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

Optimal. Leaf size=332 \[ -\frac {\sqrt {x} \left (7 a^2 d^2-6 a b c d+3 b^2 c^2\right )}{6 c^2 d \left (c+d x^2\right )}-\frac {2 a^2}{3 c x^{3/2} \left (c+d x^2\right )}-\frac {(b c-a d) (7 a d+b c) \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{8 \sqrt {2} c^{11/4} d^{5/4}}+\frac {(b c-a d) (7 a d+b c) \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{8 \sqrt {2} c^{11/4} d^{5/4}}-\frac {(b c-a d) (7 a d+b c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{4 \sqrt {2} c^{11/4} d^{5/4}}+\frac {(b c-a d) (7 a d+b c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )}{4 \sqrt {2} c^{11/4} d^{5/4}} \]

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Rubi [A]  time = 0.34, antiderivative size = 332, 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, 457, 329, 211, 1165, 628, 1162, 617, 204} \begin {gather*} -\frac {\sqrt {x} \left (7 a^2 d^2-6 a b c d+3 b^2 c^2\right )}{6 c^2 d \left (c+d x^2\right )}-\frac {2 a^2}{3 c x^{3/2} \left (c+d x^2\right )}-\frac {(b c-a d) (7 a d+b c) \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{8 \sqrt {2} c^{11/4} d^{5/4}}+\frac {(b c-a d) (7 a d+b c) \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{8 \sqrt {2} c^{11/4} d^{5/4}}-\frac {(b c-a d) (7 a d+b c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{4 \sqrt {2} c^{11/4} d^{5/4}}+\frac {(b c-a d) (7 a d+b c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )}{4 \sqrt {2} c^{11/4} d^{5/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2)/(3*c*x^(3/2)*(c + d*x^2)) - ((3*b^2*c^2 - 6*a*b*c*d + 7*a^2*d^2)*Sqrt[x])/(6*c^2*d*(c + d*x^2)) - ((b
*c - a*d)*(b*c + 7*a*d)*ArcTan[1 - (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/(4*Sqrt[2]*c^(11/4)*d^(5/4)) + ((b*c -
a*d)*(b*c + 7*a*d)*ArcTan[1 + (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/(4*Sqrt[2]*c^(11/4)*d^(5/4)) - ((b*c - a*d)*
(b*c + 7*a*d)*Log[Sqrt[c] - Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/(8*Sqrt[2]*c^(11/4)*d^(5/4)) + ((b*c
- a*d)*(b*c + 7*a*d)*Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/(8*Sqrt[2]*c^(11/4)*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 457

