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

Optimal. Leaf size=175 \[ -\frac {a^2 \left (c+d x^2\right )^{5/2}}{c x}-\frac {c \left (b^2 c^2-12 a d (2 a d+b c)\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{16 d^{3/2}}-\frac {x \left (c+d x^2\right )^{3/2} \left (b^2 c^2-12 a d (2 a d+b c)\right )}{24 c d}-\frac {x \sqrt {c+d x^2} \left (b^2 c^2-12 a d (2 a d+b c)\right )}{16 d}+\frac {b^2 x \left (c+d x^2\right )^{5/2}}{6 d} \]

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Rubi [A]  time = 0.12, antiderivative size = 172, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {462, 388, 195, 217, 206} \begin {gather*} -\frac {a^2 \left (c+d x^2\right )^{5/2}}{c x}-\frac {c \left (b^2 c^2-12 a d (2 a d+b c)\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{16 d^{3/2}}-\frac {x \sqrt {c+d x^2} \left (b^2 c^2-12 a d (2 a d+b c)\right )}{16 d}-\frac {1}{24} x \left (c+d x^2\right )^{3/2} \left (\frac {b^2 c}{d}-\frac {12 a (2 a d+b c)}{c}\right )+\frac {b^2 x \left (c+d x^2\right )^{5/2}}{6 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((b^2*c^2 - 12*a*d*(b*c + 2*a*d))*x*Sqrt[c + d*x^2])/(16*d) - (((b^2*c)/d - (12*a*(b*c + 2*a*d))/c)*x*(c + d*
x^2)^(3/2))/24 - (a^2*(c + d*x^2)^(5/2))/(c*x) + (b^2*x*(c + d*x^2)^(5/2))/(6*d) - (c*(b^2*c^2 - 12*a*d*(b*c +
2*a*d))*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/(16*d^(3/2))

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 388

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

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]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{x^2} \, dx &=-\frac {a^2 \left (c+d x^2\right )^{5/2}}{c x}+\frac {\int \left (2 a (b c+2 a d)+b^2 c x^2\right ) \left (c+d x^2\right )^{3/2} \, dx}{c}\\ &=-\frac {a^2 \left (c+d x^2\right )^{5/2}}{c x}+\frac {b^2 x \left (c+d x^2\right )^{5/2}}{6 d}-\frac {\left (b^2 c^2-12 a d (b c+2 a d)\right ) \int \left (c+d x^2\right )^{3/2} \, dx}{6 c d}\\ &=-\frac {1}{24} \left (\frac {b^2 c}{d}-\frac {12 a (b c+2 a d)}{c}\right ) x \left (c+d x^2\right )^{3/2}-\frac {a^2 \left (c+d x^2\right )^{5/2}}{c x}+\frac {b^2 x \left (c+d x^2\right )^{5/2}}{6 d}-\frac {\left (b^2 c^2-12 a d (b c+2 a d)\right ) \int \sqrt {c+d x^2} \, dx}{8 d}\\ &=-\frac {\left (b^2 c^2-12 a d (b c+2 a d)\right ) x \sqrt {c+d x^2}}{16 d}-\frac {1}{24} \left (\frac {b^2 c}{d}-\frac {12 a (b c+2 a d)}{c}\right ) x \left (c+d x^2\right )^{3/2}-\frac {a^2 \left (c+d x^2\right )^{5/2}}{c x}+\frac {b^2 x \left (c+d x^2\right )^{5/2}}{6 d}-\frac {\left (c \left (b^2 c^2-12 a d (b c+2 a d)\right )\right ) \int \frac {1}{\sqrt {c+d x^2}} \, dx}{16 d}\\ &=-\frac {\left (b^2 c^2-12 a d (b c+2 a d)\right ) x \sqrt {c+d x^2}}{16 d}-\frac {1}{24} \left (\frac {b^2 c}{d}-\frac {12 a (b c+2 a d)}{c}\right ) x \left (c+d x^2\right )^{3/2}-\frac {a^2 \left (c+d x^2\right )^{5/2}}{c x}+\frac {b^2 x \left (c+d x^2\right )^{5/2}}{6 d}-\frac {\left (c \left (b^2 c^2-12 a d (b c+2 a d)\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )}{16 d}\\ &=-\frac {\left (b^2 c^2-12 a d (b c+2 a d)\right ) x \sqrt {c+d x^2}}{16 d}-\frac {1}{24} \left (\frac {b^2 c}{d}-\frac {12 a (b c+2 a d)}{c}\right ) x \left (c+d x^2\right )^{3/2}-\frac {a^2 \left (c+d x^2\right )^{5/2}}{c x}+\frac {b^2 x \left (c+d x^2\right )^{5/2}}{6 d}-\frac {c \left (b^2 c^2-12 a d (b c+2 a d)\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{16 d^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 135, normalized size = 0.77 \begin {gather*} \sqrt {c+d x^2} \left (\frac {x \left (8 a^2 d^2+20 a b c d+b^2 c^2\right )}{16 d}-\frac {a^2 c}{x}+\frac {1}{24} b x^3 (12 a d+7 b c)+\frac {1}{6} b^2 d x^5\right )-\frac {c \left (-24 a^2 d^2-12 a b c d+b^2 c^2\right ) \log \left (\sqrt {d} \sqrt {c+d x^2}+d x\right )}{16 d^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[c + d*x^2]*(-((a^2*c)/x) + ((b^2*c^2 + 20*a*b*c*d + 8*a^2*d^2)*x)/(16*d) + (b*(7*b*c + 12*a*d)*x^3)/24 +
(b^2*d*x^5)/6) - (c*(b^2*c^2 - 12*a*b*c*d - 24*a^2*d^2)*Log[d*x + Sqrt[d]*Sqrt[c + d*x^2]])/(16*d^(3/2))

