\(\int \frac {\sqrt {x} (a+b x^2)^2}{(c+d x^2)^3} \, dx\) [436]

Optimal result

Integrand size = 24, antiderivative size = 364 \[ \int \frac {\sqrt {x} \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\frac {(b c-a d)^2 x^{3/2}}{4 c d^2 \left (c+d x^2\right )^2}-\frac {(b c-a d) (11 b c+5 a d) x^{3/2}}{16 c^2 d^2 \left (c+d x^2\right )}-\frac {\left (21 b^2 c^2+6 a b c d+5 a^2 d^2\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{32 \sqrt {2} c^{9/4} d^{11/4}}+\frac {\left (21 b^2 c^2+6 a b c d+5 a^2 d^2\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{32 \sqrt {2} c^{9/4} d^{11/4}}+\frac {\left (21 b^2 c^2+6 a b c d+5 a^2 d^2\right ) \log \left (\sqrt {c}-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {d} x\right )}{64 \sqrt {2} c^{9/4} d^{11/4}}-\frac {\left (21 b^2 c^2+6 a b c d+5 a^2 d^2\right ) \log \left (\sqrt {c}+\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {d} x\right )}{64 \sqrt {2} c^{9/4} d^{11/4}} \]

[Out]

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-d^(1/4)*2^(1/2)*x^(1/2)/c^(1/4))/c^(9/4)/d^(11/4)*2^(1/2)+1/64*(5*a^2*d^2
+6*a*b*c*d+21*b^2*c^2)*arctan(1+d^(1/4)*2^(1/2)*x^(1/2)/c^(1/4))/c^(9/4)/d^(11/4)*2^(1/2)+1/128*(5*a^2*d^2+6*a
*b*c*d+21*b^2*c^2)*ln(c^(1/2)+x*d^(1/2)-c^(1/4)*d^(1/4)*2^(1/2)*x^(1/2))/c^(9/4)/d^(11/4)*2^(1/2)-1/128*(5*a^2
*d^2+6*a*b*c*d+21*b^2*c^2)*ln(c^(1/2)+x*d^(1/2)+c^(1/4)*d^(1/4)*2^(1/2)*x^(1/2))/c^(9/4)/d^(11/4)*2^(1/2)

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 364, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {474, 468, 335, 303, 1176, 631, 210, 1179, 642} \[ \int \frac {\sqrt {x} \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=-\frac {\left (5 a^2 d^2+6 a b c d+21 b^2 c^2\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{32 \sqrt {2} c^{9/4} d^{11/4}}+\frac {\left (5 a^2 d^2+6 a b c d+21 b^2 c^2\right ) \arctan \left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )}{32 \sqrt {2} c^{9/4} d^{11/4}}+\frac {\left (5 a^2 d^2+6 a b c d+21 b^2 c^2\right ) \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{64 \sqrt {2} c^{9/4} d^{11/4}}-\frac {\left (5 a^2 d^2+6 a b c d+21 b^2 c^2\right ) \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{64 \sqrt {2} c^{9/4} d^{11/4}}-\frac {x^{3/2} (5 a d+11 b c) (b c-a d)}{16 c^2 d^2 \left (c+d x^2\right )}+\frac {x^{3/2} (b c-a d)^2}{4 c d^2 \left (c+d x^2\right )^2} \]

[In]

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

[Out]

((b*c - a*d)^2*x^(3/2))/(4*c*d^2*(c + d*x^2)^2) - ((b*c - a*d)*(11*b*c + 5*a*d)*x^(3/2))/(16*c^2*d^2*(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)])/(32*Sqrt[2]*c^(9/4)
*d^(11/4)) + ((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)])/(32*Sqrt[2]*
c^(9/4)*d^(11/4)) + ((21*b^2*c^2 + 6*a*b*c*d + 5*a^2*d^2)*Log[Sqrt[c] - Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt
[d]*x])/(64*Sqrt[2]*c^(9/4)*d^(11/4)) - ((21*b^2*c^2 + 6*a*b*c*d + 5*a^2*d^2)*Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^
(1/4)*Sqrt[x] + Sqrt[d]*x])/(64*Sqrt[2]*c^(9/4)*d^(11/4))

Rule 210

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 303

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

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 468

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

Rule 474

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

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[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])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[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 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[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 1179

