3.457 \(\int \frac{(c+d x^2)^3}{x^{3/2} (a+b x^2)^2} \, dx\)

Optimal. Leaf size=368 \[ -\frac{(b c-a d)^2 (7 a d+5 b c) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{8 \sqrt{2} a^{9/4} b^{11/4}}+\frac{(b c-a d)^2 (7 a d+5 b c) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{8 \sqrt{2} a^{9/4} b^{11/4}}+\frac{(b c-a d)^2 (7 a d+5 b c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{4 \sqrt{2} a^{9/4} b^{11/4}}-\frac{(b c-a d)^2 (7 a d+5 b c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{4 \sqrt{2} a^{9/4} b^{11/4}}-\frac{c^2 (5 b c-a d)}{2 a^2 b \sqrt{x}}-\frac{d^2 x^{3/2} (3 b c-7 a d)}{6 a b^2}+\frac{\left (c+d x^2\right )^2 (b c-a d)}{2 a b \sqrt{x} \left (a+b x^2\right )} \]

[Out]

-(c^2*(5*b*c - a*d))/(2*a^2*b*Sqrt[x]) - (d^2*(3*b*c - 7*a*d)*x^(3/2))/(6*a*b^2) + ((b*c - a*d)*(c + d*x^2)^2)
/(2*a*b*Sqrt[x]*(a + b*x^2)) + ((b*c - a*d)^2*(5*b*c + 7*a*d)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(
4*Sqrt[2]*a^(9/4)*b^(11/4)) - ((b*c - a*d)^2*(5*b*c + 7*a*d)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(4
*Sqrt[2]*a^(9/4)*b^(11/4)) - ((b*c - a*d)^2*(5*b*c + 7*a*d)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sq
rt[b]*x])/(8*Sqrt[2]*a^(9/4)*b^(11/4)) + ((b*c - a*d)^2*(5*b*c + 7*a*d)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*
Sqrt[x] + Sqrt[b]*x])/(8*Sqrt[2]*a^(9/4)*b^(11/4))

________________________________________________________________________________________

Rubi [A]  time = 0.431288, antiderivative size = 368, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {466, 468, 570, 297, 1162, 617, 204, 1165, 628} \[ -\frac{(b c-a d)^2 (7 a d+5 b c) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{8 \sqrt{2} a^{9/4} b^{11/4}}+\frac{(b c-a d)^2 (7 a d+5 b c) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{8 \sqrt{2} a^{9/4} b^{11/4}}+\frac{(b c-a d)^2 (7 a d+5 b c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{4 \sqrt{2} a^{9/4} b^{11/4}}-\frac{(b c-a d)^2 (7 a d+5 b c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{4 \sqrt{2} a^{9/4} b^{11/4}}-\frac{c^2 (5 b c-a d)}{2 a^2 b \sqrt{x}}-\frac{d^2 x^{3/2} (3 b c-7 a d)}{6 a b^2}+\frac{\left (c+d x^2\right )^2 (b c-a d)}{2 a b \sqrt{x} \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^2)^3/(x^(3/2)*(a + b*x^2)^2),x]

[Out]

-(c^2*(5*b*c - a*d))/(2*a^2*b*Sqrt[x]) - (d^2*(3*b*c - 7*a*d)*x^(3/2))/(6*a*b^2) + ((b*c - a*d)*(c + d*x^2)^2)
/(2*a*b*Sqrt[x]*(a + b*x^2)) + ((b*c - a*d)^2*(5*b*c + 7*a*d)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(
4*Sqrt[2]*a^(9/4)*b^(11/4)) - ((b*c - a*d)^2*(5*b*c + 7*a*d)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(4
*Sqrt[2]*a^(9/4)*b^(11/4)) - ((b*c - a*d)^2*(5*b*c + 7*a*d)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sq
rt[b]*x])/(8*Sqrt[2]*a^(9/4)*b^(11/4)) + ((b*c - a*d)^2*(5*b*c + 7*a*d)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*
Sqrt[x] + Sqrt[b]*x])/(8*Sqrt[2]*a^(9/4)*b^(11/4))