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d
)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b*e*n*(p + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b
*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] &&
LeQ[-1, m, -(n*(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^{5/2} \left (c+d x^2\right )^2} \, dx &=-\frac {2 a^2}{3 c x^{3/2} \left (c+d x^2\right )}+\frac {2 \int \frac {\frac {1}{2} a (6 b c-7 a d)+\frac {3}{2} b^2 c x^2}{\sqrt {x} \left (c+d x^2\right )^2} \, dx}{3 c}\\ &=-\frac {2 a^2}{3 c x^{3/2} \left (c+d x^2\right )}+\frac {\left (6 a b-\frac {3 b^2 c}{d}-\frac {7 a^2 d}{c}\right ) \sqrt {x}}{6 c \left (c+d x^2\right )}+\frac {((b c-a d) (b c+7 a d)) \int \frac {1}{\sqrt {x} \left (c+d x^2\right )} \, dx}{4 c^2 d}\\ &=-\frac {2 a^2}{3 c x^{3/2} \left (c+d x^2\right )}+\frac {\left (6 a b-\frac {3 b^2 c}{d}-\frac {7 a^2 d}{c}\right ) \sqrt {x}}{6 c \left (c+d x^2\right )}+\frac {((b c-a d) (b c+7 a d)) \operatorname {Subst}\left (\int \frac {1}{c+d x^4} \, dx,x,\sqrt {x}\right )}{2 c^2 d}\\ &=-\frac {2 a^2}{3 c x^{3/2} \left (c+d x^2\right )}+\frac {\left (6 a b-\frac {3 b^2 c}{d}-\frac {7 a^2 d}{c}\right ) \sqrt {x}}{6 c \left (c+d x^2\right )}+\frac {((b c-a d) (b c+7 a d)) \operatorname {Subst}\left (\int \frac {\sqrt {c}-\sqrt {d} x^2}{c+d x^4} \, dx,x,\sqrt {x}\right )}{4 c^{5/2} d}+\frac {((b c-a d) (b c+7 a d)) \operatorname {Subst}\left (\int \frac {\sqrt {c}+\sqrt {d} x^2}{c+d x^4} \, dx,x,\sqrt {x}\right )}{4 c^{5/2} d}\\ &=-\frac {2 a^2}{3 c x^{3/2} \left (c+d x^2\right )}+\frac {\left (6 a b-\frac {3 b^2 c}{d}-\frac {7 a^2 d}{c}\right ) \sqrt {x}}{6 c \left (c+d x^2\right )}+\frac {((b c-a d) (b c+7 a d)) \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 )}{8 c^{5/2} d^{3/2}}+\frac {((b c-a d) (b c+7 a d)) \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 )}{8 c^{5/2} d^{3/2}}-\frac {((b c-a d) (b c+7 a d)) \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 )}{8 \sqrt {2} c^{11/4} d^{5/4}}-\frac {((b c-a d) (b c+7 a d)) \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 )}{8 \sqrt {2} c^{11/4} d^{5/4}}\\ &=-\frac {2 a^2}{3 c x^{3/2} \left (c+d x^2\right )}+\frac {\left (6 a b-\frac {3 b^2 c}{d}-\frac {7 a^2 d}{c}\right ) \sqrt {x}}{6 c \left (c+d x^2\right )}-\frac {(b c-a d) (b c+7 a d) \log \left (\sqrt {c}-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {d} x\right )}{8 \sqrt {2} c^{11/4} d^{5/4}}+\frac {(b c-a d) (b c+7 a d) \log \left (\sqrt {c}+\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {d} x\right )}{8 \sqrt {2} c^{11/4} d^{5/4}}+\frac {((b c-a d) (b c+7 a d)) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{4 \sqrt {2} c^{11/4} d^{5/4}}-\frac {((b c-a d) (b c+7 a d)) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{4 \sqrt {2} c^{11/4} d^{5/4}}\\ &=-\frac {2 a^2}{3 c x^{3/2} \left (c+d x^2\right )}+\frac {\left (6 a b-\frac {3 b^2 c}{d}-\frac {7 a^2 d}{c}\right ) \sqrt {x}}{6 c \left (c+d x^2\right )}-\frac {(b c-a d) (b c+7 a d) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{4 \sqrt {2} c^{11/4} d^{5/4}}+\frac {(b c-a d) (b c+7 a d) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{4 \sqrt {2} c^{11/4} d^{5/4}}-\frac {(b c-a d) (b c+7 a d) \log \left (\sqrt {c}-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {d} x\right )}{8 \sqrt {2} c^{11/4} d^{5/4}}+\frac {(b c-a d) (b c+7 a d) \log \left (\sqrt {c}+\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {d} x\right )}{8 \sqrt {2} c^{11/4} d^{5/4}}\\ \end {align*}

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Mathematica [A]  time = 0.19, size = 315, normalized size = 0.95 \begin {gather*} \frac {-\frac {3 \sqrt {2} \left (-7 a^2 d^2+6 a b c d+b^2 c^2\right ) \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{d^{5/4}}+\frac {3 \sqrt {2} \left (-7 a^2 d^2+6 a b c d+b^2 c^2\right ) \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{d^{5/4}}-\frac {6 \sqrt {2} \left (-7 a^2 d^2+6 a b c d+b^2 c^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{d^{5/4}}+\frac {6 \sqrt {2} \left (-7 a^2 d^2+6 a b c d+b^2 c^2\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )}{d^{5/4}}-\frac {32 a^2 c^{3/4}}{x^{3/2}}-\frac {24 c^{3/4} \sqrt {x} (b c-a d)^2}{d \left (c+d x^2\right )}}{48 c^{11/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-32*a^2*c^(3/4))/x^(3/2) - (24*c^(3/4)*(b*c - a*d)^2*Sqrt[x])/(d*(c + d*x^2)) - (6*Sqrt[2]*(b^2*c^2 + 6*a*b*
c*d - 7*a^2*d^2)*ArcTan[1 - (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/d^(5/4) + (6*Sqrt[2]*(b^2*c^2 + 6*a*b*c*d - 7*
a^2*d^2)*ArcTan[1 + (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/d^(5/4) - (3*Sqrt[2]*(b^2*c^2 + 6*a*b*c*d - 7*a^2*d^2)
*Log[Sqrt[c] - Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/d^(5/4) + (3*Sqrt[2]*(b^2*c^2 + 6*a*b*c*d - 7*a^2
*d^2)*Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/d^(5/4))/(48*c^(11/4))

