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

Optimal. Leaf size=310 \[ -\frac {(b c-a d) (a d+7 b c) \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{8 \sqrt {2} c^{5/4} d^{11/4}}+\frac {(b c-a d) (a d+7 b c) \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{8 \sqrt {2} c^{5/4} d^{11/4}}+\frac {(b c-a d) (a d+7 b c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{4 \sqrt {2} c^{5/4} d^{11/4}}-\frac {(b c-a d) (a d+7 b c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )}{4 \sqrt {2} c^{5/4} d^{11/4}}+\frac {x^{3/2} (b c-a d)^2}{2 c d^2 \left (c+d x^2\right )}+\frac {2 b^2 x^{3/2}}{3 d^2} \]

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Rubi [A]  time = 0.28, antiderivative size = 310, 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 = {463, 459, 329, 297, 1162, 617, 204, 1165, 628} \begin {gather*} -\frac {(b c-a d) (a d+7 b c) \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{8 \sqrt {2} c^{5/4} d^{11/4}}+\frac {(b c-a d) (a d+7 b c) \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{8 \sqrt {2} c^{5/4} d^{11/4}}+\frac {(b c-a d) (a d+7 b c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{4 \sqrt {2} c^{5/4} d^{11/4}}-\frac {(b c-a d) (a d+7 b c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )}{4 \sqrt {2} c^{5/4} d^{11/4}}+\frac {x^{3/2} (b c-a d)^2}{2 c d^2 \left (c+d x^2\right )}+\frac {2 b^2 x^{3/2}}{3 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b^2*x^(3/2))/(3*d^2) + ((b*c - a*d)^2*x^(3/2))/(2*c*d^2*(c + d*x^2)) + ((b*c - a*d)*(7*b*c + a*d)*ArcTan[1
- (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/(4*Sqrt[2]*c^(5/4)*d^(11/4)) - ((b*c - a*d)*(7*b*c + a*d)*ArcTan[1 + (Sq
rt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/(4*Sqrt[2]*c^(5/4)*d^(11/4)) - ((b*c - a*d)*(7*b*c + a*d)*Log[Sqrt[c] - Sqrt[
2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/(8*Sqrt[2]*c^(5/4)*d^(11/4)) + ((b*c - a*d)*(7*b*c + a*d)*Log[Sqrt[c]
+ Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/(8*Sqrt[2]*c^(5/4)*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 459

