3.170 \(\int \frac{A+B x^3}{x^{7/2} (a+b x^3)^2} \, dx\)

Optimal. Leaf size=318 \[ -\frac{11 A b-5 a B}{15 a^2 b x^{5/2}}+\frac{(11 A b-5 a B) \log \left (-\sqrt{3} \sqrt [6]{a} \sqrt [6]{b} \sqrt{x}+\sqrt [3]{a}+\sqrt [3]{b} x\right )}{12 \sqrt{3} a^{17/6} \sqrt [6]{b}}-\frac{(11 A b-5 a B) \log \left (\sqrt{3} \sqrt [6]{a} \sqrt [6]{b} \sqrt{x}+\sqrt [3]{a}+\sqrt [3]{b} x\right )}{12 \sqrt{3} a^{17/6} \sqrt [6]{b}}+\frac{(11 A b-5 a B) \tan ^{-1}\left (\sqrt{3}-\frac{2 \sqrt [6]{b} \sqrt{x}}{\sqrt [6]{a}}\right )}{18 a^{17/6} \sqrt [6]{b}}-\frac{(11 A b-5 a B) \tan ^{-1}\left (\frac{2 \sqrt [6]{b} \sqrt{x}}{\sqrt [6]{a}}+\sqrt{3}\right )}{18 a^{17/6} \sqrt [6]{b}}-\frac{(11 A b-5 a B) \tan ^{-1}\left (\frac{\sqrt [6]{b} \sqrt{x}}{\sqrt [6]{a}}\right )}{9 a^{17/6} \sqrt [6]{b}}+\frac{A b-a B}{3 a b x^{5/2} \left (a+b x^3\right )} \]

[Out]

-(11*A*b - 5*a*B)/(15*a^2*b*x^(5/2)) + (A*b - a*B)/(3*a*b*x^(5/2)*(a + b*x^3)) + ((11*A*b - 5*a*B)*ArcTan[Sqrt
[3] - (2*b^(1/6)*Sqrt[x])/a^(1/6)])/(18*a^(17/6)*b^(1/6)) - ((11*A*b - 5*a*B)*ArcTan[Sqrt[3] + (2*b^(1/6)*Sqrt
[x])/a^(1/6)])/(18*a^(17/6)*b^(1/6)) - ((11*A*b - 5*a*B)*ArcTan[(b^(1/6)*Sqrt[x])/a^(1/6)])/(9*a^(17/6)*b^(1/6
)) + ((11*A*b - 5*a*B)*Log[a^(1/3) - Sqrt[3]*a^(1/6)*b^(1/6)*Sqrt[x] + b^(1/3)*x])/(12*Sqrt[3]*a^(17/6)*b^(1/6
)) - ((11*A*b - 5*a*B)*Log[a^(1/3) + Sqrt[3]*a^(1/6)*b^(1/6)*Sqrt[x] + b^(1/3)*x])/(12*Sqrt[3]*a^(17/6)*b^(1/6
))

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Rubi [A]  time = 0.50449, antiderivative size = 318, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.409, Rules used = {457, 325, 329, 209, 634, 618, 204, 628, 205} \[ -\frac{11 A b-5 a B}{15 a^2 b x^{5/2}}+\frac{(11 A b-5 a B) \log \left (-\sqrt{3} \sqrt [6]{a} \sqrt [6]{b} \sqrt{x}+\sqrt [3]{a}+\sqrt [3]{b} x\right )}{12 \sqrt{3} a^{17/6} \sqrt [6]{b}}-\frac{(11 A b-5 a B) \log \left (\sqrt{3} \sqrt [6]{a} \sqrt [6]{b} \sqrt{x}+\sqrt [3]{a}+\sqrt [3]{b} x\right )}{12 \sqrt{3} a^{17/6} \sqrt [6]{b}}+\frac{(11 A b-5 a B) \tan ^{-1}\left (\sqrt{3}-\frac{2 \sqrt [6]{b} \sqrt{x}}{\sqrt [6]{a}}\right )}{18 a^{17/6} \sqrt [6]{b}}-\frac{(11 A b-5 a B) \tan ^{-1}\left (\frac{2 \sqrt [6]{b} \sqrt{x}}{\sqrt [6]{a}}+\sqrt{3}\right )}{18 a^{17/6} \sqrt [6]{b}}-\frac{(11 A b-5 a B) \tan ^{-1}\left (\frac{\sqrt [6]{b} \sqrt{x}}{\sqrt [6]{a}}\right )}{9 a^{17/6} \sqrt [6]{b}}+\frac{A b-a B}{3 a b x^{5/2} \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^3)/(x^(7/2)*(a + b*x^3)^2),x]