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IntegrateAlgebraic [A]  time = 0.74, size = 219, normalized size = 0.66 \begin {gather*} \frac {-4 a^2 c d-7 a^2 d^2 x^2+6 a b c d x^2-3 b^2 c^2 x^2}{6 c^2 d x^{3/2} \left (c+d x^2\right )}-\frac {\left (-7 a^2 d^2+6 a b c d+b^2 c^2\right ) \tan ^{-1}\left (\frac {\sqrt {c}-\sqrt {d} x}{\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}}\right )}{4 \sqrt {2} c^{11/4} d^{5/4}}+\frac {\left (-7 a^2 d^2+6 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}}{\sqrt {c}+\sqrt {d} x}\right )}{4 \sqrt {2} c^{11/4} d^{5/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*a^2*c*d - 3*b^2*c^2*x^2 + 6*a*b*c*d*x^2 - 7*a^2*d^2*x^2)/(6*c^2*d*x^(3/2)*(c + d*x^2)) - ((b^2*c^2 + 6*a*b
*c*d - 7*a^2*d^2)*ArcTan[(Sqrt[c] - Sqrt[d]*x)/(Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x])])/(4*Sqrt[2]*c^(11/4)*d^(5/4)
) + ((b^2*c^2 + 6*a*b*c*d - 7*a^2*d^2)*ArcTanh[(Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x])/(Sqrt[c] + Sqrt[d]*x)])/(4*Sq
rt[2]*c^(11/4)*d^(5/4))

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fricas [B]  time = 1.38, size = 1340, normalized size = 4.04

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)^2,x, algorithm="fricas")

[Out]

-1/24*(12*(c^2*d^2*x^4 + c^3*d*x^2)*(-(b^8*c^8 + 24*a*b^7*c^7*d + 188*a^2*b^6*c^6*d^2 + 360*a^3*b^5*c^5*d^3 -
1434*a^4*b^4*c^4*d^4 - 2520*a^5*b^3*c^3*d^5 + 9212*a^6*b^2*c^2*d^6 - 8232*a^7*b*c*d^7 + 2401*a^8*d^8)/(c^11*d^
5))^(1/4)*arctan((sqrt(c^6*d^2*sqrt(-(b^8*c^8 + 24*a*b^7*c^7*d + 188*a^2*b^6*c^6*d^2 + 360*a^3*b^5*c^5*d^3 - 1
434*a^4*b^4*c^4*d^4 - 2520*a^5*b^3*c^3*d^5 + 9212*a^6*b^2*c^2*d^6 - 8232*a^7*b*c*d^7 + 2401*a^8*d^8)/(c^11*d^5
)) + (b^4*c^4 + 12*a*b^3*c^3*d + 22*a^2*b^2*c^2*d^2 - 84*a^3*b*c*d^3 + 49*a^4*d^4)*x)*c^8*d^4*(-(b^8*c^8 + 24*
a*b^7*c^7*d + 188*a^2*b^6*c^6*d^2 + 360*a^3*b^5*c^5*d^3 - 1434*a^4*b^4*c^4*d^4 - 2520*a^5*b^3*c^3*d^5 + 9212*a
^6*b^2*c^2*d^6 - 8232*a^7*b*c*d^7 + 2401*a^8*d^8)/(c^11*d^5))^(3/4) + (b^2*c^10*d^4 + 6*a*b*c^9*d^5 - 7*a^2*c^
8*d^6)*sqrt(x)*(-(b^8*c^8 + 24*a*b^7*c^7*d + 188*a^2*b^6*c^6*d^2 + 360*a^3*b^5*c^5*d^3 - 1434*a^4*b^4*c^4*d^4
- 2520*a^5*b^3*c^3*d^5 + 9212*a^6*b^2*c^2*d^6 - 8232*a^7*b*c*d^7 + 2401*a^8*d^8)/(c^11*d^5))^(3/4))/(b^8*c^8 +
24*a*b^7*c^7*d + 188*a^2*b^6*c^6*d^2 + 360*a^3*b^5*c^5*d^3 - 1434*a^4*b^4*c^4*d^4 - 2520*a^5*b^3*c^3*d^5 + 92
12*a^6*b^2*c^2*d^6 - 8232*a^7*b*c*d^7 + 2401*a^8*d^8)) + 3*(c^2*d^2*x^4 + c^3*d*x^2)*(-(b^8*c^8 + 24*a*b^7*c^7
*d + 188*a^2*b^6*c^6*d^2 + 360*a^3*b^5*c^5*d^3 - 1434*a^4*b^4*c^4*d^4 - 2520*a^5*b^3*c^3*d^5 + 9212*a^6*b^2*c^
2*d^6 - 8232*a^7*b*c*d^7 + 2401*a^8*d^8)/(c^11*d^5))^(1/4)*log(c^3*d*(-(b^8*c^8 + 24*a*b^7*c^7*d + 188*a^2*b^6
*c^6*d^2 + 360*a^3*b^5*c^5*d^3 - 1434*a^4*b^4*c^4*d^4 - 2520*a^5*b^3*c^3*d^5 + 9212*a^6*b^2*c^2*d^6 - 8232*a^7
*b*c*d^7 + 2401*a^8*d^8)/(c^11*d^5))^(1/4) - (b^2*c^2 + 6*a*b*c*d - 7*a^2*d^2)*sqrt(x)) - 3*(c^2*d^2*x^4 + c^3
*d*x^2)*(-(b^8*c^8 + 24*a*b^7*c^7*d + 188*a^2*b^6*c^6*d^2 + 360*a^3*b^5*c^5*d^3 - 1434*a^4*b^4*c^4*d^4 - 2520*
a^5*b^3*c^3*d^5 + 9212*a^6*b^2*c^2*d^6 - 8232*a^7*b*c*d^7 + 2401*a^8*d^8)/(c^11*d^5))^(1/4)*log(-c^3*d*(-(b^8*
c^8 + 24*a*b^7*c^7*d + 188*a^2*b^6*c^6*d^2 + 360*a^3*b^5*c^5*d^3 - 1434*a^4*b^4*c^4*d^4 - 2520*a^5*b^3*c^3*d^5
+ 9212*a^6*b^2*c^2*d^6 - 8232*a^7*b*c*d^7 + 2401*a^8*d^8)/(c^11*d^5))^(1/4) - (b^2*c^2 + 6*a*b*c*d - 7*a^2*d^
2)*sqrt(x)) + 4*(4*a^2*c*d + (3*b^2*c^2 - 6*a*b*c*d + 7*a^2*d^2)*x^2)*sqrt(x))/(c^2*d^2*x^4 + c^3*d*x^2)

