3.4.97 \(\int \frac {x^{7/2} (a+b x^2)^2}{c+d x^2} \, dx\)

Optimal. Leaf size=311 \[ -\frac {c^{5/4} (b c-a d)^2 \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{2 \sqrt {2} d^{17/4}}+\frac {c^{5/4} (b c-a d)^2 \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{2 \sqrt {2} d^{17/4}}-\frac {c^{5/4} (b c-a d)^2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{\sqrt {2} d^{17/4}}+\frac {c^{5/4} (b c-a d)^2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )}{\sqrt {2} d^{17/4}}-\frac {2 c \sqrt {x} (b c-a d)^2}{d^4}+\frac {2 x^{5/2} (b c-a d)^2}{5 d^3}-\frac {2 b x^{9/2} (b c-2 a d)}{9 d^2}+\frac {2 b^2 x^{13/2}}{13 d} \]

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Rubi [A]  time = 0.31, antiderivative size = 311, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 9, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {461, 321, 329, 211, 1165, 628, 1162, 617, 204} \begin {gather*} -\frac {c^{5/4} (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} d^{17/4}}+\frac {c^{5/4} (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} d^{17/4}}-\frac {c^{5/4} (b c-a d)^2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{\sqrt {2} d^{17/4}}+\frac {c^{5/4} (b c-a d)^2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )}{\sqrt {2} d^{17/4}}-\frac {2 b x^{9/2} (b c-2 a d)}{9 d^2}+\frac {2 x^{5/2} (b c-a d)^2}{5 d^3}-\frac {2 c \sqrt {x} (b c-a d)^2}{d^4}+\frac {2 b^2 x^{13/2}}{13 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^(7/2)*(a + b*x^2)^2)/(c + d*x^2),x]

[Out]

(-2*c*(b*c - a*d)^2*Sqrt[x])/d^4 + (2*(b*c - a*d)^2*x^(5/2))/(5*d^3) - (2*b*(b*c - 2*a*d)*x^(9/2))/(9*d^2) + (
2*b^2*x^(13/2))/(13*d) - (c^(5/4)*(b*c - a*d)^2*ArcTan[1 - (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/(Sqrt[2]*d^(17/
4)) + (c^(5/4)*(b*c - a*d)^2*ArcTan[1 + (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/(Sqrt[2]*d^(17/4)) - (c^(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]*d^(17/4)) + (c^(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]*d^(17/4))

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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && 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 461

Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegr
and[((e*x)^m*(a + b*x^n)^p)/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n
, 0] && IGtQ[p, 0] && (IntegerQ[m] || IGtQ[2*(m + 1), 0] ||  !RationalQ[m])