[Out]

-(11*A*b - 5*a*B)/(15*a^2*b*x^(5/2)) + (A*b - a*B)/(3*a*b*x^(5/2)*(a + b*x^3)) + ((11*A*b - 5*a*B)*ArcTan[Sqrt
[3] - (2*b^(1/6)*Sqrt[x])/a^(1/6)])/(18*a^(17/6)*b^(1/6)) - ((11*A*b - 5*a*B)*ArcTan[Sqrt[3] + (2*b^(1/6)*Sqrt
[x])/a^(1/6)])/(18*a^(17/6)*b^(1/6)) - ((11*A*b - 5*a*B)*ArcTan[(b^(1/6)*Sqrt[x])/a^(1/6)])/(9*a^(17/6)*b^(1/6
)) + ((11*A*b - 5*a*B)*Log[a^(1/3) - Sqrt[3]*a^(1/6)*b^(1/6)*Sqrt[x] + b^(1/3)*x])/(12*Sqrt[3]*a^(17/6)*b^(1/6
)) - ((11*A*b - 5*a*B)*Log[a^(1/3) + Sqrt[3]*a^(1/6)*b^(1/6)*Sqrt[x] + b^(1/3)*x])/(12*Sqrt[3]*a^(17/6)*b^(1/6
))

Rule 457

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d
)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b*e*n*(p + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b
*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] &&
LeQ[-1, m, -(n*(p + 1))]))

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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 209