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IntegrateAlgebraic [A]  time = 0.30, size = 147, normalized size = 0.84 \begin {gather*} \frac {\sqrt {c+d x^2} \left (-48 a^2 c d+24 a^2 d^2 x^2+60 a b c d x^2+24 a b d^2 x^4+3 b^2 c^2 x^2+14 b^2 c d x^4+8 b^2 d^2 x^6\right )}{48 d x}+\frac {\left (-24 a^2 c d^2-12 a b c^2 d+b^2 c^3\right ) \log \left (\sqrt {c+d x^2}-\sqrt {d} x\right )}{16 d^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d*x^2]*(-48*a^2*c*d + 3*b^2*c^2*x^2 + 60*a*b*c*d*x^2 + 24*a^2*d^2*x^2 + 14*b^2*c*d*x^4 + 24*a*b*d^2*
x^4 + 8*b^2*d^2*x^6))/(48*d*x) + ((b^2*c^3 - 12*a*b*c^2*d - 24*a^2*c*d^2)*Log[-(Sqrt[d]*x) + Sqrt[c + d*x^2]])
/(16*d^(3/2))

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fricas [A]  time = 1.36, size = 293, normalized size = 1.67 \begin {gather*} \left [-\frac {3 \, {\left (b^{2} c^{3} - 12 \, a b c^{2} d - 24 \, a^{2} c d^{2}\right )} \sqrt {d} x \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) - 2 \, {\left (8 \, b^{2} d^{3} x^{6} - 48 \, a^{2} c d^{2} + 2 \, {\left (7 \, b^{2} c d^{2} + 12 \, a b d^{3}\right )} x^{4} + 3 \, {\left (b^{2} c^{2} d + 20 \, a b c d^{2} + 8 \, a^{2} d^{3}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{96 \, d^{2} x}, \frac {3 \, {\left (b^{2} c^{3} - 12 \, a b c^{2} d - 24 \, a^{2} c d^{2}\right )} \sqrt {-d} x \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) + {\left (8 \, b^{2} d^{3} x^{6} - 48 \, a^{2} c d^{2} + 2 \, {\left (7 \, b^{2} c d^{2} + 12 \, a b d^{3}\right )} x^{4} + 3 \, {\left (b^{2} c^{2} d + 20 \, a b c d^{2} + 8 \, a^{2} d^{3}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{48 \, d^{2} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/96*(3*(b^2*c^3 - 12*a*b*c^2*d - 24*a^2*c*d^2)*sqrt(d)*x*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) -
2*(8*b^2*d^3*x^6 - 48*a^2*c*d^2 + 2*(7*b^2*c*d^2 + 12*a*b*d^3)*x^4 + 3*(b^2*c^2*d + 20*a*b*c*d^2 + 8*a^2*d^3)*
x^2)*sqrt(d*x^2 + c))/(d^2*x), 1/48*(3*(b^2*c^3 - 12*a*b*c^2*d - 24*a^2*c*d^2)*sqrt(-d)*x*arctan(sqrt(-d)*x/sq
rt(d*x^2 + c)) + (8*b^2*d^3*x^6 - 48*a^2*c*d^2 + 2*(7*b^2*c*d^2 + 12*a*b*d^3)*x^4 + 3*(b^2*c^2*d + 20*a*b*c*d^
2 + 8*a^2*d^3)*x^2)*sqrt(d*x^2 + c))/(d^2*x)]

