3.424 \(\int \frac {(c+d x^2)^2}{x^{11/2} (a+b x^2)} \, dx\)

Optimal. Leaf size=288 \[ -\frac {\sqrt [4]{b} (b c-a d)^2 \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{2 \sqrt {2} a^{13/4}}+\frac {\sqrt [4]{b} (b c-a d)^2 \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{2 \sqrt {2} a^{13/4}}+\frac {\sqrt [4]{b} (b c-a d)^2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} a^{13/4}}-\frac {\sqrt [4]{b} (b c-a d)^2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{\sqrt {2} a^{13/4}}-\frac {2 (b c-a d)^2}{a^3 \sqrt {x}}+\frac {2 c (b c-2 a d)}{5 a^2 x^{5/2}}-\frac {2 c^2}{9 a x^{9/2}} \]

[Out]

-2/9*c^2/a/x^(9/2)+2/5*c*(-2*a*d+b*c)/a^2/x^(5/2)+1/2*b^(1/4)*(-a*d+b*c)^2*arctan(1-b^(1/4)*2^(1/2)*x^(1/2)/a^
(1/4))/a^(13/4)*2^(1/2)-1/2*b^(1/4)*(-a*d+b*c)^2*arctan(1+b^(1/4)*2^(1/2)*x^(1/2)/a^(1/4))/a^(13/4)*2^(1/2)-1/
4*b^(1/4)*(-a*d+b*c)^2*ln(a^(1/2)+x*b^(1/2)-a^(1/4)*b^(1/4)*2^(1/2)*x^(1/2))/a^(13/4)*2^(1/2)+1/4*b^(1/4)*(-a*
d+b*c)^2*ln(a^(1/2)+x*b^(1/2)+a^(1/4)*b^(1/4)*2^(1/2)*x^(1/2))/a^(13/4)*2^(1/2)-2*(-a*d+b*c)^2/a^3/x^(1/2)

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Rubi [A]  time = 0.32, antiderivative size = 288, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {462, 453, 325, 329, 297, 1162, 617, 204, 1165, 628} \[ \frac {2 c (b c-2 a d)}{5 a^2 x^{5/2}}-\frac {2 (b c-a d)^2}{a^3 \sqrt {x}}-\frac {\sqrt [4]{b} (b c-a d)^2 \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{2 \sqrt {2} a^{13/4}}+\frac {\sqrt [4]{b} (b c-a d)^2 \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{2 \sqrt {2} a^{13/4}}+\frac {\sqrt [4]{b} (b c-a d)^2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} a^{13/4}}-\frac {\sqrt [4]{b} (b c-a d)^2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{\sqrt {2} a^{13/4}}-\frac {2 c^2}{9 a x^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^2)^2/(x^(11/2)*(a + b*x^2)),x]

[Out]

(-2*c^2)/(9*a*x^(9/2)) + (2*c*(b*c - 2*a*d))/(5*a^2*x^(5/2)) - (2*(b*c - a*d)^2)/(a^3*Sqrt[x]) + (b^(1/4)*(b*c
- a*d)^2*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(Sqrt[2]*a^(13/4)) - (b^(1/4)*(b*c - a*d)^2*ArcTan[1
+ (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(Sqrt[2]*a^(13/4)) - (b^(1/4)*(b*c - a*d)^2*Log[Sqrt[a] - Sqrt[2]*a^(1/4
)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(2*Sqrt[2]*a^(13/4)) + (b^(1/4)*(b*c - a*d)^2*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^
(1/4)*Sqrt[x] + Sqrt[b]*x])/(2*Sqrt[2]*a^(13/4))

