3.954 \(\int \frac {x^5 (a+b x^2)^{5/2}}{\sqrt {c+d x^2}} \, dx\)

Optimal. Leaf size=340 \[ \frac {\sqrt {a+b x^2} \sqrt {c+d x^2} (b c-a d)^2 \left (3 a^2 d^2+14 a b c d+63 b^2 c^2\right )}{256 b^2 d^5}-\frac {\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2} (b c-a d) \left (3 a^2 d^2+14 a b c d+63 b^2 c^2\right )}{384 b^2 d^4}+\frac {\left (a+b x^2\right )^{5/2} \sqrt {c+d x^2} \left (3 a^2 d^2+14 a b c d+63 b^2 c^2\right )}{480 b^2 d^3}-\frac {(b c-a d)^3 \left (3 a^2 d^2+14 a b c d+63 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{256 b^{5/2} d^{11/2}}-\frac {3 \left (a+b x^2\right )^{7/2} \sqrt {c+d x^2} (a d+3 b c)}{80 b^2 d^2}+\frac {x^2 \left (a+b x^2\right )^{7/2} \sqrt {c+d x^2}}{10 b d} \]

[Out]

-1/256*(-a*d+b*c)^3*(3*a^2*d^2+14*a*b*c*d+63*b^2*c^2)*arctanh(d^(1/2)*(b*x^2+a)^(1/2)/b^(1/2)/(d*x^2+c)^(1/2))
/b^(5/2)/d^(11/2)-1/384*(-a*d+b*c)*(3*a^2*d^2+14*a*b*c*d+63*b^2*c^2)*(b*x^2+a)^(3/2)*(d*x^2+c)^(1/2)/b^2/d^4+1
/480*(3*a^2*d^2+14*a*b*c*d+63*b^2*c^2)*(b*x^2+a)^(5/2)*(d*x^2+c)^(1/2)/b^2/d^3-3/80*(a*d+3*b*c)*(b*x^2+a)^(7/2
)*(d*x^2+c)^(1/2)/b^2/d^2+1/10*x^2*(b*x^2+a)^(7/2)*(d*x^2+c)^(1/2)/b/d+1/256*(-a*d+b*c)^2*(3*a^2*d^2+14*a*b*c*
d+63*b^2*c^2)*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)/b^2/d^5

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Rubi [A]  time = 0.42, antiderivative size = 340, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {446, 90, 80, 50, 63, 217, 206} \[ \frac {\left (a+b x^2\right )^{5/2} \sqrt {c+d x^2} \left (3 a^2 d^2+14 a b c d+63 b^2 c^2\right )}{480 b^2 d^3}-\frac {\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2} (b c-a d) \left (3 a^2 d^2+14 a b c d+63 b^2 c^2\right )}{384 b^2 d^4}+\frac {\sqrt {a+b x^2} \sqrt {c+d x^2} (b c-a d)^2 \left (3 a^2 d^2+14 a b c d+63 b^2 c^2\right )}{256 b^2 d^5}-\frac {(b c-a d)^3 \left (3 a^2 d^2+14 a b c d+63 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{256 b^{5/2} d^{11/2}}-\frac {3 \left (a+b x^2\right )^{7/2} \sqrt {c+d x^2} (a d+3 b c)}{80 b^2 d^2}+\frac {x^2 \left (a+b x^2\right )^{7/2} \sqrt {c+d x^2}}{10 b d} \]

Antiderivative was successfully verified.

