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

Optimal. Leaf size=267 \[ -\frac {2 a^2}{5 c x^{5/2}}+\frac {(b c-a d)^2 \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{2 \sqrt {2} c^{9/4} d^{3/4}}-\frac {(b c-a d)^2 \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{2 \sqrt {2} c^{9/4} d^{3/4}}-\frac {(b c-a d)^2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{\sqrt {2} c^{9/4} d^{3/4}}+\frac {(b c-a d)^2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )}{\sqrt {2} c^{9/4} d^{3/4}}-\frac {2 a (2 b c-a d)}{c^2 \sqrt {x}} \]

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Rubi [A]  time = 0.28, antiderivative size = 267, 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, 297, 1162, 617, 204, 1165, 628} \begin {gather*} -\frac {2 a^2}{5 c x^{5/2}}+\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^{9/4} d^{3/4}}-\frac {(b c-a d)^2 \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{2 \sqrt {2} c^{9/4} d^{3/4}}-\frac {(b c-a d)^2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{\sqrt {2} c^{9/4} d^{3/4}}+\frac {(b c-a d)^2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )}{\sqrt {2} c^{9/4} d^{3/4}}-\frac {2 a (2 b c-a d)}{c^2 \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.12, size = 254, normalized size = 0.95 \begin {gather*} \frac {-\frac {8 a^2 c^{5/4}}{x^{5/2}}+\frac {5 \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 )}{d^{3/4}}-\frac {5 \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 )}{d^{3/4}}-\frac {10 \sqrt {2} (b c-a d)^2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{d^{3/4}}+\frac {10 \sqrt {2} (b c-a d)^2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )}{d^{3/4}}+\frac {40 a \sqrt [4]{c} (a d-2 b c)}{\sqrt {x}}}{20 c^{9/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-8*a^2*c^(5/4))/x^(5/2) + (40*a*c^(1/4)*(-2*b*c + a*d))/Sqrt[x] - (10*Sqrt[2]*(b*c - a*d)^2*ArcTan[1 - (Sqrt
[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/d^(3/4) + (10*Sqrt[2]*(b*c - a*d)^2*ArcTan[1 + (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4
)])/d^(3/4) + (5*Sqrt[2]*(b*c - a*d)^2*Log[Sqrt[c] - Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/d^(3/4) - (
5*Sqrt[2]*(b*c - a*d)^2*Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/d^(3/4))/(20*c^(9/4))

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IntegrateAlgebraic [A]  time = 0.21, size = 165, normalized size = 0.62 \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^{9/4} d^{3/4}}-\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^{9/4} d^{3/4}}-\frac {2 a \left (a c-5 a d x^2+10 b c x^2\right )}{5 c^2 x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*(a*c + 10*b*c*x^2 - 5*a*d*x^2))/(5*c^2*x^(5/2)) - ((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^(9/4)*d^(3/4)) - ((b*c - a*d)^2*ArcTanh[(Sqrt[2]*c^(1/4)*d^(1
/4)*Sqrt[x])/(Sqrt[c] + Sqrt[d]*x)])/(Sqrt[2]*c^(9/4)*d^(3/4))

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fricas [B]  time = 1.22, size = 1648, normalized size = 6.17

