\(\int \frac {\sqrt {x} (a+b x^2)^2}{c+d x^2} \, dx\) [418]

Optimal result

Integrand size = 24, antiderivative size = 268 \[ \int \frac {\sqrt {x} \left (a+b x^2\right )^2}{c+d x^2} \, dx=-\frac {2 b (b c-2 a d) x^{3/2}}{3 d^2}+\frac {2 b^2 x^{7/2}}{7 d}-\frac {(b c-a d)^2 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{\sqrt {2} \sqrt [4]{c} d^{11/4}}+\frac {(b c-a d)^2 \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{\sqrt {2} \sqrt [4]{c} d^{11/4}}+\frac {(b c-a d)^2 \log \left (\sqrt {c}-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {d} x\right )}{2 \sqrt {2} \sqrt [4]{c} d^{11/4}}-\frac {(b c-a d)^2 \log \left (\sqrt {c}+\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {d} x\right )}{2 \sqrt {2} \sqrt [4]{c} d^{11/4}} \]

[Out]

-2/3*b*(-2*a*d+b*c)*x^(3/2)/d^2+2/7*b^2*x^(7/2)/d-1/2*(-a*d+b*c)^2*arctan(1-d^(1/4)*2^(1/2)*x^(1/2)/c^(1/4))/c
^(1/4)/d^(11/4)*2^(1/2)+1/2*(-a*d+b*c)^2*arctan(1+d^(1/4)*2^(1/2)*x^(1/2)/c^(1/4))/c^(1/4)/d^(11/4)*2^(1/2)+1/
4*(-a*d+b*c)^2*ln(c^(1/2)+x*d^(1/2)-c^(1/4)*d^(1/4)*2^(1/2)*x^(1/2))/c^(1/4)/d^(11/4)*2^(1/2)-1/4*(-a*d+b*c)^2
*ln(c^(1/2)+x*d^(1/2)+c^(1/4)*d^(1/4)*2^(1/2)*x^(1/2))/c^(1/4)/d^(11/4)*2^(1/2)

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {472, 335, 303, 1176, 631, 210, 1179, 642} \[ \int \frac {\sqrt {x} \left (a+b x^2\right )^2}{c+d x^2} \, dx=-\frac {(b c-a d)^2 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{\sqrt {2} \sqrt [4]{c} d^{11/4}}+\frac {(b c-a d)^2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )}{\sqrt {2} \sqrt [4]{c} d^{11/4}}+\frac {(b c-a d)^2 \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{2 \sqrt {2} \sqrt [4]{c} d^{11/4}}-\frac {(b c-a d)^2 \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{2 \sqrt {2} \sqrt [4]{c} d^{11/4}}-\frac {2 b x^{3/2} (b c-2 a d)}{3 d^2}+\frac {2 b^2 x^{7/2}}{7 d} \]

[In]

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

[Out]

(-2*b*(b*c - 2*a*d)*x^(3/2))/(3*d^2) + (2*b^2*x^(7/2))/(7*d) - ((b*c - a*d)^2*ArcTan[1 - (Sqrt[2]*d^(1/4)*Sqrt
[x])/c^(1/4)])/(Sqrt[2]*c^(1/4)*d^(11/4)) + ((b*c - a*d)^2*ArcTan[1 + (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/(Sqr
t[2]*c^(1/4)*d^(11/4)) + ((b*c - a*d)^2*Log[Sqrt[c] - Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/(2*Sqrt[2]
*c^(1/4)*d^(11/4)) - ((b*c - a*d)^2*Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/(2*Sqrt[2]*c^(
1/4)*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 472

