\(\int \frac {\sqrt {x} (a+b x^2)^2}{(c+d x^2)^3} \, dx\) [758]

Optimal result

Integrand size = 24, antiderivative size = 288 \[ \int \frac {\sqrt {x} \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\frac {(b c-a d)^2 x^{3/2}}{4 c d^2 \left (c+d x^2\right )^2}-\frac {(b c-a d) (11 b c+5 a d) x^{3/2}}{16 c^2 d^2 \left (c+d x^2\right )}-\frac {\left (21 b^2 c^2+6 a b c d+5 a^2 d^2\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{32 \sqrt {2} c^{9/4} d^{11/4}}+\frac {\left (21 b^2 c^2+6 a b c d+5 a^2 d^2\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{32 \sqrt {2} c^{9/4} d^{11/4}}-\frac {\left (21 b^2 c^2+6 a b c d+5 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}}{\sqrt {c}+\sqrt {d} x}\right )}{32 \sqrt {2} c^{9/4} d^{11/4}} \] Output:

1/4*(-a*d+b*c)^2*x^(3/2)/c/d^2/(d*x^2+c)^2-1/16*(-a*d+b*c)*(5*a*d+11*b*c)* 
x^(3/2)/c^2/d^2/(d*x^2+c)-1/64*(5*a^2*d^2+6*a*b*c*d+21*b^2*c^2)*arctan(1-2 
^(1/2)*d^(1/4)*x^(1/2)/c^(1/4))*2^(1/2)/c^(9/4)/d^(11/4)+1/64*(5*a^2*d^2+6 
*a*b*c*d+21*b^2*c^2)*arctan(1+2^(1/2)*d^(1/4)*x^(1/2)/c^(1/4))*2^(1/2)/c^( 
9/4)/d^(11/4)-1/64*(5*a^2*d^2+6*a*b*c*d+21*b^2*c^2)*arctanh(2^(1/2)*c^(1/4 
)*d^(1/4)*x^(1/2)/(c^(1/2)+d^(1/2)*x))*2^(1/2)/c^(9/4)/d^(11/4)
 

Mathematica [A] (verified)

Time = 0.66 (sec) , antiderivative size = 209, normalized size of antiderivative = 0.73 \[ \int \frac {\sqrt {x} \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\frac {-\frac {4 \sqrt [4]{c} d^{3/4} (b c-a d) x^{3/2} \left (a d \left (9 c+5 d x^2\right )+b c \left (7 c+11 d x^2\right )\right )}{\left (c+d x^2\right )^2}-\sqrt {2} \left (21 b^2 c^2+6 a b c d+5 a^2 d^2\right ) \arctan \left (\frac {\sqrt {c}-\sqrt {d} x}{\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}}\right )-\sqrt {2} \left (21 b^2 c^2+6 a b c d+5 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}}{\sqrt {c}+\sqrt {d} x}\right )}{64 c^{9/4} d^{11/4}} \] Input:

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

Output:

((-4*c^(1/4)*d^(3/4)*(b*c - a*d)*x^(3/2)*(a*d*(9*c + 5*d*x^2) + b*c*(7*c + 
 11*d*x^2)))/(c + d*x^2)^2 - Sqrt[2]*(21*b^2*c^2 + 6*a*b*c*d + 5*a^2*d^2)* 
ArcTan[(Sqrt[c] - Sqrt[d]*x)/(Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x])] - Sqrt[2]* 
(21*b^2*c^2 + 6*a*b*c*d + 5*a^2*d^2)*ArcTanh[(Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt 
[x])/(Sqrt[c] + Sqrt[d]*x)])/(64*c^(9/4)*d^(11/4))
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.13, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {366, 27, 362, 266, 826, 1476, 1082, 217, 1479, 25, 27, 1103}

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.