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 468

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

Rule 570

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_))^
(r_.), x_Symbol] :> Int[ExpandIntegrand[(g*x)^m*(a + b*x^n)^p*(c + d*x^n)^q*(e + f*x^n)^r, x], x] /; FreeQ[{a,
b, c, d, e, f, g, m, n}, x] && IGtQ[p, -2] && IGtQ[q, 0] && IGtQ[r, 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 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])

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]

Rubi steps

\begin{align*} \int \frac{\left (c+d x^2\right )^3}{x^{3/2} \left (a+b x^2\right )^2} \, dx &=2 \operatorname{Subst}\left (\int \frac{\left (c+d x^4\right )^3}{x^2 \left (a+b x^4\right )^2} \, dx,x,\sqrt{x}\right )\\ &=\frac{(b c-a d) \left (c+d x^2\right )^2}{2 a b \sqrt{x} \left (a+b x^2\right )}-\frac{\operatorname{Subst}\left (\int \frac{\left (c+d x^4\right ) \left (-c (5 b c-a d)+d (3 b c-7 a d) x^4\right )}{x^2 \left (a+b x^4\right )} \, dx,x,\sqrt{x}\right )}{2 a b}\\ &=\frac{(b c-a d) \left (c+d x^2\right )^2}{2 a b \sqrt{x} \left (a+b x^2\right )}-\frac{\operatorname{Subst}\left (\int \left (\frac{c^2 (-5 b c+a d)}{a x^2}+\frac{d^2 (3 b c-7 a d) x^2}{b}+\frac{(-b c+a d)^2 (5 b c+7 a d) x^2}{a b \left (a+b x^4\right )}\right ) \, dx,x,\sqrt{x}\right )}{2 a b}\\ &=-\frac{c^2 (5 b c-a d)}{2 a^2 b \sqrt{x}}-\frac{d^2 (3 b c-7 a d) x^{3/2}}{6 a b^2}+\frac{(b c-a d) \left (c+d x^2\right )^2}{2 a b \sqrt{x} \left (a+b x^2\right )}-\frac{\left ((b c-a d)^2 (5 b c+7 a d)\right ) \operatorname{Subst}\left (\int \frac{x^2}{a+b x^4} \, dx,x,\sqrt{x}\right )}{2 a^2 b^2}\\ &=-\frac{c^2 (5 b c-a d)}{2 a^2 b \sqrt{x}}-\frac{d^2 (3 b c-7 a d) x^{3/2}}{6 a b^2}+\frac{(b c-a d) \left (c+d x^2\right )^2}{2 a b \sqrt{x} \left (a+b x^2\right )}+\frac{\left ((b c-a d)^2 (5 b c+7 a d)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a}-\sqrt{b} x^2}{a+b x^4} \, dx,x,\sqrt{x}\right )}{4 a^2 b^{5/2}}-\frac{\left ((b c-a d)^2 (5 b c+7 a d)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a}+\sqrt{b} x^2}{a+b x^4} \, dx,x,\sqrt{x}\right )}{4 a^2 b^{5/2}}\\ &=-\frac{c^2 (5 b c-a d)}{2 a^2 b \sqrt{x}}-\frac{d^2 (3 b c-7 a d) x^{3/2}}{6 a b^2}+\frac{(b c-a d) \left (c+d x^2\right )^2}{2 a b \sqrt{x} \left (a+b x^2\right )}-\frac{\left ((b c-a d)^2 (5 b c+7 a d)\right ) \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 )}{8 a^2 b^3}-\frac{\left ((b c-a d)^2 (5 b c+7 a d)\right ) \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 )}{8 a^2 b^3}-\frac{\left ((b c-a d)^2 (5 b c+7 a d)\right ) \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 )}{8 \sqrt{2} a^{9/4} b^{11/4}}-\frac{\left ((b c-a d)^2 (5 b c+7 a d)\right ) \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 )}{8 \sqrt{2} a^{9/4} b^{11/4}}\\ &=-\frac{c^2 (5 b c-a d)}{2 a^2 b \sqrt{x}}-\frac{d^2 (3 b c-7 a d) x^{3/2}}{6 a b^2}+\frac{(b c-a d) \left (c+d x^2\right )^2}{2 a b \sqrt{x} \left (a+b x^2\right )}-\frac{(b c-a d)^2 (5 b c+7 a d) \log \left (\sqrt{a}-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{8 \sqrt{2} a^{9/4} b^{11/4}}+\frac{(b c-a d)^2 (5 b c+7 a d) \log \left (\sqrt{a}+\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{8 \sqrt{2} a^{9/4} b^{11/4}}-\frac{\left ((b c-a d)^2 (5 b c+7 a d)\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{4 \sqrt{2} a^{9/4} b^{11/4}}+\frac{\left ((b c-a d)^2 (5 b c+7 a d)\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{4 \sqrt{2} a^{9/4} b^{11/4}}\\ &=-\frac{c^2 (5 b c-a d)}{2 a^2 b \sqrt{x}}-\frac{d^2 (3 b c-7 a d) x^{3/2}}{6 a b^2}+\frac{(b c-a d) \left (c+d x^2\right )^2}{2 a b \sqrt{x} \left (a+b x^2\right )}+\frac{(b c-a d)^2 (5 b c+7 a d) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{4 \sqrt{2} a^{9/4} b^{11/4}}-\frac{(b c-a d)^2 (5 b c+7 a d) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{4 \sqrt{2} a^{9/4} b^{11/4}}-\frac{(b c-a d)^2 (5 b c+7 a d) \log \left (\sqrt{a}-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{8 \sqrt{2} a^{9/4} b^{11/4}}+\frac{(b c-a d)^2 (5 b c+7 a d) \log \left (\sqrt{a}+\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{8 \sqrt{2} a^{9/4} b^{11/4}}\\ \end{align*}

