\(\int \frac {x (c+d x)^{5/2}}{(a-b x^2)^3} \, dx\) [709]

Optimal result

Integrand size = 21, antiderivative size = 238 \[ \int \frac {x (c+d x)^{5/2}}{\left (a-b x^2\right )^3} \, dx=\frac {(c+d x)^{5/2}}{4 b \left (a-b x^2\right )^2}-\frac {5 d (a d+b c x) \sqrt {c+d x}}{16 a b^2 \left (a-b x^2\right )}+\frac {5 d \sqrt {\sqrt {b} c-\sqrt {a} d} \left (2 \sqrt {b} c+\sqrt {a} d\right ) \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {c+d x}}{\sqrt {\sqrt {b} c-\sqrt {a} d}}\right )}{32 a^{3/2} b^{9/4}}-\frac {5 d \left (2 \sqrt {b} c-\sqrt {a} d\right ) \sqrt {\sqrt {b} c+\sqrt {a} d} \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {c+d x}}{\sqrt {\sqrt {b} c+\sqrt {a} d}}\right )}{32 a^{3/2} b^{9/4}} \] Output:

1/4*(d*x+c)^(5/2)/b/(-b*x^2+a)^2-5/16*d*(b*c*x+a*d)*(d*x+c)^(1/2)/a/b^2/(- 
b*x^2+a)+5/32*d*(b^(1/2)*c-a^(1/2)*d)^(1/2)*(2*b^(1/2)*c+a^(1/2)*d)*arctan 
h(b^(1/4)*(d*x+c)^(1/2)/(b^(1/2)*c-a^(1/2)*d)^(1/2))/a^(3/2)/b^(9/4)-5/32* 
d*(2*b^(1/2)*c-a^(1/2)*d)*(b^(1/2)*c+a^(1/2)*d)^(1/2)*arctanh(b^(1/4)*(d*x 
+c)^(1/2)/(b^(1/2)*c+a^(1/2)*d)^(1/2))/a^(3/2)/b^(9/4)
 

Mathematica [A] (verified)

Time = 1.98 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.13 \[ \int \frac {x (c+d x)^{5/2}}{\left (a-b x^2\right )^3} \, dx=\frac {\frac {2 \sqrt {a} \sqrt {b} \sqrt {c+d x} \left (-5 a^2 d^2+5 b^2 c d x^3+a b \left (4 c^2+3 c d x+9 d^2 x^2\right )\right )}{\left (a-b x^2\right )^2}+5 d \left (2 \sqrt {b} c-\sqrt {a} d\right ) \sqrt {-b c-\sqrt {a} \sqrt {b} d} \arctan \left (\frac {\sqrt {-b c-\sqrt {a} \sqrt {b} d} \sqrt {c+d x}}{\sqrt {b} c+\sqrt {a} d}\right )-5 d \left (2 \sqrt {b} c+\sqrt {a} d\right ) \sqrt {-b c+\sqrt {a} \sqrt {b} d} \arctan \left (\frac {\sqrt {-b c+\sqrt {a} \sqrt {b} d} \sqrt {c+d x}}{\sqrt {b} c-\sqrt {a} d}\right )}{32 a^{3/2} b^{5/2}} \] Input:

Integrate[(x*(c + d*x)^(5/2))/(a - b*x^2)^3,x]
 

Output:

((2*Sqrt[a]*Sqrt[b]*Sqrt[c + d*x]*(-5*a^2*d^2 + 5*b^2*c*d*x^3 + a*b*(4*c^2 
 + 3*c*d*x + 9*d^2*x^2)))/(a - b*x^2)^2 + 5*d*(2*Sqrt[b]*c - Sqrt[a]*d)*Sq 
rt[-(b*c) - Sqrt[a]*Sqrt[b]*d]*ArcTan[(Sqrt[-(b*c) - Sqrt[a]*Sqrt[b]*d]*Sq 
rt[c + d*x])/(Sqrt[b]*c + Sqrt[a]*d)] - 5*d*(2*Sqrt[b]*c + Sqrt[a]*d)*Sqrt 
[-(b*c) + Sqrt[a]*Sqrt[b]*d]*ArcTan[(Sqrt[-(b*c) + Sqrt[a]*Sqrt[b]*d]*Sqrt 
[c + d*x])/(Sqrt[b]*c - Sqrt[a]*d)])/(32*a^(3/2)*b^(5/2))
 

Rubi [A] (verified)

Time = 0.96 (sec) , antiderivative size = 359, normalized size of antiderivative = 1.51, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {561, 25, 27, 1598, 27, 1440, 27, 25, 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.