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 325

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

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 453

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

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 (c+d x^2\right )^2}{x^{11/2} \left (a+b x^2\right )} \, dx &=-\frac {2 c^2}{9 a x^{9/2}}+\frac {2 \int \frac {-\frac {9}{2} c (b c-2 a d)+\frac {9}{2} a d^2 x^2}{x^{7/2} \left (a+b x^2\right )} \, dx}{9 a}\\ &=-\frac {2 c^2}{9 a x^{9/2}}+\frac {2 c (b c-2 a d)}{5 a^2 x^{5/2}}+\frac {(b c-a d)^2 \int \frac {1}{x^{3/2} \left (a+b x^2\right )} \, dx}{a^2}\\ &=-\frac {2 c^2}{9 a x^{9/2}}+\frac {2 c (b c-2 a d)}{5 a^2 x^{5/2}}-\frac {2 (b c-a d)^2}{a^3 \sqrt {x}}-\frac {\left (b (b c-a d)^2\right ) \int \frac {\sqrt {x}}{a+b x^2} \, dx}{a^3}\\ &=-\frac {2 c^2}{9 a x^{9/2}}+\frac {2 c (b c-2 a d)}{5 a^2 x^{5/2}}-\frac {2 (b c-a d)^2}{a^3 \sqrt {x}}-\frac {\left (2 b (b c-a d)^2\right ) \operatorname {Subst}\left (\int \frac {x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{a^3}\\ &=-\frac {2 c^2}{9 a x^{9/2}}+\frac {2 c (b c-2 a d)}{5 a^2 x^{5/2}}-\frac {2 (b c-a d)^2}{a^3 \sqrt {x}}+\frac {\left (\sqrt {b} (b c-a d)^2\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a}-\sqrt {b} x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{a^3}-\frac {\left (\sqrt {b} (b c-a d)^2\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a}+\sqrt {b} x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{a^3}\\ &=-\frac {2 c^2}{9 a x^{9/2}}+\frac {2 c (b c-2 a d)}{5 a^2 x^{5/2}}-\frac {2 (b c-a d)^2}{a^3 \sqrt {x}}-\frac {(b c-a d)^2 \operatorname {Subst}\left (\int \frac {1}{\frac {\sqrt {a}}{\sqrt {b}}-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}+x^2} \, dx,x,\sqrt {x}\right )}{2 a^3}-\frac {(b c-a d)^2 \operatorname {Subst}\left (\int \frac {1}{\frac {\sqrt {a}}{\sqrt {b}}+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}+x^2} \, dx,x,\sqrt {x}\right )}{2 a^3}-\frac {\left (\sqrt [4]{b} (b c-a d)^2\right ) \operatorname {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{a}}{\sqrt [4]{b}}+2 x}{-\frac {\sqrt {a}}{\sqrt {b}}-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}-x^2} \, dx,x,\sqrt {x}\right )}{2 \sqrt {2} a^{13/4}}-\frac {\left (\sqrt [4]{b} (b c-a d)^2\right ) \operatorname {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{a}}{\sqrt [4]{b}}-2 x}{-\frac {\sqrt {a}}{\sqrt {b}}+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}-x^2} \, dx,x,\sqrt {x}\right )}{2 \sqrt {2} a^{13/4}}\\ &=-\frac {2 c^2}{9 a x^{9/2}}+\frac {2 c (b c-2 a d)}{5 a^2 x^{5/2}}-\frac {2 (b c-a d)^2}{a^3 \sqrt {x}}-\frac {\sqrt [4]{b} (b c-a d)^2 \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{2 \sqrt {2} a^{13/4}}+\frac {\sqrt [4]{b} (b c-a d)^2 \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{2 \sqrt {2} a^{13/4}}-\frac {\left (\sqrt [4]{b} (b c-a d)^2\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} a^{13/4}}+\frac {\left (\sqrt [4]{b} (b c-a d)^2\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} a^{13/4}}\\ &=-\frac {2 c^2}{9 a x^{9/2}}+\frac {2 c (b c-2 a d)}{5 a^2 x^{5/2}}-\frac {2 (b c-a d)^2}{a^3 \sqrt {x}}+\frac {\sqrt [4]{b} (b c-a d)^2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} a^{13/4}}-\frac {\sqrt [4]{b} (b c-a d)^2 \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} a^{13/4}}-\frac {\sqrt [4]{b} (b c-a d)^2 \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{2 \sqrt {2} a^{13/4}}+\frac {\sqrt [4]{b} (b c-a d)^2 \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{2 \sqrt {2} a^{13/4}}\\ \end {align*}

