\(\int \frac {(a+b x^2)^2}{x^{3/2} (c+d x^2)} \, dx\) [420]

Optimal result

Integrand size = 24, antiderivative size = 260 \[ \int \frac {\left (a+b x^2\right )^2}{x^{3/2} \left (c+d x^2\right )} \, dx=-\frac {2 a^2}{c \sqrt {x}}+\frac {2 b^2 x^{3/2}}{3 d}+\frac {(b c-a d)^2 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{\sqrt {2} c^{5/4} d^{7/4}}-\frac {(b c-a d)^2 \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{\sqrt {2} c^{5/4} d^{7/4}}-\frac {(b c-a d)^2 \log \left (\sqrt {c}-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {d} x\right )}{2 \sqrt {2} c^{5/4} d^{7/4}}+\frac {(b c-a d)^2 \log \left (\sqrt {c}+\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {d} x\right )}{2 \sqrt {2} c^{5/4} d^{7/4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 260, 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 = {473, 470, 335, 303, 1176, 631, 210, 1179, 642} \[ \int \frac {\left (a+b x^2\right )^2}{x^{3/2} \left (c+d x^2\right )} \, dx=-\frac {2 a^2}{c \sqrt {x}}+\frac {(b c-a d)^2 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{\sqrt {2} c^{5/4} d^{7/4}}-\frac {(b c-a d)^2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )}{\sqrt {2} c^{5/4} d^{7/4}}-\frac {(b c-a d)^2 \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{2 \sqrt {2} c^{5/4} d^{7/4}}+\frac {(b c-a d)^2 \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{2 \sqrt {2} c^{5/4} d^{7/4}}+\frac {2 b^2 x^{3/2}}{3 d} \]

[In]

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

[Out]

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

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 470

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

Rule 473

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

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

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.59 \[ \int \frac {\left (a+b x^2\right )^2}{x^{3/2} \left (c+d x^2\right )} \, dx=\frac {\frac {4 \sqrt [4]{c} d^{3/4} \left (-3 a^2 d+b^2 c x^2\right )}{\sqrt {x}}+3 \sqrt {2} (b c-a d)^2 \arctan \left (\frac {\sqrt {c}-\sqrt {d} x}{\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}}\right )+3 \sqrt {2} (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}}{\sqrt {c}+\sqrt {d} x}\right )}{6 c^{5/4} d^{7/4}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.82 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.59

method	result	size
derivativedivides	\(\frac {2 b^{2} x^{\frac {3}{2}}}{3 d}-\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {c}{d}}}{x +\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {c}{d}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )\right )}{4 c \,d^{2} \left (\frac {c}{d}\right )^{\frac {1}{4}}}-\frac {2 a^{2}}{c \sqrt {x}}\)	\(153\)
default	\(\frac {2 b^{2} x^{\frac {3}{2}}}{3 d}-\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {c}{d}}}{x +\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {c}{d}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )\right )}{4 c \,d^{2} \left (\frac {c}{d}\right )^{\frac {1}{4}}}-\frac {2 a^{2}}{c \sqrt {x}}\)	\(153\)
risch	\(-\frac {2 \left (-b^{2} c \,x^{2}+3 a^{2} d \right )}{3 c \sqrt {x}\, d}-\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {c}{d}}}{x +\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {c}{d}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )\right )}{4 c \,d^{2} \left (\frac {c}{d}\right )^{\frac {1}{4}}}\)	\(158\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.28 (sec) , antiderivative size = 1335, normalized size of antiderivative = 5.13 \[ \int \frac {\left (a+b x^2\right )^2}{x^{3/2} \left (c+d x^2\right )} \, dx=\text {Too large to display} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 20.53 (sec) , antiderivative size = 357, normalized size of antiderivative = 1.37 \[ \int \frac {\left (a+b x^2\right )^2}{x^{3/2} \left (c+d x^2\right )} \, dx=a^{2} \left (\begin {cases} \frac {\tilde {\infty }}{x^{\frac {5}{2}}} & \text {for}\: c = 0 \wedge d = 0 \\- \frac {2}{5 d x^{\frac {5}{2}}} & \text {for}\: c = 0 \\- \frac {2}{c \sqrt {x}} & \text {for}\: d = 0 \\- \frac {\log {\left (\sqrt {x} - \sqrt [4]{- \frac {c}{d}} \right )}}{2 c \sqrt [4]{- \frac {c}{d}}} + \frac {\log {\left (\sqrt {x} + \sqrt [4]{- \frac {c}{d}} \right )}}{2 c \sqrt [4]{- \frac {c}{d}}} - \frac {\operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt [4]{- \frac {c}{d}}} \right )}}{c \sqrt [4]{- \frac {c}{d}}} - \frac {2}{c \sqrt {x}} & \text {otherwise} \end {cases}\right ) + 2 a b \left (\begin {cases} \frac {\tilde {\infty }}{\sqrt {x}} & \text {for}\: c = 0 \wedge d = 0 \\\frac {2 x^{\frac {3}{2}}}{3 c} & \text {for}\: d = 0 \\- \frac {2}{d \sqrt {x}} & \text {for}\: c = 0 \\\frac {\log {\left (\sqrt {x} - \sqrt [4]{- \frac {c}{d}} \right )}}{2 d \sqrt [4]{- \frac {c}{d}}} - \frac {\log {\left (\sqrt {x} + \sqrt [4]{- \frac {c}{d}} \right )}}{2 d \sqrt [4]{- \frac {c}{d}}} + \frac {\operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt [4]{- \frac {c}{d}}} \right )}}{d \sqrt [4]{- \frac {c}{d}}} & \text {otherwise} \end {cases}\right ) + b^{2} \left (\begin {cases} \tilde {\infty } x^{\frac {3}{2}} & \text {for}\: c = 0 \wedge d = 0 \\\frac {2 x^{\frac {7}{2}}}{7 c} & \text {for}\: d = 0 \\\frac {2 x^{\frac {3}{2}}}{3 d} & \text {for}\: c = 0 \\- \frac {c \log {\left (\sqrt {x} - \sqrt [4]{- \frac {c}{d}} \right )}}{2 d^{2} \sqrt [4]{- \frac {c}{d}}} + \frac {c \log {\left (\sqrt {x} + \sqrt [4]{- \frac {c}{d}} \right )}}{2 d^{2} \sqrt [4]{- \frac {c}{d}}} - \frac {c \operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt [4]{- \frac {c}{d}}} \right )}}{d^{2} \sqrt [4]{- \frac {c}{d}}} + \frac {2 x^{\frac {3}{2}}}{3 d} & \text {otherwise} \end {cases}\right ) \]