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

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}& = \int \left (-\frac {b (b c-2 a d) \sqrt {x}}{d^2}+\frac {b^2 x^{5/2}}{d}+\frac {\left (b^2 c^2-2 a b c d+a^2 d^2\right ) \sqrt {x}}{d^2 \left (c+d x^2\right )}\right ) \, dx \\ & = -\frac {2 b (b c-2 a d) x^{3/2}}{3 d^2}+\frac {2 b^2 x^{7/2}}{7 d}+\frac {(b c-a d)^2 \int \frac {\sqrt {x}}{c+d x^2} \, dx}{d^2} \\ & = -\frac {2 b (b c-2 a d) x^{3/2}}{3 d^2}+\frac {2 b^2 x^{7/2}}{7 d}+\frac {\left (2 (b c-a d)^2\right ) \text {Subst}\left (\int \frac {x^2}{c+d x^4} \, dx,x,\sqrt {x}\right )}{d^2} \\ & = -\frac {2 b (b c-2 a d) x^{3/2}}{3 d^2}+\frac {2 b^2 x^{7/2}}{7 d}-\frac {(b c-a d)^2 \text {Subst}\left (\int \frac {\sqrt {c}-\sqrt {d} x^2}{c+d x^4} \, dx,x,\sqrt {x}\right )}{d^{5/2}}+\frac {(b c-a d)^2 \text {Subst}\left (\int \frac {\sqrt {c}+\sqrt {d} x^2}{c+d x^4} \, dx,x,\sqrt {x}\right )}{d^{5/2}} \\ & = -\frac {2 b (b c-2 a d) x^{3/2}}{3 d^2}+\frac {2 b^2 x^{7/2}}{7 d}+\frac {(b c-a d)^2 \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 )}{2 d^3}+\frac {(b c-a d)^2 \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 )}{2 d^3}+\frac {(b c-a d)^2 \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 )}{2 \sqrt {2} \sqrt [4]{c} d^{11/4}}+\frac {(b c-a d)^2 \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 )}{2 \sqrt {2} \sqrt [4]{c} d^{11/4}} \\ & = -\frac {2 b (b c-2 a d) x^{3/2}}{3 d^2}+\frac {2 b^2 x^{7/2}}{7 d}+\frac {(b c-a d)^2 \log \left (\sqrt {c}-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {d} x\right )}{2 \sqrt {2} \sqrt [4]{c} d^{11/4}}-\frac {(b c-a d)^2 \log \left (\sqrt {c}+\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {d} x\right )}{2 \sqrt {2} \sqrt [4]{c} d^{11/4}}+\frac {(b c-a d)^2 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{\sqrt {2} \sqrt [4]{c} d^{11/4}}-\frac {(b c-a d)^2 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{\sqrt {2} \sqrt [4]{c} d^{11/4}} \\ & = -\frac {2 b (b c-2 a d) x^{3/2}}{3 d^2}+\frac {2 b^2 x^{7/2}}{7 d}-\frac {(b c-a d)^2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{\sqrt {2} \sqrt [4]{c} d^{11/4}}+\frac {(b c-a d)^2 \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{\sqrt {2} \sqrt [4]{c} d^{11/4}}+\frac {(b c-a d)^2 \log \left (\sqrt {c}-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {d} x\right )}{2 \sqrt {2} \sqrt [4]{c} d^{11/4}}-\frac {(b c-a d)^2 \log \left (\sqrt {c}+\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {d} x\right )}{2 \sqrt {2} \sqrt [4]{c} d^{11/4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.58 \[ \int \frac {\sqrt {x} \left (a+b x^2\right )^2}{c+d x^2} \, dx=\frac {4 b d^{3/4} x^{3/2} \left (-7 b c+14 a d+3 b d x^2\right )-\frac {21 \sqrt {2} (b c-a d)^2 \arctan \left (\frac {\sqrt {c}-\sqrt {d} x}{\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}}\right )}{\sqrt [4]{c}}-\frac {21 \sqrt {2} (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}}{\sqrt {c}+\sqrt {d} x}\right )}{\sqrt [4]{c}}}{42 d^{11/4}} \]

[In]

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

[Out]

(4*b*d^(3/4)*x^(3/2)*(-7*b*c + 14*a*d + 3*b*d*x^2) - (21*Sqrt[2]*(b*c - a*d)^2*ArcTan[(Sqrt[c] - Sqrt[d]*x)/(S
qrt[2]*c^(1/4)*d^(1/4)*Sqrt[x])])/c^(1/4) - (21*Sqrt[2]*(b*c - a*d)^2*ArcTanh[(Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x]
)/(Sqrt[c] + Sqrt[d]*x)])/c^(1/4))/(42*d^(11/4))

