3.829 \(\int \frac{(a+b x^2)^2 \sqrt{c+d x^2}}{x^{11/2}} \, dx\)

Optimal. Leaf size=386 \[ \frac{2 \sqrt [4]{d} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (a d (6 b c-a d)+15 b^2 c^2\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right ),\frac{1}{2}\right )}{15 c^{7/4} \sqrt{c+d x^2}}-\frac{2 a^2 \left (c+d x^2\right )^{3/2}}{9 c x^{9/2}}-\frac{2 \sqrt{c+d x^2} \left (a d (6 b c-a d)+15 b^2 c^2\right )}{15 c^2 \sqrt{x}}+\frac{4 \sqrt{d} \sqrt{x} \sqrt{c+d x^2} \left (a d (6 b c-a d)+15 b^2 c^2\right )}{15 c^2 \left (\sqrt{c}+\sqrt{d} x\right )}-\frac{4 \sqrt [4]{d} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (a d (6 b c-a d)+15 b^2 c^2\right ) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )|\frac{1}{2}\right )}{15 c^{7/4} \sqrt{c+d x^2}}-\frac{2 a \left (c+d x^2\right )^{3/2} (6 b c-a d)}{15 c^2 x^{5/2}} \]

[Out]

(-2*(15*b^2*c^2 + a*d*(6*b*c - a*d))*Sqrt[c + d*x^2])/(15*c^2*Sqrt[x]) + (4*Sqrt[d]*(15*b^2*c^2 + a*d*(6*b*c -
a*d))*Sqrt[x]*Sqrt[c + d*x^2])/(15*c^2*(Sqrt[c] + Sqrt[d]*x)) - (2*a^2*(c + d*x^2)^(3/2))/(9*c*x^(9/2)) - (2*
a*(6*b*c - a*d)*(c + d*x^2)^(3/2))/(15*c^2*x^(5/2)) - (4*d^(1/4)*(15*b^2*c^2 + a*d*(6*b*c - a*d))*(Sqrt[c] + S
qrt[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*x)^2]*EllipticE[2*ArcTan[(d^(1/4)*Sqrt[x])/c^(1/4)], 1/2])/(15*c
^(7/4)*Sqrt[c + d*x^2]) + (2*d^(1/4)*(15*b^2*c^2 + a*d*(6*b*c - a*d))*(Sqrt[c] + Sqrt[d]*x)*Sqrt[(c + d*x^2)/(
Sqrt[c] + Sqrt[d]*x)^2]*EllipticF[2*ArcTan[(d^(1/4)*Sqrt[x])/c^(1/4)], 1/2])/(15*c^(7/4)*Sqrt[c + d*x^2])

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Rubi [A]  time = 0.323593, antiderivative size = 383, normalized size of antiderivative = 0.99, number of steps used = 7, number of rules used = 7, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.269, Rules used = {462, 453, 277, 329, 305, 220, 1196} \[ -\frac{2 a^2 \left (c+d x^2\right )^{3/2}}{9 c x^{9/2}}-\frac{2 \sqrt{c+d x^2} \left (\frac{a d (6 b c-a d)}{c^2}+15 b^2\right )}{15 \sqrt{x}}+\frac{4 \sqrt{d} \sqrt{x} \sqrt{c+d x^2} \left (a d (6 b c-a d)+15 b^2 c^2\right )}{15 c^2 \left (\sqrt{c}+\sqrt{d} x\right )}+\frac{2 \sqrt [4]{d} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (a d (6 b c-a d)+15 b^2 c^2\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )|\frac{1}{2}\right )}{15 c^{7/4} \sqrt{c+d x^2}}-\frac{4 \sqrt [4]{d} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (a d (6 b c-a d)+15 b^2 c^2\right ) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )|\frac{1}{2}\right )}{15 c^{7/4} \sqrt{c+d x^2}}-\frac{2 a \left (c+d x^2\right )^{3/2} (6 b c-a d)}{15 c^2 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)^2*Sqrt[c + d*x^2])/x^(11/2),x]

[Out]

