3.446 \(\int \frac{(c+d x^2)^3}{x^{5/2} (a+b x^2)} \, dx\)
Optimal. Leaf size=284 \[ \frac{(b c-a d)^3 \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{2 \sqrt{2} a^{7/4} b^{9/4}}-\frac{(b c-a d)^3 \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{2 \sqrt{2} a^{7/4} b^{9/4}}+\frac{(b c-a d)^3 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{\sqrt{2} a^{7/4} b^{9/4}}-\frac{(b c-a d)^3 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{\sqrt{2} a^{7/4} b^{9/4}}+\frac{2 d^2 \sqrt{x} (3 b c-a d)}{b^2}-\frac{2 c^3}{3 a x^{3/2}}+\frac{2 d^3 x^{5/2}}{5 b} \]
[Out]
(-2*c^3)/(3*a*x^(3/2)) + (2*d^2*(3*b*c - a*d)*Sqrt[x])/b^2 + (2*d^3*x^(5/2))/(5*b) + ((b*c - a*d)^3*ArcTan[1 -
(Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(Sqrt[2]*a^(7/4)*b^(9/4)) - ((b*c - a*d)^3*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sq
rt[x])/a^(1/4)])/(Sqrt[2]*a^(7/4)*b^(9/4)) + ((b*c - a*d)^3*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sq
rt[b]*x])/(2*Sqrt[2]*a^(7/4)*b^(9/4)) - ((b*c - a*d)^3*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]
*x])/(2*Sqrt[2]*a^(7/4)*b^(9/4))
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Rubi [A] time = 0.26692, antiderivative size = 284, normalized size of antiderivative = 1.,
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
= {466, 461, 211, 1165, 628, 1162, 617, 204} \[ \frac{(b c-a d)^3 \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{2 \sqrt{2} a^{7/4} b^{9/4}}-\frac{(b c-a d)^3 \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{2 \sqrt{2} a^{7/4} b^{9/4}}+\frac{(b c-a d)^3 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{\sqrt{2} a^{7/4} b^{9/4}}-\frac{(b c-a d)^3 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{\sqrt{2} a^{7/4} b^{9/4}}+\frac{2 d^2 \sqrt{x} (3 b c-a d)}{b^2}-\frac{2 c^3}{3 a x^{3/2}}+\frac{2 d^3 x^{5/2}}{5 b} \]
Antiderivative was successfully verified.
[In]
Int[(c + d*x^2)^3/(x^(5/2)*(a + b*x^2)),x]
[Out]
(-2*c^3)/(3*a*x^(3/2)) + (2*d^2*(3*b*c - a*d)*Sqrt[x])/b^2 + (2*d^3*x^(5/2))/(5*b) + ((b*c - a*d)^3*ArcTan[1 -
(Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(Sqrt[2]*a^(7/4)*b^(9/4)) - ((b*c - a*d)^3*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sq
rt[x])/a^(1/4)])/(Sqrt[2]*a^(7/4)*b^(9/4)) + ((b*c - a*d)^3*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sq
rt[b]*x])/(2*Sqrt[2]*a^(7/4)*b^(9/4)) - ((b*c - a*d)^3*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]
*x])/(2*Sqrt[2]*a^(7/4)*b^(9/4))
Rule 466
Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = Deno
minator[m]}, Dist[k/e, Subst[Int[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/e^n)^p*(c + (d*x^(k*n))/e^n)^q, x], x, (e*
x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && FractionQ[m] && Intege
rQ[p]
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 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 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]
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 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 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])
Rubi steps