[In]

Int[(x^5*(a + b*x^2)^(5/2))/Sqrt[c + d*x^2],x]

[Out]

((b*c - a*d)^2*(63*b^2*c^2 + 14*a*b*c*d + 3*a^2*d^2)*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(256*b^2*d^5) - ((b*c -
a*d)*(63*b^2*c^2 + 14*a*b*c*d + 3*a^2*d^2)*(a + b*x^2)^(3/2)*Sqrt[c + d*x^2])/(384*b^2*d^4) + ((63*b^2*c^2 + 1
4*a*b*c*d + 3*a^2*d^2)*(a + b*x^2)^(5/2)*Sqrt[c + d*x^2])/(480*b^2*d^3) - (3*(3*b*c + a*d)*(a + b*x^2)^(7/2)*S
qrt[c + d*x^2])/(80*b^2*d^2) + (x^2*(a + b*x^2)^(7/2)*Sqrt[c + d*x^2])/(10*b*d) - ((b*c - a*d)^3*(63*b^2*c^2 +
14*a*b*c*d + 3*a^2*d^2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x^2])/(Sqrt[b]*Sqrt[c + d*x^2])])/(256*b^(5/2)*d^(11/2))

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 80

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

Rule 90

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a + b*
x)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 3)), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {x^5 \left (a+b x^2\right )^{5/2}}{\sqrt {c+d x^2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^2 (a+b x)^{5/2}}{\sqrt {c+d x}} \, dx,x,x^2\right )\\ &=\frac {x^2 \left (a+b x^2\right )^{7/2} \sqrt {c+d x^2}}{10 b d}+\frac {\operatorname {Subst}\left (\int \frac {(a+b x)^{5/2} \left (-a c-\frac {3}{2} (3 b c+a d) x\right )}{\sqrt {c+d x}} \, dx,x,x^2\right )}{10 b d}\\ &=-\frac {3 (3 b c+a d) \left (a+b x^2\right )^{7/2} \sqrt {c+d x^2}}{80 b^2 d^2}+\frac {x^2 \left (a+b x^2\right )^{7/2} \sqrt {c+d x^2}}{10 b d}+\frac {\left (63 b^2 c^2+14 a b c d+3 a^2 d^2\right ) \operatorname {Subst}\left (\int \frac {(a+b x)^{5/2}}{\sqrt {c+d x}} \, dx,x,x^2\right )}{160 b^2 d^2}\\ &=\frac {\left (63 b^2 c^2+14 a b c d+3 a^2 d^2\right ) \left (a+b x^2\right )^{5/2} \sqrt {c+d x^2}}{480 b^2 d^3}-\frac {3 (3 b c+a d) \left (a+b x^2\right )^{7/2} \sqrt {c+d x^2}}{80 b^2 d^2}+\frac {x^2 \left (a+b x^2\right )^{7/2} \sqrt {c+d x^2}}{10 b d}-\frac {\left ((b c-a d) \left (63 b^2 c^2+14 a b c d+3 a^2 d^2\right )\right ) \operatorname {Subst}\left (\int \frac {(a+b x)^{3/2}}{\sqrt {c+d x}} \, dx,x,x^2\right )}{192 b^2 d^3}\\ &=-\frac {(b c-a d) \left (63 b^2 c^2+14 a b c d+3 a^2 d^2\right ) \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{384 b^2 d^4}+\frac {\left (63 b^2 c^2+14 a b c d+3 a^2 d^2\right ) \left (a+b x^2\right )^{5/2} \sqrt {c+d x^2}}{480 b^2 d^3}-\frac {3 (3 b c+a d) \left (a+b x^2\right )^{7/2} \sqrt {c+d x^2}}{80 b^2 d^2}+\frac {x^2 \left (a+b x^2\right )^{7/2} \sqrt {c+d x^2}}{10 b d}+\frac {\left ((b c-a d)^2 \left (63 b^2 c^2+14 a b c d+3 a^2 d^2\right )\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a+b x}}{\sqrt {c+d x}} \, dx,x,x^2\right )}{256 b^2 d^4}\\ &=\frac {(b c-a d)^2 \left (63 b^2 c^2+14 a b c d+3 a^2 d^2\right ) \sqrt {a+b x^2} \sqrt {c+d x^2}}{256 b^2 d^5}-\frac {(b c-a d) \left (63 b^2 c^2+14 a b c d+3 a^2 d^2\right ) \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{384 b^2 d^4}+\frac {\left (63 b^2 c^2+14 a b c d+3 a^2 d^2\right ) \left (a+b x^2\right )^{5/2} \sqrt {c+d x^2}}{480 b^2 d^3}-\frac {3 (3 b c+a d) \left (a+b x^2\right )^{7/2} \sqrt {c+d x^2}}{80 b^2 d^2}+\frac {x^2 \left (a+b x^2\right )^{7/2} \sqrt {c+d x^2}}{10 b d}-\frac {\left ((b c-a d)^3 \left (63 b^2 c^2+14 a b c d+3 a^2 d^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx,x,x^2\right )}{512 b^2 d^5}\\ &=\frac {(b c-a d)^2 \left (63 b^2 c^2+14 a b c d+3 a^2 d^2\right ) \sqrt {a+b x^2} \sqrt {c+d x^2}}{256 b^2 d^5}-\frac {(b c-a d) \left (63 b^2 c^2+14 a b c d+3 a^2 d^2\right ) \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{384 b^2 d^4}+\frac {\left (63 b^2 c^2+14 a b c d+3 a^2 d^2\right ) \left (a+b x^2\right )^{5/2} \sqrt {c+d x^2}}{480 b^2 d^3}-\frac {3 (3 b c+a d) \left (a+b x^2\right )^{7/2} \sqrt {c+d x^2}}{80 b^2 d^2}+\frac {x^2 \left (a+b x^2\right )^{7/2} \sqrt {c+d x^2}}{10 b d}-\frac {\left ((b c-a d)^3 \left (63 b^2 c^2+14 a b c d+3 a^2 d^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {c-\frac {a d}{b}+\frac {d x^2}{b}}} \, dx,x,\sqrt {a+b x^2}\right )}{256 b^3 d^5}\\ &=\frac {(b c-a d)^2 \left (63 b^2 c^2+14 a b c d+3 a^2 d^2\right ) \sqrt {a+b x^2} \sqrt {c+d x^2}}{256 b^2 d^5}-\frac {(b c-a d) \left (63 b^2 c^2+14 a b c d+3 a^2 d^2\right ) \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{384 b^2 d^4}+\frac {\left (63 b^2 c^2+14 a b c d+3 a^2 d^2\right ) \left (a+b x^2\right )^{5/2} \sqrt {c+d x^2}}{480 b^2 d^3}-\frac {3 (3 b c+a d) \left (a+b x^2\right )^{7/2} \sqrt {c+d x^2}}{80 b^2 d^2}+\frac {x^2 \left (a+b x^2\right )^{7/2} \sqrt {c+d x^2}}{10 b d}-\frac {\left ((b c-a d)^3 \left (63 b^2 c^2+14 a b c d+3 a^2 d^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1-\frac {d x^2}{b}} \, dx,x,\frac {\sqrt {a+b x^2}}{\sqrt {c+d x^2}}\right )}{256 b^3 d^5}\\ &=\frac {(b c-a d)^2 \left (63 b^2 c^2+14 a b c d+3 a^2 d^2\right ) \sqrt {a+b x^2} \sqrt {c+d x^2}}{256 b^2 d^5}-\frac {(b c-a d) \left (63 b^2 c^2+14 a b c d+3 a^2 d^2\right ) \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{384 b^2 d^4}+\frac {\left (63 b^2 c^2+14 a b c d+3 a^2 d^2\right ) \left (a+b x^2\right )^{5/2} \sqrt {c+d x^2}}{480 b^2 d^3}-\frac {3 (3 b c+a d) \left (a+b x^2\right )^{7/2} \sqrt {c+d x^2}}{80 b^2 d^2}+\frac {x^2 \left (a+b x^2\right )^{7/2} \sqrt {c+d x^2}}{10 b d}-\frac {(b c-a d)^3 \left (63 b^2 c^2+14 a b c d+3 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{256 b^{5/2} d^{11/2}}\\ \end {align*}