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giac [A]  time = 0.44, size = 384, normalized size = 1.16 \begin {gather*} -\frac {2 \, a^{2}}{3 \, c^{2} x^{\frac {3}{2}}} + \frac {\sqrt {2} {\left (\left (c d^{3}\right )^{\frac {1}{4}} b^{2} c^{2} + 6 \, \left (c d^{3}\right )^{\frac {1}{4}} a b c d - 7 \, \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 )}{8 \, c^{3} d^{2}} + \frac {\sqrt {2} {\left (\left (c d^{3}\right )^{\frac {1}{4}} b^{2} c^{2} + 6 \, \left (c d^{3}\right )^{\frac {1}{4}} a b c d - 7 \, \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 )}{8 \, c^{3} d^{2}} + \frac {\sqrt {2} {\left (\left (c d^{3}\right )^{\frac {1}{4}} b^{2} c^{2} + 6 \, \left (c d^{3}\right )^{\frac {1}{4}} a b c d - 7 \, \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 )}{16 \, c^{3} d^{2}} - \frac {\sqrt {2} {\left (\left (c d^{3}\right )^{\frac {1}{4}} b^{2} c^{2} + 6 \, \left (c d^{3}\right )^{\frac {1}{4}} a b c d - 7 \, \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 )}{16 \, c^{3} d^{2}} - \frac {b^{2} c^{2} \sqrt {x} - 2 \, a b c d \sqrt {x} + a^{2} d^{2} \sqrt {x}}{2 \, {\left (d x^{2} + c\right )} c^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*a^2/(c^2*x^(3/2)) + 1/8*sqrt(2)*((c*d^3)^(1/4)*b^2*c^2 + 6*(c*d^3)^(1/4)*a*b*c*d - 7*(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^2) + 1/8*sqrt(2)*((c*d^3)^(1/4)*b^
2*c^2 + 6*(c*d^3)^(1/4)*a*b*c*d - 7*(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^2) + 1/16*sqrt(2)*((c*d^3)^(1/4)*b^2*c^2 + 6*(c*d^3)^(1/4)*a*b*c*d - 7*(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) - 1/16*sqrt(2)*((c*d^3)^(1/4)*b^2*c^2 + 6*(c
*d^3)^(1/4)*a*b*c*d - 7*(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) - 1
/2*(b^2*c^2*sqrt(x) - 2*a*b*c*d*sqrt(x) + a^2*d^2*sqrt(x))/((d*x^2 + c)*c^2*d)

