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

Optimal. Leaf size=198 \[ -\frac {(2 b c-7 a d) (b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 b^{9/2}}+\frac {\sqrt {c+d x^2} (2 b c-7 a d) (b c-a d)}{2 b^4}+\frac {\left (c+d x^2\right )^{3/2} (2 b c-7 a d)}{6 b^3}+\frac {\left (c+d x^2\right )^{5/2} (2 b c-7 a d)}{10 b^2 (b c-a d)}+\frac {a \left (c+d x^2\right )^{7/2}}{2 b \left (a+b x^2\right ) (b c-a d)} \]

[Out]

1/6*(-7*a*d+2*b*c)*(d*x^2+c)^(3/2)/b^3+1/10*(-7*a*d+2*b*c)*(d*x^2+c)^(5/2)/b^2/(-a*d+b*c)+1/2*a*(d*x^2+c)^(7/2
)/b/(-a*d+b*c)/(b*x^2+a)-1/2*(-7*a*d+2*b*c)*(-a*d+b*c)^(3/2)*arctanh(b^(1/2)*(d*x^2+c)^(1/2)/(-a*d+b*c)^(1/2))
/b^(9/2)+1/2*(-7*a*d+2*b*c)*(-a*d+b*c)*(d*x^2+c)^(1/2)/b^4

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Rubi [A]  time = 0.18, antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {446, 78, 50, 63, 208} \[ \frac {\left (c+d x^2\right )^{5/2} (2 b c-7 a d)}{10 b^2 (b c-a d)}+\frac {\left (c+d x^2\right )^{3/2} (2 b c-7 a d)}{6 b^3}+\frac {\sqrt {c+d x^2} (2 b c-7 a d) (b c-a d)}{2 b^4}-\frac {(2 b c-7 a d) (b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 b^{9/2}}+\frac {a \left (c+d x^2\right )^{7/2}}{2 b \left (a+b x^2\right ) (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*b*c - 7*a*d)*(b*c - a*d)*Sqrt[c + d*x^2])/(2*b^4) + ((2*b*c - 7*a*d)*(c + d*x^2)^(3/2))/(6*b^3) + ((2*b*c
- 7*a*d)*(c + d*x^2)^(5/2))/(10*b^2*(b*c - a*d)) + (a*(c + d*x^2)^(7/2))/(2*b*(b*c - a*d)*(a + b*x^2)) - ((2*b
*c - 7*a*d)*(b*c - a*d)^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^2])/Sqrt[b*c - a*d]])/(2*b^(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 78