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Mathematica [A]  time = 1.20, size = 271, normalized size = 0.80 \[ \frac {\sqrt {c+d x^2} \left (\frac {5 (b c-a d)^3 \left (3 a^2 d^2+14 a b c d+63 b^2 c^2\right ) \left (\frac {16 d^3 \left (a+b x^2\right )^3}{15 (b c-a d)^3}-\frac {4 d^2 \left (a+b x^2\right )^2}{3 (b c-a d)^2}+\frac {2 d \left (a+b x^2\right )}{b c-a d}-\frac {2 \sqrt {d} \sqrt {a+b x^2} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b c-a d}}\right )}{\sqrt {b c-a d} \sqrt {\frac {b \left (c+d x^2\right )}{b c-a d}}}\right )}{4 b d^5}-\frac {24 \left (a+b x^2\right )^4 (a d+3 b c)}{b d}+64 x^2 \left (a+b x^2\right )^4\right )}{640 b d \sqrt {a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^5*(a + b*x^2)^(5/2))/Sqrt[c + d*x^2],x]

[Out]

(Sqrt[c + d*x^2]*((-24*(3*b*c + a*d)*(a + b*x^2)^4)/(b*d) + 64*x^2*(a + b*x^2)^4 + (5*(b*c - a*d)^3*(63*b^2*c^
2 + 14*a*b*c*d + 3*a^2*d^2)*((2*d*(a + b*x^2))/(b*c - a*d) - (4*d^2*(a + b*x^2)^2)/(3*(b*c - a*d)^2) + (16*d^3
*(a + b*x^2)^3)/(15*(b*c - a*d)^3) - (2*Sqrt[d]*Sqrt[a + b*x^2]*ArcSinh[(Sqrt[d]*Sqrt[a + b*x^2])/Sqrt[b*c - a
*d]])/(Sqrt[b*c - a*d]*Sqrt[(b*(c + d*x^2))/(b*c - a*d)])))/(4*b*d^5)))/(640*b*d*Sqrt[a + b*x^2])

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fricas [A]  time = 1.21, size = 734, normalized size = 2.16 \[ \left [-\frac {15 \, {\left (63 \, b^{5} c^{5} - 175 \, a b^{4} c^{4} d + 150 \, a^{2} b^{3} c^{3} d^{2} - 30 \, a^{3} b^{2} c^{2} d^{3} - 5 \, a^{4} b c d^{4} - 3 \, a^{5} d^{5}\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{4} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x^{2} + 4 \, {\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {b d}\right ) - 4 \, {\left (384 \, b^{5} d^{5} x^{8} + 945 \, b^{5} c^{4} d - 2310 \, a b^{4} c^{3} d^{2} + 1564 \, a^{2} b^{3} c^{2} d^{3} - 90 \, a^{3} b^{2} c d^{4} - 45 \, a^{4} b d^{5} - 144 \, {\left (3 \, b^{5} c d^{4} - 7 \, a b^{4} d^{5}\right )} x^{6} + 8 \, {\left (63 \, b^{5} c^{2} d^{3} - 148 \, a b^{4} c d^{4} + 93 \, a^{2} b^{3} d^{5}\right )} x^{4} - 2 \, {\left (315 \, b^{5} c^{3} d^{2} - 749 \, a b^{4} c^{2} d^{3} + 481 \, a^{2} b^{3} c d^{4} - 15 \, a^{3} b^{2} d^{5}\right )} x^{2}\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{15360 \, b^{3} d^{6}}, \frac {15 \, {\left (63 \, b^{5} c^{5} - 175 \, a b^{4} c^{4} d + 150 \, a^{2} b^{3} c^{3} d^{2} - 30 \, a^{3} b^{2} c^{2} d^{3} - 5 \, a^{4} b c d^{4} - 3 \, a^{5} d^{5}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {-b d}}{2 \, {\left (b^{2} d^{2} x^{4} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x^{2}\right )}}\right ) + 2 \, {\left (384 \, b^{5} d^{5} x^{8} + 945 \, b^{5} c^{4} d - 2310 \, a b^{4} c^{3} d^{2} + 1564 \, a^{2} b^{3} c^{2} d^{3} - 90 \, a^{3} b^{2} c d^{4} - 45 \, a^{4} b d^{5} - 144 \, {\left (3 \, b^{5} c d^{4} - 7 \, a b^{4} d^{5}\right )} x^{6} + 8 \, {\left (63 \, b^{5} c^{2} d^{3} - 148 \, a b^{4} c d^{4} + 93 \, a^{2} b^{3} d^{5}\right )} x^{4} - 2 \, {\left (315 \, b^{5} c^{3} d^{2} - 749 \, a b^{4} c^{2} d^{3} + 481 \, a^{2} b^{3} c d^{4} - 15 \, a^{3} b^{2} d^{5}\right )} x^{2}\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{7680 \, b^{3} d^{6}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b*x^2+a)^(5/2)/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

