3.447 \(\int \frac{(c+d x^2)^3}{x^{7/2} (a+b x^2)} \, dx\)
Optimal. Leaf size=283 \[ \frac{(b c-a d)^3 \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{2 \sqrt{2} a^{9/4} b^{7/4}}-\frac{(b c-a d)^3 \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{2 \sqrt{2} a^{9/4} b^{7/4}}-\frac{(b c-a d)^3 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{\sqrt{2} a^{9/4} b^{7/4}}+\frac{(b c-a d)^3 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{\sqrt{2} a^{9/4} b^{7/4}}+\frac{2 c^2 (b c-3 a d)}{a^2 \sqrt{x}}-\frac{2 c^3}{5 a x^{5/2}}+\frac{2 d^3 x^{3/2}}{3 b} \]
[Out]
(-2*c^3)/(5*a*x^(5/2)) + (2*c^2*(b*c - 3*a*d))/(a^2*Sqrt[x]) + (2*d^3*x^(3/2))/(3*b) - ((b*c - a*d)^3*ArcTan[1
- (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(Sqrt[2]*a^(9/4)*b^(7/4)) + ((b*c - a*d)^3*ArcTan[1 + (Sqrt[2]*b^(1/4)*
Sqrt[x])/a^(1/4)])/(Sqrt[2]*a^(9/4)*b^(7/4)) + ((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^(9/4)*b^(7/4)) - ((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^(9/4)*b^(7/4))
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Rubi [A] time = 0.275281, antiderivative size = 283, 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, 297, 1162, 617, 204, 1165, 628} \[ \frac{(b c-a d)^3 \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{2 \sqrt{2} a^{9/4} b^{7/4}}-\frac{(b c-a d)^3 \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{2 \sqrt{2} a^{9/4} b^{7/4}}-\frac{(b c-a d)^3 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{\sqrt{2} a^{9/4} b^{7/4}}+\frac{(b c-a d)^3 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{\sqrt{2} a^{9/4} b^{7/4}}+\frac{2 c^2 (b c-3 a d)}{a^2 \sqrt{x}}-\frac{2 c^3}{5 a x^{5/2}}+\frac{2 d^3 x^{3/2}}{3 b} \]
Antiderivative was successfully verified.
[In]
Int[(c + d*x^2)^3/(x^(7/2)*(a + b*x^2)),x]
[Out]
(-2*c^3)/(5*a*x^(5/2)) + (2*c^2*(b*c - 3*a*d))/(a^2*Sqrt[x]) + (2*d^3*x^(3/2))/(3*b) - ((b*c - a*d)^3*ArcTan[1
- (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(Sqrt[2]*a^(9/4)*b^(7/4)) + ((b*c - a*d)^3*ArcTan[1 + (Sqrt[2]*b^(1/4)*
Sqrt[x])/a^(1/4)])/(Sqrt[2]*a^(9/4)*b^(7/4)) + ((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^(9/4)*b^(7/4)) - ((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^(9/4)*b^(7/4))
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 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^{7/2} \left (a+b x^2\right )} \, dx &=2 \operatorname{Subst}\left (\int \frac{\left (c+d x^4\right )^3}{x^6 \left (a+b x^4\right )} \, dx,x,\sqrt{x}\right )\\ &=2 \operatorname{Subst}\left (\int \left (\frac{c^3}{a x^6}+\frac{c^2 (-b c+3 a d)}{a^2 x^2}+\frac{d^3 x^2}{b}-\frac{(-b c+a d)^3 x^2}{a^2 b \left (a+b x^4\right )}\right ) \, dx,x,\sqrt{x}\right )\\ &=-\frac{2 c^3}{5 a x^{5/2}}+\frac{2 c^2 (b c-3 a d)}{a^2 \sqrt{x}}+\frac{2 d^3 x^{3/2}}{3 b}+\frac{\left (2 (b c-a d)^3\right ) \operatorname{Subst}\left (\int \frac{x^2}{a+b x^4} \, dx,x,\sqrt{x}\right )}{a^2 b}\\ &=-\frac{2 c^3}{5 a x^{5/2}}+\frac{2 c^2 (b c-3 a d)}{a^2 \sqrt{x}}+\frac{2 d^3 x^{3/2}}{3 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^2 b^{3/2}}+\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^2 b^{3/2}}\\ &=-\frac{2 c^3}{5 a x^{5/2}}+\frac{2 c^2 (b c-3 a d)}{a^2 \sqrt{x}}+\frac{2 d^3 x^{3/2}}{3 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^2 b^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^2 b^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^{9/4} b^{7/4}}+\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^{9/4} b^{7/4}}\\ &=-\frac{2 c^3}{5 a x^{5/2}}+\frac{2 c^2 (b c-3 a d)}{a^2 \sqrt{x}}+\frac{2 d^3 x^{3/2}}{3 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^{9/4} b^{7/4}}-\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^{9/4} b^{7/4}}+\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^{9/4} b^{7/4}}-\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^{9/4} b^{7/4}}\\ &=-\frac{2 c^3}{5 a x^{5/2}}+\frac{2 c^2 (b c-3 a d)}{a^2 \sqrt{x}}+\frac{2 d^3 x^{3/2}}{3 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^{9/4} b^{7/4}}+\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^{9/4} b^{7/4}}+\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^{9/4} b^{7/4}}-\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^{9/4} b^{7/4}}\\ \end{align*}
Mathematica [C] time = 0.352557, size = 88, normalized size = 0.31 \[ -\frac{2 \left (a \left (-5 a^2 d^3 x^4+3 a b c^2 \left (c+15 d x^2\right )-15 b^2 c^3 x^2\right )-5 x^4 (b c-a d)^3 \, _2F_1\left (\frac{3}{4},1;\frac{7}{4};-\frac{b x^2}{a}\right )\right )}{15 a^3 b x^{5/2}} \]
Antiderivative was successfully verified.
