3.5.5 \(\int \frac {(a+b x^2)^2}{x^{9/2} (c+d x^2)} \, dx\)

Optimal. Leaf size=269 \[ -\frac {2 a^2}{7 c x^{7/2}}-\frac {(b c-a d)^2 \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{2 \sqrt {2} c^{11/4} \sqrt [4]{d}}+\frac {(b c-a d)^2 \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{2 \sqrt {2} c^{11/4} \sqrt [4]{d}}-\frac {(b c-a d)^2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{\sqrt {2} c^{11/4} \sqrt [4]{d}}+\frac {(b c-a d)^2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )}{\sqrt {2} c^{11/4} \sqrt [4]{d}}-\frac {2 a (2 b c-a d)}{3 c^2 x^{3/2}} \]

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Rubi [A]  time = 0.28, antiderivative size = 269, 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, 453, 329, 211, 1165, 628, 1162, 617, 204} \begin {gather*} -\frac {2 a^2}{7 c x^{7/2}}-\frac {2 a (2 b c-a d)}{3 c^2 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^{11/4} \sqrt [4]{d}}+\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^{11/4} \sqrt [4]{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^{11/4} \sqrt [4]{d}}+\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^{11/4} \sqrt [4]{d}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^2/(x^(9/2)*(c + d*x^2)),x]

[Out]

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

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

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Mathematica [A]  time = 0.13, size = 254, normalized size = 0.94 \begin {gather*} \frac {-\frac {24 a^2 c^{7/4}}{x^{7/2}}+\frac {56 a c^{3/4} (a d-2 b c)}{x^{3/2}}-\frac {21 \sqrt {2} (b c-a d)^2 \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{\sqrt [4]{d}}+\frac {21 \sqrt {2} (b c-a d)^2 \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{\sqrt [4]{d}}-\frac {42 \sqrt {2} (b c-a d)^2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{\sqrt [4]{d}}+\frac {42 \sqrt {2} (b c-a d)^2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )}{\sqrt [4]{d}}}{84 c^{11/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^2/(x^(9/2)*(c + d*x^2)),x]

[Out]

((-24*a^2*c^(7/4))/x^(7/2) + (56*a*c^(3/4)*(-2*b*c + a*d))/x^(3/2) - (42*Sqrt[2]*(b*c - a*d)^2*ArcTan[1 - (Sqr
t[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/d^(1/4) + (42*Sqrt[2]*(b*c - a*d)^2*ArcTan[1 + (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/
4)])/d^(1/4) - (21*Sqrt[2]*(b*c - a*d)^2*Log[Sqrt[c] - Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/d^(1/4) +
(21*Sqrt[2]*(b*c - a*d)^2*Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/d^(1/4))/(84*c^(11/4))

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IntegrateAlgebraic [A]  time = 0.20, size = 165, normalized size = 0.61 \begin {gather*} -\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^{11/4} \sqrt [4]{d}}+\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^{11/4} \sqrt [4]{d}}-\frac {2 a \left (3 a c-7 a d x^2+14 b c x^2\right )}{21 c^2 x^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b*x^2)^2/(x^(9/2)*(c + d*x^2)),x]

[Out]

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

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fricas [B]  time = 1.34, size = 1252, normalized size = 4.65

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/x^(9/2)/(d*x^2+c),x, algorithm="fricas")

[Out]

1/42*(84*c^2*x^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 - 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^11*d))^(1/4)*arctan((sqrt(c^6*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^11*d)) + (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^8*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^11*d))^(3/4) - (b^2*c^10*d - 2*a*
b*c^9*d^2 + a^2*c^8*d^3)*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^11*d))^(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)) + 21*c^2*x^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^11*d))^(
1/4)*log(c^3*(-(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^11*d))^(1/4) + (b^2*c^2 - 2*a*b*c*d + a^2*d^2
)*sqrt(x)) - 21*c^2*x^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^11*d))^(1/4)*log(-c^3*(-(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^11*d))^(1/4) + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(x)) - 4*(3*a^2*c +
7*(2*a*b*c - a^2*d)*x^2)*sqrt(x))/(c^2*x^4)

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giac [A]  time = 0.48, size = 354, normalized size = 1.32 \begin {gather*} \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^{3} d} + \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^{3} d} + \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^{3} d} - \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^{3} d} - \frac {2 \, {\left (14 \, a b c x^{2} - 7 \, a^{2} d x^{2} + 3 \, a^{2} c\right )}}{21 \, c^{2} x^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/x^(9/2)/(d*x^2+c),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.02, size = 461, normalized size = 1.71 \begin {gather*} \frac {\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, a^{2} d^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )}{2 c^{3}}+\frac {\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, a^{2} d^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )}{2 c^{3}}+\frac {\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, a^{2} d^{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 c^{3}}-\frac {\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, a b d \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )}{c^{2}}-\frac {\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, a b d \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )}{c^{2}}-\frac {\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, a b 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 )}{2 c^{2}}+\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 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 c}+\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 c}+\frac {2 a^{2} d}{3 c^{2} x^{\frac {3}{2}}}-\frac {4 a b}{3 c \,x^{\frac {3}{2}}}-\frac {2 a^{2}}{7 c \,x^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2/x^(9/2)/(d*x^2+c),x)

[Out]