[-1/15360*(15*(63*b^5*c^5 - 175*a*b^4*c^4*d + 150*a^2*b^3*c^3*d^2 - 30*a^3*b^2*c^2*d^3 - 5*a^4*b*c*d^4 - 3*a^5
*d^5)*sqrt(b*d)*log(8*b^2*d^2*x^4 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 8*(b^2*c*d + a*b*d^2)*x^2 + 4*(2*b*d*x^2 +
b*c + a*d)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(b*d)) - 4*(384*b^5*d^5*x^8 + 945*b^5*c^4*d - 2310*a*b^4*c^3*d
^2 + 1564*a^2*b^3*c^2*d^3 - 90*a^3*b^2*c*d^4 - 45*a^4*b*d^5 - 144*(3*b^5*c*d^4 - 7*a*b^4*d^5)*x^6 + 8*(63*b^5*
c^2*d^3 - 148*a*b^4*c*d^4 + 93*a^2*b^3*d^5)*x^4 - 2*(315*b^5*c^3*d^2 - 749*a*b^4*c^2*d^3 + 481*a^2*b^3*c*d^4 -
15*a^3*b^2*d^5)*x^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c))/(b^3*d^6), 1/7680*(15*(63*b^5*c^5 - 175*a*b^4*c^4*d + 1
50*a^2*b^3*c^3*d^2 - 30*a^3*b^2*c^2*d^3 - 5*a^4*b*c*d^4 - 3*a^5*d^5)*sqrt(-b*d)*arctan(1/2*(2*b*d*x^2 + b*c +
a*d)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(-b*d)/(b^2*d^2*x^4 + a*b*c*d + (b^2*c*d + a*b*d^2)*x^2)) + 2*(384*b^
5*d^5*x^8 + 945*b^5*c^4*d - 2310*a*b^4*c^3*d^2 + 1564*a^2*b^3*c^2*d^3 - 90*a^3*b^2*c*d^4 - 45*a^4*b*d^5 - 144*
(3*b^5*c*d^4 - 7*a*b^4*d^5)*x^6 + 8*(63*b^5*c^2*d^3 - 148*a*b^4*c*d^4 + 93*a^2*b^3*d^5)*x^4 - 2*(315*b^5*c^3*d
^2 - 749*a*b^4*c^2*d^3 + 481*a^2*b^3*c*d^4 - 15*a^3*b^2*d^5)*x^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c))/(b^3*d^6)]

