3.8.80 \(\int \frac {x^3 (a+b x^2)^{5/2}}{\sqrt {c+d x^2}} \, dx\)

Optimal. Leaf size=237 \[ \frac {5 (b c-a d)^3 (a d+7 b c) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{128 b^{3/2} d^{9/2}}-\frac {5 \sqrt {a+b x^2} \sqrt {c+d x^2} (b c-a d)^2 (a d+7 b c)}{128 b d^4}+\frac {5 \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2} (b c-a d) (a d+7 b c)}{192 b d^3}-\frac {\left (a+b x^2\right )^{5/2} \sqrt {c+d x^2} (a d+7 b c)}{48 b d^2}+\frac {\left (a+b x^2\right )^{7/2} \sqrt {c+d x^2}}{8 b d} \]

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Rubi [A]  time = 0.23, antiderivative size = 237, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {446, 80, 50, 63, 217, 206} \begin {gather*} \frac {5 (b c-a d)^3 (a d+7 b c) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{128 b^{3/2} d^{9/2}}-\frac {\left (a+b x^2\right )^{5/2} \sqrt {c+d x^2} (a d+7 b c)}{48 b d^2}+\frac {5 \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2} (b c-a d) (a d+7 b c)}{192 b d^3}-\frac {5 \sqrt {a+b x^2} \sqrt {c+d x^2} (b c-a d)^2 (a d+7 b c)}{128 b d^4}+\frac {\left (a+b x^2\right )^{7/2} \sqrt {c+d x^2}}{8 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^3*(a + b*x^2)^(5/2))/Sqrt[c + d*x^2],x]

[Out]

(-5*(b*c - a*d)^2*(7*b*c + a*d)*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(128*b*d^4) + (5*(b*c - a*d)*(7*b*c + a*d)*(a
+ b*x^2)^(3/2)*Sqrt[c + d*x^2])/(192*b*d^3) - ((7*b*c + a*d)*(a + b*x^2)^(5/2)*Sqrt[c + d*x^2])/(48*b*d^2) +
((a + b*x^2)^(7/2)*Sqrt[c + d*x^2])/(8*b*d) + (5*(b*c - a*d)^3*(7*b*c + a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x^2])
/(Sqrt[b]*Sqrt[c + d*x^2])])/(128*b^(3/2)*d^(9/2))