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maple [A]  time = 0.02, size = 498, normalized size = 1.50 \begin {gather*} -\frac {a^{2} d \sqrt {x}}{2 \left (d \,x^{2}+c \right ) c^{2}}+\frac {a b \sqrt {x}}{\left (d \,x^{2}+c \right ) c}-\frac {b^{2} \sqrt {x}}{2 \left (d \,x^{2}+c \right ) d}-\frac {7 \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 )}{8 c^{3}}-\frac {7 \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 )}{8 c^{3}}-\frac {7 \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 )}{16 c^{3}}+\frac {3 \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 )}{4 c^{2}}+\frac {3 \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 )}{4 c^{2}}+\frac {3 \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 )}{8 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 )}{8 c 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 )}{8 c 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 )}{16 c d}-\frac {2 a^{2}}{3 c^{2} x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/c^2*d*x^(1/2)/(d*x^2+c)*a^2+1/c*x^(1/2)/(d*x^2+c)*a*b-1/2/d*x^(1/2)/(d*x^2+c)*b^2-7/8/c^3*d*(c/d)^(1/4)*2
^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)+1)*a^2+3/4/c^2*(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2
)+1)*a*b+1/8/c/d*(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)+1)*b^2-7/8/c^3*d*(c/d)^(1/4)*2^(1/2)*a
rctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)-1)*a^2+3/4/c^2*(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)-1)*a*b
+1/8/c/d*(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)-1)*b^2-7/16/c^3*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+3/8/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+1/16/c/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^2-2/3*a^2/c^2/x^(3/2)

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maxima [A]  time = 2.43, size = 326, normalized size = 0.98 \begin {gather*} -\frac {4 \, a^{2} c d + {\left (3 \, b^{2} c^{2} - 6 \, a b c d + 7 \, a^{2} d^{2}\right )} x^{2}}{6 \, {\left (c^{2} d^{2} x^{\frac {7}{2}} + c^{3} d x^{\frac {3}{2}}\right )}} + \frac {\frac {2 \, \sqrt {2} {\left (b^{2} c^{2} + 6 \, a b c d - 7 \, 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} + 6 \, a b c d - 7 \, 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} + 6 \, a b c d - 7 \, 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} + 6 \, a b c d - 7 \, 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}}}}{16 \, c^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(4*a^2*c*d + (3*b^2*c^2 - 6*a*b*c*d + 7*a^2*d^2)*x^2)/(c^2*d^2*x^(7/2) + c^3*d*x^(3/2)) + 1/16*(2*sqrt(2)
*(b^2*c^2 + 6*a*b*c*d - 7*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 + 6*a*b*c*d - 7*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 + 6*a*b*c*d - 7*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 + 6*a*b*c*d - 7*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*d)

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mupad [B]  time = 0.56, size = 1340, normalized size = 4.04

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)^2),x)

[Out]