Mathematica [C]  time = 1.99514, size = 355, normalized size = 0.96 \[ -\frac{32768 b^3 x^6 \left (c+d x^2\right )^3 \text{HypergeometricPFQ}\left (\left \{\frac{3}{4},2,2,2,2\right \},\left \{1,1,1,\frac{19}{4}\right \},-\frac{b x^2}{a}\right )+55 \left (a \left (7 a^2 \left (43923 c^2 d x^2+14641 c^3+43923 c d^2 x^4+11953 d^3 x^6\right )+18 a b x^2 \left (1083 c^2 d x^2+361 c^3+2427 c d^2 x^4+809 d^3 x^6\right )-21 b^2 x^4 \left (3 c^2 d x^2-1919 c^3+3 c d^2 x^4+d^3 x^6\right )\right )-7 \, _2F_1\left (\frac{3}{4},1;\frac{7}{4};-\frac{b x^2}{a}\right ) \left (3 a^2 b x^2 \left (7203 c^2 d x^2+2401 c^3+8355 c d^2 x^4+2401 d^3 x^6\right )+a^3 \left (43923 c^2 d x^2+14641 c^3+43923 c d^2 x^4+11953 d^3 x^6\right )+9 a b^2 x^4 \left (209 c^2 d x^2+27 c^3+81 c d^2 x^4+27 d^3 x^6\right )+b^3 x^6 \left (3 c^2 d x^2-1919 c^3+3 c d^2 x^4+d^3 x^6\right )\right )\right )}{887040 a^3 b^2 x^{9/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c + d*x^2)^3/(x^(3/2)*(a + b*x^2)^2),x]

[Out]

-(55*(a*(-21*b^2*x^4*(-1919*c^3 + 3*c^2*d*x^2 + 3*c*d^2*x^4 + d^3*x^6) + 18*a*b*x^2*(361*c^3 + 1083*c^2*d*x^2
+ 2427*c*d^2*x^4 + 809*d^3*x^6) + 7*a^2*(14641*c^3 + 43923*c^2*d*x^2 + 43923*c*d^2*x^4 + 11953*d^3*x^6)) - 7*(
b^3*x^6*(-1919*c^3 + 3*c^2*d*x^2 + 3*c*d^2*x^4 + d^3*x^6) + 9*a*b^2*x^4*(27*c^3 + 209*c^2*d*x^2 + 81*c*d^2*x^4
+ 27*d^3*x^6) + 3*a^2*b*x^2*(2401*c^3 + 7203*c^2*d*x^2 + 8355*c*d^2*x^4 + 2401*d^3*x^6) + a^3*(14641*c^3 + 43
923*c^2*d*x^2 + 43923*c*d^2*x^4 + 11953*d^3*x^6))*Hypergeometric2F1[3/4, 1, 7/4, -((b*x^2)/a)]) + 32768*b^3*x^
6*(c + d*x^2)^3*HypergeometricPFQ[{3/4, 2, 2, 2, 2}, {1, 1, 1, 19/4}, -((b*x^2)/a)])/(887040*a^3*b^2*x^(9/2))

