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

Optimal. Leaf size=260 \[ -\frac {2 a^2}{c \sqrt {x}}-\frac {(b c-a d)^2 \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{2 \sqrt {2} c^{5/4} d^{7/4}}+\frac {(b c-a d)^2 \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{2 \sqrt {2} c^{5/4} d^{7/4}}+\frac {(b c-a d)^2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{\sqrt {2} c^{5/4} d^{7/4}}-\frac {(b c-a d)^2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )}{\sqrt {2} c^{5/4} d^{7/4}}+\frac {2 b^2 x^{3/2}}{3 d} \]

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Rubi [A]  time = 0.27, antiderivative size = 260, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {462, 459, 329, 297, 1162, 617, 204, 1165, 628} \begin {gather*} -\frac {2 a^2}{c \sqrt {x}}-\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 {(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 (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )}{\sqrt {2} c^{5/4} d^{7/4}}+\frac {2 b^2 x^{3/2}}{3 d} \end {gather*}

Antiderivative was successfully verified.

[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 204

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

Rule 297

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 329

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 459

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 462

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 617

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 628

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 1162

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 1165

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*} \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 \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 ) \operatorname {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 \operatorname {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 \operatorname {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 \operatorname {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 \operatorname {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 \operatorname {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 \operatorname {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 \operatorname {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 \operatorname {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*}

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Mathematica [A]  time = 0.13, size = 261, normalized size = 1.00 \begin {gather*} \frac {-24 a^2 \sqrt [4]{c} d^{7/4}-3 \sqrt {2} \sqrt {x} (b c-a d)^2 \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )+3 \sqrt {2} \sqrt {x} (b c-a d)^2 \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )+6 \sqrt {2} \sqrt {x} (b c-a d)^2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )-6 \sqrt {2} \sqrt {x} (b c-a d)^2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )+8 b^2 c^{5/4} d^{3/4} x^2}{12 c^{5/4} d^{7/4} \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-24*a^2*c^(1/4)*d^(7/4) + 8*b^2*c^(5/4)*d^(3/4)*x^2 + 6*Sqrt[2]*(b*c - a*d)^2*Sqrt[x]*ArcTan[1 - (Sqrt[2]*d^(
1/4)*Sqrt[x])/c^(1/4)] - 6*Sqrt[2]*(b*c - a*d)^2*Sqrt[x]*ArcTan[1 + (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)] - 3*Sqr
t[2]*(b*c - a*d)^2*Sqrt[x]*Log[Sqrt[c] - Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x] + 3*Sqrt[2]*(b*c - a*d)^
2*Sqrt[x]*Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/(12*c^(5/4)*d^(7/4)*Sqrt[x])

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IntegrateAlgebraic [A]  time = 0.22, size = 162, normalized size = 0.62 \begin {gather*} \frac {2 \left (b^2 c x^2-3 a^2 d\right )}{3 c d \sqrt {x}}+\frac {(b c-a d)^2 \tan ^{-1}\left (\frac {\frac {\sqrt [4]{c}}{\sqrt {2} \sqrt [4]{d}}-\frac {\sqrt [4]{d} x}{\sqrt {2} \sqrt [4]{c}}}{\sqrt {x}}\right )}{\sqrt {2} c^{5/4} d^{7/4}}+\frac {(b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}}{\sqrt {c}+\sqrt {d} x}\right )}{\sqrt {2} c^{5/4} d^{7/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.96, size = 1636, normalized size = 6.29

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(12*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)*arctan((sqrt((b^12*c^12 - 12*a
*b^11*c^11*d + 66*a^2*b^10*c^10*d^2 - 220*a^3*b^9*c^9*d^3 + 495*a^4*b^8*c^8*d^4 - 792*a^5*b^7*c^7*d^5 + 924*a^
6*b^6*c^6*d^6 - 792*a^7*b^5*c^5*d^7 + 495*a^8*b^4*c^4*d^8 - 220*a^9*b^3*c^3*d^9 + 66*a^10*b^2*c^2*d^10 - 12*a^
11*b*c*d^11 + a^12*d^12)*x - (b^8*c^11*d^3 - 8*a*b^7*c^10*d^4 + 28*a^2*b^6*c^9*d^5 - 56*a^3*b^5*c^8*d^6 + 70*a
^4*b^4*c^7*d^7 - 56*a^5*b^3*c^6*d^8 + 28*a^6*b^2*c^5*d^9 - 8*a^7*b*c^4*d^10 + a^8*c^3*d^11)*sqrt(-(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)))*c*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^5*d^7
))^(1/4) - (b^6*c^7*d^2 - 6*a*b^5*c^6*d^3 + 15*a^2*b^4*c^5*d^4 - 20*a^3*b^3*c^4*d^5 + 15*a^4*b^2*c^3*d^6 - 6*a
^5*b*c^2*d^7 + a^6*c*d^8)*sqrt(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))/(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)) - 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 - 5
6*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)

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giac [A]  time = 0.48, size = 344, normalized size = 1.32 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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
)