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 {x^{7/2} \left (a+b x^2\right )^2}{c+d x^2} \, dx &=\int \left (-\frac {b (b c-2 a d) x^{7/2}}{d^2}+\frac {b^2 x^{11/2}}{d}+\frac {\left (b^2 c^2-2 a b c d+a^2 d^2\right ) x^{7/2}}{d^2 \left (c+d x^2\right )}\right ) \, dx\\ &=-\frac {2 b (b c-2 a d) x^{9/2}}{9 d^2}+\frac {2 b^2 x^{13/2}}{13 d}+\frac {(b c-a d)^2 \int \frac {x^{7/2}}{c+d x^2} \, dx}{d^2}\\ &=\frac {2 (b c-a d)^2 x^{5/2}}{5 d^3}-\frac {2 b (b c-2 a d) x^{9/2}}{9 d^2}+\frac {2 b^2 x^{13/2}}{13 d}-\frac {\left (c (b c-a d)^2\right ) \int \frac {x^{3/2}}{c+d x^2} \, dx}{d^3}\\ &=-\frac {2 c (b c-a d)^2 \sqrt {x}}{d^4}+\frac {2 (b c-a d)^2 x^{5/2}}{5 d^3}-\frac {2 b (b c-2 a d) x^{9/2}}{9 d^2}+\frac {2 b^2 x^{13/2}}{13 d}+\frac {\left (c^2 (b c-a d)^2\right ) \int \frac {1}{\sqrt {x} \left (c+d x^2\right )} \, dx}{d^4}\\ &=-\frac {2 c (b c-a d)^2 \sqrt {x}}{d^4}+\frac {2 (b c-a d)^2 x^{5/2}}{5 d^3}-\frac {2 b (b c-2 a d) x^{9/2}}{9 d^2}+\frac {2 b^2 x^{13/2}}{13 d}+\frac {\left (2 c^2 (b c-a d)^2\right ) \operatorname {Subst}\left (\int \frac {1}{c+d x^4} \, dx,x,\sqrt {x}\right )}{d^4}\\ &=-\frac {2 c (b c-a d)^2 \sqrt {x}}{d^4}+\frac {2 (b c-a d)^2 x^{5/2}}{5 d^3}-\frac {2 b (b c-2 a d) x^{9/2}}{9 d^2}+\frac {2 b^2 x^{13/2}}{13 d}+\frac {\left (c^{3/2} (b c-a d)^2\right ) \operatorname {Subst}\left (\int \frac {\sqrt {c}-\sqrt {d} x^2}{c+d x^4} \, dx,x,\sqrt {x}\right )}{d^4}+\frac {\left (c^{3/2} (b c-a d)^2\right ) \operatorname {Subst}\left (\int \frac {\sqrt {c}+\sqrt {d} x^2}{c+d x^4} \, dx,x,\sqrt {x}\right )}{d^4}\\ &=-\frac {2 c (b c-a d)^2 \sqrt {x}}{d^4}+\frac {2 (b c-a d)^2 x^{5/2}}{5 d^3}-\frac {2 b (b c-2 a d) x^{9/2}}{9 d^2}+\frac {2 b^2 x^{13/2}}{13 d}+\frac {\left (c^{3/2} (b c-a d)^2\right ) \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 d^{9/2}}+\frac {\left (c^{3/2} (b c-a d)^2\right ) \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 d^{9/2}}-\frac {\left (c^{5/4} (b c-a d)^2\right ) \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} d^{17/4}}-\frac {\left (c^{5/4} (b c-a d)^2\right ) \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} d^{17/4}}\\ &=-\frac {2 c (b c-a d)^2 \sqrt {x}}{d^4}+\frac {2 (b c-a d)^2 x^{5/2}}{5 d^3}-\frac {2 b (b c-2 a d) x^{9/2}}{9 d^2}+\frac {2 b^2 x^{13/2}}{13 d}-\frac {c^{5/4} (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} d^{17/4}}+\frac {c^{5/4} (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} d^{17/4}}+\frac {\left (c^{5/4} (b c-a d)^2\right ) \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} d^{17/4}}-\frac {\left (c^{5/4} (b c-a d)^2\right ) \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} d^{17/4}}\\ &=-\frac {2 c (b c-a d)^2 \sqrt {x}}{d^4}+\frac {2 (b c-a d)^2 x^{5/2}}{5 d^3}-\frac {2 b (b c-2 a d) x^{9/2}}{9 d^2}+\frac {2 b^2 x^{13/2}}{13 d}-\frac {c^{5/4} (b c-a d)^2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{\sqrt {2} d^{17/4}}+\frac {c^{5/4} (b c-a d)^2 \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{\sqrt {2} d^{17/4}}-\frac {c^{5/4} (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} d^{17/4}}+\frac {c^{5/4} (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} d^{17/4}}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 299, normalized size = 0.96 \begin {gather*} \frac {-585 \sqrt {2} c^{5/4} (b c-a d)^2 \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )+585 \sqrt {2} c^{5/4} (b c-a d)^2 \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )-1170 \sqrt {2} c^{5/4} (b c-a d)^2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )+1170 \sqrt {2} c^{5/4} (b c-a d)^2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )-520 b d^{9/4} x^{9/2} (b c-2 a d)+936 d^{5/4} x^{5/2} (b c-a d)^2-4680 c \sqrt [4]{d} \sqrt {x} (b c-a d)^2+360 b^2 d^{13/4} x^{13/2}}{2340 d^{17/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^(7/2)*(a + b*x^2)^2)/(c + d*x^2),x]