[In]
Integrate[(c + d*x^2)^3/(x^(7/2)*(a + b*x^2)),x]
[Out]
(-2*(a*(-15*b^2*c^3*x^2 - 5*a^2*d^3*x^4 + 3*a*b*c^2*(c + 15*d*x^2)) - 5*(b*c - a*d)^3*x^4*Hypergeometric2F1[3/
4, 1, 7/4, -((b*x^2)/a)]))/(15*a^3*b*x^(5/2))
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Maple [B] time = 0.015, 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^(7/2)/(b*x^2+a),x)
[Out]
2/3*d^3*x^(3/2)/b-2/5*c^3/a/x^(5/2)-6*c^2/a/x^(1/2)*d+2*c^3/a^2/x^(1/2)*b-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^(7/2)/(b*x^2+a),x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [B] time = 2.05288, size = 5407, normalized size = 19.11 \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^(7/2)/(b*x^2+a),x, algorithm="fricas")
[Out]
1/30*(60*a^2*b*x^3*(-(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^9*b^7))^(1/4)*arctan((sqrt((b^18*c^18 - 18*a
*b^17*c^17*d + 153*a^2*b^16*c^16*d^2 - 816*a^3*b^15*c^15*d^3 + 3060*a^4*b^14*c^14*d^4 - 8568*a^5*b^13*c^13*d^5
+ 18564*a^6*b^12*c^12*d^6 - 31824*a^7*b^11*c^11*d^7 + 43758*a^8*b^10*c^10*d^8 - 48620*a^9*b^9*c^9*d^9 + 43758
*a^10*b^8*c^8*d^10 - 31824*a^11*b^7*c^7*d^11 + 18564*a^12*b^6*c^6*d^12 - 8568*a^13*b^5*c^5*d^13 + 3060*a^14*b^
4*c^4*d^14 - 816*a^15*b^3*c^3*d^15 + 153*a^16*b^2*c^2*d^16 - 18*a^17*b*c*d^17 + a^18*d^18)*x - (a^5*b^15*c^12
- 12*a^6*b^14*c^11*d + 66*a^7*b^13*c^10*d^2 - 220*a^8*b^12*c^9*d^3 + 495*a^9*b^11*c^8*d^4 - 792*a^10*b^10*c^7*
d^5 + 924*a^11*b^9*c^6*d^6 - 792*a^12*b^8*c^5*d^7 + 495*a^13*b^7*c^4*d^8 - 220*a^14*b^6*c^3*d^9 + 66*a^15*b^5*
c^2*d^10 - 12*a^16*b^4*c*d^11 + a^17*b^3*d^12)*sqrt(-(b^12*c^12 - 12*a*b^11*c^11*d + 66*a^2*b^10*c^10*d^2 - 22
0*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 + 49
5*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^9*b^7)))*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^9*b^7))^(1/4) + (a^2*b^11*c^9 - 9*a^3*b^10*c^8*d + 36*a^4*
b^9*c^7*d^2 - 84*a^5*b^8*c^6*d^3 + 126*a^6*b^7*c^5*d^4 - 126*a^7*b^6*c^4*d^5 + 84*a^8*b^5*c^3*d^6 - 36*a^9*b^4
*c^2*d^7 + 9*a^10*b^3*c*d^8 - a^11*b^2*d^9)*sqrt(x)*(-(b^12*c^12 - 12*a*b^11*c^11*d + 66*a^2*b^10*c^10*d^2 - 2
20*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 + 4
95*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^9*b^7))^(1/
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^2*b*x^3*(-(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^9
*b^7))^(1/4)*log(a^7*b^5*(-(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^9*b^7))^(3/4) - (b^9*c^9 - 9*a*b^8*c^8
*d + 36*a^2*b^7*c^7*d^2 - 84*a^3*b^6*c^6*d^3 + 126*a^4*b^5*c^5*d^4 - 126*a^5*b^4*c^4*d^5 + 84*a^6*b^3*c^3*d^6
