\(\int \frac {(c+d x^2)^3}{x^{3/2} (a+b x^2)^2} \, dx\) [779]

Optimal result

Integrand size = 24, antiderivative size = 263 \[ \int \frac {\left (c+d x^2\right )^3}{x^{3/2} \left (a+b x^2\right )^2} \, dx=-\frac {2 c^3}{a^2 \sqrt {x}}+\frac {2 d^3 x^{3/2}}{3 b^2}-\frac {(b c-a d)^3 x^{3/2}}{2 a^2 b^2 \left (a+b x^2\right )}+\frac {(b c-a d)^2 (5 b c+7 a d) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{9/4} b^{11/4}}-\frac {(b c-a d)^2 (5 b c+7 a d) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{9/4} b^{11/4}}+\frac {(b c-a d)^2 (5 b c+7 a d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a}+\sqrt {b} x}\right )}{4 \sqrt {2} a^{9/4} b^{11/4}} \] Output:

-2*c^3/a^2/x^(1/2)+2/3*d^3*x^(3/2)/b^2-1/2*(-a*d+b*c)^3*x^(3/2)/a^2/b^2/(b 
*x^2+a)+1/8*(-a*d+b*c)^2*(7*a*d+5*b*c)*arctan(1-2^(1/2)*b^(1/4)*x^(1/2)/a^ 
(1/4))*2^(1/2)/a^(9/4)/b^(11/4)-1/8*(-a*d+b*c)^2*(7*a*d+5*b*c)*arctan(1+2^ 
(1/2)*b^(1/4)*x^(1/2)/a^(1/4))*2^(1/2)/a^(9/4)/b^(11/4)+1/8*(-a*d+b*c)^2*( 
7*a*d+5*b*c)*arctanh(2^(1/2)*a^(1/4)*b^(1/4)*x^(1/2)/(a^(1/2)+b^(1/2)*x))* 
2^(1/2)/a^(9/4)/b^(11/4)
 

Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 229, normalized size of antiderivative = 0.87 \[ \int \frac {\left (c+d x^2\right )^3}{x^{3/2} \left (a+b x^2\right )^2} \, dx=\frac {\frac {4 \sqrt [4]{a} b^{3/4} \left (-15 b^3 c^3 x^2+7 a^3 d^3 x^2+3 a b^2 c^2 \left (-4 c+3 d x^2\right )+a^2 b d^2 x^2 \left (-9 c+4 d x^2\right )\right )}{\sqrt {x} \left (a+b x^2\right )}+3 \sqrt {2} (b c-a d)^2 (5 b c+7 a d) \arctan \left (\frac {\sqrt {a}-\sqrt {b} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}\right )+3 \sqrt {2} (b c-a d)^2 (5 b c+7 a d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a}+\sqrt {b} x}\right )}{24 a^{9/4} b^{11/4}} \] Input:

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

Output:

((4*a^(1/4)*b^(3/4)*(-15*b^3*c^3*x^2 + 7*a^3*d^3*x^2 + 3*a*b^2*c^2*(-4*c + 
 3*d*x^2) + a^2*b*d^2*x^2*(-9*c + 4*d*x^2)))/(Sqrt[x]*(a + b*x^2)) + 3*Sqr 
t[2]*(b*c - a*d)^2*(5*b*c + 7*a*d)*ArcTan[(Sqrt[a] - Sqrt[b]*x)/(Sqrt[2]*a 
^(1/4)*b^(1/4)*Sqrt[x])] + 3*Sqrt[2]*(b*c - a*d)^2*(5*b*c + 7*a*d)*ArcTanh 
[(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])/(Sqrt[a] + Sqrt[b]*x)])/(24*a^(9/4)*b^( 
11/4))
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 373, normalized size of antiderivative = 1.42, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {368, 968, 25, 1040, 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.