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Mathematica [C]  time = 0.21, size = 101, normalized size = 0.35 \[ -\frac {2 \left (a \left (a^2 \left (5 c^2+18 c d x^2+45 d^2 x^4\right )-9 a b c x^2 \left (c+10 d x^2\right )+45 b^2 c^2 x^4\right )+15 b x^6 (b c-a d)^2 \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};-\frac {b x^2}{a}\right )\right )}{45 a^4 x^{9/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^2)^2/(x^(11/2)*(a + b*x^2)),x]

[Out]

(-2*(a*(45*b^2*c^2*x^4 - 9*a*b*c*x^2*(c + 10*d*x^2) + a^2*(5*c^2 + 18*c*d*x^2 + 45*d^2*x^4)) + 15*b*(b*c - a*d
)^2*x^6*Hypergeometric2F1[3/4, 1, 7/4, -((b*x^2)/a)]))/(45*a^4*x^(9/2))

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fricas [B]  time = 0.59, size = 1686, normalized size = 5.85 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^2/x^(11/2)/(b*x^2+a),x, algorithm="fricas")

[Out]

1/90*(180*a^3*x^5*(-(b^9*c^8 - 8*a*b^8*c^7*d + 28*a^2*b^7*c^6*d^2 - 56*a^3*b^6*c^5*d^3 + 70*a^4*b^5*c^4*d^4 -
56*a^5*b^4*c^3*d^5 + 28*a^6*b^3*c^2*d^6 - 8*a^7*b^2*c*d^7 + a^8*b*d^8)/a^13)^(1/4)*arctan((sqrt((b^14*c^12 - 1
2*a*b^13*c^11*d + 66*a^2*b^12*c^10*d^2 - 220*a^3*b^11*c^9*d^3 + 495*a^4*b^10*c^8*d^4 - 792*a^5*b^9*c^7*d^5 + 9
24*a^6*b^8*c^6*d^6 - 792*a^7*b^7*c^5*d^7 + 495*a^8*b^6*c^4*d^8 - 220*a^9*b^5*c^3*d^9 + 66*a^10*b^4*c^2*d^10 -
12*a^11*b^3*c*d^11 + a^12*b^2*d^12)*x - (a^7*b^9*c^8 - 8*a^8*b^8*c^7*d + 28*a^9*b^7*c^6*d^2 - 56*a^10*b^6*c^5*
d^3 + 70*a^11*b^5*c^4*d^4 - 56*a^12*b^4*c^3*d^5 + 28*a^13*b^3*c^2*d^6 - 8*a^14*b^2*c*d^7 + a^15*b*d^8)*sqrt(-(
b^9*c^8 - 8*a*b^8*c^7*d + 28*a^2*b^7*c^6*d^2 - 56*a^3*b^6*c^5*d^3 + 70*a^4*b^5*c^4*d^4 - 56*a^5*b^4*c^3*d^5 +
28*a^6*b^3*c^2*d^6 - 8*a^7*b^2*c*d^7 + a^8*b*d^8)/a^13))*a^3*(-(b^9*c^8 - 8*a*b^8*c^7*d + 28*a^2*b^7*c^6*d^2 -
56*a^3*b^6*c^5*d^3 + 70*a^4*b^5*c^4*d^4 - 56*a^5*b^4*c^3*d^5 + 28*a^6*b^3*c^2*d^6 - 8*a^7*b^2*c*d^7 + a^8*b*d
^8)/a^13)^(1/4) - (a^3*b^7*c^6 - 6*a^4*b^6*c^5*d + 15*a^5*b^5*c^4*d^2 - 20*a^6*b^4*c^3*d^3 + 15*a^7*b^3*c^2*d^
4 - 6*a^8*b^2*c*d^5 + a^9*b*d^6)*sqrt(x)*(-(b^9*c^8 - 8*a*b^8*c^7*d + 28*a^2*b^7*c^6*d^2 - 56*a^3*b^6*c^5*d^3
+ 70*a^4*b^5*c^4*d^4 - 56*a^5*b^4*c^3*d^5 + 28*a^6*b^3*c^2*d^6 - 8*a^7*b^2*c*d^7 + a^8*b*d^8)/a^13)^(1/4))/(b^
9*c^8 - 8*a*b^8*c^7*d + 28*a^2*b^7*c^6*d^2 - 56*a^3*b^6*c^5*d^3 + 70*a^4*b^5*c^4*d^4 - 56*a^5*b^4*c^3*d^5 + 28
*a^6*b^3*c^2*d^6 - 8*a^7*b^2*c*d^7 + a^8*b*d^8)) - 45*a^3*x^5*(-(b^9*c^8 - 8*a*b^8*c^7*d + 28*a^2*b^7*c^6*d^2
