3.421 \(\int \frac {(a+b x^2)^2}{x^{5/2} (c+d x^2)} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.26, 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, 211, 1165, 628, 1162, 617, 204} \[ -\frac {2 a^2}{3 c x^{3/2}}+\frac {(b c-a d)^2 \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{2 \sqrt {2} c^{7/4} d^{5/4}}-\frac {(b c-a d)^2 \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{2 \sqrt {2} c^{7/4} d^{5/4}}+\frac {(b c-a d)^2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{\sqrt {2} c^{7/4} d^{5/4}}-\frac {(b c-a d)^2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )}{\sqrt {2} c^{7/4} d^{5/4}}+\frac {2 b^2 \sqrt {x}}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^2/(x^(5/2)*(c + d*x^2)),x]

[Out]

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

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 211

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

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

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

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^2/(x^(5/2)*(c + d*x^2)),x]

[Out]

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

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fricas [B]  time = 0.50, size = 1253, normalized size = 4.82 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/x^(5/2)/(d*x^2+c),x, algorithm="fricas")

[Out]

-1/6*(12*c*d*x^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 - 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^7*d^5))^(1/4)*arctan((sqrt(c^4*d^2*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^7*d^5)) + (b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a
^3*b*c*d^3 + a^4*d^4)*x)*c^5*d^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)/(c^7*d^5))^(3/4) - (b^2*c^7*
d^4 - 2*a*b*c^6*d^5 + a^2*c^5*d^6)*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^7*d^5))^(3/4))/
(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^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^7*d^5))^(1/4)*log(c^2*d*(-(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^7*d^5))^(1/4) + (b^2*c^2 - 2*a*b
*c*d + a^2*d^2)*sqrt(x)) - 3*c*d*x^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^7*d^5))^(1/4)*log(-c^
2*d*(-(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^7*d^5))^(1/4) + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(x
)) - 4*(3*b^2*c*x^2 - a^2*d)*sqrt(x))/(c*d*x^2)

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giac [A]  time = 0.40, size = 344, normalized size = 1.32 \[ \frac {2 \, b^{2} \sqrt {x}}{d} - \frac {2 \, a^{2}}{3 \, c x^{\frac {3}{2}}} - \frac {\sqrt {2} {\left (\left (c d^{3}\right )^{\frac {1}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac {1}{4}} a b c d + \left (c d^{3}\right )^{\frac {1}{4}} a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {c}{d}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {c}{d}\right )^{\frac {1}{4}}}\right )}{2 \, c^{2} d^{2}} - \frac {\sqrt {2} {\left (\left (c d^{3}\right )^{\frac {1}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac {1}{4}} a b c d + \left (c d^{3}\right )^{\frac {1}{4}} a^{2} d^{2}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {c}{d}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {c}{d}\right )^{\frac {1}{4}}}\right )}{2 \, c^{2} d^{2}} - \frac {\sqrt {2} {\left (\left (c d^{3}\right )^{\frac {1}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac {1}{4}} a b c d + \left (c d^{3}\right )^{\frac {1}{4}} a^{2} d^{2}\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {c}{d}\right )^{\frac {1}{4}} + x + \sqrt {\frac {c}{d}}\right )}{4 \, c^{2} d^{2}} + \frac {\sqrt {2} {\left (\left (c d^{3}\right )^{\frac {1}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac {1}{4}} a b c d + \left (c d^{3}\right )^{\frac {1}{4}} a^{2} d^{2}\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {c}{d}\right )^{\frac {1}{4}} + x + \sqrt {\frac {c}{d}}\right )}{4 \, c^{2} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/x^(5/2)/(d*x^2+c),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.01, size = 439, normalized size = 1.69 \[ -\frac {\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, a^{2} d \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )}{2 c^{2}}-\frac {\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, a^{2} d \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )}{2 c^{2}}-\frac {\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, a^{2} d \ln \left (\frac {x +\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {c}{d}}}{x -\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {c}{d}}}\right )}{4 c^{2}}+\frac {\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, a b \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )}{c}+\frac {\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, a b \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )}{c}+\frac {\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, a b \ln \left (\frac {x +\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {c}{d}}}{x -\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {c}{d}}}\right )}{2 c}-\frac {\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, b^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )}{2 d}-\frac {\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, b^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )}{2 d}-\frac {\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, b^{2} \ln \left (\frac {x +\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {c}{d}}}{x -\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {c}{d}}}\right )}{4 d}+\frac {2 b^{2} \sqrt {x}}{d}-\frac {2 a^{2}}{3 c \,x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2/x^(5/2)/(d*x^2+c),x)