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

\(\Big \downarrow \) 368

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

\(\Big \downarrow \) 968

\(\displaystyle 2 \left (\frac {\left (c+d x^2\right )^2 (b c-a d)}{4 a b \sqrt {x} \left (a+b x^2\right )}-\frac {\int -\frac {\left (d x^2+c\right ) \left (c (5 b c-a d)-d (3 b c-7 a d) x^2\right )}{x \left (b x^2+a\right )}d\sqrt {x}}{4 a b}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle 2 \left (\frac {\int \frac {\left (d x^2+c\right ) \left (c (5 b c-a d)-d (3 b c-7 a d) x^2\right )}{x \left (b x^2+a\right )}d\sqrt {x}}{4 a b}+\frac {\left (c+d x^2\right )^2 (b c-a d)}{4 a b \sqrt {x} \left (a+b x^2\right )}\right )\)

\(\Big \downarrow \) 1040

\(\displaystyle 2 \left (\frac {\int \left (-\frac {(a d-5 b c) c^2}{a x}-\frac {d^2 (3 b c-7 a d) x}{b}-\frac {(a d-b c)^2 (5 b c+7 a d) x}{a b \left (b x^2+a\right )}\right )d\sqrt {x}}{4 a b}+\frac {\left (c+d x^2\right )^2 (b c-a d)}{4 a b \sqrt {x} \left (a+b x^2\right )}\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle 2 \left (\frac {\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right ) (b c-a d)^2 (7 a d+5 b c)}{2 \sqrt {2} a^{5/4} b^{7/4}}-\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right ) (b c-a d)^2 (7 a d+5 b c)}{2 \sqrt {2} a^{5/4} b^{7/4}}-\frac {(b c-a d)^2 (7 a d+5 b c) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{4 \sqrt {2} a^{5/4} b^{7/4}}+\frac {(b c-a d)^2 (7 a d+5 b c) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{4 \sqrt {2} a^{5/4} b^{7/4}}-\frac {c^2 (5 b c-a d)}{a \sqrt {x}}-\frac {d^2 x^{3/2} (3 b c-7 a d)}{3 b}}{4 a b}+\frac {\left (c+d x^2\right )^2 (b c-a d)}{4 a b \sqrt {x} \left (a+b x^2\right )}\right )\)

Input:

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

Output:

2*(((b*c - a*d)*(c + d*x^2)^2)/(4*a*b*Sqrt[x]*(a + b*x^2)) + (-((c^2*(5*b* 
c - a*d))/(a*Sqrt[x])) - (d^2*(3*b*c - 7*a*d)*x^(3/2))/(3*b) + ((b*c - a*d 
)^2*(5*b*c + 7*a*d)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(2*Sqrt 
[2]*a^(5/4)*b^(7/4)) - ((b*c - a*d)^2*(5*b*c + 7*a*d)*ArcTan[1 + (Sqrt[2]* 
b^(1/4)*Sqrt[x])/a^(1/4)])/(2*Sqrt[2]*a^(5/4)*b^(7/4)) - ((b*c - a*d)^2*(5 
*b*c + 7*a*d)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/ 
(4*Sqrt[2]*a^(5/4)*b^(7/4)) + ((b*c - a*d)^2*(5*b*c + 7*a*d)*Log[Sqrt[a] + 
 Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(4*Sqrt[2]*a^(5/4)*b^(7/4)) 
)/(4*a*b))
 

Defintions of rubi rules used

rule 25

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

rule 368

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_) 
, x_Symbol] :> With[{k = Denominator[m]}, Simp[k/e   Subst[Int[x^(k*(m + 1) 
 - 1)*(a + b*(x^(k*2)/e^2))^p*(c + d*(x^(k*2)/e^2))^q, x], x, (e*x)^(1/k)], 
 x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && FractionQ[m 
] && IntegerQ[p]
 

rule 968

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ 
))^(q_), x_Symbol] :> Simp[(-(c*b - a*d))*(e*x)^(m + 1)*(a + b*x^n)^(p + 1) 
*((c + d*x^n)^(q - 1)/(a*b*e*n*(p + 1))), x] + Simp[1/(a*b*n*(p + 1))   Int 
[(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 2)*Simp[c*(c*b*n*(p + 1) + (c 
*b - a*d)*(m + 1)) + d*(c*b*n*(p + 1) + (c*b - a*d)*(m + n*(q - 1) + 1))*x^ 
n, x], x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[ 
n, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, 
 x]
 

rule 1040

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n 
_))^(q_.)*((e_) + (f_.)*(x_)^(n_))^(r_.), x_Symbol] :> Int[ExpandIntegrand[ 
(g*x)^m*(a + b*x^n)^p*(c + d*x^n)^q*(e + f*x^n)^r, x], x] /; FreeQ[{a, b, c 
, d, e, f, g, m, n}, x] && IGtQ[p, -2] && IGtQ[q, 0] && IGtQ[r, 0]
 

