3.831 \(\int \frac {(a+b x^2)^2 \sqrt {c+d x^2}}{x^{15/2}} \, dx\)

Optimal. Leaf size=441 \[ -\frac {2 \sqrt {c+d x^2} \left (7 a^2 d^2-26 a b c d+39 b^2 c^2\right )}{195 c^2 x^{5/2}}+\frac {2 d^{5/4} \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (7 a^2 d^2-26 a b c d+39 b^2 c^2\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )|\frac {1}{2}\right )}{195 c^{11/4} \sqrt {c+d x^2}}-\frac {4 d^{5/4} \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (7 a^2 d^2-26 a b c d+39 b^2 c^2\right ) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )|\frac {1}{2}\right )}{195 c^{11/4} \sqrt {c+d x^2}}-\frac {4 d \sqrt {c+d x^2} \left (7 a^2 d^2-26 a b c d+39 b^2 c^2\right )}{195 c^3 \sqrt {x}}+\frac {4 d^{3/2} \sqrt {x} \sqrt {c+d x^2} \left (7 a^2 d^2-26 a b c d+39 b^2 c^2\right )}{195 c^3 \left (\sqrt {c}+\sqrt {d} x\right )}-\frac {2 a^2 \left (c+d x^2\right )^{3/2}}{13 c x^{13/2}}-\frac {2 a \left (c+d x^2\right )^{3/2} (26 b c-7 a d)}{117 c^2 x^{9/2}} \]

[Out]

-2/13*a^2*(d*x^2+c)^(3/2)/c/x^(13/2)-2/117*a*(-7*a*d+26*b*c)*(d*x^2+c)^(3/2)/c^2/x^(9/2)-2/195*(7*a^2*d^2-26*a
*b*c*d+39*b^2*c^2)*(d*x^2+c)^(1/2)/c^2/x^(5/2)-4/195*d*(7*a^2*d^2-26*a*b*c*d+39*b^2*c^2)*(d*x^2+c)^(1/2)/c^3/x
^(1/2)+4/195*d^(3/2)*(7*a^2*d^2-26*a*b*c*d+39*b^2*c^2)*x^(1/2)*(d*x^2+c)^(1/2)/c^3/(c^(1/2)+x*d^(1/2))-4/195*d
^(5/4)*(7*a^2*d^2-26*a*b*c*d+39*b^2*c^2)*(cos(2*arctan(d^(1/4)*x^(1/2)/c^(1/4)))^2)^(1/2)/cos(2*arctan(d^(1/4)
*x^(1/2)/c^(1/4)))*EllipticE(sin(2*arctan(d^(1/4)*x^(1/2)/c^(1/4))),1/2*2^(1/2))*(c^(1/2)+x*d^(1/2))*((d*x^2+c
)/(c^(1/2)+x*d^(1/2))^2)^(1/2)/c^(11/4)/(d*x^2+c)^(1/2)+2/195*d^(5/4)*(7*a^2*d^2-26*a*b*c*d+39*b^2*c^2)*(cos(2
*arctan(d^(1/4)*x^(1/2)/c^(1/4)))^2)^(1/2)/cos(2*arctan(d^(1/4)*x^(1/2)/c^(1/4)))*EllipticF(sin(2*arctan(d^(1/
4)*x^(1/2)/c^(1/4))),1/2*2^(1/2))*(c^(1/2)+x*d^(1/2))*((d*x^2+c)/(c^(1/2)+x*d^(1/2))^2)^(1/2)/c^(11/4)/(d*x^2+
c)^(1/2)

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Rubi [A]  time = 0.39, antiderivative size = 437, normalized size of antiderivative = 0.99, number of steps used = 8, number of rules used = 8, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {462, 453, 277, 325, 329, 305, 220, 1196} \[ -\frac {4 d \sqrt {c+d x^2} \left (7 a^2 d^2-26 a b c d+39 b^2 c^2\right )}{195 c^3 \sqrt {x}}+\frac {4 d^{3/2} \sqrt {x} \sqrt {c+d x^2} \left (7 a^2 d^2-26 a b c d+39 b^2 c^2\right )}{195 c^3 \left (\sqrt {c}+\sqrt {d} x\right )}+\frac {2 d^{5/4} \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (7 a^2 d^2-26 a b c d+39 b^2 c^2\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )|\frac {1}{2}\right )}{195 c^{11/4} \sqrt {c+d x^2}}-\frac {4 d^{5/4} \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (7 a^2 d^2-26 a b c d+39 b^2 c^2\right ) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )|\frac {1}{2}\right )}{195 c^{11/4} \sqrt {c+d x^2}}-\frac {2 a^2 \left (c+d x^2\right )^{3/2}}{13 c x^{13/2}}-\frac {2 \sqrt {c+d x^2} \left (39 b^2-\frac {a d (26 b c-7 a d)}{c^2}\right )}{195 x^{5/2}}-\frac {2 a \left (c+d x^2\right )^{3/2} (26 b c-7 a d)}{117 c^2 x^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)^2*Sqrt[c + d*x^2])/x^(15/2),x]