\begin{align*} \int \frac{\left (c+d x^2\right )^3}{x^{5/2} \left (a+b x^2\right )} \, dx &=2 \operatorname{Subst}\left (\int \frac{\left (c+d x^4\right )^3}{x^4 \left (a+b x^4\right )} \, dx,x,\sqrt{x}\right )\\ &=2 \operatorname{Subst}\left (\int \left (\frac{d^2 (3 b c-a d)}{b^2}+\frac{c^3}{a x^4}+\frac{d^3 x^4}{b}+\frac{(-b c+a d)^3}{a b^2 \left (a+b x^4\right )}\right ) \, dx,x,\sqrt{x}\right )\\ &=-\frac{2 c^3}{3 a x^{3/2}}+\frac{2 d^2 (3 b c-a d) \sqrt{x}}{b^2}+\frac{2 d^3 x^{5/2}}{5 b}-\frac{\left (2 (b c-a d)^3\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x^4} \, dx,x,\sqrt{x}\right )}{a b^2}\\ &=-\frac{2 c^3}{3 a x^{3/2}}+\frac{2 d^2 (3 b c-a d) \sqrt{x}}{b^2}+\frac{2 d^3 x^{5/2}}{5 b}-\frac{(b c-a d)^3 \operatorname{Subst}\left (\int \frac{\sqrt{a}-\sqrt{b} x^2}{a+b x^4} \, dx,x,\sqrt{x}\right )}{a^{3/2} b^2}-\frac{(b c-a d)^3 \operatorname{Subst}\left (\int \frac{\sqrt{a}+\sqrt{b} x^2}{a+b x^4} \, dx,x,\sqrt{x}\right )}{a^{3/2} b^2}\\ &=-\frac{2 c^3}{3 a x^{3/2}}+\frac{2 d^2 (3 b c-a d) \sqrt{x}}{b^2}+\frac{2 d^3 x^{5/2}}{5 b}-\frac{(b c-a d)^3 \operatorname{Subst}\left (\int \frac{1}{\frac{\sqrt{a}}{\sqrt{b}}-\frac{\sqrt{2} \sqrt [4]{a} x}{\sqrt [4]{b}}+x^2} \, dx,x,\sqrt{x}\right )}{2 a^{3/2} b^{5/2}}-\frac{(b c-a d)^3 \operatorname{Subst}\left (\int \frac{1}{\frac{\sqrt{a}}{\sqrt{b}}+\frac{\sqrt{2} \sqrt [4]{a} x}{\sqrt [4]{b}}+x^2} \, dx,x,\sqrt{x}\right )}{2 a^{3/2} b^{5/2}}+\frac{(b c-a d)^3 \operatorname{Subst}\left (\int \frac{\frac{\sqrt{2} \sqrt [4]{a}}{\sqrt [4]{b}}+2 x}{-\frac{\sqrt{a}}{\sqrt{b}}-\frac{\sqrt{2} \sqrt [4]{a} x}{\sqrt [4]{b}}-x^2} \, dx,x,\sqrt{x}\right )}{2 \sqrt{2} a^{7/4} b^{9/4}}+\frac{(b c-a d)^3 \operatorname{Subst}\left (\int \frac{\frac{\sqrt{2} \sqrt [4]{a}}{\sqrt [4]{b}}-2 x}{-\frac{\sqrt{a}}{\sqrt{b}}+\frac{\sqrt{2} \sqrt [4]{a} x}{\sqrt [4]{b}}-x^2} \, dx,x,\sqrt{x}\right )}{2 \sqrt{2} a^{7/4} b^{9/4}}\\ &=-\frac{2 c^3}{3 a x^{3/2}}+\frac{2 d^2 (3 b c-a d) \sqrt{x}}{b^2}+\frac{2 d^3 x^{5/2}}{5 b}+\frac{(b c-a d)^3 \log \left (\sqrt{a}-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{2 \sqrt{2} a^{7/4} b^{9/4}}-\frac{(b c-a d)^3 \log \left (\sqrt{a}+\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{2 \sqrt{2} a^{7/4} b^{9/4}}-\frac{(b c-a d)^3 \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{\sqrt{2} a^{7/4} b^{9/4}}+\frac{(b c-a d)^3 \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{\sqrt{2} a^{7/4} b^{9/4}}\\ &=-\frac{2 c^3}{3 a x^{3/2}}+\frac{2 d^2 (3 b c-a d) \sqrt{x}}{b^2}+\frac{2 d^3 x^{5/2}}{5 b}+\frac{(b c-a d)^3 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{\sqrt{2} a^{7/4} b^{9/4}}-\frac{(b c-a d)^3 \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{\sqrt{2} a^{7/4} b^{9/4}}+\frac{(b c-a d)^3 \log \left (\sqrt{a}-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{2 \sqrt{2} a^{7/4} b^{9/4}}-\frac{(b c-a d)^3 \log \left (\sqrt{a}+\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{2 \sqrt{2} a^{7/4} b^{9/4}}\\ \end{align*}
Mathematica [C] time = 0.356801, size = 89, normalized size = 0.31 \[ -\frac{2 \left (a \left (15 a^2 d^3 x^2-3 a b d^2 x^2 \left (15 c+d x^2\right )+5 b^2 c^3\right )+15 x^2 (b c-a d)^3 \, _2F_1\left (\frac{1}{4},1;\frac{5}{4};-\frac{b x^2}{a}\right )\right )}{15 a^2 b^2 x^{3/2}} \]
Antiderivative was successfully verified.