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

Rule 208

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

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 (c+d x^2\right )^{5/2}}{\left (a+b x^2\right )^2} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x (c+d x)^{5/2}}{(a+b x)^2} \, dx,x,x^2\right )\\ &=\frac {a \left (c+d x^2\right )^{7/2}}{2 b (b c-a d) \left (a+b x^2\right )}+\frac {(2 b c-7 a d) \operatorname {Subst}\left (\int \frac {(c+d x)^{5/2}}{a+b x} \, dx,x,x^2\right )}{4 b (b c-a d)}\\ &=\frac {(2 b c-7 a d) \left (c+d x^2\right )^{5/2}}{10 b^2 (b c-a d)}+\frac {a \left (c+d x^2\right )^{7/2}}{2 b (b c-a d) \left (a+b x^2\right )}+\frac {(2 b c-7 a d) \operatorname {Subst}\left (\int \frac {(c+d x)^{3/2}}{a+b x} \, dx,x,x^2\right )}{4 b^2}\\ &=\frac {(2 b c-7 a d) \left (c+d x^2\right )^{3/2}}{6 b^3}+\frac {(2 b c-7 a d) \left (c+d x^2\right )^{5/2}}{10 b^2 (b c-a d)}+\frac {a \left (c+d x^2\right )^{7/2}}{2 b (b c-a d) \left (a+b x^2\right )}+\frac {((2 b c-7 a d) (b c-a d)) \operatorname {Subst}\left (\int \frac {\sqrt {c+d x}}{a+b x} \, dx,x,x^2\right )}{4 b^3}\\ &=\frac {(2 b c-7 a d) (b c-a d) \sqrt {c+d x^2}}{2 b^4}+\frac {(2 b c-7 a d) \left (c+d x^2\right )^{3/2}}{6 b^3}+\frac {(2 b c-7 a d) \left (c+d x^2\right )^{5/2}}{10 b^2 (b c-a d)}+\frac {a \left (c+d x^2\right )^{7/2}}{2 b (b c-a d) \left (a+b x^2\right )}+\frac {\left ((2 b c-7 a d) (b c-a d)^2\right ) \operatorname {Subst}\left (\int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx,x,x^2\right )}{4 b^4}\\ &=\frac {(2 b c-7 a d) (b c-a d) \sqrt {c+d x^2}}{2 b^4}+\frac {(2 b c-7 a d) \left (c+d x^2\right )^{3/2}}{6 b^3}+\frac {(2 b c-7 a d) \left (c+d x^2\right )^{5/2}}{10 b^2 (b c-a d)}+\frac {a \left (c+d x^2\right )^{7/2}}{2 b (b c-a d) \left (a+b x^2\right )}+\frac {\left ((2 b c-7 a d) (b c-a d)^2\right ) \operatorname {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{2 b^4 d}\\ &=\frac {(2 b c-7 a d) (b c-a d) \sqrt {c+d x^2}}{2 b^4}+\frac {(2 b c-7 a d) \left (c+d x^2\right )^{3/2}}{6 b^3}+\frac {(2 b c-7 a d) \left (c+d x^2\right )^{5/2}}{10 b^2 (b c-a d)}+\frac {a \left (c+d x^2\right )^{7/2}}{2 b (b c-a d) \left (a+b x^2\right )}-\frac {(2 b c-7 a d) (b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 b^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.43, size = 164, normalized size = 0.83 \[ \frac {\left (b c-\frac {7 a d}{2}\right ) \left (\frac {2 (b c-a d) \left (\sqrt {b} \sqrt {c+d x^2} \left (-3 a d+4 b c+b d x^2\right )-3 (b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )\right )}{3 b^{7/2}}+\frac {2 \left (c+d x^2\right )^{5/2}}{5 b}\right )+\frac {a \left (c+d x^2\right )^{7/2}}{a+b x^2}}{2 b (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.15, size = 573, normalized size = 2.89 \[ \left [\frac {15 \, {\left (2 \, a b^{2} c^{2} - 9 \, a^{2} b c d + 7 \, a^{3} d^{2} + {\left (2 \, b^{3} c^{2} - 9 \, a b^{2} c d + 7 \, a^{2} b d^{2}\right )} x^{2}\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \, {\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} - 4 \, {\left (b^{2} d x^{2} + 2 \, b^{2} c - a b d\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b c - a d}{b}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 4 \, {\left (6 \, b^{3} d^{2} x^{6} + 61 \, a b^{2} c^{2} - 170 \, a^{2} b c d + 105 \, a^{3} d^{2} + 2 \, {\left (11 \, b^{3} c d - 7 \, a b^{2} d^{2}\right )} x^{4} + 2 \, {\left (23 \, b^{3} c^{2} - 59 \, a b^{2} c d + 35 \, a^{2} b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{120 \, {\left (b^{5} x^{2} + a b^{4}\right )}}, -\frac {15 \, {\left (2 \, a b^{2} c^{2} - 9 \, a^{2} b c d + 7 \, a^{3} d^{2} + {\left (2 \, b^{3} c^{2} - 9 \, a b^{2} c d + 7 \, a^{2} b d^{2}\right )} x^{2}\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b c - a d}{b}}}{2 \, {\left (b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x^{2}\right )}}\right ) - 2 \, {\left (6 \, b^{3} d^{2} x^{6} + 61 \, a b^{2} c^{2} - 170 \, a^{2} b c d + 105 \, a^{3} d^{2} + 2 \, {\left (11 \, b^{3} c d - 7 \, a b^{2} d^{2}\right )} x^{4} + 2 \, {\left (23 \, b^{3} c^{2} - 59 \, a b^{2} c d + 35 \, a^{2} b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{60 \, {\left (b^{5} x^{2} + a b^{4}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/120*(15*(2*a*b^2*c^2 - 9*a^2*b*c*d + 7*a^3*d^2 + (2*b^3*c^2 - 9*a*b^2*c*d + 7*a^2*b*d^2)*x^2)*sqrt((b*c - a
*d)/b)*log((b^2*d^2*x^4 + 8*b^2*c^2 - 8*a*b*c*d + a^2*d^2 + 2*(4*b^2*c*d - 3*a*b*d^2)*x^2 - 4*(b^2*d*x^2 + 2*b
^2*c - a*b*d)*sqrt(d*x^2 + c)*sqrt((b*c - a*d)/b))/(b^2*x^4 + 2*a*b*x^2 + a^2)) + 4*(6*b^3*d^2*x^6 + 61*a*b^2*
c^2 - 170*a^2*b*c*d + 105*a^3*d^2 + 2*(11*b^3*c*d - 7*a*b^2*d^2)*x^4 + 2*(23*b^3*c^2 - 59*a*b^2*c*d + 35*a^2*b
*d^2)*x^2)*sqrt(d*x^2 + c))/(b^5*x^2 + a*b^4), -1/60*(15*(2*a*b^2*c^2 - 9*a^2*b*c*d + 7*a^3*d^2 + (2*b^3*c^2 -
9*a*b^2*c*d + 7*a^2*b*d^2)*x^2)*sqrt(-(b*c - a*d)/b)*arctan(-1/2*(b*d*x^2 + 2*b*c - a*d)*sqrt(d*x^2 + c)*sqrt
(-(b*c - a*d)/b)/(b*c^2 - a*c*d + (b*c*d - a*d^2)*x^2)) - 2*(6*b^3*d^2*x^6 + 61*a*b^2*c^2 - 170*a^2*b*c*d + 10
5*a^3*d^2 + 2*(11*b^3*c*d - 7*a*b^2*d^2)*x^4 + 2*(23*b^3*c^2 - 59*a*b^2*c*d + 35*a^2*b*d^2)*x^2)*sqrt(d*x^2 +
c))/(b^5*x^2 + a*b^4)]