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Maple [B]  time = 0.02, size = 682, normalized size = 1.9 \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^(3/2)/(b*x^2+a)^2,x)

[Out]

2/3*d^3*x^(3/2)/b^2-2*c^3/a^2/x^(1/2)+1/2/b^2*a*x^(3/2)/(b*x^2+a)*d^3-3/2/b*x^(3/2)/(b*x^2+a)*c*d^2+3/2/a*x^(3
/2)/(b*x^2+a)*c^2*d-1/2*b/a^2*x^(3/2)/(b*x^2+a)*c^3-7/16/b^3*a/(1/b*a)^(1/4)*2^(1/2)*d^3*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)))-7/8/b^3*a/(1/b*a)^(1/4)*2^(1/2)
*d^3*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)+1)-7/8/b^3*a/(1/b*a)^(1/4)*2^(1/2)*d^3*arctan(2^(1/2)/(1/b*a)^(1/4)*
x^(1/2)-1)+3/16/b/a/(1/b*a)^(1/4)*2^(1/2)*c^2*d*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)))+3/8/b/a/(1/b*a)^(1/4)*2^(1/2)*c^2*d*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)
+1)+3/8/b/a/(1/b*a)^(1/4)*2^(1/2)*c^2*d*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)-1)-5/16/a^2/(1/b*a)^(1/4)*2^(1/2)
*c^3*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)))-5/8/a
^2/(1/b*a)^(1/4)*2^(1/2)*c^3*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)+1)-5/8/a^2/(1/b*a)^(1/4)*2^(1/2)*c^3*arctan(
2^(1/2)/(1/b*a)^(1/4)*x^(1/2)-1)+9/16/b^2/(1/b*a)^(1/4)*2^(1/2)*c*d^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)))+9/8/b^2/(1/b*a)^(1/4)*2^(1/2)*c*d^2*arctan(2^(1/2)
/(1/b*a)^(1/4)*x^(1/2)+1)+9/8/b^2/(1/b*a)^(1/4)*2^(1/2)*c*d^2*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)-1)

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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^(3/2)/(b*x^2+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.64249, size = 6132, normalized size = 16.66 \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^(3/2)/(b*x^2+a)^2,x, algorithm="fricas")

[Out]