[Out]

(-2*(39*b^2 - (a*d*(26*b*c - 7*a*d))/c^2)*Sqrt[c + d*x^2])/(195*x^(5/2)) - (4*d*(39*b^2*c^2 - 26*a*b*c*d + 7*a
^2*d^2)*Sqrt[c + d*x^2])/(195*c^3*Sqrt[x]) + (4*d^(3/2)*(39*b^2*c^2 - 26*a*b*c*d + 7*a^2*d^2)*Sqrt[x]*Sqrt[c +
d*x^2])/(195*c^3*(Sqrt[c] + Sqrt[d]*x)) - (2*a^2*(c + d*x^2)^(3/2))/(13*c*x^(13/2)) - (2*a*(26*b*c - 7*a*d)*(
c + d*x^2)^(3/2))/(117*c^2*x^(9/2)) - (4*d^(5/4)*(39*b^2*c^2 - 26*a*b*c*d + 7*a^2*d^2)*(Sqrt[c] + Sqrt[d]*x)*S
qrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*x)^2]*EllipticE[2*ArcTan[(d^(1/4)*Sqrt[x])/c^(1/4)], 1/2])/(195*c^(11/4)*Sq
rt[c + d*x^2]) + (2*d^(5/4)*(39*b^2*c^2 - 26*a*b*c*d + 7*a^2*d^2)*(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])/(195*c^(11/4)*Sqrt[c + d*x^2])