Maple [A] (verified)

Time = 2.80 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.57

method	result	size
risch	\(\frac {2 \left (3 b d \,x^{2}+14 a d -7 b c \right ) b \,x^{\frac {3}{2}}}{21 d^{2}}+\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {c}{d}}}{x +\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {c}{d}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )\right )}{4 d^{3} \left (\frac {c}{d}\right )^{\frac {1}{4}}}\)	\(153\)
derivativedivides	\(\frac {2 b \left (\frac {b d \,x^{\frac {7}{2}}}{7}+\frac {\left (2 a d -b c \right ) x^{\frac {3}{2}}}{3}\right )}{d^{2}}+\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {c}{d}}}{x +\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {c}{d}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )\right )}{4 d^{3} \left (\frac {c}{d}\right )^{\frac {1}{4}}}\)	\(156\)
default	\(\frac {2 b \left (\frac {b d \,x^{\frac {7}{2}}}{7}+\frac {\left (2 a d -b c \right ) x^{\frac {3}{2}}}{3}\right )}{d^{2}}+\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {c}{d}}}{x +\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {c}{d}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )\right )}{4 d^{3} \left (\frac {c}{d}\right )^{\frac {1}{4}}}\)	\(156\)

[In]

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

[Out]

2/21*(3*b*d*x^2+14*a*d-7*b*c)*b*x^(3/2)/d^2+1/4*(a^2*d^2-2*a*b*c*d+b^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 = 0.27 (sec) , antiderivative size = 1330, normalized size of antiderivative = 4.96 \[ \int \frac {\sqrt {x} \left (a+b x^2\right )^2}{c+d x^2} \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/42*(21*d^2*(-(b^8*c^8 - 8*a*b^7*c^7*d + 28*a^2*b^6*c^6*d^2 - 56*a^3*b^5*c^5*d^3 + 70*a^4*b^4*c^4*d^4 - 56*a^
5*b^3*c^3*d^5 + 28*a^6*b^2*c^2*d^6 - 8*a^7*b*c*d^7 + a^8*d^8)/(c*d^11))^(1/4)*log(c*d^8*(-(b^8*c^8 - 8*a*b^7*c