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/10*(20*c^2*x^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^9*d^3))^(1/4)*arctan((sqrt((b^12*c^12 -
12*a*b^11*c^11*d + 66*a^2*b^10*c^10*d^2 - 220*a^3*b^9*c^9*d^3 + 495*a^4*b^8*c^8*d^4 - 792*a^5*b^7*c^7*d^5 + 92
4*a^6*b^6*c^6*d^6 - 792*a^7*b^5*c^5*d^7 + 495*a^8*b^4*c^4*d^8 - 220*a^9*b^3*c^3*d^9 + 66*a^10*b^2*c^2*d^10 - 1
2*a^11*b*c*d^11 + a^12*d^12)*x - (b^8*c^13*d - 8*a*b^7*c^12*d^2 + 28*a^2*b^6*c^11*d^3 - 56*a^3*b^5*c^10*d^4 +
70*a^4*b^4*c^9*d^5 - 56*a^5*b^3*c^8*d^6 + 28*a^6*b^2*c^7*d^7 - 8*a^7*b*c^6*d^8 + a^8*c^5*d^9)*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^9*d^3)))*c^2*d*(-(b^8*c^8 - 8*a*b^7*c^7*d + 28*a^2*b^6*c^6*d^2 - 56*a^
3*b^5*c^5*d^3 + 70*a^4*b^4*c^4*d^4 - 56*a^5*b^3*c^3*d^5 + 28*a^6*b^2*c^2*d^6 - 8*a^7*b*c*d^7 + a^8*d^8)/(c^9*d
^3))^(1/4) - (b^6*c^8*d - 6*a*b^5*c^7*d^2 + 15*a^2*b^4*c^6*d^3 - 20*a^3*b^3*c^5*d^4 + 15*a^4*b^2*c^4*d^5 - 6*a
^5*b*c^3*d^6 + a^6*c^2*d^7)*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^9*d^3))^(1/4))/(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)) - 5*c^2*x^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^9*d^3
))^(1/4)*log(c^7*d^2*(-(b^8*c^8 - 8*a*b^7*c^7*d + 28*a^2*b^6*c^6*d^2 - 56*a^3*b^5*c^5*d^3 + 70*a^4*b^4*c^4*d^4
- 56*a^5*b^3*c^3*d^5 + 28*a^6*b^2*c^2*d^6 - 8*a^7*b*c*d^7 + a^8*d^8)/(c^9*d^3))^(3/4) + (b^6*c^6 - 6*a*b^5*c^
5*d + 15*a^2*b^4*c^4*d^2 - 20*a^3*b^3*c^3*d^3 + 15*a^4*b^2*c^2*d^4 - 6*a^5*b*c*d^5 + a^6*d^6)*sqrt(x)) + 5*c^2
*x^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^9*d^3))^(1/4)*log(-c^7*d^2*(-(b^8*c^8 - 8*a*b^7*c^7*d
+ 28*a^2*b^6*c^6*d^2 - 56*a^3*b^5*c^5*d^3 + 70*a^4*b^4*c^4*d^4 - 56*a^5*b^3*c^3*d^5 + 28*a^6*b^2*c^2*d^6 - 8*
a^7*b*c*d^7 + a^8*d^8)/(c^9*d^3))^(3/4) + (b^6*c^6 - 6*a*b^5*c^5*d + 15*a^2*b^4*c^4*d^2 - 20*a^3*b^3*c^3*d^3 +
15*a^4*b^2*c^2*d^4 - 6*a^5*b*c*d^5 + a^6*d^6)*sqrt(x)) + 4*(a^2*c + 5*(2*a*b*c - a^2*d)*x^2)*sqrt(x))/(c^2*x^
3)

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giac [A]  time = 0.47, size = 353, normalized size = 1.32 \begin {gather*} -\frac {2 \, {\left (10 \, a b c x^{2} - 5 \, a^{2} d x^{2} + a^{2} c\right )}}{5 \, c^{2} x^{\frac {5}{2}}} + \frac {\sqrt {2} {\left (\left (c d^{3}\right )^{\frac {3}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac {3}{4}} a b c d + \left (c d^{3}\right )^{\frac {3}{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^{3}} + \frac {\sqrt {2} {\left (\left (c d^{3}\right )^{\frac {3}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac {3}{4}} a b c d + \left (c d^{3}\right )^{\frac {3}{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^{3}} - \frac {\sqrt {2} {\left (\left (c d^{3}\right )^{\frac {3}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac {3}{4}} a b c d + \left (c d^{3}\right )^{\frac {3}{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^{3}} + \frac {\sqrt {2} {\left (\left (c d^{3}\right )^{\frac {3}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac {3}{4}} a b c d + \left (c d^{3}\right )^{\frac {3}{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^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*(10*a*b*c*x^2 - 5*a^2*d*x^2 + a^2*c)/(c^2*x^(5/2)) + 1/2*sqrt(2)*((c*d^3)^(3/4)*b^2*c^2 - 2*(c*d^3)^(3/4)
*a*b*c*d + (c*d^3)^(3/4)*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^3)
+ 1/2*sqrt(2)*((c*d^3)^(3/4)*b^2*c^2 - 2*(c*d^3)^(3/4)*a*b*c*d + (c*d^3)^(3/4)*a^2*d^2)*arctan(-1/2*sqrt(2)*(s
qrt(2)*(c/d)^(1/4) - 2*sqrt(x))/(c/d)^(1/4))/(c^3*d^3) - 1/4*sqrt(2)*((c*d^3)^(3/4)*b^2*c^2 - 2*(c*d^3)^(3/4)*
a*b*c*d + (c*d^3)^(3/4)*a^2*d^2)*log(sqrt(2)*sqrt(x)*(c/d)^(1/4) + x + sqrt(c/d))/(c^3*d^3) + 1/4*sqrt(2)*((c*
d^3)^(3/4)*b^2*c^2 - 2*(c*d^3)^(3/4)*a*b*c*d + (c*d^3)^(3/4)*a^2*d^2)*log(-sqrt(2)*sqrt(x)*(c/d)^(1/4) + x + s
qrt(c/d))/(c^3*d^3)