- 36*a^7*b^2*c^2*d^7 + 9*a^8*b*c*d^8 - a^9*d^9)*sqrt(x)) + 15*a^2*b*x^3*(-(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^9*b^7))^(1/4)*log(-a^7*b^5*(-(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^9*b^7))^(3/4) - (b^9*c^9 -
9*a*b^8*c^8*d + 36*a^2*b^7*c^7*d^2 - 84*a^3*b^6*c^6*d^3 + 126*a^4*b^5*c^5*d^4 - 126*a^5*b^4*c^4*d^5 + 84*a^6*b
^3*c^3*d^6 - 36*a^7*b^2*c^2*d^7 + 9*a^8*b*c*d^8 - a^9*d^9)*sqrt(x)) + 4*(5*a^2*d^3*x^4 - 3*a*b*c^3 + 15*(b^2*c
^3 - 3*a*b*c^2*d)*x^2)*sqrt(x))/(a^2*b*x^3)
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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**(7/2)/(b*x**2+a),x)
[Out]
Timed out
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Giac [B] time = 1.24746, size = 614, normalized size = 2.17 \begin{align*} \frac{2 \, d^{3} x^{\frac{3}{2}}}{3 \, b} + \frac{2 \,{\left (5 \, b c^{3} x^{2} - 15 \, a c^{2} d x^{2} - a c^{3}\right )}}{5 \, a^{2} x^{\frac{5}{2}}} + \frac{\sqrt{2}{\left (\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 + 3 \, \left (a b^{3}\right )^{\frac{3}{4}} a^{2} b c d^{2} - \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 )}{2 \, a^{3} b^{4}} + \frac{\sqrt{2}{\left (\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 + 3 \, \left (a b^{3}\right )^{\frac{3}{4}} a^{2} b c d^{2} - \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 )}{2 \, a^{3} b^{4}} - \frac{\sqrt{2}{\left (\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 + 3 \, \left (a b^{3}\right )^{\frac{3}{4}} a^{2} b c d^{2} - \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 )}{4 \, a^{3} b^{4}} + \frac{\sqrt{2}{\left (\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 + 3 \, \left (a b^{3}\right )^{\frac{3}{4}} a^{2} b c d^{2} - \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 )}{4 \, a^{3} b^{4}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x^2+c)^3/x^(7/2)/(b*x^2+a),x, algorithm="giac")
[Out]
2/3*d^3*x^(3/2)/b + 2/5*(5*b*c^3*x^2 - 15*a*c^2*d*x^2 - a*c^3)/(a^2*x^(5/2)) + 1/2*sqrt(2)*((a*b^3)^(3/4)*b^3*
c^3 - 3*(a*b^3)^(3/4)*a*b^2*c^2*d + 3*(a*b^3)^(3/4)*a^2*b*c*d^2 - (a*b^3)^(3/4)*a^3*d^3)*arctan(1/2*sqrt(2)*(s
qrt(2)*(a/b)^(1/4) + 2*sqrt(x))/(a/b)^(1/4))/(a^3*b^4) + 1/2*sqrt(2)*((a*b^3)^(3/4)*b^3*c^3 - 3*(a*b^3)^(3/4)*
a*b^2*c^2*d + 3*(a*b^3)^(3/4)*a^2*b*c*d^2 - (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^4) - 1/4*sqrt(2)*((a*b^3)^(3/4)*b^3*c^3 - 3*(a*b^3)^(3/4)*a*b^2*c^2*d + 3*(a*b^
3)^(3/4)*a^2*b*c*d^2 - (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^4) + 1/4
*sqrt(2)*((a*b^3)^(3/4)*b^3*c^3 - 3*(a*b^3)^(3/4)*a*b^2*c^2*d + 3*(a*b^3)^(3/4)*a^2*b*c*d^2 - (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^4)