^7*d + 28*a^2*b^6*c^6*d^2 - 56*a^3*b^5*c^5*d^3 + 70*a^4*b^4*c^4*d^4 - 56*a^5*b^3*c^3*d^5 + 28*a^6*b^2*c^2*d^6
- 8*a^7*b*c*d^7 + a^8*d^8)/(c*d^11))^(3/4) + (b^6*c^6 - 6*a*b^5*c^5*d + 15*a^2*b^4*c^4*d^2 - 20*a^3*b^3*c^3*d^
3 + 15*a^4*b^2*c^2*d^4 - 6*a^5*b*c*d^5 + a^6*d^6)*sqrt(x)) - 21*I*d^2*(-(b^8*c^8 - 8*a*b^7*c^7*d + 28*a^2*b^6*
c^6*d^2 - 56*a^3*b^5*c^5*d^3 + 70*a^4*b^4*c^4*d^4 - 56*a^5*b^3*c^3*d^5 + 28*a^6*b^2*c^2*d^6 - 8*a^7*b*c*d^7 +
a^8*d^8)/(c*d^11))^(1/4)*log(I*c*d^8*(-(b^8*c^8 - 8*a*b^7*c^7*d + 28*a^2*b^6*c^6*d^2 - 56*a^3*b^5*c^5*d^3 + 70
*a^4*b^4*c^4*d^4 - 56*a^5*b^3*c^3*d^5 + 28*a^6*b^2*c^2*d^6 - 8*a^7*b*c*d^7 + a^8*d^8)/(c*d^11))^(3/4) + (b^6*c
^6 - 6*a*b^5*c^5*d + 15*a^2*b^4*c^4*d^2 - 20*a^3*b^3*c^3*d^3 + 15*a^4*b^2*c^2*d^4 - 6*a^5*b*c*d^5 + a^6*d^6)*s
qrt(x)) + 21*I*d^2*(-(b^8*c^8 - 8*a*b^7*c^7*d + 28*a^2*b^6*c^6*d^2 - 56*a^3*b^5*c^5*d^3 + 70*a^4*b^4*c^4*d^4 -
56*a^5*b^3*c^3*d^5 + 28*a^6*b^2*c^2*d^6 - 8*a^7*b*c*d^7 + a^8*d^8)/(c*d^11))^(1/4)*log(-I*c*d^8*(-(b^8*c^8 -
8*a*b^7*c^7*d + 28*a^2*b^6*c^6*d^2 - 56*a^3*b^5*c^5*d^3 + 70*a^4*b^4*c^4*d^4 - 56*a^5*b^3*c^3*d^5 + 28*a^6*b^2
*c^2*d^6 - 8*a^7*b*c*d^7 + a^8*d^8)/(c*d^11))^(3/4) + (b^6*c^6 - 6*a*b^5*c^5*d + 15*a^2*b^4*c^4*d^2 - 20*a^3*b
^3*c^3*d^3 + 15*a^4*b^2*c^2*d^4 - 6*a^5*b*c*d^5 + a^6*d^6)*sqrt(x)) - 21*d^2*(-(b^8*c^8 - 8*a*b^7*c^7*d + 28*a
^2*b^6*c^6*d^2 - 56*a^3*b^5*c^5*d^3 + 70*a^4*b^4*c^4*d^4 - 56*a^5*b^3*c^3*d^5 + 28*a^6*b^2*c^2*d^6 - 8*a^7*b*c
*d^7 + a^8*d^8)/(c*d^11))^(1/4)*log(-c*d^8*(-(b^8*c^8 - 8*a*b^7*c^7*d + 28*a^2*b^6*c^6*d^2 - 56*a^3*b^5*c^5*d^
3 + 70*a^4*b^4*c^4*d^4 - 56*a^5*b^3*c^3*d^5 + 28*a^6*b^2*c^2*d^6 - 8*a^7*b*c*d^7 + a^8*d^8)/(c*d^11))^(3/4) +
(b^6*c^6 - 6*a*b^5*c^5*d + 15*a^2*b^4*c^4*d^2 - 20*a^3*b^3*c^3*d^3 + 15*a^4*b^2*c^2*d^4 - 6*a^5*b*c*d^5 + a^6*
d^6)*sqrt(x)) + 4*(3*b^2*d*x^3 - 7*(b^2*c - 2*a*b*d)*x)*sqrt(x))/d^2

Sympy [A] (verification not implemented)