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

Rule 463

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

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}{\left (c+d x^2\right )^2} \, dx &=\frac {(b c-a d)^2 x^{3/2}}{2 c d^2 \left (c+d x^2\right )}-\frac {\int \frac {\sqrt {x} \left (\frac {1}{2} \left (-4 a^2 d^2+3 (b c-a d)^2\right )-2 b^2 c d x^2\right )}{c+d x^2} \, dx}{2 c d^2}\\ &=\frac {2 b^2 x^{3/2}}{3 d^2}+\frac {(b c-a d)^2 x^{3/2}}{2 c d^2 \left (c+d x^2\right )}-\frac {((b c-a d) (7 b c+a d)) \int \frac {\sqrt {x}}{c+d x^2} \, dx}{4 c d^2}\\ &=\frac {2 b^2 x^{3/2}}{3 d^2}+\frac {(b c-a d)^2 x^{3/2}}{2 c d^2 \left (c+d x^2\right )}-\frac {((b c-a d) (7 b c+a d)) \operatorname {Subst}\left (\int \frac {x^2}{c+d x^4} \, dx,x,\sqrt {x}\right )}{2 c d^2}\\ &=\frac {2 b^2 x^{3/2}}{3 d^2}+\frac {(b c-a d)^2 x^{3/2}}{2 c d^2 \left (c+d x^2\right )}+\frac {((b c-a d) (7 b c+a d)) \operatorname {Subst}\left (\int \frac {\sqrt {c}-\sqrt {d} x^2}{c+d x^4} \, dx,x,\sqrt {x}\right )}{4 c d^{5/2}}-\frac {((b c-a d) (7 b c+a d)) \operatorname {Subst}\left (\int \frac {\sqrt {c}+\sqrt {d} x^2}{c+d x^4} \, dx,x,\sqrt {x}\right )}{4 c d^{5/2}}\\ &=\frac {2 b^2 x^{3/2}}{3 d^2}+\frac {(b c-a d)^2 x^{3/2}}{2 c d^2 \left (c+d x^2\right )}-\frac {((b c-a d) (7 b c+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 d^3}-\frac {((b c-a d) (7 b c+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 d^3}-\frac {((b c-a d) (7 b c+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^{5/4} d^{11/4}}-\frac {((b c-a d) (7 b c+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^{5/4} d^{11/4}}\\ &=\frac {2 b^2 x^{3/2}}{3 d^2}+\frac {(b c-a d)^2 x^{3/2}}{2 c d^2 \left (c+d x^2\right )}-\frac {(b c-a d) (7 b c+a d) \log \left (\sqrt {c}-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {d} x\right )}{8 \sqrt {2} c^{5/4} d^{11/4}}+\frac {(b c-a d) (7 b c+a d) \log \left (\sqrt {c}+\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {d} x\right )}{8 \sqrt {2} c^{5/4} d^{11/4}}-\frac {((b c-a d) (7 b c+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^{5/4} d^{11/4}}+\frac {((b c-a d) (7 b c+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^{5/4} d^{11/4}}\\ &=\frac {2 b^2 x^{3/2}}{3 d^2}+\frac {(b c-a d)^2 x^{3/2}}{2 c d^2 \left (c+d x^2\right )}+\frac {(b c-a d) (7 b c+a d) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{4 \sqrt {2} c^{5/4} d^{11/4}}-\frac {(b c-a d) (7 b c+a d) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{4 \sqrt {2} c^{5/4} d^{11/4}}-\frac {(b c-a d) (7 b c+a d) \log \left (\sqrt {c}-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {d} x\right )}{8 \sqrt {2} c^{5/4} d^{11/4}}+\frac {(b c-a d) (7 b c+a d) \log \left (\sqrt {c}+\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {d} x\right )}{8 \sqrt {2} c^{5/4} d^{11/4}}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 319, normalized size = 1.03 \begin {gather*} \frac {-\frac {3 \sqrt {2} \left (-a^2 d^2-6 a b c d+7 b^2 c^2\right ) \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{c^{5/4}}+\frac {3 \sqrt {2} \left (-a^2 d^2-6 a b c d+7 b^2 c^2\right ) \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{c^{5/4}}+\frac {6 \sqrt {2} \left (-a^2 d^2-6 a b c d+7 b^2 c^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{c^{5/4}}-\frac {6 \sqrt {2} \left (-a^2 d^2-6 a b c d+7 b^2 c^2\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )}{c^{5/4}}+\frac {24 d^{3/4} x^{3/2} (b c-a d)^2}{c \left (c+d x^2\right )}+32 b^2 d^{3/4} x^{3/2}}{48 d^{11/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.79, size = 215, normalized size = 0.69 \begin {gather*} \frac {x^{3/2} \left (3 a^2 d^2-6 a b c d+7 b^2 c^2+4 b^2 c d x^2\right )}{6 c d^2 \left (c+d x^2\right )}+\frac {\left (-a^2 d^2-6 a b c d+7 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^{5/4} d^{11/4}}+\frac {\left (-a^2 d^2-6 a b c d+7 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^{5/4} d^{11/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.46, size = 1723, normalized size = 5.56

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)^2,x, algorithm="fricas")

[Out]

