3.418 \(\int \frac {\sqrt {x} (a+b x^2)^2}{c+d x^2} \, dx\)

Optimal. Leaf size=268 \[ \frac {(b c-a d)^2 \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{2 \sqrt {2} \sqrt [4]{c} d^{11/4}}-\frac {(b c-a d)^2 \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{2 \sqrt {2} \sqrt [4]{c} d^{11/4}}-\frac {(b c-a d)^2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{\sqrt {2} \sqrt [4]{c} d^{11/4}}+\frac {(b c-a d)^2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )}{\sqrt {2} \sqrt [4]{c} d^{11/4}}-\frac {2 b x^{3/2} (b c-2 a d)}{3 d^2}+\frac {2 b^2 x^{7/2}}{7 d} \]

[Out]

-2/3*b*(-2*a*d+b*c)*x^(3/2)/d^2+2/7*b^2*x^(7/2)/d-1/2*(-a*d+b*c)^2*arctan(1-d^(1/4)*2^(1/2)*x^(1/2)/c^(1/4))/c
^(1/4)/d^(11/4)*2^(1/2)+1/2*(-a*d+b*c)^2*arctan(1+d^(1/4)*2^(1/2)*x^(1/2)/c^(1/4))/c^(1/4)/d^(11/4)*2^(1/2)+1/
4*(-a*d+b*c)^2*ln(c^(1/2)+x*d^(1/2)-c^(1/4)*d^(1/4)*2^(1/2)*x^(1/2))/c^(1/4)/d^(11/4)*2^(1/2)-1/4*(-a*d+b*c)^2
*ln(c^(1/2)+x*d^(1/2)+c^(1/4)*d^(1/4)*2^(1/2)*x^(1/2))/c^(1/4)/d^(11/4)*2^(1/2)

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Rubi [A]  time = 0.23, antiderivative size = 268, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {461, 329, 297, 1162, 617, 204, 1165, 628} \[ -\frac {2 b x^{3/2} (b c-2 a d)}{3 d^2}+\frac {(b c-a d)^2 \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{2 \sqrt {2} \sqrt [4]{c} d^{11/4}}-\frac {(b c-a d)^2 \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{2 \sqrt {2} \sqrt [4]{c} d^{11/4}}-\frac {(b c-a d)^2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{\sqrt {2} \sqrt [4]{c} d^{11/4}}+\frac {(b c-a d)^2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )}{\sqrt {2} \sqrt [4]{c} d^{11/4}}+\frac {2 b^2 x^{7/2}}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.12, size = 249, normalized size = 0.93 \[ \frac {-56 b \sqrt [4]{c} d^{3/4} x^{3/2} (b c-2 a d)+21 \sqrt {2} (b c-a d)^2 \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )-21 \sqrt {2} (b c-a d)^2 \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )-42 \sqrt {2} (b c-a d)^2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )+42 \sqrt {2} (b c-a d)^2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )+24 b^2 \sqrt [4]{c} d^{7/4} x^{7/2}}{84 \sqrt [4]{c} d^{11/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.56, size = 1629, normalized size = 6.08 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*x^(1/2)/(d*x^2+c),x, algorithm="fricas")

[Out]

-1/42*(84*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*d^11))^(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 + 924*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 - 12*a^1
1*b*c*d^11 + a^12*d^12)*x - (b^8*c^9*d^5 - 8*a*b^7*c^8*d^6 + 28*a^2*b^6*c^7*d^7 - 56*a^3*b^5*c^6*d^8 + 70*a^4*
b^4*c^5*d^9 - 56*a^5*b^3*c^4*d^10 + 28*a^6*b^2*c^3*d^11 - 8*a^7*b*c^2*d^12 + a^8*c*d^13)*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*d^11)))*d^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*d^11))^(1/4
) - (b^6*c^6*d^3 - 6*a*b^5*c^5*d^4 + 15*a^2*b^4*c^4*d^5 - 20*a^3*b^3*c^3*d^6 + 15*a^4*b^2*c^2*d^7 - 6*a^5*b*c*
d^8 + a^6*d^9)*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*d^11))^(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)) - 21*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*d^11))^(1/4)*log(c*d^8
*(-(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*d^11))^(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)) + 21*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*d^11))^(1/4)*log(-c*d^8*(-(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*d
^11))^(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*(3*b^2*d*x^3 - 7*(b^2*c - 2*a*b*d)*x)*sqrt(x))/d^2