Int[((a_) + (b_.)*(x_)^(n_))^(-1), x_Symbol] :> Module[{r = Numerator[Rt[a/b, n]], s = Denominator[Rt[a/b, n]]
, k, u, v}, Simp[u = Int[(r - s*Cos[((2*k - 1)*Pi)/n]*x)/(r^2 - 2*r*s*Cos[((2*k - 1)*Pi)/n]*x + s^2*x^2), x] +
Int[(r + s*Cos[((2*k - 1)*Pi)/n]*x)/(r^2 + 2*r*s*Cos[((2*k - 1)*Pi)/n]*x + s^2*x^2), x]; (2*r^2*Int[1/(r^2 +
s^2*x^2), x])/(a*n) + Dist[(2*r)/(a*n), Sum[u, {k, 1, (n - 2)/4}], x], x]] /; FreeQ[{a, b}, x] && IGtQ[(n - 2)
/4, 0] && PosQ[a/b]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; 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 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 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{A+B x^3}{x^{7/2} \left (a+b x^3\right )^2} \, dx &=\frac{A b-a B}{3 a b x^{5/2} \left (a+b x^3\right )}+\frac{\left (\frac{11 A b}{2}-\frac{5 a B}{2}\right ) \int \frac{1}{x^{7/2} \left (a+b x^3\right )} \, dx}{3 a b}\\ &=-\frac{11 A b-5 a B}{15 a^2 b x^{5/2}}+\frac{A b-a B}{3 a b x^{5/2} \left (a+b x^3\right )}-\frac{(11 A b-5 a B) \int \frac{1}{\sqrt{x} \left (a+b x^3\right )} \, dx}{6 a^2}\\ &=-\frac{11 A b-5 a B}{15 a^2 b x^{5/2}}+\frac{A b-a B}{3 a b x^{5/2} \left (a+b x^3\right )}-\frac{(11 A b-5 a B) \operatorname{Subst}\left (\int \frac{1}{a+b x^6} \, dx,x,\sqrt{x}\right )}{3 a^2}\\ &=-\frac{11 A b-5 a B}{15 a^2 b x^{5/2}}+\frac{A b-a B}{3 a b x^{5/2} \left (a+b x^3\right )}-\frac{(11 A b-5 a B) \operatorname{Subst}\left (\int \frac{\sqrt [6]{a}-\frac{1}{2} \sqrt{3} \sqrt [6]{b} x}{\sqrt [3]{a}-\sqrt{3} \sqrt [6]{a} \sqrt [6]{b} x+\sqrt [3]{b} x^2} \, dx,x,\sqrt{x}\right )}{9 a^{17/6}}-\frac{(11 A b-5 a B) \operatorname{Subst}\left (\int \frac{\sqrt [6]{a}+\frac{1}{2} \sqrt{3} \sqrt [6]{b} x}{\sqrt [3]{a}+\sqrt{3} \sqrt [6]{a} \sqrt [6]{b} x+\sqrt [3]{b} x^2} \, dx,x,\sqrt{x}\right )}{9 a^{17/6}}-\frac{(11 A b-5 a B) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{a}+\sqrt [3]{b} x^2} \, dx,x,\sqrt{x}\right )}{9 a^{8/3}}\\ &=-\frac{11 A b-5 a B}{15 a^2 b x^{5/2}}+\frac{A b-a B}{3 a b x^{5/2} \left (a+b x^3\right )}-\frac{(11 A b-5 a B) \tan ^{-1}\left (\frac{\sqrt [6]{b} \sqrt{x}}{\sqrt [6]{a}}\right )}{9 a^{17/6} \sqrt [6]{b}}-\frac{(11 A b-5 a B) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{a}-\sqrt{3} \sqrt [6]{a} \sqrt [6]{b} x+\sqrt [3]{b} x^2} \, dx,x,\sqrt{x}\right )}{36 a^{8/3}}-\frac{(11 A b-5 a B) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{a}+\sqrt{3} \sqrt [6]{a} \sqrt [6]{b} x+\sqrt [3]{b} x^2} \, dx,x,\sqrt{x}\right )}{36 a^{8/3}}+\frac{(11 A b-5 a B) \operatorname{Subst}\left (\int \frac{-\sqrt{3} \sqrt [6]{a} \sqrt [6]{b}+2 \sqrt [3]{b} x}{\sqrt [3]{a}-\sqrt{3} \sqrt [6]{a} \sqrt [6]{b} x+\sqrt [3]{b} x^2} \, dx,x,\sqrt{x}\right )}{12 \sqrt{3} a^{17/6} \sqrt [6]{b}}-\frac{(11 A b-5 a B) \operatorname{Subst}\left (\int \frac{\sqrt{3} \sqrt [6]{a} \sqrt [6]{b}+2 \sqrt [3]{b} x}{\sqrt [3]{a}+\sqrt{3} \sqrt [6]{a} \sqrt [6]{b} x+\sqrt [3]{b} x^2} \, dx,x,\sqrt{x}\right )}{12 \sqrt{3} a^{17/6} \sqrt [6]{b}}\\ &=-\frac{11 A b-5 a B}{15 a^2 b x^{5/2}}+\frac{A b-a B}{3 a b x^{5/2} \left (a+b x^3\right )}-\frac{(11 A b-5 a B) \tan ^{-1}\left (\frac{\sqrt [6]{b} \sqrt{x}}{\sqrt [6]{a}}\right )}{9 a^{17/6} \sqrt [6]{b}}+\frac{(11 A b-5 a B) \log \left (\sqrt [3]{a}-\sqrt{3} \sqrt [6]{a} \sqrt [6]{b} \sqrt{x}+\sqrt [3]{b} x\right )}{12 \sqrt{3} a^{17/6} \sqrt [6]{b}}-\frac{(11 A b-5 a B) \log \left (\sqrt [3]{a}+\sqrt{3} \sqrt [6]{a} \sqrt [6]{b} \sqrt{x}+\sqrt [3]{b} x\right )}{12 \sqrt{3} a^{17/6} \sqrt [6]{b}}-\frac{(11 A b-5 a B) \operatorname{Subst}\left (\int \frac{1}{-\frac{1}{3}-x^2} \, dx,x,1-\frac{2 \sqrt [6]{b} \sqrt{x}}{\sqrt{3} \sqrt [6]{a}}\right )}{18 \sqrt{3} a^{17/6} \sqrt [6]{b}}+\frac{(11 A b-5 a B) \operatorname{Subst}\left (\int \frac{1}{-\frac{1}{3}-x^2} \, dx,x,1+\frac{2 \sqrt [6]{b} \sqrt{x}}{\sqrt{3} \sqrt [6]{a}}\right )}{18 \sqrt{3} a^{17/6} \sqrt [6]{b}}\\ &=-\frac{11 A b-5 a B}{15 a^2 b x^{5/2}}+\frac{A b-a B}{3 a b x^{5/2} \left (a+b x^3\right )}+\frac{(11 A b-5 a B) \tan ^{-1}\left (\sqrt{3}-\frac{2 \sqrt [6]{b} \sqrt{x}}{\sqrt [6]{a}}\right )}{18 a^{17/6} \sqrt [6]{b}}-\frac{(11 A b-5 a B) \tan ^{-1}\left (\sqrt{3}+\frac{2 \sqrt [6]{b} \sqrt{x}}{\sqrt [6]{a}}\right )}{18 a^{17/6} \sqrt [6]{b}}-\frac{(11 A b-5 a B) \tan ^{-1}\left (\frac{\sqrt [6]{b} \sqrt{x}}{\sqrt [6]{a}}\right )}{9 a^{17/6} \sqrt [6]{b}}+\frac{(11 A b-5 a B) \log \left (\sqrt [3]{a}-\sqrt{3} \sqrt [6]{a} \sqrt [6]{b} \sqrt{x}+\sqrt [3]{b} x\right )}{12 \sqrt{3} a^{17/6} \sqrt [6]{b}}-\frac{(11 A b-5 a B) \log \left (\sqrt [3]{a}+\sqrt{3} \sqrt [6]{a} \sqrt [6]{b} \sqrt{x}+\sqrt [3]{b} x\right )}{12 \sqrt{3} a^{17/6} \sqrt [6]{b}}\\ \end{align*}