1/24*(12*(a^2*b^3*x^3 + a^3*b^2*x)*(-(625*b^12*c^12 - 1500*a*b^11*c^11*d - 3150*a^2*b^10*c^10*d^2 + 11060*a^3*
b^9*c^9*d^3 + 1071*a^4*b^8*c^8*d^4 - 28728*a^5*b^7*c^7*d^5 + 19068*a^6*b^6*c^6*d^6 + 27144*a^7*b^5*c^5*d^7 - 3
7665*a^8*b^4*c^4*d^8 + 2324*a^9*b^3*c^3*d^9 + 19698*a^10*b^2*c^2*d^10 - 12348*a^11*b*c*d^11 + 2401*a^12*d^12)/
(a^9*b^11))^(1/4)*arctan((sqrt((15625*b^18*c^18 - 56250*a*b^17*c^17*d - 84375*a^2*b^16*c^16*d^2 + 570000*a^3*b
^15*c^15*d^3 - 211500*a^4*b^14*c^14*d^4 - 2174040*a^5*b^13*c^13*d^5 + 2720004*a^6*b^12*c^12*d^6 + 3321072*a^7*
b^11*c^11*d^7 - 8368866*a^8*b^10*c^10*d^8 + 640420*a^9*b^9*c^9*d^9 + 11255310*a^10*b^8*c^8*d^10 - 8509968*a^11
*b^7*c^7*d^11 - 4831644*a^12*b^6*c^6*d^12 + 9537192*a^13*b^5*c^5*d^13 - 3095820*a^14*b^4*c^4*d^14 - 2551920*a^
15*b^3*c^3*d^15 + 2614689*a^16*b^2*c^2*d^16 - 907578*a^17*b*c*d^17 + 117649*a^18*d^18)*x - (625*a^5*b^17*c^12
- 1500*a^6*b^16*c^11*d - 3150*a^7*b^15*c^10*d^2 + 11060*a^8*b^14*c^9*d^3 + 1071*a^9*b^13*c^8*d^4 - 28728*a^10*
b^12*c^7*d^5 + 19068*a^11*b^11*c^6*d^6 + 27144*a^12*b^10*c^5*d^7 - 37665*a^13*b^9*c^4*d^8 + 2324*a^14*b^8*c^3*
d^9 + 19698*a^15*b^7*c^2*d^10 - 12348*a^16*b^6*c*d^11 + 2401*a^17*b^5*d^12)*sqrt(-(625*b^12*c^12 - 1500*a*b^11
*c^11*d - 3150*a^2*b^10*c^10*d^2 + 11060*a^3*b^9*c^9*d^3 + 1071*a^4*b^8*c^8*d^4 - 28728*a^5*b^7*c^7*d^5 + 1906
8*a^6*b^6*c^6*d^6 + 27144*a^7*b^5*c^5*d^7 - 37665*a^8*b^4*c^4*d^8 + 2324*a^9*b^3*c^3*d^9 + 19698*a^10*b^2*c^2*
d^10 - 12348*a^11*b*c*d^11 + 2401*a^12*d^12)/(a^9*b^11)))*a^2*b^3*(-(625*b^12*c^12 - 1500*a*b^11*c^11*d - 3150
*a^2*b^10*c^10*d^2 + 11060*a^3*b^9*c^9*d^3 + 1071*a^4*b^8*c^8*d^4 - 28728*a^5*b^7*c^7*d^5 + 19068*a^6*b^6*c^6*
d^6 + 27144*a^7*b^5*c^5*d^7 - 37665*a^8*b^4*c^4*d^8 + 2324*a^9*b^3*c^3*d^9 + 19698*a^10*b^2*c^2*d^10 - 12348*a
^11*b*c*d^11 + 2401*a^12*d^12)/(a^9*b^11))^(1/4) - (125*a^2*b^12*c^9 - 225*a^3*b^11*c^8*d - 540*a^4*b^10*c^7*d
^2 + 1308*a^5*b^9*c^6*d^3 + 342*a^6*b^8*c^5*d^4 - 2430*a^7*b^7*c^4*d^5 + 1140*a^8*b^6*c^3*d^6 + 1260*a^9*b^5*c
^2*d^7 - 1323*a^10*b^4*c*d^8 + 343*a^11*b^3*d^9)*sqrt(x)*(-(625*b^12*c^12 - 1500*a*b^11*c^11*d - 3150*a^2*b^10
*c^10*d^2 + 11060*a^3*b^9*c^9*d^3 + 1071*a^4*b^8*c^8*d^4 - 28728*a^5*b^7*c^7*d^5 + 19068*a^6*b^6*c^6*d^6 + 271
44*a^7*b^5*c^5*d^7 - 37665*a^8*b^4*c^4*d^8 + 2324*a^9*b^3*c^3*d^9 + 19698*a^10*b^2*c^2*d^10 - 12348*a^11*b*c*d
^11 + 2401*a^12*d^12)/(a^9*b^11))^(1/4))/(625*b^12*c^12 - 1500*a*b^11*c^11*d - 3150*a^2*b^10*c^10*d^2 + 11060*
a^3*b^9*c^9*d^3 + 1071*a^4*b^8*c^8*d^4 - 28728*a^5*b^7*c^7*d^5 + 19068*a^6*b^6*c^6*d^6 + 27144*a^7*b^5*c^5*d^7
- 37665*a^8*b^4*c^4*d^8 + 2324*a^9*b^3*c^3*d^9 + 19698*a^10*b^2*c^2*d^10 - 12348*a^11*b*c*d^11 + 2401*a^12*d^
12)) - 3*(a^2*b^3*x^3 + a^3*b^2*x)*(-(625*b^12*c^12 - 1500*a*b^11*c^11*d - 3150*a^2*b^10*c^10*d^2 + 11060*a^3*
b^9*c^9*d^3 + 1071*a^4*b^8*c^8*d^4 - 28728*a^5*b^7*c^7*d^5 + 19068*a^6*b^6*c^6*d^6 + 27144*a^7*b^5*c^5*d^7 - 3
7665*a^8*b^4*c^4*d^8 + 2324*a^9*b^3*c^3*d^9 + 19698*a^10*b^2*c^2*d^10 - 12348*a^11*b*c*d^11 + 2401*a^12*d^12)/
(a^9*b^11))^(1/4)*log(a^7*b^8*(-(625*b^12*c^12 - 1500*a*b^11*c^11*d - 3150*a^2*b^10*c^10*d^2 + 11060*a^3*b^9*c
^9*d^3 + 1071*a^4*b^8*c^8*d^4 - 28728*a^5*b^7*c^7*d^5 + 19068*a^6*b^6*c^6*d^6 + 27144*a^7*b^5*c^5*d^7 - 37665*