[Out]

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

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maxima [A]  time = 2.41, size = 286, normalized size = 1.10 \[ \frac {2 \, b^{2} \sqrt {x}}{d} - \frac {2 \, a^{2}}{3 \, c x^{\frac {3}{2}}} - \frac {\frac {2 \, \sqrt {2} {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} + 2 \, \sqrt {d} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {c} \sqrt {\sqrt {c} \sqrt {d}}} + \frac {2 \, \sqrt {2} {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} - 2 \, \sqrt {d} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {c} \sqrt {\sqrt {c} \sqrt {d}}} + \frac {\sqrt {2} {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} \sqrt {x} + \sqrt {d} x + \sqrt {c}\right )}{c^{\frac {3}{4}} d^{\frac {1}{4}}} - \frac {\sqrt {2} {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (-\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} \sqrt {x} + \sqrt {d} x + \sqrt {c}\right )}{c^{\frac {3}{4}} d^{\frac {1}{4}}}}{4 \, c d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/x^(5/2)/(d*x^2+c),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.38, size = 1201, normalized size = 4.62 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^2/(x^(5/2)*(c + d*x^2)),x)

[Out]

(2*b^2*x^(1/2))/d - (2*a^2)/(3*c*x^(3/2)) - (atan(((((x^(1/2)*(16*a^4*c^3*d^10 + 16*b^4*c^7*d^6 - 64*a*b^3*c^6
*d^7 - 64*a^3*b*c^4*d^9 + 96*a^2*b^2*c^5*d^8))/2 - ((a*d - b*c)^2*(16*a^2*c^5*d^9 + 16*b^2*c^7*d^7 - 32*a*b*c^
6*d^8))/(2*(-c)^(7/4)*d^(5/4)))*(a*d - b*c)^2*1i)/((-c)^(7/4)*d^(5/4)) + (((x^(1/2)*(16*a^4*c^3*d^10 + 16*b^4*
c^7*d^6 - 64*a*b^3*c^6*d^7 - 64*a^3*b*c^4*d^9 + 96*a^2*b^2*c^5*d^8))/2 + ((a*d - b*c)^2*(16*a^2*c^5*d^9 + 16*b
^2*c^7*d^7 - 32*a*b*c^6*d^8))/(2*(-c)^(7/4)*d^(5/4)))*(a*d - b*c)^2*1i)/((-c)^(7/4)*d^(5/4)))/((((x^(1/2)*(16*
a^4*c^3*d^10 + 16*b^4*c^7*d^6 - 64*a*b^3*c^6*d^7 - 64*a^3*b*c^4*d^9 + 96*a^2*b^2*c^5*d^8))/2 - ((a*d - b*c)^2*
(16*a^2*c^5*d^9 + 16*b^2*c^7*d^7 - 32*a*b*c^6*d^8))/(2*(-c)^(7/4)*d^(5/4)))*(a*d - b*c)^2)/((-c)^(7/4)*d^(5/4)
) - (((x^(1/2)*(16*a^4*c^3*d^10 + 16*b^4*c^7*d^6 - 64*a*b^3*c^6*d^7 - 64*a^3*b*c^4*d^9 + 96*a^2*b^2*c^5*d^8))/
2 + ((a*d - b*c)^2*(16*a^2*c^5*d^9 + 16*b^2*c^7*d^7 - 32*a*b*c^6*d^8))/(2*(-c)^(7/4)*d^(5/4)))*(a*d - b*c)^2)/
((-c)^(7/4)*d^(5/4))))*(a*d - b*c)^2*1i)/((-c)^(7/4)*d^(5/4)) - (atan(((((x^(1/2)*(16*a^4*c^3*d^10 + 16*b^4*c^
7*d^6 - 64*a*b^3*c^6*d^7 - 64*a^3*b*c^4*d^9 + 96*a^2*b^2*c^5*d^8))/2 - ((a*d - b*c)^2*(16*a^2*c^5*d^9 + 16*b^2
*c^7*d^7 - 32*a*b*c^6*d^8)*1i)/(2*(-c)^(7/4)*d^(5/4)))*(a*d - b*c)^2)/((-c)^(7/4)*d^(5/4)) + (((x^(1/2)*(16*a^
4*c^3*d^10 + 16*b^4*c^7*d^6 - 64*a*b^3*c^6*d^7 - 64*a^3*b*c^4*d^9 + 96*a^2*b^2*c^5*d^8))/2 + ((a*d - b*c)^2*(1
6*a^2*c^5*d^9 + 16*b^2*c^7*d^7 - 32*a*b*c^6*d^8)*1i)/(2*(-c)^(7/4)*d^(5/4)))*(a*d - b*c)^2)/((-c)^(7/4)*d^(5/4
)))/((((x^(1/2)*(16*a^4*c^3*d^10 + 16*b^4*c^7*d^6 - 64*a*b^3*c^6*d^7 - 64*a^3*b*c^4*d^9 + 96*a^2*b^2*c^5*d^8))
/2 - ((a*d - b*c)^2*(16*a^2*c^5*d^9 + 16*b^2*c^7*d^7 - 32*a*b*c^6*d^8)*1i)/(2*(-c)^(7/4)*d^(5/4)))*(a*d - b*c)
^2*1i)/((-c)^(7/4)*d^(5/4)) - (((x^(1/2)*(16*a^4*c^3*d^10 + 16*b^4*c^7*d^6 - 64*a*b^3*c^6*d^7 - 64*a^3*b*c^4*d
^9 + 96*a^2*b^2*c^5*d^8))/2 + ((a*d - b*c)^2*(16*a^2*c^5*d^9 + 16*b^2*c^7*d^7 - 32*a*b*c^6*d^8)*1i)/(2*(-c)^(7
/4)*d^(5/4)))*(a*d - b*c)^2*1i)/((-c)^(7/4)*d^(5/4))))*(a*d - b*c)^2)/((-c)^(7/4)*d^(5/4))