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maple [B]  time = 0.02, size = 439, normalized size = 1.69 \begin {gather*} \frac {2 b^{2} x^{\frac {3}{2}}}{3 d}-\frac {\sqrt {2}\, a^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )}{2 \left (\frac {c}{d}\right )^{\frac {1}{4}} c}-\frac {\sqrt {2}\, a^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )}{2 \left (\frac {c}{d}\right )^{\frac {1}{4}} c}-\frac {\sqrt {2}\, a^{2} \ln \left (\frac {x -\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {c}{d}}}{x +\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {c}{d}}}\right )}{4 \left (\frac {c}{d}\right )^{\frac {1}{4}} c}+\frac {\sqrt {2}\, a b \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )}{\left (\frac {c}{d}\right )^{\frac {1}{4}} d}+\frac {\sqrt {2}\, a b \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )}{\left (\frac {c}{d}\right )^{\frac {1}{4}} d}+\frac {\sqrt {2}\, a b \ln \left (\frac {x -\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {c}{d}}}{x +\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {c}{d}}}\right )}{2 \left (\frac {c}{d}\right )^{\frac {1}{4}} d}-\frac {\sqrt {2}\, b^{2} c \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )}{2 \left (\frac {c}{d}\right )^{\frac {1}{4}} d^{2}}-\frac {\sqrt {2}\, b^{2} c \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )}{2 \left (\frac {c}{d}\right )^{\frac {1}{4}} d^{2}}-\frac {\sqrt {2}\, b^{2} c \ln \left (\frac {x -\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {c}{d}}}{x +\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {c}{d}}}\right )}{4 \left (\frac {c}{d}\right )^{\frac {1}{4}} d^{2}}-\frac {2 a^{2}}{c \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.38, size = 223, normalized size = 0.86 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.34, size = 416, normalized size = 1.60 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 36.72, size = 394, normalized size = 1.52 \begin {gather*} a^{2} \left (\begin {cases} \frac {\tilde {\infty }}{x^{\frac {5}{2}}} & \text {for}\: c = 0 \wedge d = 0 \\- \frac {2}{c \sqrt {x}} & \text {for}\: d = 0 \\- \frac {2}{5 d x^{\frac {5}{2}}} & \text {for}\: c = 0 \\- \frac {2}{c \sqrt {x}} + \frac {\left (-1\right )^{\frac {3}{4}} \log {\left (- \sqrt [4]{-1} \sqrt [4]{c} \sqrt [4]{\frac {1}{d}} + \sqrt {x} \right )}}{2 c^{\frac {5}{4}} \sqrt [4]{\frac {1}{d}}} - \frac {\left (-1\right )^{\frac {3}{4}} \log {\left (\sqrt [4]{-1} \sqrt [4]{c} \sqrt [4]{\frac {1}{d}} + \sqrt {x} \right )}}{2 c^{\frac {5}{4}} \sqrt [4]{\frac {1}{d}}} - \frac {\left (-1\right )^{\frac {3}{4}} \operatorname {atan}{\left (\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {x}}{\sqrt [4]{c} \sqrt [4]{\frac {1}{d}}} \right )}}{c^{\frac {5}{4}} \sqrt [4]{\frac {1}{d}}} & \text {otherwise} \end {cases}\right ) + 4 a b \operatorname {RootSum} {\left (256 t^{4} c d^{3} + 1, \left (t \mapsto t \log {\left (64 t^{3} c d^{2} + \sqrt {x} \right )} \right )\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 {\left (-1\right )^{\frac {3}{4}} c^{\frac {3}{4}} \log {\left (- \sqrt [4]{-1} \sqrt [4]{c} \sqrt [4]{\frac {1}{d}} + \sqrt {x} \right )}}{2 d^{2} \sqrt [4]{\frac {1}{d}}} - \frac {\left (-1\right )^{\frac {3}{4}} c^{\frac {3}{4}} \log {\left (\sqrt [4]{-1} \sqrt [4]{c} \sqrt [4]{\frac {1}{d}} + \sqrt {x} \right )}}{2 d^{2} \sqrt [4]{\frac {1}{d}}} - \frac {\left (-1\right )^{\frac {3}{4}} c^{\frac {3}{4}} \operatorname {atan}{\left (\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {x}}{\sqrt [4]{c} \sqrt [4]{\frac {1}{d}}} \right )}}{d^{2} \sqrt [4]{\frac {1}{d}}} + \frac {2 x^{\frac {3}{2}}}{3 d} & \text {otherwise} \end {cases}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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/(c*sqrt(x)), Eq(d, 0)), (-2/(5*d*x**(5/2)), Eq(c, 0)),
(-2/(c*sqrt(x)) + (-1)**(3/4)*log(-(-1)**(1/4)*c**(1/4)*(1/d)**(1/4) + sqrt(x))/(2*c**(5/4)*(1/d)**(1/4)) - (
-1)**(3/4)*log((-1)**(1/4)*c**(1/4)*(1/d)**(1/4) + sqrt(x))/(2*c**(5/4)*(1/d)**(1/4)) - (-1)**(3/4)*atan((-1)*
*(3/4)*sqrt(x)/(c**(1/4)*(1/d)**(1/4)))/(c**(5/4)*(1/d)**(1/4)), True)) + 4*a*b*RootSum(256*_t**4*c*d**3 + 1,
Lambda(_t, _t*log(64*_t**3*c*d**2 + sqrt(x)))) + 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)), ((-1)**(3/4)*c**(3/4)*log(-(-1)**(1/4)*c**(1/4)*(1/d)**(1/4
) + sqrt(x))/(2*d**2*(1/d)**(1/4)) - (-1)**(3/4)*c**(3/4)*log((-1)**(1/4)*c**(1/4)*(1/d)**(1/4) + sqrt(x))/(2*
d**2*(1/d)**(1/4)) - (-1)**(3/4)*c**(3/4)*atan((-1)**(3/4)*sqrt(x)/(c**(1/4)*(1/d)**(1/4)))/(d**2*(1/d)**(1/4)
) + 2*x**(3/2)/(3*d), True))

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