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 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^3 \left (a+b x^2\right )^{5/2}}{\sqrt {c+d x^2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x (a+b x)^{5/2}}{\sqrt {c+d x}} \, dx,x,x^2\right )\\ &=\frac {\left (a+b x^2\right )^{7/2} \sqrt {c+d x^2}}{8 b d}-\frac {(7 b c+a d) \operatorname {Subst}\left (\int \frac {(a+b x)^{5/2}}{\sqrt {c+d x}} \, dx,x,x^2\right )}{16 b d}\\ &=-\frac {(7 b c+a d) \left (a+b x^2\right )^{5/2} \sqrt {c+d x^2}}{48 b d^2}+\frac {\left (a+b x^2\right )^{7/2} \sqrt {c+d x^2}}{8 b d}+\frac {(5 (b c-a d) (7 b c+a d)) \operatorname {Subst}\left (\int \frac {(a+b x)^{3/2}}{\sqrt {c+d x}} \, dx,x,x^2\right )}{96 b d^2}\\ &=\frac {5 (b c-a d) (7 b c+a d) \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{192 b d^3}-\frac {(7 b c+a d) \left (a+b x^2\right )^{5/2} \sqrt {c+d x^2}}{48 b d^2}+\frac {\left (a+b x^2\right )^{7/2} \sqrt {c+d x^2}}{8 b d}-\frac {\left (5 (b c-a d)^2 (7 b c+a d)\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a+b x}}{\sqrt {c+d x}} \, dx,x,x^2\right )}{128 b d^3}\\ &=-\frac {5 (b c-a d)^2 (7 b c+a d) \sqrt {a+b x^2} \sqrt {c+d x^2}}{128 b d^4}+\frac {5 (b c-a d) (7 b c+a d) \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{192 b d^3}-\frac {(7 b c+a d) \left (a+b x^2\right )^{5/2} \sqrt {c+d x^2}}{48 b d^2}+\frac {\left (a+b x^2\right )^{7/2} \sqrt {c+d x^2}}{8 b d}+\frac {\left (5 (b c-a d)^3 (7 b c+a d)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx,x,x^2\right )}{256 b d^4}\\ &=-\frac {5 (b c-a d)^2 (7 b c+a d) \sqrt {a+b x^2} \sqrt {c+d x^2}}{128 b d^4}+\frac {5 (b c-a d) (7 b c+a d) \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{192 b d^3}-\frac {(7 b c+a d) \left (a+b x^2\right )^{5/2} \sqrt {c+d x^2}}{48 b d^2}+\frac {\left (a+b x^2\right )^{7/2} \sqrt {c+d x^2}}{8 b d}+\frac {\left (5 (b c-a d)^3 (7 b c+a d)\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 )}{128 b^2 d^4}\\ &=-\frac {5 (b c-a d)^2 (7 b c+a d) \sqrt {a+b x^2} \sqrt {c+d x^2}}{128 b d^4}+\frac {5 (b c-a d) (7 b c+a d) \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{192 b d^3}-\frac {(7 b c+a d) \left (a+b x^2\right )^{5/2} \sqrt {c+d x^2}}{48 b d^2}+\frac {\left (a+b x^2\right )^{7/2} \sqrt {c+d x^2}}{8 b d}+\frac {\left (5 (b c-a d)^3 (7 b c+a d)\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 )}{128 b^2 d^4}\\ &=-\frac {5 (b c-a d)^2 (7 b c+a d) \sqrt {a+b x^2} \sqrt {c+d x^2}}{128 b d^4}+\frac {5 (b c-a d) (7 b c+a d) \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{192 b d^3}-\frac {(7 b c+a d) \left (a+b x^2\right )^{5/2} \sqrt {c+d x^2}}{48 b d^2}+\frac {\left (a+b x^2\right )^{7/2} \sqrt {c+d x^2}}{8 b d}+\frac {5 (b c-a d)^3 (7 b c+a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{128 b^{3/2} d^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.59, size = 214, normalized size = 0.90 \begin {gather*} \frac {b \sqrt {d} \sqrt {a+b x^2} \left (c+d x^2\right ) \left (15 a^3 d^3+a^2 b d^2 \left (118 d x^2-191 c\right )+a b^2 d \left (265 c^2-172 c d x^2+136 d^2 x^4\right )+b^3 \left (-105 c^3+70 c^2 d x^2-56 c d^2 x^4+48 d^3 x^6\right )\right )+15 (a d+7 b c) (b c-a d)^{7/2} \sqrt {\frac {b \left (c+d x^2\right )}{b c-a d}} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b c-a d}}\right )}{384 b^2 d^{9/2} \sqrt {c+d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(a + b*x^2)^(5/2))/Sqrt[c + d*x^2],x]

[Out]

(b*Sqrt[d]*Sqrt[a + b*x^2]*(c + d*x^2)*(15*a^3*d^3 + a^2*b*d^2*(-191*c + 118*d*x^2) + a*b^2*d*(265*c^2 - 172*c
*d*x^2 + 136*d^2*x^4) + b^3*(-105*c^3 + 70*c^2*d*x^2 - 56*c*d^2*x^4 + 48*d^3*x^6)) + 15*(b*c - a*d)^(7/2)*(7*b
*c + a*d)*Sqrt[(b*(c + d*x^2))/(b*c - a*d)]*ArcSinh[(Sqrt[d]*Sqrt[a + b*x^2])/Sqrt[b*c - a*d]])/(384*b^2*d^(9/
2)*Sqrt[c + d*x^2])