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sympy [A]  time = 31.82, size = 576, normalized size = 2.22 \[ \begin {cases} \tilde {\infty } \left (- \frac {2 a^{2}}{7 x^{\frac {7}{2}}} - \frac {4 a b}{3 x^{\frac {3}{2}}} + 2 b^{2} \sqrt {x}\right ) & \text {for}\: c = 0 \wedge d = 0 \\\frac {- \frac {2 a^{2}}{3 x^{\frac {3}{2}}} + 4 a b \sqrt {x} + \frac {2 b^{2} x^{\frac {5}{2}}}{5}}{c} & \text {for}\: d = 0 \\\frac {- \frac {2 a^{2}}{7 x^{\frac {7}{2}}} - \frac {4 a b}{3 x^{\frac {3}{2}}} + 2 b^{2} \sqrt {x}}{d} & \text {for}\: c = 0 \\- \frac {2 a^{2}}{3 c x^{\frac {3}{2}}} + \frac {\sqrt [4]{-1} a^{2} d \sqrt [4]{\frac {1}{d}} \log {\left (- \sqrt [4]{-1} \sqrt [4]{c} \sqrt [4]{\frac {1}{d}} + \sqrt {x} \right )}}{2 c^{\frac {7}{4}}} - \frac {\sqrt [4]{-1} a^{2} d \sqrt [4]{\frac {1}{d}} \log {\left (\sqrt [4]{-1} \sqrt [4]{c} \sqrt [4]{\frac {1}{d}} + \sqrt {x} \right )}}{2 c^{\frac {7}{4}}} + \frac {\sqrt [4]{-1} a^{2} d \sqrt [4]{\frac {1}{d}} \operatorname {atan}{\left (\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {x}}{\sqrt [4]{c} \sqrt [4]{\frac {1}{d}}} \right )}}{c^{\frac {7}{4}}} - \frac {\sqrt [4]{-1} a b \sqrt [4]{\frac {1}{d}} \log {\left (- \sqrt [4]{-1} \sqrt [4]{c} \sqrt [4]{\frac {1}{d}} + \sqrt {x} \right )}}{c^{\frac {3}{4}}} + \frac {\sqrt [4]{-1} a b \sqrt [4]{\frac {1}{d}} \log {\left (\sqrt [4]{-1} \sqrt [4]{c} \sqrt [4]{\frac {1}{d}} + \sqrt {x} \right )}}{c^{\frac {3}{4}}} - \frac {2 \sqrt [4]{-1} a b \sqrt [4]{\frac {1}{d}} \operatorname {atan}{\left (\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {x}}{\sqrt [4]{c} \sqrt [4]{\frac {1}{d}}} \right )}}{c^{\frac {3}{4}}} + \frac {\sqrt [4]{-1} b^{2} \sqrt [4]{c} \sqrt [4]{\frac {1}{d}} \log {\left (- \sqrt [4]{-1} \sqrt [4]{c} \sqrt [4]{\frac {1}{d}} + \sqrt {x} \right )}}{2 d} - \frac {\sqrt [4]{-1} b^{2} \sqrt [4]{c} \sqrt [4]{\frac {1}{d}} \log {\left (\sqrt [4]{-1} \sqrt [4]{c} \sqrt [4]{\frac {1}{d}} + \sqrt {x} \right )}}{2 d} + \frac {\sqrt [4]{-1} b^{2} \sqrt [4]{c} \sqrt [4]{\frac {1}{d}} \operatorname {atan}{\left (\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {x}}{\sqrt [4]{c} \sqrt [4]{\frac {1}{d}}} \right )}}{d} + \frac {2 b^{2} \sqrt {x}}{d} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2/x**(5/2)/(d*x**2+c),x)