(atan((((x^(1/2)*(1568*a^4*c^6*d^10 + 32*b^4*c^10*d^6 + 384*a*b^3*c^9*d^7 - 2688*a^3*b*c^7*d^9 + 704*a^2*b^2*c
^8*d^8) - ((a*d - b*c)*(7*a*d + b*c)*(256*b^2*c^11*d^7 - 1792*a^2*c^9*d^9 + 1536*a*b*c^10*d^8))/(8*(-c)^(11/4)
*d^(5/4)))*(a*d - b*c)*(7*a*d + b*c)*1i)/(8*(-c)^(11/4)*d^(5/4)) + ((x^(1/2)*(1568*a^4*c^6*d^10 + 32*b^4*c^10*
d^6 + 384*a*b^3*c^9*d^7 - 2688*a^3*b*c^7*d^9 + 704*a^2*b^2*c^8*d^8) + ((a*d - b*c)*(7*a*d + b*c)*(256*b^2*c^11
*d^7 - 1792*a^2*c^9*d^9 + 1536*a*b*c^10*d^8))/(8*(-c)^(11/4)*d^(5/4)))*(a*d - b*c)*(7*a*d + b*c)*1i)/(8*(-c)^(
11/4)*d^(5/4)))/(((x^(1/2)*(1568*a^4*c^6*d^10 + 32*b^4*c^10*d^6 + 384*a*b^3*c^9*d^7 - 2688*a^3*b*c^7*d^9 + 704
*a^2*b^2*c^8*d^8) - ((a*d - b*c)*(7*a*d + b*c)*(256*b^2*c^11*d^7 - 1792*a^2*c^9*d^9 + 1536*a*b*c^10*d^8))/(8*(
-c)^(11/4)*d^(5/4)))*(a*d - b*c)*(7*a*d + b*c))/(8*(-c)^(11/4)*d^(5/4)) - ((x^(1/2)*(1568*a^4*c^6*d^10 + 32*b^
4*c^10*d^6 + 384*a*b^3*c^9*d^7 - 2688*a^3*b*c^7*d^9 + 704*a^2*b^2*c^8*d^8) + ((a*d - b*c)*(7*a*d + b*c)*(256*b
^2*c^11*d^7 - 1792*a^2*c^9*d^9 + 1536*a*b*c^10*d^8))/(8*(-c)^(11/4)*d^(5/4)))*(a*d - b*c)*(7*a*d + b*c))/(8*(-
c)^(11/4)*d^(5/4))))*(a*d - b*c)*(7*a*d + b*c)*1i)/(4*(-c)^(11/4)*d^(5/4)) - ((2*a^2)/(3*c) + (x^2*(7*a^2*d^2
+ 3*b^2*c^2 - 6*a*b*c*d))/(6*c^2*d))/(c*x^(3/2) + d*x^(7/2)) + (atan((((x^(1/2)*(1568*a^4*c^6*d^10 + 32*b^4*c^
10*d^6 + 384*a*b^3*c^9*d^7 - 2688*a^3*b*c^7*d^9 + 704*a^2*b^2*c^8*d^8) - ((a*d - b*c)*(7*a*d + b*c)*(256*b^2*c
^11*d^7 - 1792*a^2*c^9*d^9 + 1536*a*b*c^10*d^8)*1i)/(8*(-c)^(11/4)*d^(5/4)))*(a*d - b*c)*(7*a*d + b*c))/(8*(-c
)^(11/4)*d^(5/4)) + ((x^(1/2)*(1568*a^4*c^6*d^10 + 32*b^4*c^10*d^6 + 384*a*b^3*c^9*d^7 - 2688*a^3*b*c^7*d^9 +
704*a^2*b^2*c^8*d^8) + ((a*d - b*c)*(7*a*d + b*c)*(256*b^2*c^11*d^7 - 1792*a^2*c^9*d^9 + 1536*a*b*c^10*d^8)*1i
)/(8*(-c)^(11/4)*d^(5/4)))*(a*d - b*c)*(7*a*d + b*c))/(8*(-c)^(11/4)*d^(5/4)))/(((x^(1/2)*(1568*a^4*c^6*d^10 +
32*b^4*c^10*d^6 + 384*a*b^3*c^9*d^7 - 2688*a^3*b*c^7*d^9 + 704*a^2*b^2*c^8*d^8) - ((a*d - b*c)*(7*a*d + b*c)*
(256*b^2*c^11*d^7 - 1792*a^2*c^9*d^9 + 1536*a*b*c^10*d^8)*1i)/(8*(-c)^(11/4)*d^(5/4)))*(a*d - b*c)*(7*a*d + b*
c)*1i)/(8*(-c)^(11/4)*d^(5/4)) - ((x^(1/2)*(1568*a^4*c^6*d^10 + 32*b^4*c^10*d^6 + 384*a*b^3*c^9*d^7 - 2688*a^3
*b*c^7*d^9 + 704*a^2*b^2*c^8*d^8) + ((a*d - b*c)*(7*a*d + b*c)*(256*b^2*c^11*d^7 - 1792*a^2*c^9*d^9 + 1536*a*b
*c^10*d^8)*1i)/(8*(-c)^(11/4)*d^(5/4)))*(a*d - b*c)*(7*a*d + b*c)*1i)/(8*(-c)^(11/4)*d^(5/4))))*(a*d - b*c)*(7
*a*d + b*c))/(4*(-c)^(11/4)*d^(5/4))

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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**(5/2)/(d*x**2+c)**2,x)

[Out]

Timed out

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