[Out]

(-4680*c*d^(1/4)*(b*c - a*d)^2*Sqrt[x] + 936*d^(5/4)*(b*c - a*d)^2*x^(5/2) - 520*b*d^(9/4)*(b*c - 2*a*d)*x^(9/
2) + 360*b^2*d^(13/4)*x^(13/2) - 1170*Sqrt[2]*c^(5/4)*(b*c - a*d)^2*ArcTan[1 - (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/
4)] + 1170*Sqrt[2]*c^(5/4)*(b*c - a*d)^2*ArcTan[1 + (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)] - 585*Sqrt[2]*c^(5/4)*(
b*c - a*d)^2*Log[Sqrt[c] - Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x] + 585*Sqrt[2]*c^(5/4)*(b*c - a*d)^2*Lo
g[Sqrt[c] + Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/(2340*d^(17/4))

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IntegrateAlgebraic [A]  time = 0.26, size = 238, normalized size = 0.77 \begin {gather*} \frac {2 \sqrt {x} \left (-585 a^2 c d^2+117 a^2 d^3 x^2+1170 a b c^2 d-234 a b c d^2 x^2+130 a b d^3 x^4-585 b^2 c^3+117 b^2 c^2 d x^2-65 b^2 c d^2 x^4+45 b^2 d^3 x^6\right )}{585 d^4}-\frac {c^{5/4} (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} d^{17/4}}+\frac {c^{5/4} (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} d^{17/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^(7/2)*(a + b*x^2)^2)/(c + d*x^2),x]

[Out]

(2*Sqrt[x]*(-585*b^2*c^3 + 1170*a*b*c^2*d - 585*a^2*c*d^2 + 117*b^2*c^2*d*x^2 - 234*a*b*c*d^2*x^2 + 117*a^2*d^
3*x^2 - 65*b^2*c*d^2*x^4 + 130*a*b*d^3*x^4 + 45*b^2*d^3*x^6))/(585*d^4) - (c^(5/4)*(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]*d^(17/4)) + (c^(5/4)*(b*c - a*d)^2*Ar
cTanh[(Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x])/(Sqrt[c] + Sqrt[d]*x)])/(Sqrt[2]*d^(17/4))

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fricas [B]  time = 1.41, size = 1334, normalized size = 4.29

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(b*x^2+a)^2/(d*x^2+c),x, algorithm="fricas")

[Out]