Mathematica [C]  time = 0.0860647, size = 74, normalized size = 0.23 \[ \frac{5 x^3 (5 a B-11 A b) \, _2F_1\left (\frac{1}{6},1;\frac{7}{6};-\frac{b x^3}{a}\right )+\frac{a \left (-6 a A+5 a B x^3-11 A b x^3\right )}{a+b x^3}}{15 a^3 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^3)/(x^(7/2)*(a + b*x^3)^2),x]

[Out]

((a*(-6*a*A - 11*A*b*x^3 + 5*a*B*x^3))/(a + b*x^3) + 5*(-11*A*b + 5*a*B)*x^3*Hypergeometric2F1[1/6, 1, 7/6, -(
(b*x^3)/a)])/(15*a^3*x^(5/2))

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Maple [A]  time = 0.039, size = 389, normalized size = 1.2 \begin{align*} -{\frac{Ab}{3\,{a}^{2} \left ( b{x}^{3}+a \right ) }\sqrt{x}}+{\frac{B}{3\,a \left ( b{x}^{3}+a \right ) }\sqrt{x}}-{\frac{11\,Ab}{9\,{a}^{3}}\sqrt [6]{{\frac{a}{b}}}\arctan \left ({\sqrt{x}{\frac{1}{\sqrt [6]{{\frac{a}{b}}}}}} \right ) }+{\frac{11\,Ab\sqrt{3}}{36\,{a}^{3}}\sqrt [6]{{\frac{a}{b}}}\ln \left ( x-\sqrt{3}\sqrt [6]{{\frac{a}{b}}}\sqrt{x}+\sqrt [3]{{\frac{a}{b}}} \right ) }-{\frac{11\,Ab}{18\,{a}^{3}}\sqrt [6]{{\frac{a}{b}}}\arctan \left ( 2\,{\sqrt{x}{\frac{1}{\sqrt [6]{{\frac{a}{b}}}}}}-\sqrt{3} \right ) }-{\frac{11\,Ab\sqrt{3}}{36\,{a}^{3}}\sqrt [6]{{\frac{a}{b}}}\ln \left ( x+\sqrt{3}\sqrt [6]{{\frac{a}{b}}}\sqrt{x}+\sqrt [3]{{\frac{a}{b}}} \right ) }-{\frac{11\,Ab}{18\,{a}^{3}}\sqrt [6]{{\frac{a}{b}}}\arctan \left ( 2\,{\sqrt{x}{\frac{1}{\sqrt [6]{{\frac{a}{b}}}}}}+\sqrt{3} \right ) }+{\frac{5\,B}{9\,{a}^{2}}\sqrt [6]{{\frac{a}{b}}}\arctan \left ({\sqrt{x}{\frac{1}{\sqrt [6]{{\frac{a}{b}}}}}} \right ) }-{\frac{5\,B\sqrt{3}}{36\,{a}^{2}}\sqrt [6]{{\frac{a}{b}}}\ln \left ( x-\sqrt{3}\sqrt [6]{{\frac{a}{b}}}\sqrt{x}+\sqrt [3]{{\frac{a}{b}}} \right ) }+{\frac{5\,B}{18\,{a}^{2}}\sqrt [6]{{\frac{a}{b}}}\arctan \left ( 2\,{\sqrt{x}{\frac{1}{\sqrt [6]{{\frac{a}{b}}}}}}-\sqrt{3} \right ) }+{\frac{5\,B\sqrt{3}}{36\,{a}^{2}}\sqrt [6]{{\frac{a}{b}}}\ln \left ( x+\sqrt{3}\sqrt [6]{{\frac{a}{b}}}\sqrt{x}+\sqrt [3]{{\frac{a}{b}}} \right ) }+{\frac{5\,B}{18\,{a}^{2}}\sqrt [6]{{\frac{a}{b}}}\arctan \left ( 2\,{\sqrt{x}{\frac{1}{\sqrt [6]{{\frac{a}{b}}}}}}+\sqrt{3} \right ) }-{\frac{2\,A}{5\,{a}^{2}}{x}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^3+A)/x^(7/2)/(b*x^3+a)^2,x)

