3.757 \(\int \frac {(c+d x^2)^{5/2}}{x^4 (a+b x^2)^2} \, dx\)

Optimal. Leaf size=176 \[ \frac {5 c (b c-a d)^{3/2} \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 a^{7/2}}-\frac {c \sqrt {c+d x^2} (5 b c-3 a d)}{6 a^2 b x^3}+\frac {\sqrt {c+d x^2} \left (3 a^2 d^2-20 a b c d+15 b^2 c^2\right )}{6 a^3 b x}+\frac {\left (c+d x^2\right )^{3/2} (b c-a d)}{2 a b x^3 \left (a+b x^2\right )} \]

[Out]

1/2*(-a*d+b*c)*(d*x^2+c)^(3/2)/a/b/x^3/(b*x^2+a)+5/2*c*(-a*d+b*c)^(3/2)*arctan(x*(-a*d+b*c)^(1/2)/a^(1/2)/(d*x
^2+c)^(1/2))/a^(7/2)-1/6*c*(-3*a*d+5*b*c)*(d*x^2+c)^(1/2)/a^2/b/x^3+1/6*(3*a^2*d^2-20*a*b*c*d+15*b^2*c^2)*(d*x
^2+c)^(1/2)/a^3/b/x

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Rubi [A]  time = 0.25, antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {468, 580, 583, 12, 377, 205} \[ \frac {\sqrt {c+d x^2} \left (3 a^2 d^2-20 a b c d+15 b^2 c^2\right )}{6 a^3 b x}-\frac {c \sqrt {c+d x^2} (5 b c-3 a d)}{6 a^2 b x^3}+\frac {5 c (b c-a d)^{3/2} \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 a^{7/2}}+\frac {\left (c+d x^2\right )^{3/2} (b c-a d)}{2 a b x^3 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^2)^(5/2)/(x^4*(a + b*x^2)^2),x]

[Out]

-(c*(5*b*c - 3*a*d)*Sqrt[c + d*x^2])/(6*a^2*b*x^3) + ((15*b^2*c^2 - 20*a*b*c*d + 3*a^2*d^2)*Sqrt[c + d*x^2])/(
6*a^3*b*x) + ((b*c - a*d)*(c + d*x^2)^(3/2))/(2*a*b*x^3*(a + b*x^2)) + (5*c*(b*c - a*d)^(3/2)*ArcTan[(Sqrt[b*c
- a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/(2*a^(7/2))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 205

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

Rule 377

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

Rule 468

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

Rule 580

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(a*g*(m + 1)), x] - Dist[1/(a*g^n*(m + 1
)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*(b*e - a*f)*(m + 1) + e*n*(b*c*(p + 1) + a*d*q)
+ d*((b*e - a*f)*(m + 1) + b*e*n*(p + q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && IGtQ[n
, 0] && GtQ[q, 0] && LtQ[m, -1] &&  !(EqQ[q, 1] && SimplerQ[e + f*x^n, c + d*x^n])

Rule 583

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

Rubi steps

\begin {align*} \int \frac {\left (c+d x^2\right )^{5/2}}{x^4 \left (a+b x^2\right )^2} \, dx &=\frac {(b c-a d) \left (c+d x^2\right )^{3/2}}{2 a b x^3 \left (a+b x^2\right )}-\frac {\int \frac {\sqrt {c+d x^2} \left (-c (5 b c-3 a d)-2 b c d x^2\right )}{x^4 \left (a+b x^2\right )} \, dx}{2 a b}\\ &=-\frac {c (5 b c-3 a d) \sqrt {c+d x^2}}{6 a^2 b x^3}+\frac {(b c-a d) \left (c+d x^2\right )^{3/2}}{2 a b x^3 \left (a+b x^2\right )}-\frac {\int \frac {c \left (15 b^2 c^2-20 a b c d+3 a^2 d^2\right )+2 b c d (5 b c-6 a d) x^2}{x^2 \left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{6 a^2 b}\\ &=-\frac {c (5 b c-3 a d) \sqrt {c+d x^2}}{6 a^2 b x^3}+\frac {\left (15 b^2 c^2-20 a b c d+3 a^2 d^2\right ) \sqrt {c+d x^2}}{6 a^3 b x}+\frac {(b c-a d) \left (c+d x^2\right )^{3/2}}{2 a b x^3 \left (a+b x^2\right )}+\frac {\int \frac {15 b c^2 (b c-a d)^2}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{6 a^3 b c}\\ &=-\frac {c (5 b c-3 a d) \sqrt {c+d x^2}}{6 a^2 b x^3}+\frac {\left (15 b^2 c^2-20 a b c d+3 a^2 d^2\right ) \sqrt {c+d x^2}}{6 a^3 b x}+\frac {(b c-a d) \left (c+d x^2\right )^{3/2}}{2 a b x^3 \left (a+b x^2\right )}+\frac {\left (5 c (b c-a d)^2\right ) \int \frac {1}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{2 a^3}\\ &=-\frac {c (5 b c-3 a d) \sqrt {c+d x^2}}{6 a^2 b x^3}+\frac {\left (15 b^2 c^2-20 a b c d+3 a^2 d^2\right ) \sqrt {c+d x^2}}{6 a^3 b x}+\frac {(b c-a d) \left (c+d x^2\right )^{3/2}}{2 a b x^3 \left (a+b x^2\right )}+\frac {\left (5 c (b c-a d)^2\right ) \operatorname {Subst}\left (\int \frac {1}{a-(-b c+a d) x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )}{2 a^3}\\ &=-\frac {c (5 b c-3 a d) \sqrt {c+d x^2}}{6 a^2 b x^3}+\frac {\left (15 b^2 c^2-20 a b c d+3 a^2 d^2\right ) \sqrt {c+d x^2}}{6 a^3 b x}+\frac {(b c-a d) \left (c+d x^2\right )^{3/2}}{2 a b x^3 \left (a+b x^2\right )}+\frac {5 c (b c-a d)^{3/2} \tan ^{-1}\left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 a^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 54, normalized size = 0.31 \[ -\frac {c \left (c+d x^2\right )^{3/2} \, _2F_1\left (-\frac {3}{2},2;-\frac {1}{2};\frac {(a d-b c) x^2}{a \left (d x^2+c\right )}\right )}{3 a^2 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^2)^(5/2)/(x^4*(a + b*x^2)^2),x]