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giac [A]  time = 0.68, size = 398, normalized size = 1.17 \[ \frac {{\left (\sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d} \sqrt {b x^{2} + a} {\left (2 \, {\left (b x^{2} + a\right )} {\left (4 \, {\left (b x^{2} + a\right )} {\left (6 \, {\left (b x^{2} + a\right )} {\left (\frac {8 \, {\left (b x^{2} + a\right )}}{b^{3} d} - \frac {9 \, b^{7} c d^{7} + 11 \, a b^{6} d^{8}}{b^{9} d^{9}}\right )} + \frac {63 \, b^{8} c^{2} d^{6} + 14 \, a b^{7} c d^{7} + 3 \, a^{2} b^{6} d^{8}}{b^{9} d^{9}}\right )} - \frac {5 \, {\left (63 \, b^{9} c^{3} d^{5} - 49 \, a b^{8} c^{2} d^{6} - 11 \, a^{2} b^{7} c d^{7} - 3 \, a^{3} b^{6} d^{8}\right )}}{b^{9} d^{9}}\right )} + \frac {15 \, {\left (63 \, b^{10} c^{4} d^{4} - 112 \, a b^{9} c^{3} d^{5} + 38 \, a^{2} b^{8} c^{2} d^{6} + 8 \, a^{3} b^{7} c d^{7} + 3 \, a^{4} b^{6} d^{8}\right )}}{b^{9} d^{9}}\right )} + \frac {15 \, {\left (63 \, b^{5} c^{5} - 175 \, a b^{4} c^{4} d + 150 \, a^{2} b^{3} c^{3} d^{2} - 30 \, a^{3} b^{2} c^{2} d^{3} - 5 \, a^{4} b c d^{4} - 3 \, a^{5} d^{5}\right )} \log \left ({\left | -\sqrt {b x^{2} + a} \sqrt {b d} + \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} b^{2} d^{5}}\right )} b}{3840 \, {\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b*x^2+a)^(5/2)/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

1/3840*(sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d)*sqrt(b*x^2 + a)*(2*(b*x^2 + a)*(4*(b*x^2 + a)*(6*(b*x^2 + a)*(8*
(b*x^2 + a)/(b^3*d) - (9*b^7*c*d^7 + 11*a*b^6*d^8)/(b^9*d^9)) + (63*b^8*c^2*d^6 + 14*a*b^7*c*d^7 + 3*a^2*b^6*d
^8)/(b^9*d^9)) - 5*(63*b^9*c^3*d^5 - 49*a*b^8*c^2*d^6 - 11*a^2*b^7*c*d^7 - 3*a^3*b^6*d^8)/(b^9*d^9)) + 15*(63*
b^10*c^4*d^4 - 112*a*b^9*c^3*d^5 + 38*a^2*b^8*c^2*d^6 + 8*a^3*b^7*c*d^7 + 3*a^4*b^6*d^8)/(b^9*d^9)) + 15*(63*b
^5*c^5 - 175*a*b^4*c^4*d + 150*a^2*b^3*c^3*d^2 - 30*a^3*b^2*c^2*d^3 - 5*a^4*b*c*d^4 - 3*a^5*d^5)*log(abs(-sqrt
(b*x^2 + a)*sqrt(b*d) + sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d)))/(sqrt(b*d)*b^2*d^5))*b/abs(b)

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maple [B]  time = 0.04, size = 1054, normalized size = 3.10 \[ \frac {\sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}\, \left (768 \sqrt {b d}\, \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, b^{4} d^{4} x^{8}+2016 \sqrt {b d}\, \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, a \,b^{3} d^{4} x^{6}-864 \sqrt {b d}\, \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, b^{4} c \,d^{3} x^{6}+1488 \sqrt {b d}\, \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, a^{2} b^{2} d^{4} x^{4}-2368 \sqrt {b d}\, \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, a \,b^{3} c \,d^{3} x^{4}+1008 \sqrt {b d}\, \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, b^{4} c^{2} d^{2} x^{4}+45 a^{5} d^{5} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+75 a^{4} b c \,d^{4} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+450 a^{3} b^{2} c^{2} d^{3} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-2250 a^{2} b^{3} c^{3} d^{2} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+2625 a \,b^{4} c^{4} d \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-945 b^{5} c^{5} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+60 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a^{3} b \,d^{4} x^{2}-1924 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a^{2} b^{2} c \,d^{3} x^{2}+2996 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a \,b^{3} c^{2} d^{2} x^{2}-1260 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, b^{4} c^{3} d \,x^{2}-90 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a^{4} d^{4}-180 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a^{3} b c \,d^{3}+3128 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a^{2} b^{2} c^{2} d^{2}-4620 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a \,b^{3} c^{3} d +1890 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, b^{4} c^{4}\right )}{7680 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, b^{2} d^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(b*x^2+a)^(5/2)/(d*x^2+c)^(1/2),x)