a^8*b^4*c^4*d^8 + 2324*a^9*b^3*c^3*d^9 + 19698*a^10*b^2*c^2*d^10 - 12348*a^11*b*c*d^11 + 2401*a^12*d^12)/(a^9*
b^11))^(3/4) + (125*b^9*c^9 - 225*a*b^8*c^8*d - 540*a^2*b^7*c^7*d^2 + 1308*a^3*b^6*c^6*d^3 + 342*a^4*b^5*c^5*d
^4 - 2430*a^5*b^4*c^4*d^5 + 1140*a^6*b^3*c^3*d^6 + 1260*a^7*b^2*c^2*d^7 - 1323*a^8*b*c*d^8 + 343*a^9*d^9)*sqrt
(x)) + 3*(a^2*b^3*x^3 + a^3*b^2*x)*(-(625*b^12*c^12 - 1500*a*b^11*c^11*d - 3150*a^2*b^10*c^10*d^2 + 11060*a^3*
b^9*c^9*d^3 + 1071*a^4*b^8*c^8*d^4 - 28728*a^5*b^7*c^7*d^5 + 19068*a^6*b^6*c^6*d^6 + 27144*a^7*b^5*c^5*d^7 - 3
7665*a^8*b^4*c^4*d^8 + 2324*a^9*b^3*c^3*d^9 + 19698*a^10*b^2*c^2*d^10 - 12348*a^11*b*c*d^11 + 2401*a^12*d^12)/
(a^9*b^11))^(1/4)*log(-a^7*b^8*(-(625*b^12*c^12 - 1500*a*b^11*c^11*d - 3150*a^2*b^10*c^10*d^2 + 11060*a^3*b^9*
c^9*d^3 + 1071*a^4*b^8*c^8*d^4 - 28728*a^5*b^7*c^7*d^5 + 19068*a^6*b^6*c^6*d^6 + 27144*a^7*b^5*c^5*d^7 - 37665
*a^8*b^4*c^4*d^8 + 2324*a^9*b^3*c^3*d^9 + 19698*a^10*b^2*c^2*d^10 - 12348*a^11*b*c*d^11 + 2401*a^12*d^12)/(a^9
*b^11))^(3/4) + (125*b^9*c^9 - 225*a*b^8*c^8*d - 540*a^2*b^7*c^7*d^2 + 1308*a^3*b^6*c^6*d^3 + 342*a^4*b^5*c^5*
d^4 - 2430*a^5*b^4*c^4*d^5 + 1140*a^6*b^3*c^3*d^6 + 1260*a^7*b^2*c^2*d^7 - 1323*a^8*b*c*d^8 + 343*a^9*d^9)*sqr
t(x)) + 4*(4*a^2*b*d^3*x^4 - 12*a*b^2*c^3 - (15*b^3*c^3 - 9*a*b^2*c^2*d + 9*a^2*b*c*d^2 - 7*a^3*d^3)*x^2)*sqrt
(x))/(a^2*b^3*x^3 + a^3*b^2*x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**2+c)**3/x**(3/2)/(b*x**2+a)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.26914, size = 680, normalized size = 1.85 \begin{align*} \frac{2 \, d^{3} x^{\frac{3}{2}}}{3 \, b^{2}} - \frac{5 \, b^{3} c^{3} x^{2} - 3 \, a b^{2} c^{2} d x^{2} + 3 \, a^{2} b c d^{2} x^{2} - a^{3} d^{3} x^{2} + 4 \, a b^{2} c^{3}}{2 \,{\left (b x^{\frac{5}{2}} + a \sqrt{x}\right )} a^{2} b^{2}} - \frac{\sqrt{2}{\left (5 \, \left (a b^{3}\right )^{\frac{3}{4}} b^{3} c^{3} - 3 \, \left (a b^{3}\right )^{\frac{3}{4}} a b^{2} c^{2} d - 9 \, \left (a b^{3}\right )^{\frac{3}{4}} a^{2} b c d^{2} + 7 \, \left (a b^{3}\right )^{\frac{3}{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 )}{8 \, a^{3} b^{5}} - \frac{\sqrt{2}{\left (5 \, \left (a b^{3}\right )^{\frac{3}{4}} b^{3} c^{3} - 3 \, \left (a b^{3}\right )^{\frac{3}{4}} a b^{2} c^{2} d - 9 \, \left (a b^{3}\right )^{\frac{3}{4}} a^{2} b c d^{2} + 7 \, \left (a b^{3}\right )^{\frac{3}{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 )}{8 \, a^{3} b^{5}} + \frac{\sqrt{2}{\left (5 \, \left (a b^{3}\right )^{\frac{3}{4}} b^{3} c^{3} - 3 \, \left (a b^{3}\right )^{\frac{3}{4}} a b^{2} c^{2} d - 9 \, \left (a b^{3}\right )^{\frac{3}{4}} a^{2} b c d^{2} + 7 \, \left (a b^{3}\right )^{\frac{3}{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 )}{16 \, a^{3} b^{5}} - \frac{\sqrt{2}{\left (5 \, \left (a b^{3}\right )^{\frac{3}{4}} b^{3} c^{3} - 3 \, \left (a b^{3}\right )^{\frac{3}{4}} a b^{2} c^{2} d - 9 \, \left (a b^{3}\right )^{\frac{3}{4}} a^{2} b c d^{2} + 7 \, \left (a b^{3}\right )^{\frac{3}{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 )}{16 \, a^{3} b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^3/x^(3/2)/(b*x^2+a)^2,x, algorithm="giac")