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 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 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 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 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 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 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^{15/2}} \, dx &=-\frac {2 a^2 \left (c+d x^2\right )^{3/2}}{13 c x^{13/2}}+\frac {2 \int \frac {\left (\frac {1}{2} a (26 b c-7 a d)+\frac {13}{2} b^2 c x^2\right ) \sqrt {c+d x^2}}{x^{11/2}} \, dx}{13 c}\\ &=-\frac {2 a^2 \left (c+d x^2\right )^{3/2}}{13 c x^{13/2}}-\frac {2 a (26 b c-7 a d) \left (c+d x^2\right )^{3/2}}{117 c^2 x^{9/2}}-\frac {1}{39} \left (-39 b^2+\frac {a d (26 b c-7 a d)}{c^2}\right ) \int \frac {\sqrt {c+d x^2}}{x^{7/2}} \, dx\\ &=-\frac {2 \left (39 b^2-\frac {a d (26 b c-7 a d)}{c^2}\right ) \sqrt {c+d x^2}}{195 x^{5/2}}-\frac {2 a^2 \left (c+d x^2\right )^{3/2}}{13 c x^{13/2}}-\frac {2 a (26 b c-7 a d) \left (c+d x^2\right )^{3/2}}{117 c^2 x^{9/2}}+\frac {1}{195} \left (2 d \left (39 b^2-\frac {a d (26 b c-7 a d)}{c^2}\right )\right ) \int \frac {1}{x^{3/2} \sqrt {c+d x^2}} \, dx\\ &=-\frac {2 \left (39 b^2-\frac {a d (26 b c-7 a d)}{c^2}\right ) \sqrt {c+d x^2}}{195 x^{5/2}}-\frac {4 d \left (39 b^2 c^2-26 a b c d+7 a^2 d^2\right ) \sqrt {c+d x^2}}{195 c^3 \sqrt {x}}-\frac {2 a^2 \left (c+d x^2\right )^{3/2}}{13 c x^{13/2}}-\frac {2 a (26 b c-7 a d) \left (c+d x^2\right )^{3/2}}{117 c^2 x^{9/2}}+\frac {\left (2 d^2 \left (39 b^2 c^2-26 a b c d+7 a^2 d^2\right )\right ) \int \frac {\sqrt {x}}{\sqrt {c+d x^2}} \, dx}{195 c^3}\\ &=-\frac {2 \left (39 b^2-\frac {a d (26 b c-7 a d)}{c^2}\right ) \sqrt {c+d x^2}}{195 x^{5/2}}-\frac {4 d \left (39 b^2 c^2-26 a b c d+7 a^2 d^2\right ) \sqrt {c+d x^2}}{195 c^3 \sqrt {x}}-\frac {2 a^2 \left (c+d x^2\right )^{3/2}}{13 c x^{13/2}}-\frac {2 a (26 b c-7 a d) \left (c+d x^2\right )^{3/2}}{117 c^2 x^{9/2}}+\frac {\left (4 d^2 \left (39 b^2 c^2-26 a b c d+7 a^2 d^2\right )\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {c+d x^4}} \, dx,x,\sqrt {x}\right )}{195 c^3}\\ &=-\frac {2 \left (39 b^2-\frac {a d (26 b c-7 a d)}{c^2}\right ) \sqrt {c+d x^2}}{195 x^{5/2}}-\frac {4 d \left (39 b^2 c^2-26 a b c d+7 a^2 d^2\right ) \sqrt {c+d x^2}}{195 c^3 \sqrt {x}}-\frac {2 a^2 \left (c+d x^2\right )^{3/2}}{13 c x^{13/2}}-\frac {2 a (26 b c-7 a d) \left (c+d x^2\right )^{3/2}}{117 c^2 x^{9/2}}+\frac {\left (4 d^{3/2} \left (39 b^2 c^2-26 a b c d+7 a^2 d^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {c+d x^4}} \, dx,x,\sqrt {x}\right )}{195 c^{5/2}}-\frac {\left (4 d^{3/2} \left (39 b^2 c^2-26 a b c d+7 a^2 d^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 )}{195 c^{5/2}}\\ &=-\frac {2 \left (39 b^2-\frac {a d (26 b c-7 a d)}{c^2}\right ) \sqrt {c+d x^2}}{195 x^{5/2}}-\frac {4 d \left (39 b^2 c^2-26 a b c d+7 a^2 d^2\right ) \sqrt {c+d x^2}}{195 c^3 \sqrt {x}}+\frac {4 d^{3/2} \left (39 b^2 c^2-26 a b c d+7 a^2 d^2\right ) \sqrt {x} \sqrt {c+d x^2}}{195 c^3 \left (\sqrt {c}+\sqrt {d} x\right )}-\frac {2 a^2 \left (c+d x^2\right )^{3/2}}{13 c x^{13/2}}-\frac {2 a (26 b c-7 a d) \left (c+d x^2\right )^{3/2}}{117 c^2 x^{9/2}}-\frac {4 d^{5/4} \left (39 b^2 c^2-26 a b c d+7 a^2 d^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 )}{195 c^{11/4} \sqrt {c+d x^2}}+\frac {2 d^{5/4} \left (39 b^2 c^2-26 a b c d+7 a^2 d^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 )}{195 c^{11/4} \sqrt {c+d x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.20, size = 182, normalized size = 0.41 \[ \frac {4 d^2 x^8 \sqrt {\frac {d x^2}{c}+1} \left (7 a^2 d^2-26 a b c d+39 b^2 c^2\right ) \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-\frac {d x^2}{c}\right )-2 \left (c+d x^2\right ) \left (a^2 \left (45 c^3+10 c^2 d x^2-14 c d^2 x^4+42 d^3 x^6\right )+26 a b c x^2 \left (5 c^2+2 c d x^2-6 d^2 x^4\right )+117 b^2 c^2 x^4 \left (c+2 d x^2\right )\right )}{585 c^3 x^{13/2} \sqrt {c+d x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)^2*Sqrt[c + d*x^2])/x^(15/2),x]

[Out]

(-2*(c + d*x^2)*(117*b^2*c^2*x^4*(c + 2*d*x^2) + 26*a*b*c*x^2*(5*c^2 + 2*c*d*x^2 - 6*d^2*x^4) + a^2*(45*c^3 +
10*c^2*d*x^2 - 14*c*d^2*x^4 + 42*d^3*x^6)) + 4*d^2*(39*b^2*c^2 - 26*a*b*c*d + 7*a^2*d^2)*x^8*Sqrt[1 + (d*x^2)/
c]*Hypergeometric2F1[1/2, 3/4, 7/4, -((d*x^2)/c)])/(585*c^3*x^(13/2)*Sqrt[c + d*x^2])