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

\(\Big \downarrow \) 561

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1598

\(\displaystyle -\frac {2 \left (\frac {d^2 \int \frac {10 a (c+d x)^2}{\left (-\frac {b c^2}{d^2}+\frac {2 b (c+d x) c}{d^2}-\frac {b (c+d x)^2}{d^2}+a\right )^2}d\sqrt {c+d x}}{16 a b}-\frac {d^2 (c+d x)^{5/2}}{8 b \left (a-\frac {b c^2}{d^2}+\frac {2 b c (c+d x)}{d^2}-\frac {b (c+d x)^2}{d^2}\right )^2}\right )}{d^2}\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1440

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1480

\(\displaystyle -\frac {2 \left (\frac {5 d^2 \left (\frac {\frac {\sqrt {b} \left (\sqrt {a} \sqrt {b} c d-a d^2+2 b c^2\right ) \int \frac {1}{\frac {\sqrt {b} \left (\sqrt {b} c+\sqrt {a} d\right )}{d^2}-\frac {b (c+d x)}{d^2}}d\sqrt {c+d x}}{2 \sqrt {a} d}-\frac {\sqrt {b} \left (-\sqrt {a} \sqrt {b} c d-a d^2+2 b c^2\right ) \int \frac {1}{\frac {\sqrt {b} \left (\sqrt {b} c-\sqrt {a} d\right )}{d^2}-\frac {b (c+d x)}{d^2}}d\sqrt {c+d x}}{2 \sqrt {a} d}}{4 a b}+\frac {\sqrt {c+d x} \left (d^2 \left (a-\frac {b c^2}{d^2}\right )+b c (c+d x)\right )}{4 a b \left (a-\frac {b c^2}{d^2}+\frac {2 b c (c+d x)}{d^2}-\frac {b (c+d x)^2}{d^2}\right )}\right )}{8 b}-\frac {d^2 (c+d x)^{5/2}}{8 b \left (a-\frac {b c^2}{d^2}+\frac {2 b c (c+d x)}{d^2}-\frac {b (c+d x)^2}{d^2}\right )^2}\right )}{d^2}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {2 \left (\frac {5 d^2 \left (\frac {\frac {d \left (\sqrt {a} \sqrt {b} c d-a d^2+2 b c^2\right ) \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {c+d x}}{\sqrt {\sqrt {a} d+\sqrt {b} c}}\right )}{2 \sqrt {a} \sqrt [4]{b} \sqrt {\sqrt {a} d+\sqrt {b} c}}-\frac {d \left (-\sqrt {a} \sqrt {b} c d-a d^2+2 b c^2\right ) \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {c+d x}}{\sqrt {\sqrt {b} c-\sqrt {a} d}}\right )}{2 \sqrt {a} \sqrt [4]{b} \sqrt {\sqrt {b} c-\sqrt {a} d}}}{4 a b}+\frac {\sqrt {c+d x} \left (d^2 \left (a-\frac {b c^2}{d^2}\right )+b c (c+d x)\right )}{4 a b \left (a-\frac {b c^2}{d^2}+\frac {2 b c (c+d x)}{d^2}-\frac {b (c+d x)^2}{d^2}\right )}\right )}{8 b}-\frac {d^2 (c+d x)^{5/2}}{8 b \left (a-\frac {b c^2}{d^2}+\frac {2 b c (c+d x)}{d^2}-\frac {b (c+d x)^2}{d^2}\right )^2}\right )}{d^2}\)

Input:

Int[(x*(c + d*x)^(5/2))/(a - b*x^2)^3,x]
 

Output:

(-2*(-1/8*(d^2*(c + d*x)^(5/2))/(b*(a - (b*c^2)/d^2 + (2*b*c*(c + d*x))/d^ 
2 - (b*(c + d*x)^2)/d^2)^2) + (5*d^2*((Sqrt[c + d*x]*((a - (b*c^2)/d^2)*d^ 
2 + b*c*(c + d*x)))/(4*a*b*(a - (b*c^2)/d^2 + (2*b*c*(c + d*x))/d^2 - (b*( 
c + d*x)^2)/d^2)) + (-1/2*(d*(2*b*c^2 - Sqrt[a]*Sqrt[b]*c*d - a*d^2)*ArcTa 
nh[(b^(1/4)*Sqrt[c + d*x])/Sqrt[Sqrt[b]*c - Sqrt[a]*d]])/(Sqrt[a]*b^(1/4)* 
Sqrt[Sqrt[b]*c - Sqrt[a]*d]) + (d*(2*b*c^2 + Sqrt[a]*Sqrt[b]*c*d - a*d^2)* 
ArcTanh[(b^(1/4)*Sqrt[c + d*x])/Sqrt[Sqrt[b]*c + Sqrt[a]*d]])/(2*Sqrt[a]*b 
^(1/4)*Sqrt[Sqrt[b]*c + Sqrt[a]*d]))/(4*a*b)))/(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 1440

Int[((d_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] 
:> Simp[(-d^3)*(d*x)^(m - 3)*(2*a + b*x^2)*((a + b*x^2 + c*x^4)^(p + 1)/(2* 
(p + 1)*(b^2 - 4*a*c))), x] + Simp[d^4/(2*(p + 1)*(b^2 - 4*a*c))   Int[(d*x 
)^(m - 4)*(2*a*(m - 3) + b*(m + 4*p + 3)*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] && Gt 
Q[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 = 0.56 (sec) , antiderivative size = 297, normalized size of antiderivative = 1.25

method	result	size
pseudoelliptic	\(-\frac {5 \left (-\frac {d^{2} \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}\, b \left (a \,d^{2}-2 b \,c^{2}+\sqrt {a b \,d^{2}}\, c \right ) \left (-b \,x^{2}+a \right )^{2} \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{2}+\left (-\frac {b \,d^{2} \left (-b \,x^{2}+a \right )^{2} \left (a \,d^{2}-2 b \,c^{2}-\sqrt {a b \,d^{2}}\, c \right ) \operatorname {arctanh}\left (\frac {b \sqrt {d x +c}}{\sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{2}+\sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}\, \sqrt {d x +c}\, \sqrt {a b \,d^{2}}\, \left (-b^{2} c d \,x^{3}-\frac {4 \left (\frac {9}{4} d^{2} x^{2}+\frac {3}{4} c d x +c^{2}\right ) a b}{5}+a^{2} d^{2}\right )\right ) \sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}\right )}{16 \sqrt {a b \,d^{2}}\, \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}\, \sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}\, a \,b^{2} \left (-b \,x^{2}+a \right )^{2}}\)	\(297\)
derivativedivides	\(-2 d^{4} \left (\frac {-\frac {5 c \left (d x +c \right )^{\frac {7}{2}}}{32 a \,d^{2}}-\frac {3 \left (3 a \,d^{2}-5 b \,c^{2}\right ) \left (d x +c \right )^{\frac {5}{2}}}{32 a b \,d^{2}}+\frac {15 c \left (a \,d^{2}-b \,c^{2}\right ) \left (d x +c \right )^{\frac {3}{2}}}{32 a b \,d^{2}}+\frac {5 \left (a \,d^{2}-b \,c^{2}\right )^{2} \sqrt {d x +c}}{32 a \,b^{2} d^{2}}}{\left (-b \left (d x +c \right )^{2}+2 b c \left (d x +c \right )+a \,d^{2}-b \,c^{2}\right )^{2}}+\frac {\frac {5 \left (-a \,d^{2}+2 b \,c^{2}-\sqrt {a b \,d^{2}}\, c \right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{64 \sqrt {a b \,d^{2}}\, \sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}-\frac {5 \left (a \,d^{2}-2 b \,c^{2}-\sqrt {a b \,d^{2}}\, c \right ) \operatorname {arctanh}\left (\frac {b \sqrt {d x +c}}{\sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{64 \sqrt {a b \,d^{2}}\, \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}}{a b \,d^{2}}\right )\)	\(317\)
default	\(2 d^{4} \left (-\frac {-\frac {5 c \left (d x +c \right )^{\frac {7}{2}}}{32 a \,d^{2}}-\frac {3 \left (3 a \,d^{2}-5 b \,c^{2}\right ) \left (d x +c \right )^{\frac {5}{2}}}{32 a b \,d^{2}}+\frac {15 c \left (a \,d^{2}-b \,c^{2}\right ) \left (d x +c \right )^{\frac {3}{2}}}{32 a b \,d^{2}}+\frac {5 \left (a \,d^{2}-b \,c^{2}\right )^{2} \sqrt {d x +c}}{32 a \,b^{2} d^{2}}}{\left (-b \left (d x +c \right )^{2}+2 b c \left (d x +c \right )+a \,d^{2}-b \,c^{2}\right )^{2}}-\frac {5 \left (\frac {\left (-a \,d^{2}+2 b \,c^{2}-\sqrt {a b \,d^{2}}\, c \right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{2 \sqrt {a b \,d^{2}}\, \sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}-\frac {\left (a \,d^{2}-2 b \,c^{2}-\sqrt {a b \,d^{2}}\, c \right ) \operatorname {arctanh}\left (\frac {b \sqrt {d x +c}}{\sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{2 \sqrt {a b \,d^{2}}\, \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{32 a b \,d^{2}}\right )\)	\(318\)