[In]

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

[Out]

a**2*Piecewise((zoo/x**(5/2), Eq(c, 0) & Eq(d, 0)), (-2/(5*d*x**(5/2)), Eq(c, 0)), (-2/(c*sqrt(x)), Eq(d, 0)),
(-log(sqrt(x) - (-c/d)**(1/4))/(2*c*(-c/d)**(1/4)) + log(sqrt(x) + (-c/d)**(1/4))/(2*c*(-c/d)**(1/4)) - atan(
sqrt(x)/(-c/d)**(1/4))/(c*(-c/d)**(1/4)) - 2/(c*sqrt(x)), True)) + 2*a*b*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(sqrt(x)/(-c/d)**(1/4))/(d*(-c/d)**(1/4)), Tru
e)) + b**2*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))

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 223, normalized size of antiderivative = 0.86 \[ \int \frac {\left (a+b x^2\right )^2}{x^{3/2} \left (c+d x^2\right )} \, dx=\frac {2 \, b^{2} x^{\frac {3}{2}}}{3 \, d} - \frac {2 \, a^{2}}{c \sqrt {x}} - \frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} + 2 \, \sqrt {d} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {\sqrt {c} \sqrt {d}} \sqrt {d}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} - 2 \, \sqrt {d} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {\sqrt {c} \sqrt {d}} \sqrt {d}} - \frac {\sqrt {2} \log \left (\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} \sqrt {x} + \sqrt {d} x + \sqrt {c}\right )}{c^{\frac {1}{4}} d^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} \sqrt {x} + \sqrt {d} x + \sqrt {c}\right )}{c^{\frac {1}{4}} d^{\frac {3}{4}}}\right )}}{4 \, c d} \]

[In]

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

[Out]

2/3*b^2*x^(3/2)/d - 2*a^2/(c*sqrt(x)) - 1/4*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqr
t(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)*a
rctan(-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)))/(c*d)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 344, normalized size of antiderivative = 1.32 \[ \int \frac {\left (a+b x^2\right )^2}{x^{3/2} \left (c+d x^2\right )} \, dx=\frac {2 \, b^{2} x^{\frac {3}{2}}}{3 \, d} - \frac {2 \, a^{2}}{c \sqrt {x}} - \frac {\sqrt {2} {\left (\left (c d^{3}\right )^{\frac {3}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac {3}{4}} a b c d + \left (c d^{3}\right )^{\frac {3}{4}} a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {c}{d}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {c}{d}\right )^{\frac {1}{4}}}\right )}{2 \, c^{2} d^{4}} - \frac {\sqrt {2} {\left (\left (c d^{3}\right )^{\frac {3}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac {3}{4}} a b c d + \left (c d^{3}\right )^{\frac {3}{4}} a^{2} d^{2}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {c}{d}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {c}{d}\right )^{\frac {1}{4}}}\right )}{2 \, c^{2} d^{4}} + \frac {\sqrt {2} {\left (\left (c d^{3}\right )^{\frac {3}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac {3}{4}} a b c d + \left (c d^{3}\right )^{\frac {3}{4}} a^{2} d^{2}\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {c}{d}\right )^{\frac {1}{4}} + x + \sqrt {\frac {c}{d}}\right )}{4 \, c^{2} d^{4}} - \frac {\sqrt {2} {\left (\left (c d^{3}\right )^{\frac {3}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac {3}{4}} a b c d + \left (c d^{3}\right )^{\frac {3}{4}} a^{2} d^{2}\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {c}{d}\right )^{\frac {1}{4}} + x + \sqrt {\frac {c}{d}}\right )}{4 \, c^{2} d^{4}} \]