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fricas [F]  time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \sqrt {d x^{2} + c}}{x^{\frac {15}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(1/2)/x^(15/2),x, algorithm="fricas")

[Out]

integral((b^2*x^4 + 2*a*b*x^2 + a^2)*sqrt(d*x^2 + c)/x^(15/2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{2} + a\right )}^{2} \sqrt {d x^{2} + c}}{x^{\frac {15}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(1/2)/x^(15/2),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^2*sqrt(d*x^2 + c)/x^(15/2), x)

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maple [A]  time = 0.06, size = 706, normalized size = 1.60 \[ \frac {-\frac {28 a^{2} d^{4} x^{8}}{195}+\frac {8 a b c \,d^{3} x^{8}}{15}-\frac {4 b^{2} c^{2} d^{2} x^{8}}{5}+\frac {28 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, a^{2} c \,d^{3} x^{6} \EllipticE \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{195}-\frac {14 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, a^{2} c \,d^{3} x^{6} \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{195}-\frac {8 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, a b \,c^{2} d^{2} x^{6} \EllipticE \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{15}+\frac {4 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, a b \,c^{2} d^{2} x^{6} \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{15}+\frac {4 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, b^{2} c^{3} d \,x^{6} \EllipticE \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{5}-\frac {2 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, b^{2} c^{3} d \,x^{6} \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{5}-\frac {56 a^{2} c \,d^{3} x^{6}}{585}+\frac {16 a b \,c^{2} d^{2} x^{6}}{45}-\frac {6 b^{2} c^{3} d \,x^{6}}{5}+\frac {8 a^{2} c^{2} d^{2} x^{4}}{585}-\frac {28 a b \,c^{3} d \,x^{4}}{45}-\frac {2 b^{2} c^{4} x^{4}}{5}-\frac {22 a^{2} c^{3} d \,x^{2}}{117}-\frac {4 a b \,c^{4} x^{2}}{9}-\frac {2 a^{2} c^{4}}{13}}{\sqrt {d \,x^{2}+c}\, c^{3} x^{\frac {13}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2*(d*x^2+c)^(1/2)/x^(15/2),x)

[Out]

2/585*(42*((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)*(-1/(-c*d)^
(1/2)*d*x)^(1/2)*EllipticE(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*x^6*a^2*c*d^3-156*((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)*(-1/(-c*d)^(1/2)*d*x)^(1/2)*Ellip
ticE(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*x^6*a*b*c^2*d^2+234*((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)*(-1/(-c*d)^(1/2)*d*x)^(1/2)*EllipticE(((d*x+(-c*d)^(1
/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*x^6*b^2*c^3*d-21*((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)*(-1/(-c*d)^(1/2)*d*x)^(1/2)*EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2
),1/2*2^(1/2))*x^6*a^2*c*d^3+78*((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)*(-1/(-c*d)^(1/2)*d*x)^(1/2)*EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*x^6*a*b*
c^2*d^2-117*((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)*(-1/(-c*d
)^(1/2)*d*x)^(1/2)*EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*x^6*b^2*c^3*d-42*x^8*a^2*d^4
+156*x^8*a*b*c*d^3-234*x^8*b^2*c^2*d^2-28*x^6*a^2*c*d^3+104*x^6*a*b*c^2*d^2-351*x^6*b^2*c^3*d+4*x^4*a^2*c^2*d^
2-182*x^4*a*b*c^3*d-117*x^4*b^2*c^4-55*a^2*c^3*d*x^2-130*x^2*a*b*c^4-45*a^2*c^4)/(d*x^2+c)^(1/2)/x^(13/2)/c^3

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{2} + a\right )}^{2} \sqrt {d x^{2} + c}}{x^{\frac {15}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(1/2)/x^(15/2),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^2*sqrt(d*x^2 + c)/x^(15/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (b\,x^2+a\right )}^2\,\sqrt {d\,x^2+c}}{x^{15/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x^2)^2*(c + d*x^2)^(1/2))/x^(15/2),x)

[Out]

int(((a + b*x^2)^2*(c + d*x^2)^(1/2))/x^(15/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2*(d*x**2+c)**(1/2)/x**(15/2),x)

[Out]

Timed out

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