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giac [A]  time = 0.54, size = 361, normalized size = 1.35 \[ \frac {\sqrt {2} {\left (\left (c d^{3}\right )^{\frac {3}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac {3}{4}} a b c d + \left (c d^{3}\right )^{\frac {3}{4}} a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {c}{d}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {c}{d}\right )^{\frac {1}{4}}}\right )}{2 \, c d^{5}} + \frac {\sqrt {2} {\left (\left (c d^{3}\right )^{\frac {3}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac {3}{4}} a b c d + \left (c d^{3}\right )^{\frac {3}{4}} a^{2} d^{2}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {c}{d}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {c}{d}\right )^{\frac {1}{4}}}\right )}{2 \, c d^{5}} - \frac {\sqrt {2} {\left (\left (c d^{3}\right )^{\frac {3}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac {3}{4}} a b c d + \left (c d^{3}\right )^{\frac {3}{4}} a^{2} d^{2}\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {c}{d}\right )^{\frac {1}{4}} + x + \sqrt {\frac {c}{d}}\right )}{4 \, c d^{5}} + \frac {\sqrt {2} {\left (\left (c d^{3}\right )^{\frac {3}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac {3}{4}} a b c d + \left (c d^{3}\right )^{\frac {3}{4}} a^{2} d^{2}\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {c}{d}\right )^{\frac {1}{4}} + x + \sqrt {\frac {c}{d}}\right )}{4 \, c d^{5}} + \frac {2 \, {\left (3 \, b^{2} d^{6} x^{\frac {7}{2}} - 7 \, b^{2} c d^{5} x^{\frac {3}{2}} + 14 \, a b d^{6} x^{\frac {3}{2}}\right )}}{21 \, d^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*x^(1/2)/(d*x^2+c),x, algorithm="giac")

[Out]

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*d^5) + 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*d^5) - 1/4*s
qrt(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*d^5) + 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*d^5) + 2/21*(3*b^2*d^6*x^(7/2) - 7*b^2*c*d^5*x^(3/2
) + 14*a*b*d^6*x^(3/2))/d^7

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maple [B]  time = 0.01, size = 461, normalized size = 1.72 \[ \frac {2 b^{2} x^{\frac {7}{2}}}{7 d}+\frac {4 a b \,x^{\frac {3}{2}}}{3 d}-\frac {2 b^{2} c \,x^{\frac {3}{2}}}{3 d^{2}}+\frac {\sqrt {2}\, a^{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}\, a^{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}\, a^{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 {\sqrt {2}\, a b c \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}} d^{2}}-\frac {\sqrt {2}\, a b c \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}} d^{2}}-\frac {\sqrt {2}\, a b 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 )}{2 \left (\frac {c}{d}\right )^{\frac {1}{4}} d^{2}}+\frac {\sqrt {2}\, b^{2} c^{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^{3}}+\frac {\sqrt {2}\, b^{2} c^{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^{3}}+\frac {\sqrt {2}\, b^{2} 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 )}{4 \left (\frac {c}{d}\right )^{\frac {1}{4}} d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2*x^(1/2)/(d*x^2+c),x)

[Out]