[In]
Integrate[(c + d*x^2)^3/(x^(5/2)*(a + b*x^2)),x]
[Out]
(-2*(a*(5*b^2*c^3 + 15*a^2*d^3*x^2 - 3*a*b*d^2*x^2*(15*c + d*x^2)) + 15*(b*c - a*d)^3*x^2*Hypergeometric2F1[1/
4, 1, 5/4, -((b*x^2)/a)]))/(15*a^2*b^2*x^(3/2))
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Maple [B] time = 0.013, size = 616, normalized size = 2.2 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((d*x^2+c)^3/x^(5/2)/(b*x^2+a),x)
[Out]
2/5*d^3*x^(5/2)/b-2*d^3/b^2*a*x^(1/2)+6*d^2/b*x^(1/2)*c-2/3*c^3/a/x^(3/2)+1/2*a/b^2*(1/b*a)^(1/4)*2^(1/2)*arct
an(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)+1)*d^3-3/2/b*(1/b*a)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)+1)*c*
d^2+3/2/a*(1/b*a)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)+1)*c^2*d-1/2/a^2*b*(1/b*a)^(1/4)*2^(1/2)*
arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)+1)*c^3+1/2*a/b^2*(1/b*a)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/
2)-1)*d^3-3/2/b*(1/b*a)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)-1)*c*d^2+3/2/a*(1/b*a)^(1/4)*2^(1/2
)*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)-1)*c^2*d-1/2/a^2*b*(1/b*a)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/b*a)^(1/4)*x
^(1/2)-1)*c^3+1/4*a/b^2*(1/b*a)^(1/4)*2^(1/2)*ln((x+(1/b*a)^(1/4)*x^(1/2)*2^(1/2)+(1/b*a)^(1/2))/(x-(1/b*a)^(1
/4)*x^(1/2)*2^(1/2)+(1/b*a)^(1/2)))*d^3-3/4/b*(1/b*a)^(1/4)*2^(1/2)*ln((x+(1/b*a)^(1/4)*x^(1/2)*2^(1/2)+(1/b*a
)^(1/2))/(x-(1/b*a)^(1/4)*x^(1/2)*2^(1/2)+(1/b*a)^(1/2)))*c*d^2+3/4/a*(1/b*a)^(1/4)*2^(1/2)*ln((x+(1/b*a)^(1/4
)*x^(1/2)*2^(1/2)+(1/b*a)^(1/2))/(x-(1/b*a)^(1/4)*x^(1/2)*2^(1/2)+(1/b*a)^(1/2)))*c^2*d-1/4/a^2*b*(1/b*a)^(1/4
)*2^(1/2)*ln((x+(1/b*a)^(1/4)*x^(1/2)*2^(1/2)+(1/b*a)^(1/2))/(x-(1/b*a)^(1/4)*x^(1/2)*2^(1/2)+(1/b*a)^(1/2)))*
c^3
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x^2+c)^3/x^(5/2)/(b*x^2+a),x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [B] time = 1.37986, size = 4078, normalized size = 14.36 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x^2+c)^3/x^(5/2)/(b*x^2+a),x, algorithm="fricas")
[Out]
1/30*(60*a*b^2*x^2*(-(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^11*b*c*d^11 + a^12*d^12)/(a^7*b^9))^(1/4)*arctan((sqrt(a^4*b^4*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^11*b*c*d^11 + a^12*d^12)/(a^7*b^9)) + (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)*x)*a^5*b^7*(-(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^11*b*c*d^11 + a^12*d^12
)/(a^7*b^9))^(3/4) + (a^5*b^10*c^3 - 3*a^6*b^9*c^2*d + 3*a^7*b^8*c*d^2 - a^8*b^7*d^3)*sqrt(x)*(-(b^12*c^12 - 1
2*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^11*b*c*d^11 + a^12*d^12)/(a^7*b^9))^(3/4))/(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^11*b*c*d^11 + a^12*d^12)) + 15*a*b^2*x^2*(-(b^1