[Out]

-1/3/a^2*x^(1/2)/(b*x^3+a)*A*b+1/3/a*x^(1/2)/(b*x^3+a)*B-11/9/a^3*A*b*(a/b)^(1/6)*arctan(x^(1/2)/(a/b)^(1/6))+
11/36/a^3*A*b*3^(1/2)*(a/b)^(1/6)*ln(x-3^(1/2)*(a/b)^(1/6)*x^(1/2)+(a/b)^(1/3))-11/18/a^3*A*b*(a/b)^(1/6)*arct
an(2*x^(1/2)/(a/b)^(1/6)-3^(1/2))-11/36/a^3*A*b*3^(1/2)*(a/b)^(1/6)*ln(x+3^(1/2)*(a/b)^(1/6)*x^(1/2)+(a/b)^(1/
3))-11/18/a^3*A*b*(a/b)^(1/6)*arctan(2*x^(1/2)/(a/b)^(1/6)+3^(1/2))+5/9/a^2*B*(a/b)^(1/6)*arctan(x^(1/2)/(a/b)
^(1/6))-5/36/a^2*B*3^(1/2)*(a/b)^(1/6)*ln(x-3^(1/2)*(a/b)^(1/6)*x^(1/2)+(a/b)^(1/3))+5/18/a^2*B*(a/b)^(1/6)*ar
ctan(2*x^(1/2)/(a/b)^(1/6)-3^(1/2))+5/36/a^2*B*3^(1/2)*(a/b)^(1/6)*ln(x+3^(1/2)*(a/b)^(1/6)*x^(1/2)+(a/b)^(1/3
))+5/18/a^2*B*(a/b)^(1/6)*arctan(2*x^(1/2)/(a/b)^(1/6)+3^(1/2))-2/5*A/a^2/x^(5/2)

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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((B*x^3+A)/x^(7/2)/(b*x^3+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.32402, size = 6660, normalized size = 20.94 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/x^(7/2)/(b*x^3+a)^2,x, algorithm="fricas")

[Out]