- 56*a^3*b^6*c^5*d^3 + 70*a^4*b^5*c^4*d^4 - 56*a^5*b^4*c^3*d^5 + 28*a^6*b^3*c^2*d^6 - 8*a^7*b^2*c*d^7 + a^8*b*
d^8)/a^13)^(1/4)*log(a^10*(-(b^9*c^8 - 8*a*b^8*c^7*d + 28*a^2*b^7*c^6*d^2 - 56*a^3*b^6*c^5*d^3 + 70*a^4*b^5*c^
4*d^4 - 56*a^5*b^4*c^3*d^5 + 28*a^6*b^3*c^2*d^6 - 8*a^7*b^2*c*d^7 + a^8*b*d^8)/a^13)^(3/4) + (b^7*c^6 - 6*a*b^
6*c^5*d + 15*a^2*b^5*c^4*d^2 - 20*a^3*b^4*c^3*d^3 + 15*a^4*b^3*c^2*d^4 - 6*a^5*b^2*c*d^5 + a^6*b*d^6)*sqrt(x))
+ 45*a^3*x^5*(-(b^9*c^8 - 8*a*b^8*c^7*d + 28*a^2*b^7*c^6*d^2 - 56*a^3*b^6*c^5*d^3 + 70*a^4*b^5*c^4*d^4 - 56*a
^5*b^4*c^3*d^5 + 28*a^6*b^3*c^2*d^6 - 8*a^7*b^2*c*d^7 + a^8*b*d^8)/a^13)^(1/4)*log(-a^10*(-(b^9*c^8 - 8*a*b^8*
c^7*d + 28*a^2*b^7*c^6*d^2 - 56*a^3*b^6*c^5*d^3 + 70*a^4*b^5*c^4*d^4 - 56*a^5*b^4*c^3*d^5 + 28*a^6*b^3*c^2*d^6
- 8*a^7*b^2*c*d^7 + a^8*b*d^8)/a^13)^(3/4) + (b^7*c^6 - 6*a*b^6*c^5*d + 15*a^2*b^5*c^4*d^2 - 20*a^3*b^4*c^3*d
^3 + 15*a^4*b^3*c^2*d^4 - 6*a^5*b^2*c*d^5 + a^6*b*d^6)*sqrt(x)) - 4*(45*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*x^4 +
5*a^2*c^2 - 9*(a*b*c^2 - 2*a^2*c*d)*x^2)*sqrt(x))/(a^3*x^5)

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giac [A]  time = 0.47, size = 390, normalized size = 1.35 \[ -\frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {3}{4}} b^{2} c^{2} - 2 \, \left (a b^{3}\right )^{\frac {3}{4}} a b c d + \left (a b^{3}\right )^{\frac {3}{4}} a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{2 \, a^{4} b^{2}} - \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {3}{4}} b^{2} c^{2} - 2 \, \left (a b^{3}\right )^{\frac {3}{4}} a b c d + \left (a b^{3}\right )^{\frac {3}{4}} a^{2} d^{2}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{2 \, a^{4} b^{2}} + \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {3}{4}} b^{2} c^{2} - 2 \, \left (a b^{3}\right )^{\frac {3}{4}} a b c d + \left (a b^{3}\right )^{\frac {3}{4}} a^{2} d^{2}\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{4 \, a^{4} b^{2}} - \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {3}{4}} b^{2} c^{2} - 2 \, \left (a b^{3}\right )^{\frac {3}{4}} a b c d + \left (a b^{3}\right )^{\frac {3}{4}} a^{2} d^{2}\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{4 \, a^{4} b^{2}} - \frac {2 \, {\left (45 \, b^{2} c^{2} x^{4} - 90 \, a b c d x^{4} + 45 \, a^{2} d^{2} x^{4} - 9 \, a b c^{2} x^{2} + 18 \, a^{2} c d x^{2} + 5 \, a^{2} c^{2}\right )}}{45 \, a^{3} x^{\frac {9}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^2/x^(11/2)/(b*x^2+a),x, algorithm="giac")