[Out]

2/3*d^3*x^(3/2)/b^2 - 1/2*(5*b^3*c^3*x^2 - 3*a*b^2*c^2*d*x^2 + 3*a^2*b*c*d^2*x^2 - a^3*d^3*x^2 + 4*a*b^2*c^3)/
((b*x^(5/2) + a*sqrt(x))*a^2*b^2) - 1/8*sqrt(2)*(5*(a*b^3)^(3/4)*b^3*c^3 - 3*(a*b^3)^(3/4)*a*b^2*c^2*d - 9*(a*
b^3)^(3/4)*a^2*b*c*d^2 + 7*(a*b^3)^(3/4)*a^3*d^3)*arctan(1/2*sqrt(2)*(sqrt(2)*(a/b)^(1/4) + 2*sqrt(x))/(a/b)^(
1/4))/(a^3*b^5) - 1/8*sqrt(2)*(5*(a*b^3)^(3/4)*b^3*c^3 - 3*(a*b^3)^(3/4)*a*b^2*c^2*d - 9*(a*b^3)^(3/4)*a^2*b*c
*d^2 + 7*(a*b^3)^(3/4)*a^3*d^3)*arctan(-1/2*sqrt(2)*(sqrt(2)*(a/b)^(1/4) - 2*sqrt(x))/(a/b)^(1/4))/(a^3*b^5) +
1/16*sqrt(2)*(5*(a*b^3)^(3/4)*b^3*c^3 - 3*(a*b^3)^(3/4)*a*b^2*c^2*d - 9*(a*b^3)^(3/4)*a^2*b*c*d^2 + 7*(a*b^3)
^(3/4)*a^3*d^3)*log(sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/(a^3*b^5) - 1/16*sqrt(2)*(5*(a*b^3)^(3/4)*b^3
*c^3 - 3*(a*b^3)^(3/4)*a*b^2*c^2*d - 9*(a*b^3)^(3/4)*a^2*b*c*d^2 + 7*(a*b^3)^(3/4)*a^3*d^3)*log(-sqrt(2)*sqrt(
x)*(a/b)^(1/4) + x + sqrt(a/b))/(a^3*b^5)