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IntegrateAlgebraic [F]  time = 3.10, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^3 \left (a+b x^2\right )^{5/2}}{\sqrt {c+d x^2}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(x^3*(a + b*x^2)^(5/2))/Sqrt[c + d*x^2],x]

[Out]

Defer[IntegrateAlgebraic][(x^3*(a + b*x^2)^(5/2))/Sqrt[c + d*x^2], x]

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fricas [A]  time = 1.35, size = 574, normalized size = 2.42 \begin {gather*} \left [-\frac {15 \, {\left (7 \, b^{4} c^{4} - 20 \, a b^{3} c^{3} d + 18 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} - a^{4} d^{4}\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 (48 \, b^{4} d^{4} x^{6} - 105 \, b^{4} c^{3} d + 265 \, a b^{3} c^{2} d^{2} - 191 \, a^{2} b^{2} c d^{3} + 15 \, a^{3} b d^{4} - 8 \, {\left (7 \, b^{4} c d^{3} - 17 \, a b^{3} d^{4}\right )} x^{4} + 2 \, {\left (35 \, b^{4} c^{2} d^{2} - 86 \, a b^{3} c d^{3} + 59 \, a^{2} b^{2} d^{4}\right )} x^{2}\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{1536 \, b^{2} d^{5}}, -\frac {15 \, {\left (7 \, b^{4} c^{4} - 20 \, a b^{3} c^{3} d + 18 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} - a^{4} d^{4}\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 (48 \, b^{4} d^{4} x^{6} - 105 \, b^{4} c^{3} d + 265 \, a b^{3} c^{2} d^{2} - 191 \, a^{2} b^{2} c d^{3} + 15 \, a^{3} b d^{4} - 8 \, {\left (7 \, b^{4} c d^{3} - 17 \, a b^{3} d^{4}\right )} x^{4} + 2 \, {\left (35 \, b^{4} c^{2} d^{2} - 86 \, a b^{3} c d^{3} + 59 \, a^{2} b^{2} d^{4}\right )} x^{2}\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{768 \, b^{2} d^{5}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b*x^2+a)^(5/2)/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

[-1/1536*(15*(7*b^4*c^4 - 20*a*b^3*c^3*d + 18*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 - a^4*d^4)*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*(48*b^4*d^4*x^6 - 105*b^4*c^3*d + 265*a*b^3*c^2*d^2 - 191*a^2*b^2*c*d^3 + 15*a
^3*b*d^4 - 8*(7*b^4*c*d^3 - 17*a*b^3*d^4)*x^4 + 2*(35*b^4*c^2*d^2 - 86*a*b^3*c*d^3 + 59*a^2*b^2*d^4)*x^2)*sqrt
(b*x^2 + a)*sqrt(d*x^2 + c))/(b^2*d^5), -1/768*(15*(7*b^4*c^4 - 20*a*b^3*c^3*d + 18*a^2*b^2*c^2*d^2 - 4*a^3*b*
c*d^3 - a^4*d^4)*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*(48*b^4*d^4*x^6 - 105*b^4*c^3*d + 265*a*b^3*c^2*d^2 - 191*a
^2*b^2*c*d^3 + 15*a^3*b*d^4 - 8*(7*b^4*c*d^3 - 17*a*b^3*d^4)*x^4 + 2*(35*b^4*c^2*d^2 - 86*a*b^3*c*d^3 + 59*a^2
*b^2*d^4)*x^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c))/(b^2*d^5)]