Time = 48.90 (sec) , antiderivative size = 386, normalized size of antiderivative = 1.44 \[ \int \frac {\sqrt {x} \left (a+b x^2\right )^2}{c+d x^2} \, dx=a^{2} \left (\begin {cases} \frac {\tilde {\infty }}{\sqrt {x}} & \text {for}\: c = 0 \wedge d = 0 \\\frac {2 x^{\frac {3}{2}}}{3 c} & \text {for}\: d = 0 \\- \frac {2}{d \sqrt {x}} & \text {for}\: c = 0 \\\frac {\log {\left (\sqrt {x} - \sqrt [4]{- \frac {c}{d}} \right )}}{2 d \sqrt [4]{- \frac {c}{d}}} - \frac {\log {\left (\sqrt {x} + \sqrt [4]{- \frac {c}{d}} \right )}}{2 d \sqrt [4]{- \frac {c}{d}}} + \frac {\operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt [4]{- \frac {c}{d}}} \right )}}{d \sqrt [4]{- \frac {c}{d}}} & \text {otherwise} \end {cases}\right ) + 2 a b \left (\begin {cases} \tilde {\infty } x^{\frac {3}{2}} & \text {for}\: c = 0 \wedge d = 0 \\\frac {2 x^{\frac {7}{2}}}{7 c} & \text {for}\: d = 0 \\\frac {2 x^{\frac {3}{2}}}{3 d} & \text {for}\: c = 0 \\- \frac {c \log {\left (\sqrt {x} - \sqrt [4]{- \frac {c}{d}} \right )}}{2 d^{2} \sqrt [4]{- \frac {c}{d}}} + \frac {c \log {\left (\sqrt {x} + \sqrt [4]{- \frac {c}{d}} \right )}}{2 d^{2} \sqrt [4]{- \frac {c}{d}}} - \frac {c \operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt [4]{- \frac {c}{d}}} \right )}}{d^{2} \sqrt [4]{- \frac {c}{d}}} + \frac {2 x^{\frac {3}{2}}}{3 d} & \text {otherwise} \end {cases}\right ) + b^{2} \left (\begin {cases} \tilde {\infty } x^{\frac {7}{2}} & \text {for}\: c = 0 \wedge d = 0 \\\frac {2 x^{\frac {11}{2}}}{11 c} & \text {for}\: d = 0 \\\frac {2 x^{\frac {7}{2}}}{7 d} & \text {for}\: c = 0 \\\frac {c^{2} \log {\left (\sqrt {x} - \sqrt [4]{- \frac {c}{d}} \right )}}{2 d^{3} \sqrt [4]{- \frac {c}{d}}} - \frac {c^{2} \log {\left (\sqrt {x} + \sqrt [4]{- \frac {c}{d}} \right )}}{2 d^{3} \sqrt [4]{- \frac {c}{d}}} + \frac {c^{2} \operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt [4]{- \frac {c}{d}}} \right )}}{d^{3} \sqrt [4]{- \frac {c}{d}}} - \frac {2 c x^{\frac {3}{2}}}{3 d^{2}} + \frac {2 x^{\frac {7}{2}}}{7 d} & \text {otherwise} \end {cases}\right ) \]

[In]

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

[Out]

a**2*Piecewise((zoo/sqrt(x), Eq(c, 0) & Eq(d, 0)), (2*x**(3/2)/(3*c), Eq(d, 0)), (-2/(d*sqrt(x)), Eq(c, 0)), (
log(sqrt(x) - (-c/d)**(1/4))/(2*d*(-c/d)**(1/4)) - log(sqrt(x) + (-c/d)**(1/4))/(2*d*(-c/d)**(1/4)) + atan(sqr
t(x)/(-c/d)**(1/4))/(d*(-c/d)**(1/4)), True)) + 2*a*b*Piecewise((zoo*x**(3/2), Eq(c, 0) & Eq(d, 0)), (2*x**(7/
2)/(7*c), Eq(d, 0)), (2*x**(3/2)/(3*d), Eq(c, 0)), (-c*log(sqrt(x) - (-c/d)**(1/4))/(2*d**2*(-c/d)**(1/4)) + c
*log(sqrt(x) + (-c/d)**(1/4))/(2*d**2*(-c/d)**(1/4)) - c*atan(sqrt(x)/(-c/d)**(1/4))/(d**2*(-c/d)**(1/4)) + 2*
x**(3/2)/(3*d), True)) + b**2*Piecewise((zoo*x**(7/2), Eq(c, 0) & Eq(d, 0)), (2*x**(11/2)/(11*c), Eq(d, 0)), (
2*x**(7/2)/(7*d), Eq(c, 0)), (c**2*log(sqrt(x) - (-c/d)**(1/4))/(2*d**3*(-c/d)**(1/4)) - c**2*log(sqrt(x) + (-
c/d)**(1/4))/(2*d**3*(-c/d)**(1/4)) + c**2*atan(sqrt(x)/(-c/d)**(1/4))/(d**3*(-c/d)**(1/4)) - 2*c*x**(3/2)/(3*
d**2) + 2*x**(7/2)/(7*d), True))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 229, normalized size of antiderivative = 0.85 \[ \int \frac {\sqrt {x} \left (a+b x^2\right )^2}{c+d x^2} \, dx=\frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} + 2 \, \sqrt {d} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {\sqrt {c} \sqrt {d}} \sqrt {d}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} - 2 \, \sqrt {d} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {\sqrt {c} \sqrt {d}} \sqrt {d}} - \frac {\sqrt {2} \log \left (\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} \sqrt {x} + \sqrt {d} x + \sqrt {c}\right )}{c^{\frac {1}{4}} d^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} \sqrt {x} + \sqrt {d} x + \sqrt {c}\right )}{c^{\frac {1}{4}} d^{\frac {3}{4}}}\right )}}{4 \, d^{2}} + \frac {2 \, {\left (3 \, b^{2} d x^{\frac {7}{2}} - 7 \, {\left (b^{2} c - 2 \, a b d\right )} x^{\frac {3}{2}}\right )}}{21 \, d^{2}} \]