1/1170*(2340*d^4*(-(b^8*c^13 - 8*a*b^7*c^12*d + 28*a^2*b^6*c^11*d^2 - 56*a^3*b^5*c^10*d^3 + 70*a^4*b^4*c^9*d^4
- 56*a^5*b^3*c^8*d^5 + 28*a^6*b^2*c^7*d^6 - 8*a^7*b*c^6*d^7 + a^8*c^5*d^8)/d^17)^(1/4)*arctan((sqrt(d^8*sqrt(
-(b^8*c^13 - 8*a*b^7*c^12*d + 28*a^2*b^6*c^11*d^2 - 56*a^3*b^5*c^10*d^3 + 70*a^4*b^4*c^9*d^4 - 56*a^5*b^3*c^8*
d^5 + 28*a^6*b^2*c^7*d^6 - 8*a^7*b*c^6*d^7 + a^8*c^5*d^8)/d^17) + (b^4*c^6 - 4*a*b^3*c^5*d + 6*a^2*b^2*c^4*d^2
- 4*a^3*b*c^3*d^3 + a^4*c^2*d^4)*x)*d^13*(-(b^8*c^13 - 8*a*b^7*c^12*d + 28*a^2*b^6*c^11*d^2 - 56*a^3*b^5*c^10
*d^3 + 70*a^4*b^4*c^9*d^4 - 56*a^5*b^3*c^8*d^5 + 28*a^6*b^2*c^7*d^6 - 8*a^7*b*c^6*d^7 + a^8*c^5*d^8)/d^17)^(3/
4) - (b^2*c^3*d^13 - 2*a*b*c^2*d^14 + a^2*c*d^15)*sqrt(x)*(-(b^8*c^13 - 8*a*b^7*c^12*d + 28*a^2*b^6*c^11*d^2 -
56*a^3*b^5*c^10*d^3 + 70*a^4*b^4*c^9*d^4 - 56*a^5*b^3*c^8*d^5 + 28*a^6*b^2*c^7*d^6 - 8*a^7*b*c^6*d^7 + a^8*c^
5*d^8)/d^17)^(3/4))/(b^8*c^13 - 8*a*b^7*c^12*d + 28*a^2*b^6*c^11*d^2 - 56*a^3*b^5*c^10*d^3 + 70*a^4*b^4*c^9*d^
4 - 56*a^5*b^3*c^8*d^5 + 28*a^6*b^2*c^7*d^6 - 8*a^7*b*c^6*d^7 + a^8*c^5*d^8)) + 585*d^4*(-(b^8*c^13 - 8*a*b^7*
c^12*d + 28*a^2*b^6*c^11*d^2 - 56*a^3*b^5*c^10*d^3 + 70*a^4*b^4*c^9*d^4 - 56*a^5*b^3*c^8*d^5 + 28*a^6*b^2*c^7*
d^6 - 8*a^7*b*c^6*d^7 + a^8*c^5*d^8)/d^17)^(1/4)*log(d^4*(-(b^8*c^13 - 8*a*b^7*c^12*d + 28*a^2*b^6*c^11*d^2 -
56*a^3*b^5*c^10*d^3 + 70*a^4*b^4*c^9*d^4 - 56*a^5*b^3*c^8*d^5 + 28*a^6*b^2*c^7*d^6 - 8*a^7*b*c^6*d^7 + a^8*c^5
*d^8)/d^17)^(1/4) + (b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2)*sqrt(x)) - 585*d^4*(-(b^8*c^13 - 8*a*b^7*c^12*d + 28*a
^2*b^6*c^11*d^2 - 56*a^3*b^5*c^10*d^3 + 70*a^4*b^4*c^9*d^4 - 56*a^5*b^3*c^8*d^5 + 28*a^6*b^2*c^7*d^6 - 8*a^7*b
*c^6*d^7 + a^8*c^5*d^8)/d^17)^(1/4)*log(-d^4*(-(b^8*c^13 - 8*a*b^7*c^12*d + 28*a^2*b^6*c^11*d^2 - 56*a^3*b^5*c
^10*d^3 + 70*a^4*b^4*c^9*d^4 - 56*a^5*b^3*c^8*d^5 + 28*a^6*b^2*c^7*d^6 - 8*a^7*b*c^6*d^7 + a^8*c^5*d^8)/d^17)^
(1/4) + (b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2)*sqrt(x)) + 4*(45*b^2*d^3*x^6 - 585*b^2*c^3 + 1170*a*b*c^2*d - 585*
a^2*c*d^2 - 65*(b^2*c*d^2 - 2*a*b*d^3)*x^4 + 117*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*x^2)*sqrt(x))/d^4