[Out]

-1/3*(c*(c + d*x^2)^(3/2)*Hypergeometric2F1[-3/2, 2, -1/2, ((-(b*c) + a*d)*x^2)/(a*(c + d*x^2))])/(a^2*x^3)

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fricas [A]  time = 0.73, size = 483, normalized size = 2.74 \[ \left [-\frac {15 \, {\left ({\left (b^{2} c^{2} - a b c d\right )} x^{5} + {\left (a b c^{2} - a^{2} c d\right )} x^{3}\right )} \sqrt {-\frac {b c - a d}{a}} \log \left (\frac {{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} - 2 \, {\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{2} + 4 \, {\left (a^{2} c x - {\left (a b c - 2 \, a^{2} d\right )} x^{3}\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b c - a d}{a}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) - 4 \, {\left ({\left (15 \, b^{2} c^{2} - 20 \, a b c d + 3 \, a^{2} d^{2}\right )} x^{4} - 2 \, a^{2} c^{2} + 2 \, {\left (5 \, a b c^{2} - 7 \, a^{2} c d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{24 \, {\left (a^{3} b x^{5} + a^{4} x^{3}\right )}}, \frac {15 \, {\left ({\left (b^{2} c^{2} - a b c d\right )} x^{5} + {\left (a b c^{2} - a^{2} c d\right )} x^{3}\right )} \sqrt {\frac {b c - a d}{a}} \arctan \left (\frac {{\left ({\left (b c - 2 \, a d\right )} x^{2} - a c\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b c - a d}{a}}}{2 \, {\left ({\left (b c d - a d^{2}\right )} x^{3} + {\left (b c^{2} - a c d\right )} x\right )}}\right ) + 2 \, {\left ({\left (15 \, b^{2} c^{2} - 20 \, a b c d + 3 \, a^{2} d^{2}\right )} x^{4} - 2 \, a^{2} c^{2} + 2 \, {\left (5 \, a b c^{2} - 7 \, a^{2} c d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{12 \, {\left (a^{3} b x^{5} + a^{4} x^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^(5/2)/x^4/(b*x^2+a)^2,x, algorithm="fricas")

[Out]

[-1/24*(15*((b^2*c^2 - a*b*c*d)*x^5 + (a*b*c^2 - a^2*c*d)*x^3)*sqrt(-(b*c - a*d)/a)*log(((b^2*c^2 - 8*a*b*c*d
+ 8*a^2*d^2)*x^4 + a^2*c^2 - 2*(3*a*b*c^2 - 4*a^2*c*d)*x^2 + 4*(a^2*c*x - (a*b*c - 2*a^2*d)*x^3)*sqrt(d*x^2 +
c)*sqrt(-(b*c - a*d)/a))/(b^2*x^4 + 2*a*b*x^2 + a^2)) - 4*((15*b^2*c^2 - 20*a*b*c*d + 3*a^2*d^2)*x^4 - 2*a^2*c
^2 + 2*(5*a*b*c^2 - 7*a^2*c*d)*x^2)*sqrt(d*x^2 + c))/(a^3*b*x^5 + a^4*x^3), 1/12*(15*((b^2*c^2 - a*b*c*d)*x^5
+ (a*b*c^2 - a^2*c*d)*x^3)*sqrt((b*c - a*d)/a)*arctan(1/2*((b*c - 2*a*d)*x^2 - a*c)*sqrt(d*x^2 + c)*sqrt((b*c
- a*d)/a)/((b*c*d - a*d^2)*x^3 + (b*c^2 - a*c*d)*x)) + 2*((15*b^2*c^2 - 20*a*b*c*d + 3*a^2*d^2)*x^4 - 2*a^2*c^
2 + 2*(5*a*b*c^2 - 7*a^2*c*d)*x^2)*sqrt(d*x^2 + c))/(a^3*b*x^5 + a^4*x^3)]