[Out]

-1/2*sqrt(2)*((a*b^3)^(3/4)*b^2*c^2 - 2*(a*b^3)^(3/4)*a*b*c*d + (a*b^3)^(3/4)*a^2*d^2)*arctan(1/2*sqrt(2)*(sqr
t(2)*(a/b)^(1/4) + 2*sqrt(x))/(a/b)^(1/4))/(a^4*b^2) - 1/2*sqrt(2)*((a*b^3)^(3/4)*b^2*c^2 - 2*(a*b^3)^(3/4)*a*
b*c*d + (a*b^3)^(3/4)*a^2*d^2)*arctan(-1/2*sqrt(2)*(sqrt(2)*(a/b)^(1/4) - 2*sqrt(x))/(a/b)^(1/4))/(a^4*b^2) +
1/4*sqrt(2)*((a*b^3)^(3/4)*b^2*c^2 - 2*(a*b^3)^(3/4)*a*b*c*d + (a*b^3)^(3/4)*a^2*d^2)*log(sqrt(2)*sqrt(x)*(a/b
)^(1/4) + x + sqrt(a/b))/(a^4*b^2) - 1/4*sqrt(2)*((a*b^3)^(3/4)*b^2*c^2 - 2*(a*b^3)^(3/4)*a*b*c*d + (a*b^3)^(3
/4)*a^2*d^2)*log(-sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/(a^4*b^2) - 2/45*(45*b^2*c^2*x^4 - 90*a*b*c*d*x
^4 + 45*a^2*d^2*x^4 - 9*a*b*c^2*x^2 + 18*a^2*c*d*x^2 + 5*a^2*c^2)/(a^3*x^(9/2))

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maple [B]  time = 0.02, size = 495, normalized size = 1.72 \[ -\frac {\sqrt {2}\, d^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )}{2 \left (\frac {a}{b}\right )^{\frac {1}{4}} a}-\frac {\sqrt {2}\, d^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )}{2 \left (\frac {a}{b}\right )^{\frac {1}{4}} a}-\frac {\sqrt {2}\, d^{2} \ln \left (\frac {x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{b}}}{x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{b}}}\right )}{4 \left (\frac {a}{b}\right )^{\frac {1}{4}} a}+\frac {\sqrt {2}\, b c d \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )}{\left (\frac {a}{b}\right )^{\frac {1}{4}} a^{2}}+\frac {\sqrt {2}\, b c d \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )}{\left (\frac {a}{b}\right )^{\frac {1}{4}} a^{2}}+\frac {\sqrt {2}\, b c d \ln \left (\frac {x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{b}}}{x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{b}}}\right )}{2 \left (\frac {a}{b}\right )^{\frac {1}{4}} a^{2}}-\frac {\sqrt {2}\, b^{2} c^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )}{2 \left (\frac {a}{b}\right )^{\frac {1}{4}} a^{3}}-\frac {\sqrt {2}\, b^{2} c^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )}{2 \left (\frac {a}{b}\right )^{\frac {1}{4}} a^{3}}-\frac {\sqrt {2}\, b^{2} c^{2} \ln \left (\frac {x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{b}}}{x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{b}}}\right )}{4 \left (\frac {a}{b}\right )^{\frac {1}{4}} a^{3}}-\frac {2 d^{2}}{a \sqrt {x}}+\frac {4 b c d}{a^{2} \sqrt {x}}-\frac {2 b^{2} c^{2}}{a^{3} \sqrt {x}}-\frac {4 c d}{5 a \,x^{\frac {5}{2}}}+\frac {2 b \,c^{2}}{5 a^{2} x^{\frac {5}{2}}}-\frac {2 c^{2}}{9 a \,x^{\frac {9}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^2+c)^2/x^(11/2)/(b*x^2+a),x)