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giac [A]  time = 0.46, size = 436, normalized size = 1.40 \begin {gather*} \frac {\sqrt {2} {\left (\left (c d^{3}\right )^{\frac {1}{4}} b^{2} c^{3} - 2 \, \left (c d^{3}\right )^{\frac {1}{4}} a b c^{2} d + \left (c d^{3}\right )^{\frac {1}{4}} a^{2} c 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 \, d^{5}} + \frac {\sqrt {2} {\left (\left (c d^{3}\right )^{\frac {1}{4}} b^{2} c^{3} - 2 \, \left (c d^{3}\right )^{\frac {1}{4}} a b c^{2} d + \left (c d^{3}\right )^{\frac {1}{4}} a^{2} c 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 \, d^{5}} + \frac {\sqrt {2} {\left (\left (c d^{3}\right )^{\frac {1}{4}} b^{2} c^{3} - 2 \, \left (c d^{3}\right )^{\frac {1}{4}} a b c^{2} d + \left (c d^{3}\right )^{\frac {1}{4}} a^{2} c d^{2}\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {c}{d}\right )^{\frac {1}{4}} + x + \sqrt {\frac {c}{d}}\right )}{4 \, d^{5}} - \frac {\sqrt {2} {\left (\left (c d^{3}\right )^{\frac {1}{4}} b^{2} c^{3} - 2 \, \left (c d^{3}\right )^{\frac {1}{4}} a b c^{2} d + \left (c d^{3}\right )^{\frac {1}{4}} a^{2} c d^{2}\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {c}{d}\right )^{\frac {1}{4}} + x + \sqrt {\frac {c}{d}}\right )}{4 \, d^{5}} + \frac {2 \, {\left (45 \, b^{2} d^{12} x^{\frac {13}{2}} - 65 \, b^{2} c d^{11} x^{\frac {9}{2}} + 130 \, a b d^{12} x^{\frac {9}{2}} + 117 \, b^{2} c^{2} d^{10} x^{\frac {5}{2}} - 234 \, a b c d^{11} x^{\frac {5}{2}} + 117 \, a^{2} d^{12} x^{\frac {5}{2}} - 585 \, b^{2} c^{3} d^{9} \sqrt {x} + 1170 \, a b c^{2} d^{10} \sqrt {x} - 585 \, a^{2} c d^{11} \sqrt {x}\right )}}{585 \, d^{13}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(b*x^2+a)^2/(d*x^2+c),x, algorithm="giac")

[Out]

1/2*sqrt(2)*((c*d^3)^(1/4)*b^2*c^3 - 2*(c*d^3)^(1/4)*a*b*c^2*d + (c*d^3)^(1/4)*a^2*c*d^2)*arctan(1/2*sqrt(2)*(
sqrt(2)*(c/d)^(1/4) + 2*sqrt(x))/(c/d)^(1/4))/d^5 + 1/2*sqrt(2)*((c*d^3)^(1/4)*b^2*c^3 - 2*(c*d^3)^(1/4)*a*b*c
^2*d + (c*d^3)^(1/4)*a^2*c*d^2)*arctan(-1/2*sqrt(2)*(sqrt(2)*(c/d)^(1/4) - 2*sqrt(x))/(c/d)^(1/4))/d^5 + 1/4*s
qrt(2)*((c*d^3)^(1/4)*b^2*c^3 - 2*(c*d^3)^(1/4)*a*b*c^2*d + (c*d^3)^(1/4)*a^2*c*d^2)*log(sqrt(2)*sqrt(x)*(c/d)
^(1/4) + x + sqrt(c/d))/d^5 - 1/4*sqrt(2)*((c*d^3)^(1/4)*b^2*c^3 - 2*(c*d^3)^(1/4)*a*b*c^2*d + (c*d^3)^(1/4)*a
^2*c*d^2)*log(-sqrt(2)*sqrt(x)*(c/d)^(1/4) + x + sqrt(c/d))/d^5 + 2/585*(45*b^2*d^12*x^(13/2) - 65*b^2*c*d^11*
x^(9/2) + 130*a*b*d^12*x^(9/2) + 117*b^2*c^2*d^10*x^(5/2) - 234*a*b*c*d^11*x^(5/2) + 117*a^2*d^12*x^(5/2) - 58
5*b^2*c^3*d^9*sqrt(x) + 1170*a*b*c^2*d^10*sqrt(x) - 585*a^2*c*d^11*sqrt(x))/d^13