[In]

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

[Out]

1/4*(b^2*c^2 - 2*a*b*c*d + 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 + 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)))/d^2 + 2/21*(3*b^2*d*x^(7/2) - 7*(b^2*c - 2*a*b*d)*x^(3/2))/d^2

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 361, normalized size of antiderivative = 1.35 \[ \int \frac {\sqrt {x} \left (a+b x^2\right )^2}{c+d x^2} \, dx=\frac {\sqrt {2} {\left (\left (c d^{3}\right )^{\frac {3}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac {3}{4}} a b c d + \left (c d^{3}\right )^{\frac {3}{4}} a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {c}{d}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {c}{d}\right )^{\frac {1}{4}}}\right )}{2 \, c d^{5}} + \frac {\sqrt {2} {\left (\left (c d^{3}\right )^{\frac {3}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac {3}{4}} a b c d + \left (c d^{3}\right )^{\frac {3}{4}} a^{2} d^{2}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {c}{d}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {c}{d}\right )^{\frac {1}{4}}}\right )}{2 \, c d^{5}} - \frac {\sqrt {2} {\left (\left (c d^{3}\right )^{\frac {3}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac {3}{4}} a b c d + \left (c d^{3}\right )^{\frac {3}{4}} a^{2} d^{2}\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {c}{d}\right )^{\frac {1}{4}} + x + \sqrt {\frac {c}{d}}\right )}{4 \, c d^{5}} + \frac {\sqrt {2} {\left (\left (c d^{3}\right )^{\frac {3}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac {3}{4}} a b c d + \left (c d^{3}\right )^{\frac {3}{4}} a^{2} d^{2}\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {c}{d}\right )^{\frac {1}{4}} + x + \sqrt {\frac {c}{d}}\right )}{4 \, c d^{5}} + \frac {2 \, {\left (3 \, b^{2} d^{6} x^{\frac {7}{2}} - 7 \, b^{2} c d^{5} x^{\frac {3}{2}} + 14 \, a b d^{6} x^{\frac {3}{2}}\right )}}{21 \, d^{7}} \]

[In]

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

[Out]

1/2*sqrt(2)*((c*d^3)^(3/4)*b^2*c^2 - 2*(c*d^3)^(3/4)*a*b*c*d + (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*d^5) + 1/2*sqrt(2)*((c*d^3)^(3/4)*b^2*c^2 - 2*(c*d^3)^(3/4)*a*b*c
*d + (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*d^5) - 1/4*s
qrt(2)*((c*d^3)^(3/4)*b^2*c^2 - 2*(c*d^3)^(3/4)*a*b*c*d + (c*d^3)^(3/4)*a^2*d^2)*log(sqrt(2)*sqrt(x)*(c/d)^(1/
4) + x + sqrt(c/d))/(c*d^5) + 1/4*sqrt(2)*((c*d^3)^(3/4)*b^2*c^2 - 2*(c*d^3)^(3/4)*a*b*c*d + (c*d^3)^(3/4)*a^2
*d^2)*log(-sqrt(2)*sqrt(x)*(c/d)^(1/4) + x + sqrt(c/d))/(c*d^5) + 2/21*(3*b^2*d^6*x^(7/2) - 7*b^2*c*d^5*x^(3/2
) + 14*a*b*d^6*x^(3/2))/d^7