-1/180*(20*sqrt(3)*(a^2*b*x^6 + a^3*x^3)*(-(15625*B^6*a^6 - 206250*A*B^5*a^5*b + 1134375*A^2*B^4*a^4*b^2 - 332
7500*A^3*B^3*a^3*b^3 + 5490375*A^4*B^2*a^2*b^4 - 4831530*A^5*B*a*b^5 + 1771561*A^6*b^6)/(a^17*b))^(1/6)*arctan
(1/3*(2*sqrt(3)*sqrt(a^6*(-(15625*B^6*a^6 - 206250*A*B^5*a^5*b + 1134375*A^2*B^4*a^4*b^2 - 3327500*A^3*B^3*a^3
*b^3 + 5490375*A^4*B^2*a^2*b^4 - 4831530*A^5*B*a*b^5 + 1771561*A^6*b^6)/(a^17*b))^(1/3) + (25*B^2*a^2 - 110*A*
B*a*b + 121*A^2*b^2)*x + (5*B*a^4 - 11*A*a^3*b)*sqrt(x)*(-(15625*B^6*a^6 - 206250*A*B^5*a^5*b + 1134375*A^2*B^
4*a^4*b^2 - 3327500*A^3*B^3*a^3*b^3 + 5490375*A^4*B^2*a^2*b^4 - 4831530*A^5*B*a*b^5 + 1771561*A^6*b^6)/(a^17*b
))^(1/6))*a^14*b*(-(15625*B^6*a^6 - 206250*A*B^5*a^5*b + 1134375*A^2*B^4*a^4*b^2 - 3327500*A^3*B^3*a^3*b^3 + 5
490375*A^4*B^2*a^2*b^4 - 4831530*A^5*B*a*b^5 + 1771561*A^6*b^6)/(a^17*b))^(5/6) + 2*sqrt(3)*(5*B*a^15*b - 11*A
*a^14*b^2)*sqrt(x)*(-(15625*B^6*a^6 - 206250*A*B^5*a^5*b + 1134375*A^2*B^4*a^4*b^2 - 3327500*A^3*B^3*a^3*b^3 +
5490375*A^4*B^2*a^2*b^4 - 4831530*A^5*B*a*b^5 + 1771561*A^6*b^6)/(a^17*b))^(5/6) - sqrt(3)*(15625*B^6*a^6 - 2
06250*A*B^5*a^5*b + 1134375*A^2*B^4*a^4*b^2 - 3327500*A^3*B^3*a^3*b^3 + 5490375*A^4*B^2*a^2*b^4 - 4831530*A^5*
B*a*b^5 + 1771561*A^6*b^6))/(15625*B^6*a^6 - 206250*A*B^5*a^5*b + 1134375*A^2*B^4*a^4*b^2 - 3327500*A^3*B^3*a^
3*b^3 + 5490375*A^4*B^2*a^2*b^4 - 4831530*A^5*B*a*b^5 + 1771561*A^6*b^6)) + 20*sqrt(3)*(a^2*b*x^6 + a^3*x^3)*(
-(15625*B^6*a^6 - 206250*A*B^5*a^5*b + 1134375*A^2*B^4*a^4*b^2 - 3327500*A^3*B^3*a^3*b^3 + 5490375*A^4*B^2*a^2
*b^4 - 4831530*A^5*B*a*b^5 + 1771561*A^6*b^6)/(a^17*b))^(1/6)*arctan(1/3*(2*sqrt(3)*sqrt(a^6*(-(15625*B^6*a^6
- 206250*A*B^5*a^5*b + 1134375*A^2*B^4*a^4*b^2 - 3327500*A^3*B^3*a^3*b^3 + 5490375*A^4*B^2*a^2*b^4 - 4831530*A
^5*B*a*b^5 + 1771561*A^6*b^6)/(a^17*b))^(1/3) + (25*B^2*a^2 - 110*A*B*a*b + 121*A^2*b^2)*x - (5*B*a^4 - 11*A*a
^3*b)*sqrt(x)*(-(15625*B^6*a^6 - 206250*A*B^5*a^5*b + 1134375*A^2*B^4*a^4*b^2 - 3327500*A^3*B^3*a^3*b^3 + 5490
375*A^4*B^2*a^2*b^4 - 4831530*A^5*B*a*b^5 + 1771561*A^6*b^6)/(a^17*b))^(1/6))*a^14*b*(-(15625*B^6*a^6 - 206250
*A*B^5*a^5*b + 1134375*A^2*B^4*a^4*b^2 - 3327500*A^3*B^3*a^3*b^3 + 5490375*A^4*B^2*a^2*b^4 - 4831530*A^5*B*a*b
^5 + 1771561*A^6*b^6)/(a^17*b))^(5/6) + 2*sqrt(3)*(5*B*a^15*b - 11*A*a^14*b^2)*sqrt(x)*(-(15625*B^6*a^6 - 2062
50*A*B^5*a^5*b + 1134375*A^2*B^4*a^4*b^2 - 3327500*A^3*B^3*a^3*b^3 + 5490375*A^4*B^2*a^2*b^4 - 4831530*A^5*B*a
*b^5 + 1771561*A^6*b^6)/(a^17*b))^(5/6) + sqrt(3)*(15625*B^6*a^6 - 206250*A*B^5*a^5*b + 1134375*A^2*B^4*a^4*b^
2 - 3327500*A^3*B^3*a^3*b^3 + 5490375*A^4*B^2*a^2*b^4 - 4831530*A^5*B*a*b^5 + 1771561*A^6*b^6))/(15625*B^6*a^6
- 206250*A*B^5*a^5*b + 1134375*A^2*B^4*a^4*b^2 - 3327500*A^3*B^3*a^3*b^3 + 5490375*A^4*B^2*a^2*b^4 - 4831530*
A^5*B*a*b^5 + 1771561*A^6*b^6)) - 5*(a^2*b*x^6 + a^3*x^3)*(-(15625*B^6*a^6 - 206250*A*B^5*a^5*b + 1134375*A^2*
B^4*a^4*b^2 - 3327500*A^3*B^3*a^3*b^3 + 5490375*A^4*B^2*a^2*b^4 - 4831530*A^5*B*a*b^5 + 1771561*A^6*b^6)/(a^17
*b))^(1/6)*log(a^6*(-(15625*B^6*a^6 - 206250*A*B^5*a^5*b + 1134375*A^2*B^4*a^4*b^2 - 3327500*A^3*B^3*a^3*b^3 +
5490375*A^4*B^2*a^2*b^4 - 4831530*A^5*B*a*b^5 + 1771561*A^6*b^6)/(a^17*b))^(1/3) + (25*B^2*a^2 - 110*A*B*a*b
+ 121*A^2*b^2)*x + (5*B*a^4 - 11*A*a^3*b)*sqrt(x)*(-(15625*B^6*a^6 - 206250*A*B^5*a^5*b + 1134375*A^2*B^4*a^4*