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

\(\Big \downarrow \) 366

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 362

\(\displaystyle \frac {\frac {\left (5 a^2 d^2+6 a b c d+21 b^2 c^2\right ) \int \frac {\sqrt {x}}{d x^2+c}dx}{4 c}-\frac {x^{3/2} (b c-a d) (5 a d+11 b c)}{2 c \left (c+d x^2\right )}}{8 c d^2}+\frac {x^{3/2} (b c-a d)^2}{4 c d^2 \left (c+d x^2\right )^2}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {\frac {\left (5 a^2 d^2+6 a b c d+21 b^2 c^2\right ) \int \frac {x}{d x^2+c}d\sqrt {x}}{2 c}-\frac {x^{3/2} (b c-a d) (5 a d+11 b c)}{2 c \left (c+d x^2\right )}}{8 c d^2}+\frac {x^{3/2} (b c-a d)^2}{4 c d^2 \left (c+d x^2\right )^2}\)

\(\Big \downarrow \) 826

\(\displaystyle \frac {\frac {\left (5 a^2 d^2+6 a b c d+21 b^2 c^2\right ) \left (\frac {\int \frac {\sqrt {d} x+\sqrt {c}}{d x^2+c}d\sqrt {x}}{2 \sqrt {d}}-\frac {\int \frac {\sqrt {c}-\sqrt {d} x}{d x^2+c}d\sqrt {x}}{2 \sqrt {d}}\right )}{2 c}-\frac {x^{3/2} (b c-a d) (5 a d+11 b c)}{2 c \left (c+d x^2\right )}}{8 c d^2}+\frac {x^{3/2} (b c-a d)^2}{4 c d^2 \left (c+d x^2\right )^2}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {\frac {\left (5 a^2 d^2+6 a b c d+21 b^2 c^2\right ) \left (\frac {\frac {\int \frac {1}{x-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{d}}+\frac {\sqrt {c}}{\sqrt {d}}}d\sqrt {x}}{2 \sqrt {d}}+\frac {\int \frac {1}{x+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{d}}+\frac {\sqrt {c}}{\sqrt {d}}}d\sqrt {x}}{2 \sqrt {d}}}{2 \sqrt {d}}-\frac {\int \frac {\sqrt {c}-\sqrt {d} x}{d x^2+c}d\sqrt {x}}{2 \sqrt {d}}\right )}{2 c}-\frac {x^{3/2} (b c-a d) (5 a d+11 b c)}{2 c \left (c+d x^2\right )}}{8 c d^2}+\frac {x^{3/2} (b c-a d)^2}{4 c d^2 \left (c+d x^2\right )^2}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {\frac {\left (5 a^2 d^2+6 a b c d+21 b^2 c^2\right ) \left (\frac {\frac {\int \frac {1}{-x-1}d\left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{\sqrt {2} \sqrt [4]{c} \sqrt [4]{d}}-\frac {\int \frac {1}{-x-1}d\left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )}{\sqrt {2} \sqrt [4]{c} \sqrt [4]{d}}}{2 \sqrt {d}}-\frac {\int \frac {\sqrt {c}-\sqrt {d} x}{d x^2+c}d\sqrt {x}}{2 \sqrt {d}}\right )}{2 c}-\frac {x^{3/2} (b c-a d) (5 a d+11 b c)}{2 c \left (c+d x^2\right )}}{8 c d^2}+\frac {x^{3/2} (b c-a d)^2}{4 c d^2 \left (c+d x^2\right )^2}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\frac {\left (5 a^2 d^2+6 a b c d+21 b^2 c^2\right ) \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )}{\sqrt {2} \sqrt [4]{c} \sqrt [4]{d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{\sqrt {2} \sqrt [4]{c} \sqrt [4]{d}}}{2 \sqrt {d}}-\frac {\int \frac {\sqrt {c}-\sqrt {d} x}{d x^2+c}d\sqrt {x}}{2 \sqrt {d}}\right )}{2 c}-\frac {x^{3/2} (b c-a d) (5 a d+11 b c)}{2 c \left (c+d x^2\right )}}{8 c d^2}+\frac {x^{3/2} (b c-a d)^2}{4 c d^2 \left (c+d x^2\right )^2}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {\frac {\left (5 a^2 d^2+6 a b c d+21 b^2 c^2\right ) \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )}{\sqrt {2} \sqrt [4]{c} \sqrt [4]{d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{\sqrt {2} \sqrt [4]{c} \sqrt [4]{d}}}{2 \sqrt {d}}-\frac {-\frac {\int -\frac {\sqrt {2} \sqrt [4]{c}-2 \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{d} \left (x-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{d}}+\frac {\sqrt {c}}{\sqrt {d}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{c} \sqrt [4]{d}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{d} \sqrt {x}+\sqrt [4]{c}\right )}{\sqrt [4]{d} \left (x+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{d}}+\frac {\sqrt {c}}{\sqrt {d}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{c} \sqrt [4]{d}}}{2 \sqrt {d}}\right )}{2 c}-\frac {x^{3/2} (b c-a d) (5 a d+11 b c)}{2 c \left (c+d x^2\right )}}{8 c d^2}+\frac {x^{3/2} (b c-a d)^2}{4 c d^2 \left (c+d x^2\right )^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\left (5 a^2 d^2+6 a b c d+21 b^2 c^2\right ) \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )}{\sqrt {2} \sqrt [4]{c} \sqrt [4]{d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{\sqrt {2} \sqrt [4]{c} \sqrt [4]{d}}}{2 \sqrt {d}}-\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{c}-2 \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{d} \left (x-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{d}}+\frac {\sqrt {c}}{\sqrt {d}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{c} \sqrt [4]{d}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{d} \sqrt {x}+\sqrt [4]{c}\right )}{\sqrt [4]{d} \left (x+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{d}}+\frac {\sqrt {c}}{\sqrt {d}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{c} \sqrt [4]{d}}}{2 \sqrt {d}}\right )}{2 c}-\frac {x^{3/2} (b c-a d) (5 a d+11 b c)}{2 c \left (c+d x^2\right )}}{8 c d^2}+\frac {x^{3/2} (b c-a d)^2}{4 c d^2 \left (c+d x^2\right )^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\left (5 a^2 d^2+6 a b c d+21 b^2 c^2\right ) \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )}{\sqrt {2} \sqrt [4]{c} \sqrt [4]{d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{\sqrt {2} \sqrt [4]{c} \sqrt [4]{d}}}{2 \sqrt {d}}-\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{c}-2 \sqrt [4]{d} \sqrt {x}}{x-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{d}}+\frac {\sqrt {c}}{\sqrt {d}}}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{c} \sqrt {d}}+\frac {\int \frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}+\sqrt [4]{c}}{x+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{d}}+\frac {\sqrt {c}}{\sqrt {d}}}d\sqrt {x}}{2 \sqrt [4]{c} \sqrt {d}}}{2 \sqrt {d}}\right )}{2 c}-\frac {x^{3/2} (b c-a d) (5 a d+11 b c)}{2 c \left (c+d x^2\right )}}{8 c d^2}+\frac {x^{3/2} (b c-a d)^2}{4 c d^2 \left (c+d x^2\right )^2}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {\frac {\left (5 a^2 d^2+6 a b c d+21 b^2 c^2\right ) \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )}{\sqrt {2} \sqrt [4]{c} \sqrt [4]{d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{\sqrt {2} \sqrt [4]{c} \sqrt [4]{d}}}{2 \sqrt {d}}-\frac {\frac {\log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{2 \sqrt {2} \sqrt [4]{c} \sqrt [4]{d}}-\frac {\log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{2 \sqrt {2} \sqrt [4]{c} \sqrt [4]{d}}}{2 \sqrt {d}}\right )}{2 c}-\frac {x^{3/2} (b c-a d) (5 a d+11 b c)}{2 c \left (c+d x^2\right )}}{8 c d^2}+\frac {x^{3/2} (b c-a d)^2}{4 c d^2 \left (c+d x^2\right )^2}\)

Input:

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

Output:

((b*c - a*d)^2*x^(3/2))/(4*c*d^2*(c + d*x^2)^2) + (-1/2*((b*c - a*d)*(11*b 
*c + 5*a*d)*x^(3/2))/(c*(c + d*x^2)) + ((21*b^2*c^2 + 6*a*b*c*d + 5*a^2*d^ 
2)*((-(ArcTan[1 - (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)]/(Sqrt[2]*c^(1/4)*d^(1 
/4))) + ArcTan[1 + (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)]/(Sqrt[2]*c^(1/4)*d^( 
1/4)))/(2*Sqrt[d]) - (-1/2*Log[Sqrt[c] - Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + 
 Sqrt[d]*x]/(Sqrt[2]*c^(1/4)*d^(1/4)) + Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^(1 
/4)*Sqrt[x] + Sqrt[d]*x]/(2*Sqrt[2]*c^(1/4)*d^(1/4)))/(2*Sqrt[d])))/(2*c)) 
/(8*c*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 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 362