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

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maxima [A]  time = 2.36, size = 293, normalized size = 1.09 \begin {gather*} \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^{2}} - \frac {2 \, {\left (3 \, a^{2} c + 7 \, {\left (2 \, a b c - a^{2} d\right )} x^{2}\right )}}{21 \, c^{2} x^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/x^(9/2)/(d*x^2+c),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.44, size = 1209, normalized size = 4.49

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^2/(x^(9/2)*(c + d*x^2)),x)

[Out]

(atan(((((x^(1/2)*(16*a^4*c^6*d^7 + 16*b^4*c^10*d^3 - 64*a*b^3*c^9*d^4 - 64*a^3*b*c^7*d^6 + 96*a^2*b^2*c^8*d^5
))/2 - ((a*d - b*c)^2*(16*a^2*c^9*d^5 + 16*b^2*c^11*d^3 - 32*a*b*c^10*d^4))/(2*(-c)^(11/4)*d^(1/4)))*(a*d - b*
c)^2*1i)/((-c)^(11/4)*d^(1/4)) + (((x^(1/2)*(16*a^4*c^6*d^7 + 16*b^4*c^10*d^3 - 64*a*b^3*c^9*d^4 - 64*a^3*b*c^
7*d^6 + 96*a^2*b^2*c^8*d^5))/2 + ((a*d - b*c)^2*(16*a^2*c^9*d^5 + 16*b^2*c^11*d^3 - 32*a*b*c^10*d^4))/(2*(-c)^
(11/4)*d^(1/4)))*(a*d - b*c)^2*1i)/((-c)^(11/4)*d^(1/4)))/((((x^(1/2)*(16*a^4*c^6*d^7 + 16*b^4*c^10*d^3 - 64*a
*b^3*c^9*d^4 - 64*a^3*b*c^7*d^6 + 96*a^2*b^2*c^8*d^5))/2 - ((a*d - b*c)^2*(16*a^2*c^9*d^5 + 16*b^2*c^11*d^3 -
32*a*b*c^10*d^4))/(2*(-c)^(11/4)*d^(1/4)))*(a*d - b*c)^2)/((-c)^(11/4)*d^(1/4)) - (((x^(1/2)*(16*a^4*c^6*d^7 +
16*b^4*c^10*d^3 - 64*a*b^3*c^9*d^4 - 64*a^3*b*c^7*d^6 + 96*a^2*b^2*c^8*d^5))/2 + ((a*d - b*c)^2*(16*a^2*c^9*d
^5 + 16*b^2*c^11*d^3 - 32*a*b*c^10*d^4))/(2*(-c)^(11/4)*d^(1/4)))*(a*d - b*c)^2)/((-c)^(11/4)*d^(1/4))))*(a*d
- b*c)^2*1i)/((-c)^(11/4)*d^(1/4)) - ((2*a^2)/(7*c) - (2*a*x^2*(a*d - 2*b*c))/(3*c^2))/x^(7/2) + (atan(((((x^(
1/2)*(16*a^4*c^6*d^7 + 16*b^4*c^10*d^3 - 64*a*b^3*c^9*d^4 - 64*a^3*b*c^7*d^6 + 96*a^2*b^2*c^8*d^5))/2 - ((a*d
- b*c)^2*(16*a^2*c^9*d^5 + 16*b^2*c^11*d^3 - 32*a*b*c^10*d^4)*1i)/(2*(-c)^(11/4)*d^(1/4)))*(a*d - b*c)^2)/((-c
)^(11/4)*d^(1/4)) + (((x^(1/2)*(16*a^4*c^6*d^7 + 16*b^4*c^10*d^3 - 64*a*b^3*c^9*d^4 - 64*a^3*b*c^7*d^6 + 96*a^
2*b^2*c^8*d^5))/2 + ((a*d - b*c)^2*(16*a^2*c^9*d^5 + 16*b^2*c^11*d^3 - 32*a*b*c^10*d^4)*1i)/(2*(-c)^(11/4)*d^(
1/4)))*(a*d - b*c)^2)/((-c)^(11/4)*d^(1/4)))/((((x^(1/2)*(16*a^4*c^6*d^7 + 16*b^4*c^10*d^3 - 64*a*b^3*c^9*d^4
- 64*a^3*b*c^7*d^6 + 96*a^2*b^2*c^8*d^5))/2 - ((a*d - b*c)^2*(16*a^2*c^9*d^5 + 16*b^2*c^11*d^3 - 32*a*b*c^10*d
^4)*1i)/(2*(-c)^(11/4)*d^(1/4)))*(a*d - b*c)^2*1i)/((-c)^(11/4)*d^(1/4)) - (((x^(1/2)*(16*a^4*c^6*d^7 + 16*b^4
*c^10*d^3 - 64*a*b^3*c^9*d^4 - 64*a^3*b*c^7*d^6 + 96*a^2*b^2*c^8*d^5))/2 + ((a*d - b*c)^2*(16*a^2*c^9*d^5 + 16
*b^2*c^11*d^3 - 32*a*b*c^10*d^4)*1i)/(2*(-c)^(11/4)*d^(1/4)))*(a*d - b*c)^2*1i)/((-c)^(11/4)*d^(1/4))))*(a*d -
b*c)^2)/((-c)^(11/4)*d^(1/4))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2/x**(9/2)/(d*x**2+c),x)

[Out]

Timed out

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