rule 2009

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

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.76

method	result	size
risch	\(\frac {-2 b^{2} c^{3}+\frac {2 a^{2} d^{3} x^{2}}{3}}{a^{2} \sqrt {x}\, b^{2}}-\frac {\left (2 a^{2} d^{2}-4 a b c d +2 b^{2} c^{2}\right ) \left (\frac {\left (-\frac {a d}{4}+\frac {b c}{4}\right ) x^{\frac {3}{2}}}{b \,x^{2}+a}+\frac {\left (\frac {7 a d}{4}+\frac {5 b c}{4}\right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}{x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{8 b \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{b^{2} a^{2}}\)	\(200\)
derivativedivides	\(\frac {2 d^{3} x^{\frac {3}{2}}}{3 b^{2}}-\frac {2 c^{3}}{a^{2} \sqrt {x}}-\frac {2 \left (\frac {\left (-\frac {1}{4} a^{3} d^{3}+\frac {3}{4} a^{2} b c \,d^{2}-\frac {3}{4} a \,b^{2} c^{2} d +\frac {1}{4} b^{3} c^{3}\right ) x^{\frac {3}{2}}}{b \,x^{2}+a}+\frac {\left (\frac {7}{4} a^{3} d^{3}-\frac {3}{4} a \,b^{2} c^{2} d +\frac {5}{4} b^{3} c^{3}-\frac {9}{4} a^{2} b c \,d^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}{x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{8 b \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{b^{2} a^{2}}\)	\(225\)
default	\(\frac {2 d^{3} x^{\frac {3}{2}}}{3 b^{2}}-\frac {2 c^{3}}{a^{2} \sqrt {x}}-\frac {2 \left (\frac {\left (-\frac {1}{4} a^{3} d^{3}+\frac {3}{4} a^{2} b c \,d^{2}-\frac {3}{4} a \,b^{2} c^{2} d +\frac {1}{4} b^{3} c^{3}\right ) x^{\frac {3}{2}}}{b \,x^{2}+a}+\frac {\left (\frac {7}{4} a^{3} d^{3}-\frac {3}{4} a \,b^{2} c^{2} d +\frac {5}{4} b^{3} c^{3}-\frac {9}{4} a^{2} b c \,d^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}{x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{8 b \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{b^{2} a^{2}}\)	\(225\)

Input:

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

Output:

2/3*(a^2*d^3*x^2-3*b^2*c^3)/a^2/x^(1/2)/b^2-1/b^2/a^2*(2*a^2*d^2-4*a*b*c*d 
+2*b^2*c^2)*((-1/4*a*d+1/4*b*c)*x^(3/2)/(b*x^2+a)+1/8*(7/4*a*d+5/4*b*c)/b/ 
(a/b)^(1/4)*2^(1/2)*(ln((x-(a/b)^(1/4)*x^(1/2)*2^(1/2)+(a/b)^(1/2))/(x+(a/ 
b)^(1/4)*x^(1/2)*2^(1/2)+(a/b)^(1/2)))+2*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2 
)+1)+2*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)-1)))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.16 (sec) , antiderivative size = 2104, normalized size of antiderivative = 8.00 \[ \int \frac {\left (c+d x^2\right )^3}{x^{3/2} \left (a+b x^2\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/24*(3*(a^2*b^3*x^3 + a^3*b^2*x)*(-(625*b^12*c^12 - 1500*a*b^11*c^11*d - 
 3150*a^2*b^10*c^10*d^2 + 11060*a^3*b^9*c^9*d^3 + 1071*a^4*b^8*c^8*d^4 - 2 
8728*a^5*b^7*c^7*d^5 + 19068*a^6*b^6*c^6*d^6 + 27144*a^7*b^5*c^5*d^7 - 376 
65*a^8*b^4*c^4*d^8 + 2324*a^9*b^3*c^3*d^9 + 19698*a^10*b^2*c^2*d^10 - 1234 
8*a^11*b*c*d^11 + 2401*a^12*d^12)/(a^9*b^11))^(1/4)*log(a^7*b^8*(-(625*b^1 
2*c^12 - 1500*a*b^11*c^11*d - 3150*a^2*b^10*c^10*d^2 + 11060*a^3*b^9*c^9*d 
^3 + 1071*a^4*b^8*c^8*d^4 - 28728*a^5*b^7*c^7*d^5 + 19068*a^6*b^6*c^6*d^6 
+ 27144*a^7*b^5*c^5*d^7 - 37665*a^8*b^4*c^4*d^8 + 2324*a^9*b^3*c^3*d^9 + 1 
9698*a^10*b^2*c^2*d^10 - 12348*a^11*b*c*d^11 + 2401*a^12*d^12)/(a^9*b^11)) 
^(3/4) + (125*b^9*c^9 - 225*a*b^8*c^8*d - 540*a^2*b^7*c^7*d^2 + 1308*a^3*b 
^6*c^6*d^3 + 342*a^4*b^5*c^5*d^4 - 2430*a^5*b^4*c^4*d^5 + 1140*a^6*b^3*c^3 
*d^6 + 1260*a^7*b^2*c^2*d^7 - 1323*a^8*b*c*d^8 + 343*a^9*d^9)*sqrt(x)) + 3 
*(-I*a^2*b^3*x^3 - I*a^3*b^2*x)*(-(625*b^12*c^12 - 1500*a*b^11*c^11*d - 31 
50*a^2*b^10*c^10*d^2 + 11060*a^3*b^9*c^9*d^3 + 1071*a^4*b^8*c^8*d^4 - 2872 
8*a^5*b^7*c^7*d^5 + 19068*a^6*b^6*c^6*d^6 + 27144*a^7*b^5*c^5*d^7 - 37665* 
a^8*b^4*c^4*d^8 + 2324*a^9*b^3*c^3*d^9 + 19698*a^10*b^2*c^2*d^10 - 12348*a 
^11*b*c*d^11 + 2401*a^12*d^12)/(a^9*b^11))^(1/4)*log(I*a^7*b^8*(-(625*b^12 
*c^12 - 1500*a*b^11*c^11*d - 3150*a^2*b^10*c^10*d^2 + 11060*a^3*b^9*c^9*d^ 
3 + 1071*a^4*b^8*c^8*d^4 - 28728*a^5*b^7*c^7*d^5 + 19068*a^6*b^6*c^6*d^6 + 
 27144*a^7*b^5*c^5*d^7 - 37665*a^8*b^4*c^4*d^8 + 2324*a^9*b^3*c^3*d^9 +...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.16 \[ \int \frac {\left (c+d x^2\right )^3}{x^{3/2} \left (a+b x^2\right )^2} \, dx=\frac {2 \, d^{3} x^{\frac {3}{2}}}{3 \, b^{2}} - \frac {4 \, a b^{2} c^{3} + {\left (5 \, b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} x^{2}}{2 \, {\left (a^{2} b^{3} x^{\frac {5}{2}} + a^{3} b^{2} \sqrt {x}\right )}} - \frac {{\left (5 \, b^{3} c^{3} - 3 \, a b^{2} c^{2} d - 9 \, a^{2} b c d^{2} + 7 \, a^{3} d^{3}\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} - \frac {\sqrt {2} \log \left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}}\right )}}{16 \, a^{2} b^{2}} \] Input:

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 504 vs. \(2 (201) = 402\).

Time = 0.14 (sec) , antiderivative size = 504, normalized size of antiderivative = 1.92 \[ \int \frac {\left (c+d x^2\right )^3}{x^{3/2} \left (a+b x^2\right )^2} \, dx=\frac {2 \, d^{3} x^{\frac {3}{2}}}{3 \, b^{2}} - \frac {5 \, b^{3} c^{3} x^{2} - 3 \, a b^{2} c^{2} d x^{2} + 3 \, a^{2} b c d^{2} x^{2} - a^{3} d^{3} x^{2} + 4 \, a b^{2} c^{3}}{2 \, {\left (b x^{\frac {5}{2}} + a \sqrt {x}\right )} a^{2} b^{2}} - \frac {\sqrt {2} {\left (5 \, \left (a b^{3}\right )^{\frac {3}{4}} b^{3} c^{3} - 3 \, \left (a b^{3}\right )^{\frac {3}{4}} a b^{2} c^{2} d - 9 \, \left (a b^{3}\right )^{\frac {3}{4}} a^{2} b c d^{2} + 7 \, \left (a b^{3}\right )^{\frac {3}{4}} a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{8 \, a^{3} b^{5}} - \frac {\sqrt {2} {\left (5 \, \left (a b^{3}\right )^{\frac {3}{4}} b^{3} c^{3} - 3 \, \left (a b^{3}\right )^{\frac {3}{4}} a b^{2} c^{2} d - 9 \, \left (a b^{3}\right )^{\frac {3}{4}} a^{2} b c d^{2} + 7 \, \left (a b^{3}\right )^{\frac {3}{4}} a^{3} d^{3}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{8 \, a^{3} b^{5}} + \frac {\sqrt {2} {\left (5 \, \left (a b^{3}\right )^{\frac {3}{4}} b^{3} c^{3} - 3 \, \left (a b^{3}\right )^{\frac {3}{4}} a b^{2} c^{2} d - 9 \, \left (a b^{3}\right )^{\frac {3}{4}} a^{2} b c d^{2} + 7 \, \left (a b^{3}\right )^{\frac {3}{4}} a^{3} d^{3}\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{16 \, a^{3} b^{5}} - \frac {\sqrt {2} {\left (5 \, \left (a b^{3}\right )^{\frac {3}{4}} b^{3} c^{3} - 3 \, \left (a b^{3}\right )^{\frac {3}{4}} a b^{2} c^{2} d - 9 \, \left (a b^{3}\right )^{\frac {3}{4}} a^{2} b c d^{2} + 7 \, \left (a b^{3}\right )^{\frac {3}{4}} a^{3} d^{3}\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{16 \, a^{3} b^{5}} \] Input:

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

Output:

2/3*d^3*x^(3/2)/b^2 - 1/2*(5*b^3*c^3*x^2 - 3*a*b^2*c^2*d*x^2 + 3*a^2*b*c*d 
^2*x^2 - a^3*d^3*x^2 + 4*a*b^2*c^3)/((b*x^(5/2) + a*sqrt(x))*a^2*b^2) - 1/ 
8*sqrt(2)*(5*(a*b^3)^(3/4)*b^3*c^3 - 3*(a*b^3)^(3/4)*a*b^2*c^2*d - 9*(a*b^ 
3)^(3/4)*a^2*b*c*d^2 + 7*(a*b^3)^(3/4)*a^3*d^3)*arctan(1/2*sqrt(2)*(sqrt(2 
)*(a/b)^(1/4) + 2*sqrt(x))/(a/b)^(1/4))/(a^3*b^5) - 1/8*sqrt(2)*(5*(a*b^3) 
^(3/4)*b^3*c^3 - 3*(a*b^3)^(3/4)*a*b^2*c^2*d - 9*(a*b^3)^(3/4)*a^2*b*c*d^2 
 + 7*(a*b^3)^(3/4)*a^3*d^3)*arctan(-1/2*sqrt(2)*(sqrt(2)*(a/b)^(1/4) - 2*s 
qrt(x))/(a/b)^(1/4))/(a^3*b^5) + 1/16*sqrt(2)*(5*(a*b^3)^(3/4)*b^3*c^3 - 3 
*(a*b^3)^(3/4)*a*b^2*c^2*d - 9*(a*b^3)^(3/4)*a^2*b*c*d^2 + 7*(a*b^3)^(3/4) 
*a^3*d^3)*log(sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/(a^3*b^5) - 1/1 
6*sqrt(2)*(5*(a*b^3)^(3/4)*b^3*c^3 - 3*(a*b^3)^(3/4)*a*b^2*c^2*d - 9*(a*b^ 
3)^(3/4)*a^2*b*c*d^2 + 7*(a*b^3)^(3/4)*a^3*d^3)*log(-sqrt(2)*sqrt(x)*(a/b) 
^(1/4) + x + sqrt(a/b))/(a^3*b^5)
 