2*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^11*b*c*d^11 + a^12*d^12)/(a^7*b^9))^(1/4)*log(a^2*b^2*(-(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^11*b*c*d^11 + a^12*d^12)
/(a^7*b^9))^(1/4) - (b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqrt(x)) - 15*a*b^2*x^2*(-(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 + 9
24*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^11*b*c*d^11 + a^12*d^12)/(a^7*b^9))^(1/4)*log(-a^2*b^2*(-(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^11*b*c*d^11 + a^12*d^12)/(a^7*b
^9))^(1/4) - (b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqrt(x)) + 4*(3*a*b*d^3*x^4 - 5*b^2*c^3 + 15*
(3*a*b*c*d^2 - a^2*d^3)*x^2)*sqrt(x))/(a*b^2*x^2)
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Sympy [A] time = 157.665, size = 842, normalized size = 2.96 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x**2+c)**3/x**(5/2)/(b*x**2+a),x)
[Out]
Piecewise((zoo*(-2*c**3/(7*x**(7/2)) - 2*c**2*d/x**(3/2) + 6*c*d**2*sqrt(x) + 2*d**3*x**(5/2)/5), Eq(a, 0) & E
q(b, 0)), ((-2*c**3/(3*x**(3/2)) + 6*c**2*d*sqrt(x) + 6*c*d**2*x**(5/2)/5 + 2*d**3*x**(9/2)/9)/a, Eq(b, 0)), (
(-2*c**3/(7*x**(7/2)) - 2*c**2*d/x**(3/2) + 6*c*d**2*sqrt(x) + 2*d**3*x**(5/2)/5)/b, Eq(a, 0)), (-(-1)**(1/4)*
a**(5/4)*b**30*d**3*(1/b)**(129/4)*log(-(-1)**(1/4)*a**(1/4)*(1/b)**(1/4) + sqrt(x))/2 + (-1)**(1/4)*a**(5/4)*
b**30*d**3*(1/b)**(129/4)*log((-1)**(1/4)*a**(1/4)*(1/b)**(1/4) + sqrt(x))/2 - (-1)**(1/4)*a**(5/4)*d**3*(1/b)
**(1/4)*atan((-1)**(3/4)*sqrt(x)/(a**(1/4)*(1/b)**(1/4)))/b**2 + 3*(-1)**(1/4)*a**(1/4)*b**31*c*d**2*(1/b)**(1
29/4)*log(-(-1)**(1/4)*a**(1/4)*(1/b)**(1/4) + sqrt(x))/2 - 3*(-1)**(1/4)*a**(1/4)*b**31*c*d**2*(1/b)**(129/4)
*log((-1)**(1/4)*a**(1/4)*(1/b)**(1/4) + sqrt(x))/2 + 3*(-1)**(1/4)*a**(1/4)*c*d**2*(1/b)**(1/4)*atan((-1)**(3
/4)*sqrt(x)/(a**(1/4)*(1/b)**(1/4)))/b - 2*a*d**3*sqrt(x)/b**2 + 6*c*d**2*sqrt(x)/b + 2*d**3*x**(5/2)/(5*b) -
2*c**3/(3*a*x**(3/2)) - 3*(-1)**(1/4)*b**32*c**2*d*(1/b)**(129/4)*log(-(-1)**(1/4)*a**(1/4)*(1/b)**(1/4) + sqr
t(x))/(2*a**(3/4)) + 3*(-1)**(1/4)*b**32*c**2*d*(1/b)**(129/4)*log((-1)**(1/4)*a**(1/4)*(1/b)**(1/4) + sqrt(x)
)/(2*a**(3/4)) - 3*(-1)**(1/4)*c**2*d*(1/b)**(1/4)*atan((-1)**(3/4)*sqrt(x)/(a**(1/4)*(1/b)**(1/4)))/a**(3/4)
+ (-1)**(1/4)*b**33*c**3*(1/b)**(129/4)*log(-(-1)**(1/4)*a**(1/4)*(1/b)**(1/4) + sqrt(x))/(2*a**(7/4)) - (-1)*
*(1/4)*b**33*c**3*(1/b)**(129/4)*log((-1)**(1/4)*a**(1/4)*(1/b)**(1/4) + sqrt(x))/(2*a**(7/4)) + (-1)**(1/4)*b
*c**3*(1/b)**(1/4)*atan((-1)**(3/4)*sqrt(x)/(a**(1/4)*(1/b)**(1/4)))/a**(7/4), True))