[Out]

Piecewise((zoo*(-2*a**2/(7*x**(7/2)) - 4*a*b/(3*x**(3/2)) + 2*b**2*sqrt(x)), Eq(c, 0) & Eq(d, 0)), ((-2*a**2/(
3*x**(3/2)) + 4*a*b*sqrt(x) + 2*b**2*x**(5/2)/5)/c, Eq(d, 0)), ((-2*a**2/(7*x**(7/2)) - 4*a*b/(3*x**(3/2)) + 2
*b**2*sqrt(x))/d, Eq(c, 0)), (-2*a**2/(3*c*x**(3/2)) + (-1)**(1/4)*a**2*d*(1/d)**(1/4)*log(-(-1)**(1/4)*c**(1/
4)*(1/d)**(1/4) + sqrt(x))/(2*c**(7/4)) - (-1)**(1/4)*a**2*d*(1/d)**(1/4)*log((-1)**(1/4)*c**(1/4)*(1/d)**(1/4
) + sqrt(x))/(2*c**(7/4)) + (-1)**(1/4)*a**2*d*(1/d)**(1/4)*atan((-1)**(3/4)*sqrt(x)/(c**(1/4)*(1/d)**(1/4)))/
c**(7/4) - (-1)**(1/4)*a*b*(1/d)**(1/4)*log(-(-1)**(1/4)*c**(1/4)*(1/d)**(1/4) + sqrt(x))/c**(3/4) + (-1)**(1/
4)*a*b*(1/d)**(1/4)*log((-1)**(1/4)*c**(1/4)*(1/d)**(1/4) + sqrt(x))/c**(3/4) - 2*(-1)**(1/4)*a*b*(1/d)**(1/4)
*atan((-1)**(3/4)*sqrt(x)/(c**(1/4)*(1/d)**(1/4)))/c**(3/4) + (-1)**(1/4)*b**2*c**(1/4)*(1/d)**(1/4)*log(-(-1)
**(1/4)*c**(1/4)*(1/d)**(1/4) + sqrt(x))/(2*d) - (-1)**(1/4)*b**2*c**(1/4)*(1/d)**(1/4)*log((-1)**(1/4)*c**(1/
4)*(1/d)**(1/4) + sqrt(x))/(2*d) + (-1)**(1/4)*b**2*c**(1/4)*(1/d)**(1/4)*atan((-1)**(3/4)*sqrt(x)/(c**(1/4)*(
1/d)**(1/4)))/d + 2*b**2*sqrt(x)/d, True))

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