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maple [B]  time = 0.02, size = 545, normalized size = 1.75 \begin {gather*} \frac {2 b^{2} x^{\frac {13}{2}}}{13 d}+\frac {4 a b \,x^{\frac {9}{2}}}{9 d}-\frac {2 b^{2} c \,x^{\frac {9}{2}}}{9 d^{2}}+\frac {2 a^{2} x^{\frac {5}{2}}}{5 d}-\frac {4 a b c \,x^{\frac {5}{2}}}{5 d^{2}}+\frac {2 b^{2} c^{2} x^{\frac {5}{2}}}{5 d^{3}}+\frac {\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, a^{2} c \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )}{2 d^{2}}+\frac {\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, a^{2} c \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )}{2 d^{2}}+\frac {\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, a^{2} c \ln \left (\frac {x +\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {c}{d}}}{x -\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {c}{d}}}\right )}{4 d^{2}}-\frac {\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, a b \,c^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )}{d^{3}}-\frac {\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, a b \,c^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )}{d^{3}}-\frac {\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, a b \,c^{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 )}{2 d^{3}}+\frac {\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, b^{2} c^{3} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )}{2 d^{4}}+\frac {\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, b^{2} c^{3} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )}{2 d^{4}}+\frac {\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, b^{2} c^{3} \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^{4}}-\frac {2 a^{2} c \sqrt {x}}{d^{2}}+\frac {4 a b \,c^{2} \sqrt {x}}{d^{3}}-\frac {2 b^{2} c^{3} \sqrt {x}}{d^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)*(b*x^2+a)^2/(d*x^2+c),x)

[Out]

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

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maxima [A]  time = 2.44, size = 360, normalized size = 1.16 \begin {gather*} \frac {{\left (\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}}}\right )} c^{2}}{4 \, d^{4}} + \frac {2 \, {\left (45 \, b^{2} d^{3} x^{\frac {13}{2}} - 65 \, {\left (b^{2} c d^{2} - 2 \, a b d^{3}\right )} x^{\frac {9}{2}} + 117 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} x^{\frac {5}{2}} - 585 \, {\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\right )} \sqrt {x}\right )}}{585 \, d^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(b*x^2+a)^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/d^4 + 2/585*(45*b^2*d^3*x^(13/2) - 65*(b^2*c*d^2 - 2*a*b*d^3)*x^(9/2) + 117*(b^2*c^2
*d - 2*a*b*c*d^2 + a^2*d^3)*x^(5/2) - 585*(b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2)*sqrt(x))/d^4

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mupad [B]  time = 0.38, size = 1202, normalized size = 3.86

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(7/2)*(a + b*x^2)^2)/(c + d*x^2),x)

[Out]