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

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maxima [A]  time = 2.51, size = 229, normalized size = 0.85 \[ \frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} + 2 \, \sqrt {d} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {\sqrt {c} \sqrt {d}} \sqrt {d}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} - 2 \, \sqrt {d} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {\sqrt {c} \sqrt {d}} \sqrt {d}} - \frac {\sqrt {2} \log \left (\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} \sqrt {x} + \sqrt {d} x + \sqrt {c}\right )}{c^{\frac {1}{4}} d^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} \sqrt {x} + \sqrt {d} x + \sqrt {c}\right )}{c^{\frac {1}{4}} d^{\frac {3}{4}}}\right )}}{4 \, d^{2}} + \frac {2 \, {\left (3 \, b^{2} d x^{\frac {7}{2}} - 7 \, {\left (b^{2} c - 2 \, a b d\right )} x^{\frac {3}{2}}\right )}}{21 \, d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [B]  time = 0.18, size = 390, normalized size = 1.46 \[ \frac {2\,b^2\,x^{7/2}}{7\,d}-x^{3/2}\,\left (\frac {2\,b^2\,c}{3\,d^2}-\frac {4\,a\,b}{3\,d}\right )+\frac {\mathrm {atan}\left (\frac {d^{1/4}\,\sqrt {x}\,{\left (a\,d-b\,c\right )}^2\,\left (a^4\,c\,d^4-4\,a^3\,b\,c^2\,d^3+6\,a^2\,b^2\,c^3\,d^2-4\,a\,b^3\,c^4\,d+b^4\,c^5\right )}{{\left (-c\right )}^{1/4}\,\left (a^6\,c\,d^6-6\,a^5\,b\,c^2\,d^5+15\,a^4\,b^2\,c^3\,d^4-20\,a^3\,b^3\,c^4\,d^3+15\,a^2\,b^4\,c^5\,d^2-6\,a\,b^5\,c^6\,d+b^6\,c^7\right )}\right )\,{\left (a\,d-b\,c\right )}^2}{{\left (-c\right )}^{1/4}\,d^{11/4}}+\frac {\mathrm {atan}\left (\frac {d^{1/4}\,\sqrt {x}\,{\left (a\,d-b\,c\right )}^2\,\left (a^4\,c\,d^4-4\,a^3\,b\,c^2\,d^3+6\,a^2\,b^2\,c^3\,d^2-4\,a\,b^3\,c^4\,d+b^4\,c^5\right )\,1{}\mathrm {i}}{{\left (-c\right )}^{1/4}\,\left (a^6\,c\,d^6-6\,a^5\,b\,c^2\,d^5+15\,a^4\,b^2\,c^3\,d^4-20\,a^3\,b^3\,c^4\,d^3+15\,a^2\,b^4\,c^5\,d^2-6\,a\,b^5\,c^6\,d+b^6\,c^7\right )}\right )\,{\left (a\,d-b\,c\right )}^2\,1{}\mathrm {i}}{{\left (-c\right )}^{1/4}\,d^{11/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(1/2)*(a + b*x^2)^2)/(c + d*x^2),x)

[Out]

(2*b^2*x^(7/2))/(7*d) - x^(3/2)*((2*b^2*c)/(3*d^2) - (4*a*b)/(3*d)) + (atan((d^(1/4)*x^(1/2)*(a*d - b*c)^2*(b^
4*c^5 + a^4*c*d^4 - 4*a^3*b*c^2*d^3 + 6*a^2*b^2*c^3*d^2 - 4*a*b^3*c^4*d))/((-c)^(1/4)*(b^6*c^7 + a^6*c*d^6 - 6
*a^5*b*c^2*d^5 + 15*a^2*b^4*c^5*d^2 - 20*a^3*b^3*c^4*d^3 + 15*a^4*b^2*c^3*d^4 - 6*a*b^5*c^6*d)))*(a*d - b*c)^2
)/((-c)^(1/4)*d^(11/4)) + (atan((d^(1/4)*x^(1/2)*(a*d - b*c)^2*(b^4*c^5 + a^4*c*d^4 - 4*a^3*b*c^2*d^3 + 6*a^2*
b^2*c^3*d^2 - 4*a*b^3*c^4*d)*1i)/((-c)^(1/4)*(b^6*c^7 + a^6*c*d^6 - 6*a^5*b*c^2*d^5 + 15*a^2*b^4*c^5*d^2 - 20*
a^3*b^3*c^4*d^3 + 15*a^4*b^2*c^3*d^4 - 6*a*b^5*c^6*d)))*(a*d - b*c)^2*1i)/((-c)^(1/4)*d^(11/4))

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sympy [A]  time = 11.40, size = 87, normalized size = 0.32 \[ \frac {4 a b x^{\frac {3}{2}}}{3 d} - \frac {2 b^{2} c x^{\frac {3}{2}}}{3 d^{2}} + \frac {2 b^{2} x^{\frac {7}{2}}}{7 d} + \frac {2 \left (a d - b c\right )^{2} \operatorname {RootSum} {\left (256 t^{4} c d^{3} + 1, \left (t \mapsto t \log {\left (64 t^{3} c d^{2} + \sqrt {x} \right )} \right )\right )}}{d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2*x**(1/2)/(d*x**2+c),x)

[Out]

4*a*b*x**(3/2)/(3*d) - 2*b**2*c*x**(3/2)/(3*d**2) + 2*b**2*x**(7/2)/(7*d) + 2*(a*d - b*c)**2*RootSum(256*_t**4
*c*d**3 + 1, Lambda(_t, _t*log(64*_t**3*c*d**2 + sqrt(x))))/d**2

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