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[(-(b*c - a*d))*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(2*a*b*e 
*(p + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(2*a*b*(p + 1))   I 
nt[(e*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && N 
eQ[b*c - a*d, 0] && LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) || 
  !RationalQ[m] || (ILtQ[p + 1/2, 0] && LeQ[-1, m, -2*(p + 1)]))
 

rule 366

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

rule 826

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 
2]], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*s)   Int[(r + s*x^2)/(a + b*x^ 
4), x], x] - Simp[1/(2*s)   Int[(r - s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{ 
a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] 
 && AtomQ[SplitProduct[SumBaseQ, b]]))
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1476

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

rule 1479

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
-2*(d/e), 2]}, Simp[e/(2*c*q)   Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], 
 x] + Simp[e/(2*c*q)   Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F 
reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
 

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 213, normalized size of antiderivative = 0.74

method	result	size
derivativedivides	\(\frac {\frac {\left (5 a^{2} d^{2}+6 a b c d -11 b^{2} c^{2}\right ) x^{\frac {7}{2}}}{16 c^{2} d}+\frac {\left (9 a^{2} d^{2}-2 a b c d -7 b^{2} c^{2}\right ) x^{\frac {3}{2}}}{16 c \,d^{2}}}{\left (x^{2} d +c \right )^{2}}+\frac {\left (5 a^{2} d^{2}+6 a b c d +21 b^{2} c^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {c}{d}}}{x +\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {c}{d}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )\right )}{128 c^{2} d^{3} \left (\frac {c}{d}\right )^{\frac {1}{4}}}\)	\(213\)
default	\(\frac {\frac {\left (5 a^{2} d^{2}+6 a b c d -11 b^{2} c^{2}\right ) x^{\frac {7}{2}}}{16 c^{2} d}+\frac {\left (9 a^{2} d^{2}-2 a b c d -7 b^{2} c^{2}\right ) x^{\frac {3}{2}}}{16 c \,d^{2}}}{\left (x^{2} d +c \right )^{2}}+\frac {\left (5 a^{2} d^{2}+6 a b c d +21 b^{2} c^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {c}{d}}}{x +\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {c}{d}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )\right )}{128 c^{2} d^{3} \left (\frac {c}{d}\right )^{\frac {1}{4}}}\)	\(213\)

Input:

int(x^(1/2)*(b*x^2+a)^2/(d*x^2+c)^3,x,method=_RETURNVERBOSE)
 

Output:

2*(1/32*(5*a^2*d^2+6*a*b*c*d-11*b^2*c^2)/c^2/d*x^(7/2)+1/32*(9*a^2*d^2-2*a 
*b*c*d-7*b^2*c^2)/c/d^2*x^(3/2))/(d*x^2+c)^2+1/128*(5*a^2*d^2+6*a*b*c*d+21 
*b^2*c^2)/c^2/d^3/(c/d)^(1/4)*2^(1/2)*(ln((x-(c/d)^(1/4)*x^(1/2)*2^(1/2)+( 
c/d)^(1/2))/(x+(c/d)^(1/4)*x^(1/2)*2^(1/2)+(c/d)^(1/2)))+2*arctan(2^(1/2)/ 
(c/d)^(1/4)*x^(1/2)+1)+2*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)-1))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.12 (sec) , antiderivative size = 1532, normalized size of antiderivative = 5.32 \[ \int \frac {\sqrt {x} \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\text {Too large to display} \] Input:

integrate(x^(1/2)*(b*x^2+a)^2/(d*x^2+c)^3,x, algorithm="fricas")
 

Output:

1/64*((c^2*d^4*x^4 + 2*c^3*d^3*x^2 + c^4*d^2)*(-(194481*b^8*c^8 + 222264*a 
*b^7*c^7*d + 280476*a^2*b^6*c^6*d^2 + 176904*a^3*b^5*c^5*d^3 + 112806*a^4* 
b^4*c^4*d^4 + 42120*a^5*b^3*c^3*d^5 + 15900*a^6*b^2*c^2*d^6 + 3000*a^7*b*c 
*d^7 + 625*a^8*d^8)/(c^9*d^11))^(1/4)*log(c^7*d^8*(-(194481*b^8*c^8 + 2222 
64*a*b^7*c^7*d + 280476*a^2*b^6*c^6*d^2 + 176904*a^3*b^5*c^5*d^3 + 112806* 
a^4*b^4*c^4*d^4 + 42120*a^5*b^3*c^3*d^5 + 15900*a^6*b^2*c^2*d^6 + 3000*a^7 
*b*c*d^7 + 625*a^8*d^8)/(c^9*d^11))^(3/4) + (9261*b^6*c^6 + 7938*a*b^5*c^5 
*d + 8883*a^2*b^4*c^4*d^2 + 3996*a^3*b^3*c^3*d^3 + 2115*a^4*b^2*c^2*d^4 + 
450*a^5*b*c*d^5 + 125*a^6*d^6)*sqrt(x)) - (I*c^2*d^4*x^4 + 2*I*c^3*d^3*x^2 
 + I*c^4*d^2)*(-(194481*b^8*c^8 + 222264*a*b^7*c^7*d + 280476*a^2*b^6*c^6* 
d^2 + 176904*a^3*b^5*c^5*d^3 + 112806*a^4*b^4*c^4*d^4 + 42120*a^5*b^3*c^3* 
d^5 + 15900*a^6*b^2*c^2*d^6 + 3000*a^7*b*c*d^7 + 625*a^8*d^8)/(c^9*d^11))^ 
(1/4)*log(I*c^7*d^8*(-(194481*b^8*c^8 + 222264*a*b^7*c^7*d + 280476*a^2*b^ 
6*c^6*d^2 + 176904*a^3*b^5*c^5*d^3 + 112806*a^4*b^4*c^4*d^4 + 42120*a^5*b^ 
3*c^3*d^5 + 15900*a^6*b^2*c^2*d^6 + 3000*a^7*b*c*d^7 + 625*a^8*d^8)/(c^9*d 
^11))^(3/4) + (9261*b^6*c^6 + 7938*a*b^5*c^5*d + 8883*a^2*b^4*c^4*d^2 + 39 
96*a^3*b^3*c^3*d^3 + 2115*a^4*b^2*c^2*d^4 + 450*a^5*b*c*d^5 + 125*a^6*d^6) 
*sqrt(x)) - (-I*c^2*d^4*x^4 - 2*I*c^3*d^3*x^2 - I*c^4*d^2)*(-(194481*b^8*c 
^8 + 222264*a*b^7*c^7*d + 280476*a^2*b^6*c^6*d^2 + 176904*a^3*b^5*c^5*d^3 
+ 112806*a^4*b^4*c^4*d^4 + 42120*a^5*b^3*c^3*d^5 + 15900*a^6*b^2*c^2*d^...
 

Sympy [F(-1)]

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

integrate(x**(1/2)*(b*x**2+a)**2/(d*x**2+c)**3,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 297, normalized size of antiderivative = 1.03 \[ \int \frac {\sqrt {x} \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=-\frac {{\left (11 \, b^{2} c^{2} d - 6 \, a b c d^{2} - 5 \, a^{2} d^{3}\right )} x^{\frac {7}{2}} + {\left (7 \, b^{2} c^{3} + 2 \, a b c^{2} d - 9 \, a^{2} c d^{2}\right )} x^{\frac {3}{2}}}{16 \, {\left (c^{2} d^{4} x^{4} + 2 \, c^{3} d^{3} x^{2} + c^{4} d^{2}\right )}} + \frac {{\left (21 \, b^{2} c^{2} + 6 \, a b c d + 5 \, a^{2} d^{2}\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} + 2 \, \sqrt {d} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {\sqrt {c} \sqrt {d}} \sqrt {d}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} - 2 \, \sqrt {d} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {\sqrt {c} \sqrt {d}} \sqrt {d}} - \frac {\sqrt {2} \log \left (\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} \sqrt {x} + \sqrt {d} x + \sqrt {c}\right )}{c^{\frac {1}{4}} d^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} \sqrt {x} + \sqrt {d} x + \sqrt {c}\right )}{c^{\frac {1}{4}} d^{\frac {3}{4}}}\right )}}{128 \, c^{2} d^{2}} \] Input:

integrate(x^(1/2)*(b*x^2+a)^2/(d*x^2+c)^3,x, algorithm="maxima")
 

Output:

-1/16*((11*b^2*c^2*d - 6*a*b*c*d^2 - 5*a^2*d^3)*x^(7/2) + (7*b^2*c^3 + 2*a 
*b*c^2*d - 9*a^2*c*d^2)*x^(3/2))/(c^2*d^4*x^4 + 2*c^3*d^3*x^2 + c^4*d^2) + 
 1/128*(21*b^2*c^2 + 6*a*b*c*d + 5*a^2*d^2)*(2*sqrt(2)*arctan(1/2*sqrt(2)* 
(sqrt(2)*c^(1/4)*d^(1/4) + 2*sqrt(d)*sqrt(x))/sqrt(sqrt(c)*sqrt(d)))/(sqrt 
(sqrt(c)*sqrt(d))*sqrt(d)) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*c^(1/4 
)*d^(1/4) - 2*sqrt(d)*sqrt(x))/sqrt(sqrt(c)*sqrt(d)))/(sqrt(sqrt(c)*sqrt(d 
))*sqrt(d)) - sqrt(2)*log(sqrt(2)*c^(1/4)*d^(1/4)*sqrt(x) + sqrt(d)*x + sq 
rt(c))/(c^(1/4)*d^(3/4)) + sqrt(2)*log(-sqrt(2)*c^(1/4)*d^(1/4)*sqrt(x) + 
sqrt(d)*x + sqrt(c))/(c^(1/4)*d^(3/4)))/(c^2*d^2)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 416, normalized size of antiderivative = 1.44 \[ \int \frac {\sqrt {x} \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=-\frac {11 \, b^{2} c^{2} d x^{\frac {7}{2}} - 6 \, a b c d^{2} x^{\frac {7}{2}} - 5 \, a^{2} d^{3} x^{\frac {7}{2}} + 7 \, b^{2} c^{3} x^{\frac {3}{2}} + 2 \, a b c^{2} d x^{\frac {3}{2}} - 9 \, a^{2} c d^{2} x^{\frac {3}{2}}}{16 \, {\left (d x^{2} + c\right )}^{2} c^{2} d^{2}} + \frac {\sqrt {2} {\left (21 \, \left (c d^{3}\right )^{\frac {3}{4}} b^{2} c^{2} + 6 \, \left (c d^{3}\right )^{\frac {3}{4}} a b c d + 5 \, \left (c d^{3}\right )^{\frac {3}{4}} a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {c}{d}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {c}{d}\right )^{\frac {1}{4}}}\right )}{64 \, c^{3} d^{5}} + \frac {\sqrt {2} {\left (21 \, \left (c d^{3}\right )^{\frac {3}{4}} b^{2} c^{2} + 6 \, \left (c d^{3}\right )^{\frac {3}{4}} a b c d + 5 \, \left (c d^{3}\right )^{\frac {3}{4}} a^{2} d^{2}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {c}{d}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {c}{d}\right )^{\frac {1}{4}}}\right )}{64 \, c^{3} d^{5}} - \frac {\sqrt {2} {\left (21 \, \left (c d^{3}\right )^{\frac {3}{4}} b^{2} c^{2} + 6 \, \left (c d^{3}\right )^{\frac {3}{4}} a b c d + 5 \, \left (c d^{3}\right )^{\frac {3}{4}} a^{2} d^{2}\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {c}{d}\right )^{\frac {1}{4}} + x + \sqrt {\frac {c}{d}}\right )}{128 \, c^{3} d^{5}} + \frac {\sqrt {2} {\left (21 \, \left (c d^{3}\right )^{\frac {3}{4}} b^{2} c^{2} + 6 \, \left (c d^{3}\right )^{\frac {3}{4}} a b c d + 5 \, \left (c d^{3}\right )^{\frac {3}{4}} a^{2} d^{2}\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {c}{d}\right )^{\frac {1}{4}} + x + \sqrt {\frac {c}{d}}\right )}{128 \, c^{3} d^{5}} \] Input:

integrate(x^(1/2)*(b*x^2+a)^2/(d*x^2+c)^3,x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