Mupad [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 390, normalized size of antiderivative = 1.46 \[ \int \frac {\sqrt {x} \left (a+b x^2\right )^2}{c+d x^2} \, dx=\frac {2\,b^2\,x^{7/2}}{7\,d}-x^{3/2}\,\left (\frac {2\,b^2\,c}{3\,d^2}-\frac {4\,a\,b}{3\,d}\right )+\frac {\mathrm {atan}\left (\frac {d^{1/4}\,\sqrt {x}\,{\left (a\,d-b\,c\right )}^2\,\left (a^4\,c\,d^4-4\,a^3\,b\,c^2\,d^3+6\,a^2\,b^2\,c^3\,d^2-4\,a\,b^3\,c^4\,d+b^4\,c^5\right )}{{\left (-c\right )}^{1/4}\,\left (a^6\,c\,d^6-6\,a^5\,b\,c^2\,d^5+15\,a^4\,b^2\,c^3\,d^4-20\,a^3\,b^3\,c^4\,d^3+15\,a^2\,b^4\,c^5\,d^2-6\,a\,b^5\,c^6\,d+b^6\,c^7\right )}\right )\,{\left (a\,d-b\,c\right )}^2}{{\left (-c\right )}^{1/4}\,d^{11/4}}+\frac {\mathrm {atan}\left (\frac {d^{1/4}\,\sqrt {x}\,{\left (a\,d-b\,c\right )}^2\,\left (a^4\,c\,d^4-4\,a^3\,b\,c^2\,d^3+6\,a^2\,b^2\,c^3\,d^2-4\,a\,b^3\,c^4\,d+b^4\,c^5\right )\,1{}\mathrm {i}}{{\left (-c\right )}^{1/4}\,\left (a^6\,c\,d^6-6\,a^5\,b\,c^2\,d^5+15\,a^4\,b^2\,c^3\,d^4-20\,a^3\,b^3\,c^4\,d^3+15\,a^2\,b^4\,c^5\,d^2-6\,a\,b^5\,c^6\,d+b^6\,c^7\right )}\right )\,{\left (a\,d-b\,c\right )}^2\,1{}\mathrm {i}}{{\left (-c\right )}^{1/4}\,d^{11/4}} \]

[In]

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

[Out]

(2*b^2*x^(7/2))/(7*d) - x^(3/2)*((2*b^2*c)/(3*d^2) - (4*a*b)/(3*d)) + (atan((d^(1/4)*x^(1/2)*(a*d - b*c)^2*(b^
4*c^5 + a^4*c*d^4 - 4*a^3*b*c^2*d^3 + 6*a^2*b^2*c^3*d^2 - 4*a*b^3*c^4*d))/((-c)^(1/4)*(b^6*c^7 + a^6*c*d^6 - 6
*a^5*b*c^2*d^5 + 15*a^2*b^4*c^5*d^2 - 20*a^3*b^3*c^4*d^3 + 15*a^4*b^2*c^3*d^4 - 6*a*b^5*c^6*d)))*(a*d - b*c)^2
)/((-c)^(1/4)*d^(11/4)) + (atan((d^(1/4)*x^(1/2)*(a*d - b*c)^2*(b^4*c^5 + a^4*c*d^4 - 4*a^3*b*c^2*d^3 + 6*a^2*
b^2*c^3*d^2 - 4*a*b^3*c^4*d)*1i)/((-c)^(1/4)*(b^6*c^7 + a^6*c*d^6 - 6*a^5*b*c^2*d^5 + 15*a^2*b^4*c^5*d^2 - 20*
a^3*b^3*c^4*d^3 + 15*a^4*b^2*c^3*d^4 - 6*a*b^5*c^6*d)))*(a*d - b*c)^2*1i)/((-c)^(1/4)*d^(11/4))