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giac [A]  time = 0.40, size = 264, normalized size = 1.33 \[ \frac {{\left (2 \, b^{3} c^{3} - 11 \, a b^{2} c^{2} d + 16 \, a^{2} b c d^{2} - 7 \, a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{2 \, \sqrt {-b^{2} c + a b d} b^{4}} + \frac {\sqrt {d x^{2} + c} a b^{2} c^{2} d - 2 \, \sqrt {d x^{2} + c} a^{2} b c d^{2} + \sqrt {d x^{2} + c} a^{3} d^{3}}{2 \, {\left ({\left (d x^{2} + c\right )} b - b c + a d\right )} b^{4}} + \frac {3 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{8} + 5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{8} c + 15 \, \sqrt {d x^{2} + c} b^{8} c^{2} - 10 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b^{7} d - 60 \, \sqrt {d x^{2} + c} a b^{7} c d + 45 \, \sqrt {d x^{2} + c} a^{2} b^{6} d^{2}}{15 \, b^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*b^3*c^3 - 11*a*b^2*c^2*d + 16*a^2*b*c*d^2 - 7*a^3*d^3)*arctan(sqrt(d*x^2 + c)*b/sqrt(-b^2*c + a*b*d))/(
sqrt(-b^2*c + a*b*d)*b^4) + 1/2*(sqrt(d*x^2 + c)*a*b^2*c^2*d - 2*sqrt(d*x^2 + c)*a^2*b*c*d^2 + sqrt(d*x^2 + c)
*a^3*d^3)/(((d*x^2 + c)*b - b*c + a*d)*b^4) + 1/15*(3*(d*x^2 + c)^(5/2)*b^8 + 5*(d*x^2 + c)^(3/2)*b^8*c + 15*s
qrt(d*x^2 + c)*b^8*c^2 - 10*(d*x^2 + c)^(3/2)*a*b^7*d - 60*sqrt(d*x^2 + c)*a*b^7*c*d + 45*sqrt(d*x^2 + c)*a^2*
b^6*d^2)/b^10

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maple [B]  time = 0.02, size = 7443, normalized size = 37.59 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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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^3*(d*x^2+c)^(5/2)/(b*x^2+a)^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 positive or negative?

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mupad [B]  time = 1.17, size = 276, normalized size = 1.39 \[ \frac {{\left (d\,x^2+c\right )}^{5/2}}{5\,b^2}-\sqrt {d\,x^2+c}\,\left (\frac {{\left (a\,d-b\,c\right )}^2}{b^4}+\frac {\left (2\,b^2\,c-2\,a\,b\,d\right )\,\left (\frac {c}{b^2}-\frac {2\,b^2\,c-2\,a\,b\,d}{b^4}\right )}{b^2}\right )-{\left (d\,x^2+c\right )}^{3/2}\,\left (\frac {c}{3\,b^2}-\frac {2\,b^2\,c-2\,a\,b\,d}{3\,b^4}\right )+\frac {\sqrt {d\,x^2+c}\,\left (\frac {a^3\,d^3}{2}-a^2\,b\,c\,d^2+\frac {a\,b^2\,c^2\,d}{2}\right )}{b^5\,\left (d\,x^2+c\right )-b^5\,c+a\,b^4\,d}-\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d\,x^2+c}\,{\left (a\,d-b\,c\right )}^{3/2}\,\left (7\,a\,d-2\,b\,c\right )}{7\,a^3\,d^3-16\,a^2\,b\,c\,d^2+11\,a\,b^2\,c^2\,d-2\,b^3\,c^3}\right )\,{\left (a\,d-b\,c\right )}^{3/2}\,\left (7\,a\,d-2\,b\,c\right )}{2\,b^{9/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c + d*x^2)^(5/2)/(5*b^2) - (c + d*x^2)^(1/2)*((a*d - b*c)^2/b^4 + ((2*b^2*c - 2*a*b*d)*(c/b^2 - (2*b^2*c - 2*
a*b*d)/b^4))/b^2) - (c + d*x^2)^(3/2)*(c/(3*b^2) - (2*b^2*c - 2*a*b*d)/(3*b^4)) + ((c + d*x^2)^(1/2)*((a^3*d^3
)/2 + (a*b^2*c^2*d)/2 - a^2*b*c*d^2))/(b^5*(c + d*x^2) - b^5*c + a*b^4*d) - (atan((b^(1/2)*(c + d*x^2)^(1/2)*(
a*d - b*c)^(3/2)*(7*a*d - 2*b*c))/(7*a^3*d^3 - 2*b^3*c^3 + 11*a*b^2*c^2*d - 16*a^2*b*c*d^2))*(a*d - b*c)^(3/2)
*(7*a*d - 2*b*c))/(2*b^(9/2))

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

[Out]

Timed out

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