[In]

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

[Out]

2/3*b^2*x^(3/2)/d - 2*a^2/(c*sqrt(x)) - 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^2*d^4) - 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^2*d^4) + 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^2*d^4) - 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^2*d^4
)

Mupad [B] (verification not implemented)

Time = 4.96 (sec) , antiderivative size = 416, normalized size of antiderivative = 1.60 \[ \int \frac {\left (a+b x^2\right )^2}{x^{3/2} \left (c+d x^2\right )} \, dx=\frac {2\,b^2\,x^{3/2}}{3\,d}-\frac {2\,a^2}{c\,\sqrt {x}}-\frac {\mathrm {atan}\left (\frac {\sqrt {x}\,{\left (a\,d-b\,c\right )}^2\,\left (16\,a^4\,c^4\,d^9-64\,a^3\,b\,c^5\,d^8+96\,a^2\,b^2\,c^6\,d^7-64\,a\,b^3\,c^7\,d^6+16\,b^4\,c^8\,d^5\right )}{{\left (-c\right )}^{5/4}\,d^{7/4}\,\left (16\,a^6\,c^3\,d^9-96\,a^5\,b\,c^4\,d^8+240\,a^4\,b^2\,c^5\,d^7-320\,a^3\,b^3\,c^6\,d^6+240\,a^2\,b^4\,c^7\,d^5-96\,a\,b^5\,c^8\,d^4+16\,b^6\,c^9\,d^3\right )}\right )\,{\left (a\,d-b\,c\right )}^2}{{\left (-c\right )}^{5/4}\,d^{7/4}}-\frac {\mathrm {atan}\left (\frac {\sqrt {x}\,{\left (a\,d-b\,c\right )}^2\,\left (16\,a^4\,c^4\,d^9-64\,a^3\,b\,c^5\,d^8+96\,a^2\,b^2\,c^6\,d^7-64\,a\,b^3\,c^7\,d^6+16\,b^4\,c^8\,d^5\right )\,1{}\mathrm {i}}{{\left (-c\right )}^{5/4}\,d^{7/4}\,\left (16\,a^6\,c^3\,d^9-96\,a^5\,b\,c^4\,d^8+240\,a^4\,b^2\,c^5\,d^7-320\,a^3\,b^3\,c^6\,d^6+240\,a^2\,b^4\,c^7\,d^5-96\,a\,b^5\,c^8\,d^4+16\,b^6\,c^9\,d^3\right )}\right )\,{\left (a\,d-b\,c\right )}^2\,1{}\mathrm {i}}{{\left (-c\right )}^{5/4}\,d^{7/4}} \]

[In]

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

[Out]

(2*b^2*x^(3/2))/(3*d) - (2*a^2)/(c*x^(1/2)) - (atan((x^(1/2)*(a*d - b*c)^2*(16*a^4*c^4*d^9 + 16*b^4*c^8*d^5 -
64*a*b^3*c^7*d^6 - 64*a^3*b*c^5*d^8 + 96*a^2*b^2*c^6*d^7))/((-c)^(5/4)*d^(7/4)*(16*a^6*c^3*d^9 + 16*b^6*c^9*d^
3 - 96*a*b^5*c^8*d^4 - 96*a^5*b*c^4*d^8 + 240*a^2*b^4*c^7*d^5 - 320*a^3*b^3*c^6*d^6 + 240*a^4*b^2*c^5*d^7)))*(
a*d - b*c)^2)/((-c)^(5/4)*d^(7/4)) - (atan((x^(1/2)*(a*d - b*c)^2*(16*a^4*c^4*d^9 + 16*b^4*c^8*d^5 - 64*a*b^3*
c^7*d^6 - 64*a^3*b*c^5*d^8 + 96*a^2*b^2*c^6*d^7)*1i)/((-c)^(5/4)*d^(7/4)*(16*a^6*c^3*d^9 + 16*b^6*c^9*d^3 - 96
*a*b^5*c^8*d^4 - 96*a^5*b*c^4*d^8 + 240*a^2*b^4*c^7*d^5 - 320*a^3*b^3*c^6*d^6 + 240*a^4*b^2*c^5*d^7)))*(a*d -
b*c)^2*1i)/((-c)^(5/4)*d^(7/4))