[Out]

1/7680*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)*(768*x^8*b^4*d^4*(b*d)^(1/2)*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)+2016*x
^6*a*b^3*d^4*(b*d)^(1/2)*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)-864*x^6*b^4*c*d^3*(b*d)^(1/2)*(b*d*x^4+a*d*x^2+b*
c*x^2+a*c)^(1/2)+1488*x^4*a^2*b^2*d^4*(b*d)^(1/2)*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)-2368*x^4*a*b^3*c*d^3*(b*
d)^(1/2)*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)+1008*x^4*b^4*c^2*d^2*(b*d)^(1/2)*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1
/2)+60*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*x^2*a^3*b*d^4*(b*d)^(1/2)-1924*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*
x^2*a^2*c*b^2*d^3*(b*d)^(1/2)+2996*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*x^2*a*c^2*b^3*d^2*(b*d)^(1/2)-1260*(b*d
*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*x^2*c^3*b^4*d*(b*d)^(1/2)+45*d^5*ln(1/2*(2*b*d*x^2+a*d+b*c+2*(b*d*x^4+a*d*x^2+
b*c*x^2+a*c)^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*a^5+75*ln(1/2*(2*b*d*x^2+a*d+b*c+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^
(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*a^4*c*b*d^4+450*ln(1/2*(2*b*d*x^2+a*d+b*c+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2
)*(b*d)^(1/2))/(b*d)^(1/2))*a^3*c^2*b^2*d^3-2250*ln(1/2*(2*b*d*x^2+a*d+b*c+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/
2)*(b*d)^(1/2))/(b*d)^(1/2))*a^2*c^3*b^3*d^2+2625*ln(1/2*(2*b*d*x^2+a*d+b*c+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1
/2)*(b*d)^(1/2))/(b*d)^(1/2))*c^4*a*b^4*d-945*b^5*ln(1/2*(2*b*d*x^2+a*d+b*c+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1
/2)*(b*d)^(1/2))/(b*d)^(1/2))*c^5-90*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*a^4*d^4*(b*d)^(1/2)-180*(b*d*x^4+a*d*
x^2+b*c*x^2+a*c)^(1/2)*a^3*c*b*d^3*(b*d)^(1/2)+3128*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*a^2*c^2*b^2*d^2*(b*d)^
(1/2)-4620*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*a*c^3*b^3*d*(b*d)^(1/2)+1890*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2
)*c^4*b^4*(b*d)^(1/2))/b^2/d^5/(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)/(b*d)^(1/2)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b*x^2+a)^(5/2)/(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more details)Is a*d-b*c zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^5\,{\left (b\,x^2+a\right )}^{5/2}}{\sqrt {d\,x^2+c}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^5*(a + b*x^2)^(5/2))/(c + d*x^2)^(1/2),x)

[Out]

int((x^5*(a + b*x^2)^(5/2))/(c + d*x^2)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{5} \left (a + b x^{2}\right )^{\frac {5}{2}}}{\sqrt {c + d x^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(b*x**2+a)**(5/2)/(d*x**2+c)**(1/2),x)

[Out]

Integral(x**5*(a + b*x**2)**(5/2)/sqrt(c + d*x**2), x)

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