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maple [B]  time = 0.02, size = 452, normalized size = 1.69 \begin {gather*} \frac {\sqrt {2}\, a^{2} d \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )}{2 \left (\frac {c}{d}\right )^{\frac {1}{4}} c^{2}}+\frac {\sqrt {2}\, a^{2} d \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )}{2 \left (\frac {c}{d}\right )^{\frac {1}{4}} c^{2}}+\frac {\sqrt {2}\, a^{2} d \ln \left (\frac {x -\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {c}{d}}}{x +\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {c}{d}}}\right )}{4 \left (\frac {c}{d}\right )^{\frac {1}{4}} c^{2}}-\frac {\sqrt {2}\, a b \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )}{\left (\frac {c}{d}\right )^{\frac {1}{4}} c}-\frac {\sqrt {2}\, a b \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )}{\left (\frac {c}{d}\right )^{\frac {1}{4}} c}-\frac {\sqrt {2}\, a b \ln \left (\frac {x -\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {c}{d}}}{x +\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {c}{d}}}\right )}{2 \left (\frac {c}{d}\right )^{\frac {1}{4}} c}+\frac {\sqrt {2}\, b^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )}{2 \left (\frac {c}{d}\right )^{\frac {1}{4}} d}+\frac {\sqrt {2}\, b^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )}{2 \left (\frac {c}{d}\right )^{\frac {1}{4}} d}+\frac {\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 \left (\frac {c}{d}\right )^{\frac {1}{4}} d}+\frac {2 a^{2} d}{c^{2} \sqrt {x}}-\frac {4 a b}{c \sqrt {x}}-\frac {2 a^{2}}{5 c \,x^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.47, size = 229, normalized size = 0.86 \begin {gather*} \frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} {\left (\frac {2 \, \sqrt {2} \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 {\sqrt {c} \sqrt {d}} \sqrt {d}} + \frac {2 \, \sqrt {2} \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 {\sqrt {c} \sqrt {d}} \sqrt {d}} - \frac {\sqrt {2} \log \left (\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} \sqrt {x} + \sqrt {d} x + \sqrt {c}\right )}{c^{\frac {1}{4}} d^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} \sqrt {x} + \sqrt {d} x + \sqrt {c}\right )}{c^{\frac {1}{4}} d^{\frac {3}{4}}}\right )}}{4 \, c^{2}} - \frac {2 \, {\left (a^{2} c + 5 \, {\left (2 \, a b c - a^{2} d\right )} x^{2}\right )}}{5 \, c^{2} x^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.34, size = 417, normalized size = 1.56 \begin {gather*} \frac {\mathrm {atan}\left (\frac {\sqrt {x}\,{\left (a\,d-b\,c\right )}^2\,\left (16\,a^4\,c^7\,d^6-64\,a^3\,b\,c^8\,d^5+96\,a^2\,b^2\,c^9\,d^4-64\,a\,b^3\,c^{10}\,d^3+16\,b^4\,c^{11}\,d^2\right )}{{\left (-c\right )}^{9/4}\,d^{3/4}\,\left (16\,a^6\,c^5\,d^7-96\,a^5\,b\,c^6\,d^6+240\,a^4\,b^2\,c^7\,d^5-320\,a^3\,b^3\,c^8\,d^4+240\,a^2\,b^4\,c^9\,d^3-96\,a\,b^5\,c^{10}\,d^2+16\,b^6\,c^{11}\,d\right )}\right )\,{\left (a\,d-b\,c\right )}^2}{{\left (-c\right )}^{9/4}\,d^{3/4}}-\frac {\frac {2\,a^2}{5\,c}-\frac {2\,a\,x^2\,\left (a\,d-2\,b\,c\right )}{c^2}}{x^{5/2}}-\frac {\mathrm {atanh}\left (\frac {\sqrt {x}\,{\left (a\,d-b\,c\right )}^2\,\left (16\,a^4\,c^7\,d^6-64\,a^3\,b\,c^8\,d^5+96\,a^2\,b^2\,c^9\,d^4-64\,a\,b^3\,c^{10}\,d^3+16\,b^4\,c^{11}\,d^2\right )}{{\left (-c\right )}^{9/4}\,d^{3/4}\,\left (16\,a^6\,c^5\,d^7-96\,a^5\,b\,c^6\,d^6+240\,a^4\,b^2\,c^7\,d^5-320\,a^3\,b^3\,c^8\,d^4+240\,a^2\,b^4\,c^9\,d^3-96\,a\,b^5\,c^{10}\,d^2+16\,b^6\,c^{11}\,d\right )}\right )\,{\left (a\,d-b\,c\right )}^2}{{\left (-c\right )}^{9/4}\,d^{3/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atan((x^(1/2)*(a*d - b*c)^2*(16*a^4*c^7*d^6 + 16*b^4*c^11*d^2 - 64*a*b^3*c^10*d^3 - 64*a^3*b*c^8*d^5 + 96*a^2
*b^2*c^9*d^4))/((-c)^(9/4)*d^(3/4)*(16*b^6*c^11*d + 16*a^6*c^5*d^7 - 96*a*b^5*c^10*d^2 - 96*a^5*b*c^6*d^6 + 24
0*a^2*b^4*c^9*d^3 - 320*a^3*b^3*c^8*d^4 + 240*a^4*b^2*c^7*d^5)))*(a*d - b*c)^2)/((-c)^(9/4)*d^(3/4)) - ((2*a^2
)/(5*c) - (2*a*x^2*(a*d - 2*b*c))/c^2)/x^(5/2) - (atanh((x^(1/2)*(a*d - b*c)^2*(16*a^4*c^7*d^6 + 16*b^4*c^11*d
^2 - 64*a*b^3*c^10*d^3 - 64*a^3*b*c^8*d^5 + 96*a^2*b^2*c^9*d^4))/((-c)^(9/4)*d^(3/4)*(16*b^6*c^11*d + 16*a^6*c
^5*d^7 - 96*a*b^5*c^10*d^2 - 96*a^5*b*c^6*d^6 + 240*a^2*b^4*c^9*d^3 - 320*a^3*b^3*c^8*d^4 + 240*a^4*b^2*c^7*d^
5)))*(a*d - b*c)^2)/((-c)^(9/4)*d^(3/4))