-1/24*(12*(c*d^3*x^2 + c^2*d^2)*(-(2401*b^8*c^8 - 8232*a*b^7*c^7*d + 9212*a^2*b^6*c^6*d^2 - 2520*a^3*b^5*c^5*d
^3 - 1434*a^4*b^4*c^4*d^4 + 360*a^5*b^3*c^3*d^5 + 188*a^6*b^2*c^2*d^6 + 24*a^7*b*c*d^7 + a^8*d^8)/(c^5*d^11))^
(1/4)*arctan((sqrt((117649*b^12*c^12 - 605052*a*b^11*c^11*d + 1195698*a^2*b^10*c^10*d^2 - 1049580*a^3*b^9*c^9*
d^3 + 247695*a^4*b^8*c^8*d^4 + 184968*a^5*b^7*c^7*d^5 - 73604*a^6*b^6*c^6*d^6 - 26424*a^7*b^5*c^5*d^7 + 5055*a
^8*b^4*c^4*d^8 + 3060*a^9*b^3*c^3*d^9 + 498*a^10*b^2*c^2*d^10 + 36*a^11*b*c*d^11 + a^12*d^12)*x - (2401*b^8*c^
11*d^5 - 8232*a*b^7*c^10*d^6 + 9212*a^2*b^6*c^9*d^7 - 2520*a^3*b^5*c^8*d^8 - 1434*a^4*b^4*c^7*d^9 + 360*a^5*b^
3*c^6*d^10 + 188*a^6*b^2*c^5*d^11 + 24*a^7*b*c^4*d^12 + a^8*c^3*d^13)*sqrt(-(2401*b^8*c^8 - 8232*a*b^7*c^7*d +
9212*a^2*b^6*c^6*d^2 - 2520*a^3*b^5*c^5*d^3 - 1434*a^4*b^4*c^4*d^4 + 360*a^5*b^3*c^3*d^5 + 188*a^6*b^2*c^2*d^
6 + 24*a^7*b*c*d^7 + a^8*d^8)/(c^5*d^11)))*c*d^3*(-(2401*b^8*c^8 - 8232*a*b^7*c^7*d + 9212*a^2*b^6*c^6*d^2 - 2
520*a^3*b^5*c^5*d^3 - 1434*a^4*b^4*c^4*d^4 + 360*a^5*b^3*c^3*d^5 + 188*a^6*b^2*c^2*d^6 + 24*a^7*b*c*d^7 + a^8*
d^8)/(c^5*d^11))^(1/4) + (343*b^6*c^7*d^3 - 882*a*b^5*c^6*d^4 + 609*a^2*b^4*c^5*d^5 + 36*a^3*b^3*c^4*d^6 - 87*
a^4*b^2*c^3*d^7 - 18*a^5*b*c^2*d^8 - a^6*c*d^9)*sqrt(x)*(-(2401*b^8*c^8 - 8232*a*b^7*c^7*d + 9212*a^2*b^6*c^6*
d^2 - 2520*a^3*b^5*c^5*d^3 - 1434*a^4*b^4*c^4*d^4 + 360*a^5*b^3*c^3*d^5 + 188*a^6*b^2*c^2*d^6 + 24*a^7*b*c*d^7
+ a^8*d^8)/(c^5*d^11))^(1/4))/(2401*b^8*c^8 - 8232*a*b^7*c^7*d + 9212*a^2*b^6*c^6*d^2 - 2520*a^3*b^5*c^5*d^3
- 1434*a^4*b^4*c^4*d^4 + 360*a^5*b^3*c^3*d^5 + 188*a^6*b^2*c^2*d^6 + 24*a^7*b*c*d^7 + a^8*d^8)) - 3*(c*d^3*x^2
+ c^2*d^2)*(-(2401*b^8*c^8 - 8232*a*b^7*c^7*d + 9212*a^2*b^6*c^6*d^2 - 2520*a^3*b^5*c^5*d^3 - 1434*a^4*b^4*c^
4*d^4 + 360*a^5*b^3*c^3*d^5 + 188*a^6*b^2*c^2*d^6 + 24*a^7*b*c*d^7 + a^8*d^8)/(c^5*d^11))^(1/4)*log(c^4*d^8*(-
(2401*b^8*c^8 - 8232*a*b^7*c^7*d + 9212*a^2*b^6*c^6*d^2 - 2520*a^3*b^5*c^5*d^3 - 1434*a^4*b^4*c^4*d^4 + 360*a^
5*b^3*c^3*d^5 + 188*a^6*b^2*c^2*d^6 + 24*a^7*b*c*d^7 + a^8*d^8)/(c^5*d^11))^(3/4) - (343*b^6*c^6 - 882*a*b^5*c
^5*d + 609*a^2*b^4*c^4*d^2 + 36*a^3*b^3*c^3*d^3 - 87*a^4*b^2*c^2*d^4 - 18*a^5*b*c*d^5 - a^6*d^6)*sqrt(x)) + 3*
(c*d^3*x^2 + c^2*d^2)*(-(2401*b^8*c^8 - 8232*a*b^7*c^7*d + 9212*a^2*b^6*c^6*d^2 - 2520*a^3*b^5*c^5*d^3 - 1434*
a^4*b^4*c^4*d^4 + 360*a^5*b^3*c^3*d^5 + 188*a^6*b^2*c^2*d^6 + 24*a^7*b*c*d^7 + a^8*d^8)/(c^5*d^11))^(1/4)*log(
-c^4*d^8*(-(2401*b^8*c^8 - 8232*a*b^7*c^7*d + 9212*a^2*b^6*c^6*d^2 - 2520*a^3*b^5*c^5*d^3 - 1434*a^4*b^4*c^4*d
^4 + 360*a^5*b^3*c^3*d^5 + 188*a^6*b^2*c^2*d^6 + 24*a^7*b*c*d^7 + a^8*d^8)/(c^5*d^11))^(3/4) - (343*b^6*c^6 -
882*a*b^5*c^5*d + 609*a^2*b^4*c^4*d^2 + 36*a^3*b^3*c^3*d^3 - 87*a^4*b^2*c^2*d^4 - 18*a^5*b*c*d^5 - a^6*d^6)*sq
rt(x)) - 4*(4*b^2*c*d*x^3 + (7*b^2*c^2 - 6*a*b*c*d + 3*a^2*d^2)*x)*sqrt(x))/(c*d^3*x^2 + c^2*d^2)