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Giac [B] time = 1.44698, size = 622, normalized size = 2.19 \begin{align*} -\frac{2 \, c^{3}}{3 \, a x^{\frac{3}{2}}} - \frac{\sqrt{2}{\left (\left (a b^{3}\right )^{\frac{1}{4}} b^{3} c^{3} - 3 \, \left (a b^{3}\right )^{\frac{1}{4}} a b^{2} c^{2} d + 3 \, \left (a b^{3}\right )^{\frac{1}{4}} a^{2} b c d^{2} - \left (a b^{3}\right )^{\frac{1}{4}} a^{3} d^{3}\right )} \arctan \left (\frac{\sqrt{2}{\left (\sqrt{2} \left (\frac{a}{b}\right )^{\frac{1}{4}} + 2 \, \sqrt{x}\right )}}{2 \, \left (\frac{a}{b}\right )^{\frac{1}{4}}}\right )}{2 \, a^{2} b^{3}} - \frac{\sqrt{2}{\left (\left (a b^{3}\right )^{\frac{1}{4}} b^{3} c^{3} - 3 \, \left (a b^{3}\right )^{\frac{1}{4}} a b^{2} c^{2} d + 3 \, \left (a b^{3}\right )^{\frac{1}{4}} a^{2} b c d^{2} - \left (a b^{3}\right )^{\frac{1}{4}} a^{3} d^{3}\right )} \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{2} \left (\frac{a}{b}\right )^{\frac{1}{4}} - 2 \, \sqrt{x}\right )}}{2 \, \left (\frac{a}{b}\right )^{\frac{1}{4}}}\right )}{2 \, a^{2} b^{3}} - \frac{\sqrt{2}{\left (\left (a b^{3}\right )^{\frac{1}{4}} b^{3} c^{3} - 3 \, \left (a b^{3}\right )^{\frac{1}{4}} a b^{2} c^{2} d + 3 \, \left (a b^{3}\right )^{\frac{1}{4}} a^{2} b c d^{2} - \left (a b^{3}\right )^{\frac{1}{4}} a^{3} d^{3}\right )} \log \left (\sqrt{2} \sqrt{x} \left (\frac{a}{b}\right )^{\frac{1}{4}} + x + \sqrt{\frac{a}{b}}\right )}{4 \, a^{2} b^{3}} + \frac{\sqrt{2}{\left (\left (a b^{3}\right )^{\frac{1}{4}} b^{3} c^{3} - 3 \, \left (a b^{3}\right )^{\frac{1}{4}} a b^{2} c^{2} d + 3 \, \left (a b^{3}\right )^{\frac{1}{4}} a^{2} b c d^{2} - \left (a b^{3}\right )^{\frac{1}{4}} a^{3} d^{3}\right )} \log \left (-\sqrt{2} \sqrt{x} \left (\frac{a}{b}\right )^{\frac{1}{4}} + x + \sqrt{\frac{a}{b}}\right )}{4 \, a^{2} b^{3}} + \frac{2 \,{\left (b^{4} d^{3} x^{\frac{5}{2}} + 15 \, b^{4} c d^{2} \sqrt{x} - 5 \, a b^{3} d^{3} \sqrt{x}\right )}}{5 \, b^{5}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x^2+c)^3/x^(5/2)/(b*x^2+a),x, algorithm="giac")
[Out]
-2/3*c^3/(a*x^(3/2)) - 1/2*sqrt(2)*((a*b^3)^(1/4)*b^3*c^3 - 3*(a*b^3)^(1/4)*a*b^2*c^2*d + 3*(a*b^3)^(1/4)*a^2*
b*c*d^2 - (a*b^3)^(1/4)*a^3*d^3)*arctan(1/2*sqrt(2)*(sqrt(2)*(a/b)^(1/4) + 2*sqrt(x))/(a/b)^(1/4))/(a^2*b^3) -
1/2*sqrt(2)*((a*b^3)^(1/4)*b^3*c^3 - 3*(a*b^3)^(1/4)*a*b^2*c^2*d + 3*(a*b^3)^(1/4)*a^2*b*c*d^2 - (a*b^3)^(1/4
)*a^3*d^3)*arctan(-1/2*sqrt(2)*(sqrt(2)*(a/b)^(1/4) - 2*sqrt(x))/(a/b)^(1/4))/(a^2*b^3) - 1/4*sqrt(2)*((a*b^3)
^(1/4)*b^3*c^3 - 3*(a*b^3)^(1/4)*a*b^2*c^2*d + 3*(a*b^3)^(1/4)*a^2*b*c*d^2 - (a*b^3)^(1/4)*a^3*d^3)*log(sqrt(2
)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/(a^2*b^3) + 1/4*sqrt(2)*((a*b^3)^(1/4)*b^3*c^3 - 3*(a*b^3)^(1/4)*a*b^2*
c^2*d + 3*(a*b^3)^(1/4)*a^2*b*c*d^2 - (a*b^3)^(1/4)*a^3*d^3)*log(-sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))
/(a^2*b^3) + 2/5*(b^4*d^3*x^(5/2) + 15*b^4*c*d^2*sqrt(x) - 5*a*b^3*d^3*sqrt(x))/b^5