b^2 - 3327500*A^3*B^3*a^3*b^3 + 5490375*A^4*B^2*a^2*b^4 - 4831530*A^5*B*a*b^5 + 1771561*A^6*b^6)/(a^17*b))^(1/
6)) + 5*(a^2*b*x^6 + a^3*x^3)*(-(15625*B^6*a^6 - 206250*A*B^5*a^5*b + 1134375*A^2*B^4*a^4*b^2 - 3327500*A^3*B^
3*a^3*b^3 + 5490375*A^4*B^2*a^2*b^4 - 4831530*A^5*B*a*b^5 + 1771561*A^6*b^6)/(a^17*b))^(1/6)*log(a^6*(-(15625*
B^6*a^6 - 206250*A*B^5*a^5*b + 1134375*A^2*B^4*a^4*b^2 - 3327500*A^3*B^3*a^3*b^3 + 5490375*A^4*B^2*a^2*b^4 - 4
831530*A^5*B*a*b^5 + 1771561*A^6*b^6)/(a^17*b))^(1/3) + (25*B^2*a^2 - 110*A*B*a*b + 121*A^2*b^2)*x - (5*B*a^4
- 11*A*a^3*b)*sqrt(x)*(-(15625*B^6*a^6 - 206250*A*B^5*a^5*b + 1134375*A^2*B^4*a^4*b^2 - 3327500*A^3*B^3*a^3*b^
3 + 5490375*A^4*B^2*a^2*b^4 - 4831530*A^5*B*a*b^5 + 1771561*A^6*b^6)/(a^17*b))^(1/6)) + 10*(a^2*b*x^6 + a^3*x^
3)*(-(15625*B^6*a^6 - 206250*A*B^5*a^5*b + 1134375*A^2*B^4*a^4*b^2 - 3327500*A^3*B^3*a^3*b^3 + 5490375*A^4*B^2
*a^2*b^4 - 4831530*A^5*B*a*b^5 + 1771561*A^6*b^6)/(a^17*b))^(1/6)*log(a^3*(-(15625*B^6*a^6 - 206250*A*B^5*a^5*
b + 1134375*A^2*B^4*a^4*b^2 - 3327500*A^3*B^3*a^3*b^3 + 5490375*A^4*B^2*a^2*b^4 - 4831530*A^5*B*a*b^5 + 177156
1*A^6*b^6)/(a^17*b))^(1/6) - (5*B*a - 11*A*b)*sqrt(x)) - 10*(a^2*b*x^6 + a^3*x^3)*(-(15625*B^6*a^6 - 206250*A*
B^5*a^5*b + 1134375*A^2*B^4*a^4*b^2 - 3327500*A^3*B^3*a^3*b^3 + 5490375*A^4*B^2*a^2*b^4 - 4831530*A^5*B*a*b^5
+ 1771561*A^6*b^6)/(a^17*b))^(1/6)*log(-a^3*(-(15625*B^6*a^6 - 206250*A*B^5*a^5*b + 1134375*A^2*B^4*a^4*b^2 -
3327500*A^3*B^3*a^3*b^3 + 5490375*A^4*B^2*a^2*b^4 - 4831530*A^5*B*a*b^5 + 1771561*A^6*b^6)/(a^17*b))^(1/6) - (
5*B*a - 11*A*b)*sqrt(x)) - 12*((5*B*a - 11*A*b)*x^3 - 6*A*a)*sqrt(x))/(a^2*b*x^6 + a^3*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((B*x**3+A)/x**(7/2)/(b*x**3+a)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.17935, size = 423, normalized size = 1.33 \begin{align*} \frac{\sqrt{3}{\left (5 \, \left (a b^{5}\right )^{\frac{1}{6}} B a - 11 \, \left (a b^{5}\right )^{\frac{1}{6}} A b\right )} \log \left (\sqrt{3} \sqrt{x} \left (\frac{a}{b}\right )^{\frac{1}{6}} + x + \left (\frac{a}{b}\right )^{\frac{1}{3}}\right )}{36 \, a^{3} b} - \frac{\sqrt{3}{\left (5 \, \left (a b^{5}\right )^{\frac{1}{6}} B a - 11 \, \left (a b^{5}\right )^{\frac{1}{6}} A b\right )} \log \left (-\sqrt{3} \sqrt{x} \left (\frac{a}{b}\right )^{\frac{1}{6}} + x + \left (\frac{a}{b}\right )^{\frac{1}{3}}\right )}{36 \, a^{3} b} + \frac{B a \sqrt{x} - A b \sqrt{x}}{3 \,{\left (b x^{3} + a\right )} a^{2}} + \frac{{\left (5 \, \left (a b^{5}\right )^{\frac{1}{6}} B a - 11 \, \left (a b^{5}\right )^{\frac{1}{6}} A b\right )} \arctan \left (\frac{\sqrt{3} \left (\frac{a}{b}\right )^{\frac{1}{6}} + 2 \, \sqrt{x}}{\left (\frac{a}{b}\right )^{\frac{1}{6}}}\right )}{18 \, a^{3} b} + \frac{{\left (5 \, \left (a b^{5}\right )^{\frac{1}{6}} B a - 11 \, \left (a b^{5}\right )^{\frac{1}{6}} A b\right )} \arctan \left (-\frac{\sqrt{3} \left (\frac{a}{b}\right )^{\frac{1}{6}} - 2 \, \sqrt{x}}{\left (\frac{a}{b}\right )^{\frac{1}{6}}}\right )}{18 \, a^{3} b} + \frac{{\left (5 \, \left (a b^{5}\right )^{\frac{1}{6}} B a - 11 \, \left (a b^{5}\right )^{\frac{1}{6}} A b\right )} \arctan \left (\frac{\sqrt{x}}{\left (\frac{a}{b}\right )^{\frac{1}{6}}}\right )}{9 \, a^{3} b} - \frac{2 \, A}{5 \, a^{2} x^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/x^(7/2)/(b*x^3+a)^2,x, algorithm="giac")