Input:

int(x*(d*x+c)^(5/2)/(-b*x^2+a)^3,x,method=_RETURNVERBOSE)
 

Output:

-5/16/(a*b*d^2)^(1/2)*(-1/2*d^2*((b*c+(a*b*d^2)^(1/2))*b)^(1/2)*b*(a*d^2-2 
*b*c^2+(a*b*d^2)^(1/2)*c)*(-b*x^2+a)^2*arctan(b*(d*x+c)^(1/2)/((-b*c+(a*b* 
d^2)^(1/2))*b)^(1/2))+(-1/2*b*d^2*(-b*x^2+a)^2*(a*d^2-2*b*c^2-(a*b*d^2)^(1 
/2)*c)*arctanh(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)*(d*x+c)^(1/2)*(a*b*d^2)^(1/2)*(-b^2*c*d*x^3-4/5*(9/4 
*d^2*x^2+3/4*c*d*x+c^2)*a*b+a^2*d^2))*((-b*c+(a*b*d^2)^(1/2))*b)^(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^2/(-b*x 
^2+a)^2
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 810 vs. \(2 (180) = 360\).

Time = 0.11 (sec) , antiderivative size = 810, normalized size of antiderivative = 3.40 \[ \int \frac {x (c+d x)^{5/2}}{\left (a-b x^2\right )^3} \, dx =\text {Too large to display} \] Input:

integrate(x*(d*x+c)^(5/2)/(-b*x^2+a)^3,x, algorithm="fricas")
 

Output:

1/64*(5*(a*b^4*x^4 - 2*a^2*b^3*x^2 + a^3*b^2)*sqrt((a^3*b^4*sqrt(d^10/(a^3 
*b^9)) + 4*b*c^3*d^2 - 3*a*c*d^4)/(a^3*b^4))*log(-125*(4*b*c^2*d^6 - a*d^8 
)*sqrt(d*x + c) + 125*(2*a^3*b^7*c*sqrt(d^10/(a^3*b^9)) + a^2*b^2*d^6)*sqr 
t((a^3*b^4*sqrt(d^10/(a^3*b^9)) + 4*b*c^3*d^2 - 3*a*c*d^4)/(a^3*b^4))) - 5 
*(a*b^4*x^4 - 2*a^2*b^3*x^2 + a^3*b^2)*sqrt((a^3*b^4*sqrt(d^10/(a^3*b^9)) 
+ 4*b*c^3*d^2 - 3*a*c*d^4)/(a^3*b^4))*log(-125*(4*b*c^2*d^6 - a*d^8)*sqrt( 
d*x + c) - 125*(2*a^3*b^7*c*sqrt(d^10/(a^3*b^9)) + a^2*b^2*d^6)*sqrt((a^3* 
b^4*sqrt(d^10/(a^3*b^9)) + 4*b*c^3*d^2 - 3*a*c*d^4)/(a^3*b^4))) - 5*(a*b^4 
*x^4 - 2*a^2*b^3*x^2 + a^3*b^2)*sqrt(-(a^3*b^4*sqrt(d^10/(a^3*b^9)) - 4*b* 
c^3*d^2 + 3*a*c*d^4)/(a^3*b^4))*log(-125*(4*b*c^2*d^6 - a*d^8)*sqrt(d*x + 
c) + 125*(2*a^3*b^7*c*sqrt(d^10/(a^3*b^9)) - a^2*b^2*d^6)*sqrt(-(a^3*b^4*s 
qrt(d^10/(a^3*b^9)) - 4*b*c^3*d^2 + 3*a*c*d^4)/(a^3*b^4))) + 5*(a*b^4*x^4 
- 2*a^2*b^3*x^2 + a^3*b^2)*sqrt(-(a^3*b^4*sqrt(d^10/(a^3*b^9)) - 4*b*c^3*d 
^2 + 3*a*c*d^4)/(a^3*b^4))*log(-125*(4*b*c^2*d^6 - a*d^8)*sqrt(d*x + c) - 
125*(2*a^3*b^7*c*sqrt(d^10/(a^3*b^9)) - a^2*b^2*d^6)*sqrt(-(a^3*b^4*sqrt(d 
^10/(a^3*b^9)) - 4*b*c^3*d^2 + 3*a*c*d^4)/(a^3*b^4))) + 4*(5*b^2*c*d*x^3 + 
 9*a*b*d^2*x^2 + 3*a*b*c*d*x + 4*a*b*c^2 - 5*a^2*d^2)*sqrt(d*x + c))/(a*b^ 
4*x^4 - 2*a^2*b^3*x^2 + a^3*b^2)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x (c+d x)^{5/2}}{\left (a-b x^2\right )^3} \, dx=\text {Timed out} \] Input:

integrate(x*(d*x+c)**(5/2)/(-b*x**2+a)**3,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {x (c+d x)^{5/2}}{\left (a-b x^2\right )^3} \, dx=\int { -\frac {{\left (d x + c\right )}^{\frac {5}{2}} x}{{\left (b x^{2} - a\right )}^{3}} \,d x } \] Input:

integrate(x*(d*x+c)^(5/2)/(-b*x^2+a)^3,x, algorithm="maxima")
 

Output:

-integrate((d*x + c)^(5/2)*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. 554 vs. \(2 (180) = 360\).

Time = 0.27 (sec) , antiderivative size = 554, normalized size of antiderivative = 2.33 \[ \int \frac {x (c+d x)^{5/2}}{\left (a-b x^2\right )^3} \, dx=-\frac {5 \, {\left (\sqrt {a b} a^{2} c d^{4} {\left | b \right |} + {\left (a b c^{2} d^{2} - a^{2} d^{4}\right )} {\left | a \right |} {\left | b \right |} {\left | d \right |} - {\left (2 \, \sqrt {a b} a b c^{3} d^{2} - \sqrt {a b} a^{2} c d^{4}\right )} {\left | b \right |}\right )} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-\frac {a b^{3} c + \sqrt {a^{2} b^{6} c^{2} - {\left (a b^{3} c^{2} - a^{2} b^{2} d^{2}\right )} a b^{3}}}{a b^{3}}}}\right )}{32 \, {\left (a^{2} b^{3} c - \sqrt {a b} a^{2} b^{2} d\right )} \sqrt {-b^{2} c - \sqrt {a b} b d} {\left | a \right |} {\left | d \right |}} + \frac {5 \, {\left (a^{2} b c d^{4} {\left | b \right |} - {\left (\sqrt {a b} b c^{2} d^{2} - \sqrt {a b} a d^{4}\right )} {\left | a \right |} {\left | b \right |} {\left | d \right |} - {\left (2 \, a b^{2} c^{3} d^{2} - a^{2} b c d^{4}\right )} {\left | b \right |}\right )} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-\frac {a b^{3} c - \sqrt {a^{2} b^{6} c^{2} - {\left (a b^{3} c^{2} - a^{2} b^{2} d^{2}\right )} a b^{3}}}{a b^{3}}}}\right )}{32 \, {\left (a^{2} b^{3} d + \sqrt {a b} a b^{3} c\right )} \sqrt {-b^{2} c + \sqrt {a b} b d} {\left | a \right |} {\left | d \right |}} + \frac {5 \, {\left (d x + c\right )}^{\frac {7}{2}} b^{2} c d^{2} - 15 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{2} c^{2} d^{2} + 15 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{2} c^{3} d^{2} - 5 \, \sqrt {d x + c} b^{2} c^{4} d^{2} + 9 \, {\left (d x + c\right )}^{\frac {5}{2}} a b d^{4} - 15 \, {\left (d x + c\right )}^{\frac {3}{2}} a b c d^{4} + 10 \, \sqrt {d x + c} a b c^{2} d^{4} - 5 \, \sqrt {d x + c} a^{2} d^{6}}{16 \, {\left ({\left (d x + c\right )}^{2} b - 2 \, {\left (d x + c\right )} b c + b c^{2} - a d^{2}\right )}^{2} a b^{2}} \] Input:

integrate(x*(d*x+c)^(5/2)/(-b*x^2+a)^3,x, algorithm="giac")
 

Output:

-5/32*(sqrt(a*b)*a^2*c*d^4*abs(b) + (a*b*c^2*d^2 - a^2*d^4)*abs(a)*abs(b)* 
abs(d) - (2*sqrt(a*b)*a*b*c^3*d^2 - sqrt(a*b)*a^2*c*d^4)*abs(b))*arctan(sq 
rt(d*x + c)/sqrt(-(a*b^3*c + sqrt(a^2*b^6*c^2 - (a*b^3*c^2 - a^2*b^2*d^2)* 
a*b^3))/(a*b^3)))/((a^2*b^3*c - sqrt(a*b)*a^2*b^2*d)*sqrt(-b^2*c - sqrt(a* 
b)*b*d)*abs(a)*abs(d)) + 5/32*(a^2*b*c*d^4*abs(b) - (sqrt(a*b)*b*c^2*d^2 - 
 sqrt(a*b)*a*d^4)*abs(a)*abs(b)*abs(d) - (2*a*b^2*c^3*d^2 - a^2*b*c*d^4)*a 
bs(b))*arctan(sqrt(d*x + c)/sqrt(-(a*b^3*c - sqrt(a^2*b^6*c^2 - (a*b^3*c^2 
 - a^2*b^2*d^2)*a*b^3))/(a*b^3)))/((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*(5*(d*x + c)^(7/2)*b^2*c*d^2 - 
 15*(d*x + c)^(5/2)*b^2*c^2*d^2 + 15*(d*x + c)^(3/2)*b^2*c^3*d^2 - 5*sqrt( 
d*x + c)*b^2*c^4*d^2 + 9*(d*x + c)^(5/2)*a*b*d^4 - 15*(d*x + c)^(3/2)*a*b* 
c*d^4 + 10*sqrt(d*x + c)*a*b*c^2*d^4 - 5*sqrt(d*x + c)*a^2*d^6)/(((d*x + c 
)^2*b - 2*(d*x + c)*b*c + b*c^2 - a*d^2)^2*a*b^2)
 

Mupad [B] (verification not implemented)

Time = 8.87 (sec) , antiderivative size = 882, normalized size of antiderivative = 3.71 \[ \int \frac {x (c+d x)^{5/2}}{\left (a-b x^2\right )^3} \, dx =\text {Too large to display} \] Input:

int((x*(c + d*x)^(5/2))/(a - b*x^2)^3,x)
 

Output:

((5*c*d^2*(c + d*x)^(7/2))/(16*a) + (15*(b*c^3*d^2 - a*c*d^4)*(c + d*x)^(3 
/2))/(16*a*b) - (5*(c + d*x)^(1/2)*(a^2*d^6 + b^2*c^4*d^2 - 2*a*b*c^2*d^4) 
)/(16*a*b^2) + (3*d^2*(3*a*d^2 - 5*b*c^2)*(c + d*x)^(5/2))/(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) - 
 2*atanh((25*d^8*(c + d*x)^(1/2)*((25*c^3*d^2)/(1024*a^3*b^3) - (75*c*d^4) 
/(4096*a^2*b^4) + (25*d^5*(a^9*b^9)^(1/2))/(4096*a^6*b^9))^(1/2))/(32*((12 
5*c*d^10)/(1024*a*b^2) - (125*c^3*d^8)/(1024*a^2*b) - (125*d^11*(a^9*b^9)^ 
(1/2))/(2048*a^5*b^7) + (125*c^2*d^9*(a^9*b^9)^(1/2))/(2048*a^6*b^6))) + ( 
25*c*d^7*(a^9*b^9)^(1/2)*(c + d*x)^(1/2)*((25*c^3*d^2)/(1024*a^3*b^3) - (7 
5*c*d^4)/(4096*a^2*b^4) + (25*d^5*(a^9*b^9)^(1/2))/(4096*a^6*b^9))^(1/2))/ 
(32*((125*d^11*(a^9*b^9)^(1/2))/(2048*b^3) - (125*a^4*b^2*c*d^10)/1024 + ( 
125*a^3*b^3*c^3*d^8)/1024 - (125*c^2*d^9*(a^9*b^9)^(1/2))/(2048*a*b^2))))* 
((25*(d^5*(a^9*b^9)^(1/2) - 3*a^4*b^5*c*d^4 + 4*a^3*b^6*c^3*d^2))/(4096*a^ 
6*b^9))^(1/2) - 2*atanh((25*d^8*(c + d*x)^(1/2)*((25*c^3*d^2)/(1024*a^3*b^ 
3) - (75*c*d^4)/(4096*a^2*b^4) - (25*d^5*(a^9*b^9)^(1/2))/(4096*a^6*b^9))^ 
(1/2))/(32*((125*c*d^10)/(1024*a*b^2) - (125*c^3*d^8)/(1024*a^2*b) + (125* 
d^11*(a^9*b^9)^(1/2))/(2048*a^5*b^7) - (125*c^2*d^9*(a^9*b^9)^(1/2))/(2048 
*a^6*b^6))) + (25*c*d^7*(a^9*b^9)^(1/2)*(c + d*x)^(1/2)*((25*c^3*d^2)/(102 
4*a^3*b^3) - (75*c*d^4)/(4096*a^2*b^4) - (25*d^5*(a^9*b^9)^(1/2))/(4096...
 

Reduce [B] (verification not implemented)

Time = 3.19 (sec) , antiderivative size = 976, normalized size of antiderivative = 4.10 \[ \int \frac {x (c+d x)^{5/2}}{\left (a-b x^2\right )^3} \, dx =\text {Too large to display} \] Input:

int(x*(d*x+c)^(5/2)/(-b*x^2+a)^3,x)
 

Output:

(20*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*d - 40*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*d*x**2 + 20*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*d*x**4 + 10*sqrt 
(b)*sqrt(sqrt(b)*sqrt(a)*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(sqr 
t(b)*sqrt(a)*d - b*c)))*a**3*d**2 - 20*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* 
d**2*x**2 + 10*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*d**2*x**4 + 10*sqrt(a)*s 
qrt(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*d - 20*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*b**2*c*d*x**2 
 + 10*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))*b**3*c*d*x**4 - 10*sqrt(a)*sqrt(sqrt(b)*sq 
rt(a)*d + b*c)*log(sqrt(sqrt(b)*sqrt(a)*d + b*c) + sqrt(b)*sqrt(c + d*x))* 
a**2*b*c*d + 20*sqrt(a)*sqrt(sqrt(b)*sqrt(a)*d + b*c)*log(sqrt(sqrt(b)*sqr 
t(a)*d + b*c) + sqrt(b)*sqrt(c + d*x))*a*b**2*c*d*x**2 - 10*sqrt(a)*sqrt(s 
qrt(b)*sqrt(a)*d + b*c)*log(sqrt(sqrt(b)*sqrt(a)*d + b*c) + sqrt(b)*sqrt(c 
 + d*x))*b**3*c*d*x**4 - 5*sqrt(b)*sqrt(sqrt(b)*sqrt(a)*d + b*c)*log( -...