[Out]

-1/2/a/(a/b)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)-1)*d^2+1/a^2/(a/b)^(1/4)*2^(1/2)*arctan(2^(1/2)/
(a/b)^(1/4)*x^(1/2)-1)*b*c*d-1/2/a^3/(a/b)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)-1)*b^2*c^2-1/4/a/(
a/b)^(1/4)*2^(1/2)*ln((x-(a/b)^(1/4)*2^(1/2)*x^(1/2)+(a/b)^(1/2))/(x+(a/b)^(1/4)*2^(1/2)*x^(1/2)+(a/b)^(1/2)))
*d^2+1/2/a^2/(a/b)^(1/4)*2^(1/2)*ln((x-(a/b)^(1/4)*2^(1/2)*x^(1/2)+(a/b)^(1/2))/(x+(a/b)^(1/4)*2^(1/2)*x^(1/2)
+(a/b)^(1/2)))*b*c*d-1/4/a^3/(a/b)^(1/4)*2^(1/2)*ln((x-(a/b)^(1/4)*2^(1/2)*x^(1/2)+(a/b)^(1/2))/(x+(a/b)^(1/4)
*2^(1/2)*x^(1/2)+(a/b)^(1/2)))*b^2*c^2-1/2/a/(a/b)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)+1)*d^2+1/a
^2/(a/b)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)+1)*b*c*d-1/2/a^3/(a/b)^(1/4)*2^(1/2)*arctan(2^(1/2)/
(a/b)^(1/4)*x^(1/2)+1)*b^2*c^2-2/9*c^2/a/x^(9/2)-2/a/x^(1/2)*d^2+4/a^2/x^(1/2)*b*c*d-2/a^3/x^(1/2)*b^2*c^2-4/5
*c/a/x^(5/2)*d+2/5*c^2/a^2/x^(5/2)*b

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maxima [A]  time = 2.60, size = 263, normalized size = 0.91 \[ -\frac {{\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} - \frac {\sqrt {2} \log \left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}}\right )}}{4 \, a^{3}} - \frac {2 \, {\left (45 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x^{4} + 5 \, a^{2} c^{2} - 9 \, {\left (a b c^{2} - 2 \, a^{2} c d\right )} x^{2}\right )}}{45 \, a^{3} x^{\frac {9}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^2/x^(11/2)/(b*x^2+a),x, algorithm="maxima")

[Out]

-1/4*(b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2)*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*a^(1/4)*b^(1/4) + 2*sqrt(b)*sq
rt(x))/sqrt(sqrt(a)*sqrt(b)))/(sqrt(sqrt(a)*sqrt(b))*sqrt(b)) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*a^(1/4)
*b^(1/4) - 2*sqrt(b)*sqrt(x))/sqrt(sqrt(a)*sqrt(b)))/(sqrt(sqrt(a)*sqrt(b))*sqrt(b)) - sqrt(2)*log(sqrt(2)*a^(
1/4)*b^(1/4)*sqrt(x) + sqrt(b)*x + sqrt(a))/(a^(1/4)*b^(3/4)) + sqrt(2)*log(-sqrt(2)*a^(1/4)*b^(1/4)*sqrt(x) +
sqrt(b)*x + sqrt(a))/(a^(1/4)*b^(3/4)))/a^3 - 2/45*(45*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*x^4 + 5*a^2*c^2 - 9*(a
*b*c^2 - 2*a^2*c*d)*x^2)/(a^3*x^(9/2))