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giac [A]  time = 0.46, size = 388, normalized size = 1.25 \begin {gather*} \frac {2 \, b^{2} x^{\frac {3}{2}}}{3 \, d^{2}} + \frac {b^{2} c^{2} x^{\frac {3}{2}} - 2 \, a b c d x^{\frac {3}{2}} + a^{2} d^{2} x^{\frac {3}{2}}}{2 \, {\left (d x^{2} + c\right )} c d^{2}} - \frac {\sqrt {2} {\left (7 \, \left (c d^{3}\right )^{\frac {3}{4}} b^{2} c^{2} - 6 \, \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 )}{8 \, c^{2} d^{5}} - \frac {\sqrt {2} {\left (7 \, \left (c d^{3}\right )^{\frac {3}{4}} b^{2} c^{2} - 6 \, \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 )}{8 \, c^{2} d^{5}} + \frac {\sqrt {2} {\left (7 \, \left (c d^{3}\right )^{\frac {3}{4}} b^{2} c^{2} - 6 \, \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 )}{16 \, c^{2} d^{5}} - \frac {\sqrt {2} {\left (7 \, \left (c d^{3}\right )^{\frac {3}{4}} b^{2} c^{2} - 6 \, \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 )}{16 \, c^{2} d^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*b^2*x^(3/2)/d^2 + 1/2*(b^2*c^2*x^(3/2) - 2*a*b*c*d*x^(3/2) + a^2*d^2*x^(3/2))/((d*x^2 + c)*c*d^2) - 1/8*sq
rt(2)*(7*(c*d^3)^(3/4)*b^2*c^2 - 6*(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^2*d^5) - 1/8*sqrt(2)*(7*(c*d^3)^(3/4)*b^2*c^2 - 6*(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^2*d^5) + 1/1
6*sqrt(2)*(7*(c*d^3)^(3/4)*b^2*c^2 - 6*(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^2*d^5) - 1/16*sqrt(2)*(7*(c*d^3)^(3/4)*b^2*c^2 - 6*(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^2*d^5)