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

Rubi steps \begin{align*} \text {integral}& = \frac {(b c-a d)^2 x^{3/2}}{4 c d^2 \left (c+d x^2\right )^2}-\frac {\int \frac {\sqrt {x} \left (\frac {1}{2} \left (-8 a^2 d^2+3 (b c-a d)^2\right )-4 b^2 c d x^2\right )}{\left (c+d x^2\right )^2} \, dx}{4 c d^2} \\ & = \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 ) \int \frac {\sqrt {x}}{c+d x^2} \, dx}{32 c^2 d^2} \\ & = \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 ) \text {Subst}\left (\int \frac {x^2}{c+d x^4} \, dx,x,\sqrt {x}\right )}{16 c^2 d^2} \\ & = \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 ) \text {Subst}\left (\int \frac {\sqrt {c}-\sqrt {d} x^2}{c+d x^4} \, dx,x,\sqrt {x}\right )}{32 c^2 d^{5/2}}+\frac {\left (21 b^2 c^2+6 a b c d+5 a^2 d^2\right ) \text {Subst}\left (\int \frac {\sqrt {c}+\sqrt {d} x^2}{c+d x^4} \, dx,x,\sqrt {x}\right )}{32 c^2 d^{5/2}} \\ & = \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 ) \text {Subst}\left (\int \frac {1}{\frac {\sqrt {c}}{\sqrt {d}}-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{d}}+x^2} \, dx,x,\sqrt {x}\right )}{64 c^2 d^3}+\frac {\left (21 b^2 c^2+6 a b c d+5 a^2 d^2\right ) \text {Subst}\left (\int \frac {1}{\frac {\sqrt {c}}{\sqrt {d}}+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{d}}+x^2} \, dx,x,\sqrt {x}\right )}{64 c^2 d^3}+\frac {\left (21 b^2 c^2+6 a b c d+5 a^2 d^2\right ) \text {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{c}}{\sqrt [4]{d}}+2 x}{-\frac {\sqrt {c}}{\sqrt {d}}-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{d}}-x^2} \, dx,x,\sqrt {x}\right )}{64 \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 {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{c}}{\sqrt [4]{d}}-2 x}{-\frac {\sqrt {c}}{\sqrt {d}}+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{d}}-x^2} \, dx,x,\sqrt {x}\right )}{64 \sqrt {2} c^{9/4} d^{11/4}} \\ & = \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 ) \log \left (\sqrt {c}-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {d} x\right )}{64 \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 ) \log \left (\sqrt {c}+\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {d} x\right )}{64 \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 {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,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 {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{32 \sqrt {2} c^{9/4} d^{11/4}} \\ & = \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 ) \tan ^{-1}\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 ) \tan ^{-1}\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 ) \log \left (\sqrt {c}-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {d} x\right )}{64 \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 ) \log \left (\sqrt {c}+\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {d} x\right )}{64 \sqrt {2} c^{9/4} d^{11/4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.68 (sec) , antiderivative size = 209, normalized size of antiderivative = 0.57 \[ \int \frac {\sqrt {x} \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\frac {-\frac {4 \sqrt [4]{c} d^{3/4} (b c-a d) x^{3/2} \left (a d \left (9 c+5 d x^2\right )+b c \left (7 c+11 d x^2\right )\right )}{\left (c+d x^2\right )^2}-\sqrt {2} \left (21 b^2 c^2+6 a b c d+5 a^2 d^2\right ) \arctan \left (\frac {\sqrt {c}-\sqrt {d} x}{\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}}\right )-\sqrt {2} \left (21 b^2 c^2+6 a b c d+5 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}}{\sqrt {c}+\sqrt {d} x}\right )}{64 c^{9/4} d^{11/4}} \]

[In]

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

[Out]

((-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))

Maple [A] (verified)

Time = 2.72 (sec) , antiderivative size = 213, normalized size of antiderivative = 0.59

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

[In]

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

[Out]

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)/d^3/c^2/(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 = 0.31 (sec) , antiderivative size = 1532, normalized size of antiderivative = 4.21 \[ \int \frac {\sqrt {x} \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\text {Too large to display} \]

[In]

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

[Out]

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 + 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 + 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 + 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 + 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)) - (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 + 15
900*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 + 22226
4*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)) - 4*((11*b^2*c^2*d - 6*a*b*c*d^2 - 5*a^2*d^3)*x^3 + (7*b^2*c^3 + 2*a*b*c^2*d - 9*a^2*c*d^2)*x)*sqrt(x))/(
c^2*d^4*x^4 + 2*c^3*d^3*x^2 + c^4*d^2)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 297, normalized size of antiderivative = 0.82 \[ \int \frac {\sqrt {x} \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=-\frac {{\left (11 \, b^{2} c^{2} d - 6 \, a b c d^{2} - 5 \, a^{2} d^{3}\right )} x^{\frac {7}{2}} + {\left (7 \, b^{2} c^{3} + 2 \, a b c^{2} d - 9 \, a^{2} c d^{2}\right )} x^{\frac {3}{2}}}{16 \, {\left (c^{2} d^{4} x^{4} + 2 \, c^{3} d^{3} x^{2} + c^{4} d^{2}\right )}} + \frac {{\left (21 \, b^{2} c^{2} + 6 \, a b c d + 5 \, a^{2} d^{2}\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} + 2 \, \sqrt {d} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {\sqrt {c} \sqrt {d}} \sqrt {d}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} - 2 \, \sqrt {d} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {\sqrt {c} \sqrt {d}} \sqrt {d}} - \frac {\sqrt {2} \log \left (\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} \sqrt {x} + \sqrt {d} x + \sqrt {c}\right )}{c^{\frac {1}{4}} d^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} \sqrt {x} + \sqrt {d} x + \sqrt {c}\right )}{c^{\frac {1}{4}} d^{\frac {3}{4}}}\right )}}{128 \, c^{2} d^{2}} \]