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sympy [A]  time = 125.08, size = 406, normalized size = 1.52 \begin {gather*} a^{2} \left (\begin {cases} \frac {\tilde {\infty }}{x^{\frac {9}{2}}} & \text {for}\: c = 0 \wedge d = 0 \\- \frac {2}{9 d x^{\frac {9}{2}}} & \text {for}\: c = 0 \\- \frac {2}{5 c x^{\frac {5}{2}}} & \text {for}\: d = 0 \\- \frac {2}{5 c x^{\frac {5}{2}}} + \frac {2 d}{c^{2} \sqrt {x}} - \frac {\left (-1\right )^{\frac {3}{4}} d \log {\left (- \sqrt [4]{-1} \sqrt [4]{c} \sqrt [4]{\frac {1}{d}} + \sqrt {x} \right )}}{2 c^{\frac {9}{4}} \sqrt [4]{\frac {1}{d}}} + \frac {\left (-1\right )^{\frac {3}{4}} d \log {\left (\sqrt [4]{-1} \sqrt [4]{c} \sqrt [4]{\frac {1}{d}} + \sqrt {x} \right )}}{2 c^{\frac {9}{4}} \sqrt [4]{\frac {1}{d}}} + \frac {\left (-1\right )^{\frac {3}{4}} d \operatorname {atan}{\left (\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {x}}{\sqrt [4]{c} \sqrt [4]{\frac {1}{d}}} \right )}}{c^{\frac {9}{4}} \sqrt [4]{\frac {1}{d}}} & \text {otherwise} \end {cases}\right ) + 2 a b \left (\begin {cases} \frac {\tilde {\infty }}{x^{\frac {5}{2}}} & \text {for}\: c = 0 \wedge d = 0 \\- \frac {2}{c \sqrt {x}} & \text {for}\: d = 0 \\- \frac {2}{5 d x^{\frac {5}{2}}} & \text {for}\: c = 0 \\- \frac {2}{c \sqrt {x}} + \frac {\left (-1\right )^{\frac {3}{4}} \log {\left (- \sqrt [4]{-1} \sqrt [4]{c} \sqrt [4]{\frac {1}{d}} + \sqrt {x} \right )}}{2 c^{\frac {5}{4}} \sqrt [4]{\frac {1}{d}}} - \frac {\left (-1\right )^{\frac {3}{4}} \log {\left (\sqrt [4]{-1} \sqrt [4]{c} \sqrt [4]{\frac {1}{d}} + \sqrt {x} \right )}}{2 c^{\frac {5}{4}} \sqrt [4]{\frac {1}{d}}} - \frac {\left (-1\right )^{\frac {3}{4}} \operatorname {atan}{\left (\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {x}}{\sqrt [4]{c} \sqrt [4]{\frac {1}{d}}} \right )}}{c^{\frac {5}{4}} \sqrt [4]{\frac {1}{d}}} & \text {otherwise} \end {cases}\right ) + 2 b^{2} \operatorname {RootSum} {\left (256 t^{4} c d^{3} + 1, \left (t \mapsto t \log {\left (64 t^{3} c d^{2} + \sqrt {x} \right )} \right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*Piecewise((zoo/x**(9/2), Eq(c, 0) & Eq(d, 0)), (-2/(9*d*x**(9/2)), Eq(c, 0)), (-2/(5*c*x**(5/2)), Eq(d, 0
)), (-2/(5*c*x**(5/2)) + 2*d/(c**2*sqrt(x)) - (-1)**(3/4)*d*log(-(-1)**(1/4)*c**(1/4)*(1/d)**(1/4) + sqrt(x))/
(2*c**(9/4)*(1/d)**(1/4)) + (-1)**(3/4)*d*log((-1)**(1/4)*c**(1/4)*(1/d)**(1/4) + sqrt(x))/(2*c**(9/4)*(1/d)**
(1/4)) + (-1)**(3/4)*d*atan((-1)**(3/4)*sqrt(x)/(c**(1/4)*(1/d)**(1/4)))/(c**(9/4)*(1/d)**(1/4)), True)) + 2*a
*b*Piecewise((zoo/x**(5/2), Eq(c, 0) & Eq(d, 0)), (-2/(c*sqrt(x)), Eq(d, 0)), (-2/(5*d*x**(5/2)), Eq(c, 0)), (
-2/(c*sqrt(x)) + (-1)**(3/4)*log(-(-1)**(1/4)*c**(1/4)*(1/d)**(1/4) + sqrt(x))/(2*c**(5/4)*(1/d)**(1/4)) - (-1
)**(3/4)*log((-1)**(1/4)*c**(1/4)*(1/d)**(1/4) + sqrt(x))/(2*c**(5/4)*(1/d)**(1/4)) - (-1)**(3/4)*atan((-1)**(
3/4)*sqrt(x)/(c**(1/4)*(1/d)**(1/4)))/(c**(5/4)*(1/d)**(1/4)), True)) + 2*b**2*RootSum(256*_t**4*c*d**3 + 1, L
ambda(_t, _t*log(64*_t**3*c*d**2 + sqrt(x))))

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