(-2*(15*b^2 + (a*d*(6*b*c - a*d))/c^2)*Sqrt[c + d*x^2])/(15*Sqrt[x]) + (4*Sqrt[d]*(15*b^2*c^2 + a*d*(6*b*c - a
*d))*Sqrt[x]*Sqrt[c + d*x^2])/(15*c^2*(Sqrt[c] + Sqrt[d]*x)) - (2*a^2*(c + d*x^2)^(3/2))/(9*c*x^(9/2)) - (2*a*
(6*b*c - a*d)*(c + d*x^2)^(3/2))/(15*c^2*x^(5/2)) - (4*d^(1/4)*(15*b^2*c^2 + a*d*(6*b*c - a*d))*(Sqrt[c] + Sqr
t[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*x)^2]*EllipticE[2*ArcTan[(d^(1/4)*Sqrt[x])/c^(1/4)], 1/2])/(15*c^(
7/4)*Sqrt[c + d*x^2]) + (2*d^(1/4)*(15*b^2*c^2 + a*d*(6*b*c - a*d))*(Sqrt[c] + Sqrt[d]*x)*Sqrt[(c + d*x^2)/(Sq
rt[c] + Sqrt[d]*x)^2]*EllipticF[2*ArcTan[(d^(1/4)*Sqrt[x])/c^(1/4)], 1/2])/(15*c^(7/4)*Sqrt[c + d*x^2])

Rule 462

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

Rule 453

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

Rule 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && 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 305

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, Int[1/Sqrt[a + b*x^4], x],
x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

\begin{align*} \int \frac{\left (a+b x^2\right )^2 \sqrt{c+d x^2}}{x^{11/2}} \, dx &=-\frac{2 a^2 \left (c+d x^2\right )^{3/2}}{9 c x^{9/2}}+\frac{2 \int \frac{\left (\frac{3}{2} a (6 b c-a d)+\frac{9}{2} b^2 c x^2\right ) \sqrt{c+d x^2}}{x^{7/2}} \, dx}{9 c}\\ &=-\frac{2 a^2 \left (c+d x^2\right )^{3/2}}{9 c x^{9/2}}-\frac{2 a (6 b c-a d) \left (c+d x^2\right )^{3/2}}{15 c^2 x^{5/2}}+\frac{1}{15} \left (15 b^2+\frac{a d (6 b c-a d)}{c^2}\right ) \int \frac{\sqrt{c+d x^2}}{x^{3/2}} \, dx\\ &=-\frac{2 \left (15 b^2+\frac{a d (6 b c-a d)}{c^2}\right ) \sqrt{c+d x^2}}{15 \sqrt{x}}-\frac{2 a^2 \left (c+d x^2\right )^{3/2}}{9 c x^{9/2}}-\frac{2 a (6 b c-a d) \left (c+d x^2\right )^{3/2}}{15 c^2 x^{5/2}}+\frac{1}{15} \left (2 d \left (15 b^2+\frac{a d (6 b c-a d)}{c^2}\right )\right ) \int \frac{\sqrt{x}}{\sqrt{c+d x^2}} \, dx\\ &=-\frac{2 \left (15 b^2+\frac{a d (6 b c-a d)}{c^2}\right ) \sqrt{c+d x^2}}{15 \sqrt{x}}-\frac{2 a^2 \left (c+d x^2\right )^{3/2}}{9 c x^{9/2}}-\frac{2 a (6 b c-a d) \left (c+d x^2\right )^{3/2}}{15 c^2 x^{5/2}}+\frac{1}{15} \left (4 d \left (15 b^2+\frac{a d (6 b c-a d)}{c^2}\right )\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{c+d x^4}} \, dx,x,\sqrt{x}\right )\\ &=-\frac{2 \left (15 b^2+\frac{a d (6 b c-a d)}{c^2}\right ) \sqrt{c+d x^2}}{15 \sqrt{x}}-\frac{2 a^2 \left (c+d x^2\right )^{3/2}}{9 c x^{9/2}}-\frac{2 a (6 b c-a d) \left (c+d x^2\right )^{3/2}}{15 c^2 x^{5/2}}+\frac{1}{15} \left (4 \sqrt{c} \sqrt{d} \left (15 b^2+\frac{a d (6 b c-a d)}{c^2}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c+d x^4}} \, dx,x,\sqrt{x}\right )-\frac{1}{15} \left (4 \sqrt{c} \sqrt{d} \left (15 b^2+\frac{a d (6 b c-a d)}{c^2}\right )\right ) \operatorname{Subst}\left (\int \frac{1-\frac{\sqrt{d} x^2}{\sqrt{c}}}{\sqrt{c+d x^4}} \, dx,x,\sqrt{x}\right )\\ &=-\frac{2 \left (15 b^2+\frac{a d (6 b c-a d)}{c^2}\right ) \sqrt{c+d x^2}}{15 \sqrt{x}}+\frac{4 \sqrt{d} \left (15 b^2+\frac{a d (6 b c-a d)}{c^2}\right ) \sqrt{x} \sqrt{c+d x^2}}{15 \left (\sqrt{c}+\sqrt{d} x\right )}-\frac{2 a^2 \left (c+d x^2\right )^{3/2}}{9 c x^{9/2}}-\frac{2 a (6 b c-a d) \left (c+d x^2\right )^{3/2}}{15 c^2 x^{5/2}}-\frac{4 \sqrt [4]{c} \sqrt [4]{d} \left (15 b^2+\frac{a d (6 b c-a d)}{c^2}\right ) \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )|\frac{1}{2}\right )}{15 \sqrt{c+d x^2}}+\frac{2 \sqrt [4]{c} \sqrt [4]{d} \left (15 b^2+\frac{a d (6 b c-a d)}{c^2}\right ) \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )|\frac{1}{2}\right )}{15 \sqrt{c+d x^2}}\\ \end{align*}