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mupad [B]  time = 0.35, size = 451, normalized size = 1.57 \[ \frac {{\left (-b\right )}^{1/4}\,\mathrm {atanh}\left (\frac {{\left (-b\right )}^{1/4}\,\sqrt {x}\,{\left (a\,d-b\,c\right )}^2\,\left (16\,a^{14}\,b^4\,d^4-64\,a^{13}\,b^5\,c\,d^3+96\,a^{12}\,b^6\,c^2\,d^2-64\,a^{11}\,b^7\,c^3\,d+16\,a^{10}\,b^8\,c^4\right )}{a^{13/4}\,\left (16\,a^{13}\,b^4\,d^6-96\,a^{12}\,b^5\,c\,d^5+240\,a^{11}\,b^6\,c^2\,d^4-320\,a^{10}\,b^7\,c^3\,d^3+240\,a^9\,b^8\,c^4\,d^2-96\,a^8\,b^9\,c^5\,d+16\,a^7\,b^{10}\,c^6\right )}\right )\,{\left (a\,d-b\,c\right )}^2}{a^{13/4}}-\frac {{\left (-b\right )}^{1/4}\,\mathrm {atan}\left (\frac {{\left (-b\right )}^{1/4}\,\sqrt {x}\,{\left (a\,d-b\,c\right )}^2\,\left (16\,a^{14}\,b^4\,d^4-64\,a^{13}\,b^5\,c\,d^3+96\,a^{12}\,b^6\,c^2\,d^2-64\,a^{11}\,b^7\,c^3\,d+16\,a^{10}\,b^8\,c^4\right )}{a^{13/4}\,\left (16\,a^{13}\,b^4\,d^6-96\,a^{12}\,b^5\,c\,d^5+240\,a^{11}\,b^6\,c^2\,d^4-320\,a^{10}\,b^7\,c^3\,d^3+240\,a^9\,b^8\,c^4\,d^2-96\,a^8\,b^9\,c^5\,d+16\,a^7\,b^{10}\,c^6\right )}\right )\,{\left (a\,d-b\,c\right )}^2}{a^{13/4}}-\frac {\frac {2\,c^2}{9\,a}+\frac {2\,x^4\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}{a^3}+\frac {2\,c\,x^2\,\left (2\,a\,d-b\,c\right )}{5\,a^2}}{x^{9/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x^2)^2/(x^(11/2)*(a + b*x^2)),x)

[Out]

((-b)^(1/4)*atanh(((-b)^(1/4)*x^(1/2)*(a*d - b*c)^2*(16*a^10*b^8*c^4 + 16*a^14*b^4*d^4 - 64*a^11*b^7*c^3*d - 6
4*a^13*b^5*c*d^3 + 96*a^12*b^6*c^2*d^2))/(a^(13/4)*(16*a^7*b^10*c^6 + 16*a^13*b^4*d^6 - 96*a^8*b^9*c^5*d - 96*
a^12*b^5*c*d^5 + 240*a^9*b^8*c^4*d^2 - 320*a^10*b^7*c^3*d^3 + 240*a^11*b^6*c^2*d^4)))*(a*d - b*c)^2)/a^(13/4)
- ((-b)^(1/4)*atan(((-b)^(1/4)*x^(1/2)*(a*d - b*c)^2*(16*a^10*b^8*c^4 + 16*a^14*b^4*d^4 - 64*a^11*b^7*c^3*d -
64*a^13*b^5*c*d^3 + 96*a^12*b^6*c^2*d^2))/(a^(13/4)*(16*a^7*b^10*c^6 + 16*a^13*b^4*d^6 - 96*a^8*b^9*c^5*d - 96
*a^12*b^5*c*d^5 + 240*a^9*b^8*c^4*d^2 - 320*a^10*b^7*c^3*d^3 + 240*a^11*b^6*c^2*d^4)))*(a*d - b*c)^2)/a^(13/4)
- ((2*c^2)/(9*a) + (2*x^4*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d))/a^3 + (2*c*x^2*(2*a*d - b*c))/(5*a^2))/x^(9/2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**2+c)**2/x**(11/2)/(b*x**2+a),x)

[Out]

Timed out

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