Mupad [B] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 657, normalized size of antiderivative = 2.50 \[ \int \frac {\left (c+d x^2\right )^3}{x^{3/2} \left (a+b x^2\right )^2} \, dx=\frac {\frac {x^2\,\left (a^3\,d^3-3\,a^2\,b\,c\,d^2+3\,a\,b^2\,c^2\,d-5\,b^3\,c^3\right )}{2\,a^2}-\frac {2\,b^2\,c^3}{a}}{b^3\,x^{5/2}+a\,b^2\,\sqrt {x}}+\frac {2\,d^3\,x^{3/2}}{3\,b^2}-\frac {\mathrm {atan}\left (\frac {\sqrt {x}\,{\left (a\,d-b\,c\right )}^2\,\left (7\,a\,d+5\,b\,c\right )\,\left (1568\,a^{13}\,b^8\,d^6-4032\,a^{12}\,b^9\,c\,d^5+1248\,a^{11}\,b^{10}\,c^2\,d^4+3968\,a^{10}\,b^{11}\,c^3\,d^3-2592\,a^9\,b^{12}\,c^4\,d^2-960\,a^8\,b^{13}\,c^5\,d+800\,a^7\,b^{14}\,c^6\right )}{4\,{\left (-a\right )}^{9/4}\,b^{11/4}\,\left (2744\,a^{14}\,b^5\,d^9-10584\,a^{13}\,b^6\,c\,d^8+10080\,a^{12}\,b^7\,c^2\,d^7+9120\,a^{11}\,b^8\,c^3\,d^6-19440\,a^{10}\,b^9\,c^4\,d^5+2736\,a^9\,b^{10}\,c^5\,d^4+10464\,a^8\,b^{11}\,c^6\,d^3-4320\,a^7\,b^{12}\,c^7\,d^2-1800\,a^6\,b^{13}\,c^8\,d+1000\,a^5\,b^{14}\,c^9\right )}\right )\,{\left (a\,d-b\,c\right )}^2\,\left (7\,a\,d+5\,b\,c\right )}{4\,{\left (-a\right )}^{9/4}\,b^{11/4}}-\frac {\mathrm {atan}\left (\frac {\sqrt {x}\,{\left (a\,d-b\,c\right )}^2\,\left (7\,a\,d+5\,b\,c\right )\,\left (1568\,a^{13}\,b^8\,d^6-4032\,a^{12}\,b^9\,c\,d^5+1248\,a^{11}\,b^{10}\,c^2\,d^4+3968\,a^{10}\,b^{11}\,c^3\,d^3-2592\,a^9\,b^{12}\,c^4\,d^2-960\,a^8\,b^{13}\,c^5\,d+800\,a^7\,b^{14}\,c^6\right )\,1{}\mathrm {i}}{4\,{\left (-a\right )}^{9/4}\,b^{11/4}\,\left (2744\,a^{14}\,b^5\,d^9-10584\,a^{13}\,b^6\,c\,d^8+10080\,a^{12}\,b^7\,c^2\,d^7+9120\,a^{11}\,b^8\,c^3\,d^6-19440\,a^{10}\,b^9\,c^4\,d^5+2736\,a^9\,b^{10}\,c^5\,d^4+10464\,a^8\,b^{11}\,c^6\,d^3-4320\,a^7\,b^{12}\,c^7\,d^2-1800\,a^6\,b^{13}\,c^8\,d+1000\,a^5\,b^{14}\,c^9\right )}\right )\,{\left (a\,d-b\,c\right )}^2\,\left (7\,a\,d+5\,b\,c\right )\,1{}\mathrm {i}}{4\,{\left (-a\right )}^{9/4}\,b^{11/4}} \] Input:

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

Output:

((x^2*(a^3*d^3 - 5*b^3*c^3 + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2))/(2*a^2) - (2* 
b^2*c^3)/a)/(b^3*x^(5/2) + a*b^2*x^(1/2)) + (2*d^3*x^(3/2))/(3*b^2) - (ata 
n((x^(1/2)*(a*d - b*c)^2*(7*a*d + 5*b*c)*(800*a^7*b^14*c^6 + 1568*a^13*b^8 
*d^6 - 960*a^8*b^13*c^5*d - 4032*a^12*b^9*c*d^5 - 2592*a^9*b^12*c^4*d^2 + 
3968*a^10*b^11*c^3*d^3 + 1248*a^11*b^10*c^2*d^4))/(4*(-a)^(9/4)*b^(11/4)*( 
1000*a^5*b^14*c^9 + 2744*a^14*b^5*d^9 - 1800*a^6*b^13*c^8*d - 10584*a^13*b 
^6*c*d^8 - 4320*a^7*b^12*c^7*d^2 + 10464*a^8*b^11*c^6*d^3 + 2736*a^9*b^10* 
c^5*d^4 - 19440*a^10*b^9*c^4*d^5 + 9120*a^11*b^8*c^3*d^6 + 10080*a^12*b^7* 
c^2*d^7)))*(a*d - b*c)^2*(7*a*d + 5*b*c))/(4*(-a)^(9/4)*b^(11/4)) - (atan( 
(x^(1/2)*(a*d - b*c)^2*(7*a*d + 5*b*c)*(800*a^7*b^14*c^6 + 1568*a^13*b^8*d 
^6 - 960*a^8*b^13*c^5*d - 4032*a^12*b^9*c*d^5 - 2592*a^9*b^12*c^4*d^2 + 39 
68*a^10*b^11*c^3*d^3 + 1248*a^11*b^10*c^2*d^4)*1i)/(4*(-a)^(9/4)*b^(11/4)* 
(1000*a^5*b^14*c^9 + 2744*a^14*b^5*d^9 - 1800*a^6*b^13*c^8*d - 10584*a^13* 
b^6*c*d^8 - 4320*a^7*b^12*c^7*d^2 + 10464*a^8*b^11*c^6*d^3 + 2736*a^9*b^10 
*c^5*d^4 - 19440*a^10*b^9*c^4*d^5 + 9120*a^11*b^8*c^3*d^6 + 10080*a^12*b^7 
*c^2*d^7)))*(a*d - b*c)^2*(7*a*d + 5*b*c)*1i)/(4*(-a)^(9/4)*b^(11/4))
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 1405, normalized size of antiderivative = 5.34 \[ \int \frac {\left (c+d x^2\right )^3}{x^{3/2} \left (a+b x^2\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(42*sqrt(x)*b**(1/4)*a**(3/4)*sqrt(2)*atan((b**(1/4)*a**(1/4)*sqrt(2) - 2* 
sqrt(x)*sqrt(b))/(b**(1/4)*a**(1/4)*sqrt(2)))*a**4*d**3 - 54*sqrt(x)*b**(1 
/4)*a**(3/4)*sqrt(2)*atan((b**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(x)*sqrt(b))/ 
(b**(1/4)*a**(1/4)*sqrt(2)))*a**3*b*c*d**2 + 42*sqrt(x)*b**(1/4)*a**(3/4)* 
sqrt(2)*atan((b**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(x)*sqrt(b))/(b**(1/4)*a** 
(1/4)*sqrt(2)))*a**3*b*d**3*x**2 - 18*sqrt(x)*b**(1/4)*a**(3/4)*sqrt(2)*at 
an((b**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(x)*sqrt(b))/(b**(1/4)*a**(1/4)*sqrt 
(2)))*a**2*b**2*c**2*d - 54*sqrt(x)*b**(1/4)*a**(3/4)*sqrt(2)*atan((b**(1/ 
4)*a**(1/4)*sqrt(2) - 2*sqrt(x)*sqrt(b))/(b**(1/4)*a**(1/4)*sqrt(2)))*a**2 
*b**2*c*d**2*x**2 + 30*sqrt(x)*b**(1/4)*a**(3/4)*sqrt(2)*atan((b**(1/4)*a* 
*(1/4)*sqrt(2) - 2*sqrt(x)*sqrt(b))/(b**(1/4)*a**(1/4)*sqrt(2)))*a*b**3*c* 
*3 - 18*sqrt(x)*b**(1/4)*a**(3/4)*sqrt(2)*atan((b**(1/4)*a**(1/4)*sqrt(2) 
- 2*sqrt(x)*sqrt(b))/(b**(1/4)*a**(1/4)*sqrt(2)))*a*b**3*c**2*d*x**2 + 30* 
sqrt(x)*b**(1/4)*a**(3/4)*sqrt(2)*atan((b**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt 
(x)*sqrt(b))/(b**(1/4)*a**(1/4)*sqrt(2)))*b**4*c**3*x**2 - 42*sqrt(x)*b**( 
1/4)*a**(3/4)*sqrt(2)*atan((b**(1/4)*a**(1/4)*sqrt(2) + 2*sqrt(x)*sqrt(b)) 
/(b**(1/4)*a**(1/4)*sqrt(2)))*a**4*d**3 + 54*sqrt(x)*b**(1/4)*a**(3/4)*sqr 
t(2)*atan((b**(1/4)*a**(1/4)*sqrt(2) + 2*sqrt(x)*sqrt(b))/(b**(1/4)*a**(1/ 
4)*sqrt(2)))*a**3*b*c*d**2 - 42*sqrt(x)*b**(1/4)*a**(3/4)*sqrt(2)*atan((b* 
*(1/4)*a**(1/4)*sqrt(2) + 2*sqrt(x)*sqrt(b))/(b**(1/4)*a**(1/4)*sqrt(2)...