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giac [B]  time = 4.94, size = 496, normalized size = 2.82 \[ -\frac {5 \, {\left (b^{2} c^{3} \sqrt {d} - 2 \, a b c^{2} d^{\frac {3}{2}} + a^{2} c d^{\frac {5}{2}}\right )} \arctan \left (\frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt {a b c d - a^{2} d^{2}}}\right )}{2 \, \sqrt {a b c d - a^{2} d^{2}} a^{3}} - \frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b^{3} c^{3} \sqrt {d} - 4 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a b^{2} c^{2} d^{\frac {3}{2}} + 5 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a^{2} b c d^{\frac {5}{2}} - 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a^{3} d^{\frac {7}{2}} - b^{3} c^{4} \sqrt {d} + 2 \, a b^{2} c^{3} d^{\frac {3}{2}} - a^{2} b c^{2} d^{\frac {5}{2}}}{{\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} b - 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b c + 4 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a d + b c^{2}\right )} a^{3} b} - \frac {2 \, {\left (6 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} b c^{3} \sqrt {d} - 9 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a c^{2} d^{\frac {3}{2}} - 12 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b c^{4} \sqrt {d} + 12 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a c^{3} d^{\frac {3}{2}} + 6 \, b c^{5} \sqrt {d} - 7 \, a c^{4} d^{\frac {3}{2}}\right )}}{3 \, {\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c\right )}^{3} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^(5/2)/x^4/(b*x^2+a)^2,x, algorithm="giac")

[Out]

-5/2*(b^2*c^3*sqrt(d) - 2*a*b*c^2*d^(3/2) + a^2*c*d^(5/2))*arctan(1/2*((sqrt(d)*x - sqrt(d*x^2 + c))^2*b - b*c
+ 2*a*d)/sqrt(a*b*c*d - a^2*d^2))/(sqrt(a*b*c*d - a^2*d^2)*a^3) - ((sqrt(d)*x - sqrt(d*x^2 + c))^2*b^3*c^3*sq
rt(d) - 4*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*b^2*c^2*d^(3/2) + 5*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a^2*b*c*d^(5/2
) - 2*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a^3*d^(7/2) - b^3*c^4*sqrt(d) + 2*a*b^2*c^3*d^(3/2) - a^2*b*c^2*d^(5/2))
/(((sqrt(d)*x - sqrt(d*x^2 + c))^4*b - 2*(sqrt(d)*x - sqrt(d*x^2 + c))^2*b*c + 4*(sqrt(d)*x - sqrt(d*x^2 + c))
^2*a*d + b*c^2)*a^3*b) - 2/3*(6*(sqrt(d)*x - sqrt(d*x^2 + c))^4*b*c^3*sqrt(d) - 9*(sqrt(d)*x - sqrt(d*x^2 + c)
)^4*a*c^2*d^(3/2) - 12*(sqrt(d)*x - sqrt(d*x^2 + c))^2*b*c^4*sqrt(d) + 12*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*c^
3*d^(3/2) + 6*b*c^5*sqrt(d) - 7*a*c^4*d^(3/2))/(((sqrt(d)*x - sqrt(d*x^2 + c))^2 - c)^3*a^3)

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maple [B]  time = 0.02, size = 7705, normalized size = 43.78 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^2+c)^(5/2)/x^4/(b*x^2+a)^2,x)

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}}}{{\left (b x^{2} + a\right )}^{2} x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^(5/2)/x^4/(b*x^2+a)^2,x, algorithm="maxima")

[Out]

integrate((d*x^2 + c)^(5/2)/((b*x^2 + a)^2*x^4), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (d\,x^2+c\right )}^{5/2}}{x^4\,{\left (b\,x^2+a\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x^2)^(5/2)/(x^4*(a + b*x^2)^2),x)

[Out]

int((c + d*x^2)^(5/2)/(x^4*(a + b*x^2)^2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c + d x^{2}\right )^{\frac {5}{2}}}{x^{4} \left (a + b x^{2}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**2+c)**(5/2)/x**4/(b*x**2+a)**2,x)

[Out]

Integral((c + d*x**2)**(5/2)/(x**4*(a + b*x**2)**2), x)

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