*b^4*c*d^2*(c + d*x)^(1/2)*((9*(d^5*(a^9*b^7)^(1 
/2) - 3*a^4*b^4*c*d^4 + 4*a^3*b^5*c^3*d^2))/(4096*(a^6*b^8*c^2 - a^7*b^7*d 
^2)))^(1/2))*((9*(d^5*(a^9*b^7)^(1/2) - 3*a^4*b^4*c*d^4 + 4*a^3*b^5*c^3*d^ 
2))/(4096*(a^6*b^8*c^2 - a^7*b^7*d^2)))^(1/2) + ((9*a*d^6 + 36*b*c^2*d^...
 

Reduce [B] (verification not implemented)

Time = 3.19 (sec) , antiderivative size = 1504, normalized size of antiderivative = 6.51 \[ \int \frac {x (c+d x)^{3/2}}{\left (a-b x^2\right )^3} \, dx =\text {Too large to display} \] Input:

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

Output:

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