[Out]

1/36*sqrt(3)*(5*(a*b^5)^(1/6)*B*a - 11*(a*b^5)^(1/6)*A*b)*log(sqrt(3)*sqrt(x)*(a/b)^(1/6) + x + (a/b)^(1/3))/(
a^3*b) - 1/36*sqrt(3)*(5*(a*b^5)^(1/6)*B*a - 11*(a*b^5)^(1/6)*A*b)*log(-sqrt(3)*sqrt(x)*(a/b)^(1/6) + x + (a/b
)^(1/3))/(a^3*b) + 1/3*(B*a*sqrt(x) - A*b*sqrt(x))/((b*x^3 + a)*a^2) + 1/18*(5*(a*b^5)^(1/6)*B*a - 11*(a*b^5)^
(1/6)*A*b)*arctan((sqrt(3)*(a/b)^(1/6) + 2*sqrt(x))/(a/b)^(1/6))/(a^3*b) + 1/18*(5*(a*b^5)^(1/6)*B*a - 11*(a*b
^5)^(1/6)*A*b)*arctan(-(sqrt(3)*(a/b)^(1/6) - 2*sqrt(x))/(a/b)^(1/6))/(a^3*b) + 1/9*(5*(a*b^5)^(1/6)*B*a - 11*
(a*b^5)^(1/6)*A*b)*arctan(sqrt(x)/(a/b)^(1/6))/(a^3*b) - 2/5*A/(a^2*x^(5/2))