Mathematica [C]  time = 0.141972, size = 148, normalized size = 0.38 \[ \frac{12 d x^6 \sqrt{\frac{c}{d x^2}+1} \left (-a^2 d^2+6 a b c d+15 b^2 c^2\right ) \, _2F_1\left (-\frac{1}{4},\frac{1}{2};\frac{3}{4};-\frac{c}{d x^2}\right )-2 \left (c+d x^2\right ) \left (a^2 \left (5 c^2+2 c d x^2-6 d^2 x^4\right )+18 a b c x^2 \left (c+2 d x^2\right )+45 b^2 c^2 x^4\right )}{45 c^2 x^{9/2} \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)^2*Sqrt[c + d*x^2])/x^(11/2),x]

[Out]

(-2*(c + d*x^2)*(45*b^2*c^2*x^4 + 18*a*b*c*x^2*(c + 2*d*x^2) + a^2*(5*c^2 + 2*c*d*x^2 - 6*d^2*x^4)) + 12*d*(15
*b^2*c^2 + 6*a*b*c*d - a^2*d^2)*Sqrt[1 + c/(d*x^2)]*x^6*Hypergeometric2F1[-1/4, 1/2, 3/4, -(c/(d*x^2))])/(45*c
^2*x^(9/2)*Sqrt[c + d*x^2])