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maple [B]  time = 0.02, size = 499, normalized size = 1.61 \begin {gather*} \frac {a^{2} x^{\frac {3}{2}}}{2 \left (d \,x^{2}+c \right ) c}-\frac {a b \,x^{\frac {3}{2}}}{\left (d \,x^{2}+c \right ) d}+\frac {b^{2} c \,x^{\frac {3}{2}}}{2 \left (d \,x^{2}+c \right ) d^{2}}+\frac {2 b^{2} 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 )}{8 \left (\frac {c}{d}\right )^{\frac {1}{4}} c d}+\frac {\sqrt {2}\, a^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )}{8 \left (\frac {c}{d}\right )^{\frac {1}{4}} c 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 )}{16 \left (\frac {c}{d}\right )^{\frac {1}{4}} c d}+\frac {3 \sqrt {2}\, a b \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )}{4 \left (\frac {c}{d}\right )^{\frac {1}{4}} d^{2}}+\frac {3 \sqrt {2}\, a b \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )}{4 \left (\frac {c}{d}\right )^{\frac {1}{4}} d^{2}}+\frac {3 \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 \left (\frac {c}{d}\right )^{\frac {1}{4}} d^{2}}-\frac {7 \sqrt {2}\, b^{2} c \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )}{8 \left (\frac {c}{d}\right )^{\frac {1}{4}} d^{3}}-\frac {7 \sqrt {2}\, b^{2} c \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )}{8 \left (\frac {c}{d}\right )^{\frac {1}{4}} d^{3}}-\frac {7 \sqrt {2}\, b^{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 )}{16 \left (\frac {c}{d}\right )^{\frac {1}{4}} d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*b^2*x^(3/2)/d^2+1/2/c*x^(3/2)/(d*x^2+c)*a^2-1/d*x^(3/2)/(d*x^2+c)*a*b+1/2/d^2*c*x^(3/2)/(d*x^2+c)*b^2+1/16
/d/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^2+3/8/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-7/16/d^3*c/(c/d)^(1/4)*2^(1/2)*ln((x-(c/d)^(1/4)*2^(1/2)*x^(1/2)+(c/d)^(1/2))/(x+(c/d
)^(1/4)*2^(1/2)*x^(1/2)+(c/d)^(1/2)))*b^2+1/8/d/c/(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)+1)*a^
2+3/4/d^2/(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)+1)*a*b-7/8/d^3*c/(c/d)^(1/4)*2^(1/2)*arctan(2
^(1/2)/(c/d)^(1/4)*x^(1/2)+1)*b^2+1/8/d/c/(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)-1)*a^2+3/4/d^
2/(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)-1)*a*b-7/8/d^3*c/(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 = 2.51, size = 258, normalized size = 0.83 \begin {gather*} \frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x^{\frac {3}{2}}}{2 \, {\left (c d^{3} x^{2} + c^{2} d^{2}\right )}} + \frac {2 \, b^{2} x^{\frac {3}{2}}}{3 \, d^{2}} - \frac {{\left (7 \, b^{2} c^{2} - 6 \, 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 )}}{16 \, c d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*x^(3/2)/(c*d^3*x^2 + c^2*d^2) + 2/3*b^2*x^(3/2)/d^2 - 1/16*(7*b^2*c^2 - 6*
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)*sq
rt(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)*s
qrt(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*d^2)

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mupad [B]  time = 0.40, size = 137, normalized size = 0.44 \begin {gather*} \frac {2\,b^2\,x^{3/2}}{3\,d^2}+\frac {x^{3/2}\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}{2\,c\,\left (d^3\,x^2+c\,d^2\right )}-\frac {\mathrm {atan}\left (\frac {d^{1/4}\,\sqrt {x}}{{\left (-c\right )}^{1/4}}\right )\,\left (a\,d-b\,c\right )\,\left (a\,d+7\,b\,c\right )}{4\,{\left (-c\right )}^{5/4}\,d^{11/4}}-\frac {\mathrm {atan}\left (\frac {d^{1/4}\,\sqrt {x}\,1{}\mathrm {i}}{{\left (-c\right )}^{1/4}}\right )\,\left (a\,d-b\,c\right )\,\left (a\,d+7\,b\,c\right )\,1{}\mathrm {i}}{4\,{\left (-c\right )}^{5/4}\,d^{11/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 42.06, size = 173, normalized size = 0.56 \begin {gather*} \frac {4 a b \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} - \frac {4 b^{2} c \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}} + \frac {2 b^{2} x^{\frac {3}{2}}}{3 d^{2}} + \frac {2 x^{\frac {3}{2}} \left (a d - b c\right )^{2}}{4 c^{2} d^{2} + 4 c d^{3} x^{2}} + \frac {2 \left (a d - b c\right )^{2} \operatorname {RootSum} {\left (65536 t^{4} c^{5} d^{3} + 1, \left (t \mapsto t \log {\left (4096 t^{3} c^{4} d^{2} + \sqrt {x} \right )} \right )\right )}}{d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*a*b*RootSum(256*_t**4*c*d**3 + 1, Lambda(_t, _t*log(64*_t**3*c*d**2 + sqrt(x))))/d - 4*b**2*c*RootSum(256*_t
**4*c*d**3 + 1, Lambda(_t, _t*log(64*_t**3*c*d**2 + sqrt(x))))/d**2 + 2*b**2*x**(3/2)/(3*d**2) + 2*x**(3/2)*(a
*d - b*c)**2/(4*c**2*d**2 + 4*c*d**3*x**2) + 2*(a*d - b*c)**2*RootSum(65536*_t**4*c**5*d**3 + 1, Lambda(_t, _t
*log(4096*_t**3*c**4*d**2 + sqrt(x))))/d**2

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