-1/16*(11*b^2*c^2*d*x^(7/2) - 6*a*b*c*d^2*x^(7/2) - 5*a^2*d^3*x^(7/2) + 7* 
b^2*c^3*x^(3/2) + 2*a*b*c^2*d*x^(3/2) - 9*a^2*c*d^2*x^(3/2))/((d*x^2 + c)^ 
2*c^2*d^2) + 1/64*sqrt(2)*(21*(c*d^3)^(3/4)*b^2*c^2 + 6*(c*d^3)^(3/4)*a*b* 
c*d + 5*(c*d^3)^(3/4)*a^2*d^2)*arctan(1/2*sqrt(2)*(sqrt(2)*(c/d)^(1/4) + 2 
*sqrt(x))/(c/d)^(1/4))/(c^3*d^5) + 1/64*sqrt(2)*(21*(c*d^3)^(3/4)*b^2*c^2 
+ 6*(c*d^3)^(3/4)*a*b*c*d + 5*(c*d^3)^(3/4)*a^2*d^2)*arctan(-1/2*sqrt(2)*( 
sqrt(2)*(c/d)^(1/4) - 2*sqrt(x))/(c/d)^(1/4))/(c^3*d^5) - 1/128*sqrt(2)*(2 
1*(c*d^3)^(3/4)*b^2*c^2 + 6*(c*d^3)^(3/4)*a*b*c*d + 5*(c*d^3)^(3/4)*a^2*d^ 
2)*log(sqrt(2)*sqrt(x)*(c/d)^(1/4) + x + sqrt(c/d))/(c^3*d^5) + 1/128*sqrt 
(2)*(21*(c*d^3)^(3/4)*b^2*c^2 + 6*(c*d^3)^(3/4)*a*b*c*d + 5*(c*d^3)^(3/4)* 
a^2*d^2)*log(-sqrt(2)*sqrt(x)*(c/d)^(1/4) + x + sqrt(c/d))/(c^3*d^5)
 

Mupad [B] (verification not implemented)

Time = 0.83 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.64 \[ \int \frac {\sqrt {x} \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\frac {\mathrm {atan}\left (\frac {d^{1/4}\,\sqrt {x}}{{\left (-c\right )}^{1/4}}\right )\,\left (5\,a^2\,d^2+6\,a\,b\,c\,d+21\,b^2\,c^2\right )}{32\,{\left (-c\right )}^{9/4}\,d^{11/4}}-\frac {\frac {x^{3/2}\,\left (-9\,a^2\,d^2+2\,a\,b\,c\,d+7\,b^2\,c^2\right )}{16\,c\,d^2}-\frac {x^{7/2}\,\left (5\,a^2\,d^2+6\,a\,b\,c\,d-11\,b^2\,c^2\right )}{16\,c^2\,d}}{c^2+2\,c\,d\,x^2+d^2\,x^4}-\frac {\mathrm {atanh}\left (\frac {d^{1/4}\,\sqrt {x}}{{\left (-c\right )}^{1/4}}\right )\,\left (5\,a^2\,d^2+6\,a\,b\,c\,d+21\,b^2\,c^2\right )}{32\,{\left (-c\right )}^{9/4}\,d^{11/4}} \] Input:

int((x^(1/2)*(a + b*x^2)^2)/(c + d*x^2)^3,x)
 

Output:

(atan((d^(1/4)*x^(1/2))/(-c)^(1/4))*(5*a^2*d^2 + 21*b^2*c^2 + 6*a*b*c*d))/ 
(32*(-c)^(9/4)*d^(11/4)) - ((x^(3/2)*(7*b^2*c^2 - 9*a^2*d^2 + 2*a*b*c*d))/ 
(16*c*d^2) - (x^(7/2)*(5*a^2*d^2 - 11*b^2*c^2 + 6*a*b*c*d))/(16*c^2*d))/(c 
^2 + d^2*x^4 + 2*c*d*x^2) - (atanh((d^(1/4)*x^(1/2))/(-c)^(1/4))*(5*a^2*d^ 
2 + 21*b^2*c^2 + 6*a*b*c*d))/(32*(-c)^(9/4)*d^(11/4))
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 1496, normalized size of antiderivative = 5.19 \[ \int \frac {\sqrt {x} \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx =\text {Too large to display} \] Input:

int(x^(1/2)*(b*x^2+a)^2/(d*x^2+c)^3,x)
 

Output:

( - 10*d**(1/4)*c**(3/4)*sqrt(2)*atan((d**(1/4)*c**(1/4)*sqrt(2) - 2*sqrt( 
x)*sqrt(d))/(d**(1/4)*c**(1/4)*sqrt(2)))*a**2*c**2*d**2 - 20*d**(1/4)*c**( 
3/4)*sqrt(2)*atan((d**(1/4)*c**(1/4)*sqrt(2) - 2*sqrt(x)*sqrt(d))/(d**(1/4 
)*c**(1/4)*sqrt(2)))*a**2*c*d**3*x**2 - 10*d**(1/4)*c**(3/4)*sqrt(2)*atan( 
(d**(1/4)*c**(1/4)*sqrt(2) - 2*sqrt(x)*sqrt(d))/(d**(1/4)*c**(1/4)*sqrt(2) 
))*a**2*d**4*x**4 - 12*d**(1/4)*c**(3/4)*sqrt(2)*atan((d**(1/4)*c**(1/4)*s 
qrt(2) - 2*sqrt(x)*sqrt(d))/(d**(1/4)*c**(1/4)*sqrt(2)))*a*b*c**3*d - 24*d 
**(1/4)*c**(3/4)*sqrt(2)*atan((d**(1/4)*c**(1/4)*sqrt(2) - 2*sqrt(x)*sqrt( 
d))/(d**(1/4)*c**(1/4)*sqrt(2)))*a*b*c**2*d**2*x**2 - 12*d**(1/4)*c**(3/4) 
*sqrt(2)*atan((d**(1/4)*c**(1/4)*sqrt(2) - 2*sqrt(x)*sqrt(d))/(d**(1/4)*c* 
*(1/4)*sqrt(2)))*a*b*c*d**3*x**4 - 42*d**(1/4)*c**(3/4)*sqrt(2)*atan((d**( 
1/4)*c**(1/4)*sqrt(2) - 2*sqrt(x)*sqrt(d))/(d**(1/4)*c**(1/4)*sqrt(2)))*b* 
*2*c**4 - 84*d**(1/4)*c**(3/4)*sqrt(2)*atan((d**(1/4)*c**(1/4)*sqrt(2) - 2 
*sqrt(x)*sqrt(d))/(d**(1/4)*c**(1/4)*sqrt(2)))*b**2*c**3*d*x**2 - 42*d**(1 
/4)*c**(3/4)*sqrt(2)*atan((d**(1/4)*c**(1/4)*sqrt(2) - 2*sqrt(x)*sqrt(d))/ 
(d**(1/4)*c**(1/4)*sqrt(2)))*b**2*c**2*d**2*x**4 + 10*d**(1/4)*c**(3/4)*sq 
rt(2)*atan((d**(1/4)*c**(1/4)*sqrt(2) + 2*sqrt(x)*sqrt(d))/(d**(1/4)*c**(1 
/4)*sqrt(2)))*a**2*c**2*d**2 + 20*d**(1/4)*c**(3/4)*sqrt(2)*atan((d**(1/4) 
*c**(1/4)*sqrt(2) + 2*sqrt(x)*sqrt(d))/(d**(1/4)*c**(1/4)*sqrt(2)))*a**2*c 
*d**3*x**2 + 10*d**(1/4)*c**(3/4)*sqrt(2)*atan((d**(1/4)*c**(1/4)*sqrt(...