[In]

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

[Out]

-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*sqr
t(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)*sq
rt(d))*sqrt(d)) - sqrt(2)*log(sqrt(2)*c^(1/4)*d^(1/4)*sqrt(x) + sqrt(d)*x + sqrt(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)

none

Time = 0.29 (sec) , antiderivative size = 416, normalized size of antiderivative = 1.14 \[ \int \frac {\sqrt {x} \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=-\frac {11 \, b^{2} c^{2} d x^{\frac {7}{2}} - 6 \, a b c d^{2} x^{\frac {7}{2}} - 5 \, a^{2} d^{3} x^{\frac {7}{2}} + 7 \, b^{2} c^{3} x^{\frac {3}{2}} + 2 \, a b c^{2} d x^{\frac {3}{2}} - 9 \, a^{2} c d^{2} x^{\frac {3}{2}}}{16 \, {\left (d x^{2} + c\right )}^{2} c^{2} d^{2}} + \frac {\sqrt {2} {\left (21 \, \left (c d^{3}\right )^{\frac {3}{4}} b^{2} c^{2} + 6 \, \left (c d^{3}\right )^{\frac {3}{4}} a b c d + 5 \, \left (c d^{3}\right )^{\frac {3}{4}} a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {c}{d}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {c}{d}\right )^{\frac {1}{4}}}\right )}{64 \, c^{3} d^{5}} + \frac {\sqrt {2} {\left (21 \, \left (c d^{3}\right )^{\frac {3}{4}} b^{2} c^{2} + 6 \, \left (c d^{3}\right )^{\frac {3}{4}} a b c d + 5 \, \left (c d^{3}\right )^{\frac {3}{4}} a^{2} d^{2}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {c}{d}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {c}{d}\right )^{\frac {1}{4}}}\right )}{64 \, c^{3} d^{5}} - \frac {\sqrt {2} {\left (21 \, \left (c d^{3}\right )^{\frac {3}{4}} b^{2} c^{2} + 6 \, \left (c d^{3}\right )^{\frac {3}{4}} a b c d + 5 \, \left (c d^{3}\right )^{\frac {3}{4}} a^{2} d^{2}\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {c}{d}\right )^{\frac {1}{4}} + x + \sqrt {\frac {c}{d}}\right )}{128 \, c^{3} d^{5}} + \frac {\sqrt {2} {\left (21 \, \left (c d^{3}\right )^{\frac {3}{4}} b^{2} c^{2} + 6 \, \left (c d^{3}\right )^{\frac {3}{4}} a b c d + 5 \, \left (c d^{3}\right )^{\frac {3}{4}} a^{2} d^{2}\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {c}{d}\right )^{\frac {1}{4}} + x + \sqrt {\frac {c}{d}}\right )}{128 \, c^{3} d^{5}} \]

[In]

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

[Out]

-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)*(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) + 1/12
8*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 = 5.23 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.51 \[ \int \frac {\sqrt {x} \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\frac {\mathrm {atan}\left (\frac {d^{1/4}\,\sqrt {x}}{{\left (-c\right )}^{1/4}}\right )\,\left (5\,a^2\,d^2+6\,a\,b\,c\,d+21\,b^2\,c^2\right )}{32\,{\left (-c\right )}^{9/4}\,d^{11/4}}-\frac {\frac {x^{3/2}\,\left (-9\,a^2\,d^2+2\,a\,b\,c\,d+7\,b^2\,c^2\right )}{16\,c\,d^2}-\frac {x^{7/2}\,\left (5\,a^2\,d^2+6\,a\,b\,c\,d-11\,b^2\,c^2\right )}{16\,c^2\,d}}{c^2+2\,c\,d\,x^2+d^2\,x^4}-\frac {\mathrm {atanh}\left (\frac {d^{1/4}\,\sqrt {x}}{{\left (-c\right )}^{1/4}}\right )\,\left (5\,a^2\,d^2+6\,a\,b\,c\,d+21\,b^2\,c^2\right )}{32\,{\left (-c\right )}^{9/4}\,d^{11/4}} \]

[In]

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

[Out]

(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))