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Maple [A]  time = 0.05, size = 659, normalized size = 1.7 \begin{align*} -{\frac{2}{45\,{c}^{2}} \left ( 6\,\sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{2}\sqrt{{\frac{-dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{-{\frac{dx}{\sqrt{-cd}}}}{\it EllipticE} \left ( \sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}},1/2\,\sqrt{2} \right ){x}^{4}{a}^{2}c{d}^{2}-36\,\sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{2}\sqrt{{\frac{-dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{-{\frac{dx}{\sqrt{-cd}}}}{\it EllipticE} \left ( \sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}},1/2\,\sqrt{2} \right ){x}^{4}ab{c}^{2}d-90\,\sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{2}\sqrt{{\frac{-dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{-{\frac{dx}{\sqrt{-cd}}}}{\it EllipticE} \left ( \sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}},1/2\,\sqrt{2} \right ){x}^{4}{b}^{2}{c}^{3}-3\,\sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{2}\sqrt{{\frac{-dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{-{\frac{dx}{\sqrt{-cd}}}}{\it EllipticF} \left ( \sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}},1/2\,\sqrt{2} \right ){x}^{4}{a}^{2}c{d}^{2}+18\,\sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{2}\sqrt{{\frac{-dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{-{\frac{dx}{\sqrt{-cd}}}}{\it EllipticF} \left ( \sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}},1/2\,\sqrt{2} \right ){x}^{4}ab{c}^{2}d+45\,\sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{2}\sqrt{{\frac{-dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{-{\frac{dx}{\sqrt{-cd}}}}{\it EllipticF} \left ( \sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}},1/2\,\sqrt{2} \right ){x}^{4}{b}^{2}{c}^{3}-6\,{x}^{6}{a}^{2}{d}^{3}+36\,{x}^{6}abc{d}^{2}+45\,{x}^{6}{b}^{2}{c}^{2}d-4\,{x}^{4}{a}^{2}c{d}^{2}+54\,{x}^{4}ab{c}^{2}d+45\,{x}^{4}{b}^{2}{c}^{3}+7\,{x}^{2}{a}^{2}{c}^{2}d+18\,{x}^{2}ab{c}^{3}+5\,{a}^{2}{c}^{3} \right ){\frac{1}{\sqrt{d{x}^{2}+c}}}{x}^{-{\frac{9}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2*(d*x^2+c)^(1/2)/x^(11/2),x)

[Out]

-2/45*(6*((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*2^(1/2)*((-d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*(-x/(-c*d)^(
1/2)*d)^(1/2)*EllipticE(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*x^4*a^2*c*d^2-36*((d*x+(-c*d)^(1/
2))/(-c*d)^(1/2))^(1/2)*2^(1/2)*((-d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*(-x/(-c*d)^(1/2)*d)^(1/2)*EllipticE((
(d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*x^4*a*b*c^2*d-90*((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*2
^(1/2)*((-d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*(-x/(-c*d)^(1/2)*d)^(1/2)*EllipticE(((d*x+(-c*d)^(1/2))/(-c*d)
^(1/2))^(1/2),1/2*2^(1/2))*x^4*b^2*c^3-3*((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*2^(1/2)*((-d*x+(-c*d)^(1/2))/
(-c*d)^(1/2))^(1/2)*(-x/(-c*d)^(1/2)*d)^(1/2)*EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*x
^4*a^2*c*d^2+18*((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*2^(1/2)*((-d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*(-x/(
-c*d)^(1/2)*d)^(1/2)*EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*x^4*a*b*c^2*d+45*((d*x+(-c
*d)^(1/2))/(-c*d)^(1/2))^(1/2)*2^(1/2)*((-d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*(-x/(-c*d)^(1/2)*d)^(1/2)*Elli
pticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*x^4*b^2*c^3-6*x^6*a^2*d^3+36*x^6*a*b*c*d^2+45*x^6*b
^2*c^2*d-4*x^4*a^2*c*d^2+54*x^4*a*b*c^2*d+45*x^4*b^2*c^3+7*x^2*a^2*c^2*d+18*x^2*a*b*c^3+5*a^2*c^3)/(d*x^2+c)^(
1/2)/x^(9/2)/c^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{2} + a\right )}^{2} \sqrt{d x^{2} + c}}{x^{\frac{11}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(1/2)/x^(11/2),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^2*sqrt(d*x^2 + c)/x^(11/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \sqrt{d x^{2} + c}}{x^{\frac{11}{2}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(1/2)/x^(11/2),x, algorithm="fricas")

[Out]

integral((b^2*x^4 + 2*a*b*x^2 + a^2)*sqrt(d*x^2 + c)/x^(11/2), 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((b*x**2+a)**2*(d*x**2+c)**(1/2)/x**(11/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{2} + a\right )}^{2} \sqrt{d x^{2} + c}}{x^{\frac{11}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(1/2)/x^(11/2),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^2*sqrt(d*x^2 + c)/x^(11/2), x)