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giac [A]  time = 0.57, size = 304, normalized size = 1.28 \begin {gather*} \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 (\frac {6 \, {\left (b x^{2} + a\right )}}{b^{2} d} - \frac {7 \, b^{3} c d^{5} + a b^{2} d^{6}}{b^{4} d^{7}}\right )} + \frac {5 \, {\left (7 \, b^{4} c^{2} d^{4} - 6 \, a b^{3} c d^{5} - a^{2} b^{2} d^{6}\right )}}{b^{4} d^{7}}\right )} - \frac {15 \, {\left (7 \, b^{5} c^{3} d^{3} - 13 \, a b^{4} c^{2} d^{4} + 5 \, a^{2} b^{3} c d^{5} + a^{3} b^{2} d^{6}\right )}}{b^{4} d^{7}}\right )} - \frac {15 \, {\left (7 \, b^{4} c^{4} - 20 \, a b^{3} c^{3} d + 18 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} - a^{4} d^{4}\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 d^{4}}\right )} b}{384 \, {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b*x^2+a)^(5/2)/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

1/384*(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)/(b^2
*d) - (7*b^3*c*d^5 + a*b^2*d^6)/(b^4*d^7)) + 5*(7*b^4*c^2*d^4 - 6*a*b^3*c*d^5 - a^2*b^2*d^6)/(b^4*d^7)) - 15*(
7*b^5*c^3*d^3 - 13*a*b^4*c^2*d^4 + 5*a^2*b^3*c*d^5 + a^3*b^2*d^6)/(b^4*d^7)) - 15*(7*b^4*c^4 - 20*a*b^3*c^3*d
+ 18*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 - a^4*d^4)*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*d^4))*b/abs(b)

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maple [B]  time = 0.02, size = 770, normalized size = 3.25 \begin {gather*} -\frac {\sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}\, \left (-96 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, b^{3} d^{3} x^{6}-272 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a \,b^{2} d^{3} x^{4}+112 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, b^{3} c \,d^{2} x^{4}+15 a^{4} 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 )+60 a^{3} b c \,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 )-270 a^{2} b^{2} c^{2} 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 )+300 a \,b^{3} c^{3} 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 )-105 b^{4} c^{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 )-236 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a^{2} b \,d^{3} x^{2}+344 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a \,b^{2} c \,d^{2} x^{2}-140 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, b^{3} c^{2} d \,x^{2}-30 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a^{3} d^{3}+382 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a^{2} b c \,d^{2}-530 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a \,b^{2} c^{2} d +210 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, b^{3} c^{3}\right )}{768 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, b \,d^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(b*x^2+a)^(5/2)/(d*x^2+c)^(1/2),x)

[Out]

-1/768*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)*(-96*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*b^3*d^3*x^6-272*(b
*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*a*b^2*d^3*x^4+112*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2
)*b^3*c*d^2*x^4-236*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*a^2*b*d^3*x^2+344*(b*d*x^4+a*d*x^2+b*c*x^2
+a*c)^(1/2)*(b*d)^(1/2)*a*b^2*c*d^2*x^2-140*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*b^3*c^2*d*x^2+15*a
^4*d^4*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))+60*a^3*b*c*d^
3*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))-270*a^2*b^2*c^2*d^
2*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))+300*a*b^3*c^3*d*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))-105*b^4*c^4*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))-30*(b*d*x^4+a*d*x^2+b*c*x^2+a*
c)^(1/2)*(b*d)^(1/2)*a^3*d^3+382*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*a^2*b*c*d^2-530*(b*d*x^4+a*d*
x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*a*b^2*c^2*d+210*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*b^3*c^3)/b/
d^4/(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 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(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 \begin {gather*} \int \frac {x^3\,{\left (b\,x^2+a\right )}^{5/2}}{\sqrt {d\,x^2+c}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(a + b*x^2)^(5/2))/(c + d*x^2)^(1/2),x)

[Out]

int((x^3*(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 \begin {gather*} \int \frac {x^{3} \left (a + b x^{2}\right )^{\frac {5}{2}}}{\sqrt {c + d x^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(b*x**2+a)**(5/2)/(d*x**2+c)**(1/2),x)

[Out]

Integral(x**3*(a + b*x**2)**(5/2)/sqrt(c + d*x**2), x)

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