x^(5/2)*((2*a^2)/(5*d) + (c*((2*b^2*c)/d^2 - (4*a*b)/d))/(5*d)) - x^(9/2)*((2*b^2*c)/(9*d^2) - (4*a*b)/(9*d))
+ (2*b^2*x^(13/2))/(13*d) - (c*x^(1/2)*((2*a^2)/d + (c*((2*b^2*c)/d^2 - (4*a*b)/d))/d))/d + ((-c)^(5/4)*atan((
((-c)^(5/4)*(a*d - b*c)^2*((16*x^(1/2)*(b^4*c^8 + a^4*c^4*d^4 - 4*a^3*b*c^5*d^3 + 6*a^2*b^2*c^6*d^2 - 4*a*b^3*
c^7*d))/d^5 - (16*(-c)^(5/4)*(a*d - b*c)^2*(b^2*c^5 + a^2*c^3*d^2 - 2*a*b*c^4*d))/d^(21/4))*1i)/(2*d^(17/4)) +
((-c)^(5/4)*(a*d - b*c)^2*((16*x^(1/2)*(b^4*c^8 + a^4*c^4*d^4 - 4*a^3*b*c^5*d^3 + 6*a^2*b^2*c^6*d^2 - 4*a*b^3
*c^7*d))/d^5 + (16*(-c)^(5/4)*(a*d - b*c)^2*(b^2*c^5 + a^2*c^3*d^2 - 2*a*b*c^4*d))/d^(21/4))*1i)/(2*d^(17/4)))
/(((-c)^(5/4)*(a*d - b*c)^2*((16*x^(1/2)*(b^4*c^8 + a^4*c^4*d^4 - 4*a^3*b*c^5*d^3 + 6*a^2*b^2*c^6*d^2 - 4*a*b^
3*c^7*d))/d^5 - (16*(-c)^(5/4)*(a*d - b*c)^2*(b^2*c^5 + a^2*c^3*d^2 - 2*a*b*c^4*d))/d^(21/4)))/(2*d^(17/4)) -
((-c)^(5/4)*(a*d - b*c)^2*((16*x^(1/2)*(b^4*c^8 + a^4*c^4*d^4 - 4*a^3*b*c^5*d^3 + 6*a^2*b^2*c^6*d^2 - 4*a*b^3*
c^7*d))/d^5 + (16*(-c)^(5/4)*(a*d - b*c)^2*(b^2*c^5 + a^2*c^3*d^2 - 2*a*b*c^4*d))/d^(21/4)))/(2*d^(17/4))))*(a
*d - b*c)^2*1i)/d^(17/4) + ((-c)^(5/4)*atan((((-c)^(5/4)*(a*d - b*c)^2*((16*x^(1/2)*(b^4*c^8 + a^4*c^4*d^4 - 4
*a^3*b*c^5*d^3 + 6*a^2*b^2*c^6*d^2 - 4*a*b^3*c^7*d))/d^5 - ((-c)^(5/4)*(a*d - b*c)^2*(b^2*c^5 + a^2*c^3*d^2 -
2*a*b*c^4*d)*16i)/d^(21/4)))/(2*d^(17/4)) + ((-c)^(5/4)*(a*d - b*c)^2*((16*x^(1/2)*(b^4*c^8 + a^4*c^4*d^4 - 4*
a^3*b*c^5*d^3 + 6*a^2*b^2*c^6*d^2 - 4*a*b^3*c^7*d))/d^5 + ((-c)^(5/4)*(a*d - b*c)^2*(b^2*c^5 + a^2*c^3*d^2 - 2
*a*b*c^4*d)*16i)/d^(21/4)))/(2*d^(17/4)))/(((-c)^(5/4)*(a*d - b*c)^2*((16*x^(1/2)*(b^4*c^8 + a^4*c^4*d^4 - 4*a
^3*b*c^5*d^3 + 6*a^2*b^2*c^6*d^2 - 4*a*b^3*c^7*d))/d^5 - ((-c)^(5/4)*(a*d - b*c)^2*(b^2*c^5 + a^2*c^3*d^2 - 2*
a*b*c^4*d)*16i)/d^(21/4))*1i)/(2*d^(17/4)) - ((-c)^(5/4)*(a*d - b*c)^2*((16*x^(1/2)*(b^4*c^8 + a^4*c^4*d^4 - 4
*a^3*b*c^5*d^3 + 6*a^2*b^2*c^6*d^2 - 4*a*b^3*c^7*d))/d^5 + ((-c)^(5/4)*(a*d - b*c)^2*(b^2*c^5 + a^2*c^3*d^2 -
2*a*b*c^4*d)*16i)/d^(21/4))*1i)/(2*d^(17/4))))*(a*d - b*c)^2)/d^(17/4)

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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(x**(7/2)*(b*x**2+a)**2/(d*x**2+c),x)

[Out]

Timed out

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