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giac [A]  time = 0.50, size = 173, normalized size = 0.99 \begin {gather*} \frac {2 \, a^{2} c^{2} \sqrt {d}}{{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c} + \frac {1}{48} \, {\left (2 \, {\left (4 \, b^{2} d x^{2} + \frac {7 \, b^{2} c d^{4} + 12 \, a b d^{5}}{d^{4}}\right )} x^{2} + \frac {3 \, {\left (b^{2} c^{2} d^{3} + 20 \, a b c d^{4} + 8 \, a^{2} d^{5}\right )}}{d^{4}}\right )} \sqrt {d x^{2} + c} x + \frac {{\left (b^{2} c^{3} \sqrt {d} - 12 \, a b c^{2} d^{\frac {3}{2}} - 24 \, a^{2} c d^{\frac {5}{2}}\right )} \log \left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2}\right )}{32 \, d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a^2*c^2*sqrt(d)/((sqrt(d)*x - sqrt(d*x^2 + c))^2 - c) + 1/48*(2*(4*b^2*d*x^2 + (7*b^2*c*d^4 + 12*a*b*d^5)/d^
4)*x^2 + 3*(b^2*c^2*d^3 + 20*a*b*c*d^4 + 8*a^2*d^5)/d^4)*sqrt(d*x^2 + c)*x + 1/32*(b^2*c^3*sqrt(d) - 12*a*b*c^
2*d^(3/2) - 24*a^2*c*d^(5/2))*log((sqrt(d)*x - sqrt(d*x^2 + c))^2)/d^2

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maple [A]  time = 0.01, size = 221, normalized size = 1.26 \begin {gather*} \frac {3 a^{2} c \sqrt {d}\, \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{2}+\frac {3 a b \,c^{2} \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{4 \sqrt {d}}-\frac {b^{2} c^{3} \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{16 d^{\frac {3}{2}}}+\frac {3 \sqrt {d \,x^{2}+c}\, a^{2} d x}{2}+\frac {3 \sqrt {d \,x^{2}+c}\, a b c x}{4}-\frac {\sqrt {d \,x^{2}+c}\, b^{2} c^{2} x}{16 d}+\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}} a^{2} d x}{c}+\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}} a b x}{2}-\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}} b^{2} c x}{24 d}+\frac {\left (d \,x^{2}+c \right )^{\frac {5}{2}} b^{2} x}{6 d}-\frac {\left (d \,x^{2}+c \right )^{\frac {5}{2}} a^{2}}{c x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*b^2*x*(d*x^2+c)^(5/2)/d-1/24*b^2*c/d*x*(d*x^2+c)^(3/2)-1/16*b^2*c^2/d*x*(d*x^2+c)^(1/2)-1/16*b^2*c^3/d^(3/
2)*ln(d^(1/2)*x+(d*x^2+c)^(1/2))+1/2*a*b*x*(d*x^2+c)^(3/2)+3/4*a*b*c*x*(d*x^2+c)^(1/2)+3/4*a*b*c^2/d^(1/2)*ln(
d^(1/2)*x+(d*x^2+c)^(1/2))-a^2*(d*x^2+c)^(5/2)/c/x+a^2*d/c*x*(d*x^2+c)^(3/2)+3/2*a^2*d*x*(d*x^2+c)^(1/2)+3/2*a
^2*d^(1/2)*c*ln(d^(1/2)*x+(d*x^2+c)^(1/2))

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maxima [A]  time = 1.14, size = 178, normalized size = 1.02 \begin {gather*} \frac {1}{2} \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b x + \frac {3}{4} \, \sqrt {d x^{2} + c} a b c x + \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{2} x}{6 \, d} - \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} c x}{24 \, d} - \frac {\sqrt {d x^{2} + c} b^{2} c^{2} x}{16 \, d} + \frac {3}{2} \, \sqrt {d x^{2} + c} a^{2} d x - \frac {b^{2} c^{3} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{16 \, d^{\frac {3}{2}}} + \frac {3 \, a b c^{2} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{4 \, \sqrt {d}} + \frac {3}{2} \, a^{2} c \sqrt {d} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right ) - \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(d*x^2 + c)^(3/2)*a*b*x + 3/4*sqrt(d*x^2 + c)*a*b*c*x + 1/6*(d*x^2 + c)^(5/2)*b^2*x/d - 1/24*(d*x^2 + c)^(
3/2)*b^2*c*x/d - 1/16*sqrt(d*x^2 + c)*b^2*c^2*x/d + 3/2*sqrt(d*x^2 + c)*a^2*d*x - 1/16*b^2*c^3*arcsinh(d*x/sqr
t(c*d))/d^(3/2) + 3/4*a*b*c^2*arcsinh(d*x/sqrt(c*d))/sqrt(d) + 3/2*a^2*c*sqrt(d)*arcsinh(d*x/sqrt(c*d)) - (d*x
^2 + c)^(3/2)*a^2/x

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (b\,x^2+a\right )}^2\,{\left (d\,x^2+c\right )}^{3/2}}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 22.41, size = 367, normalized size = 2.10 \begin {gather*} - \frac {a^{2} c^{\frac {3}{2}}}{x \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {a^{2} \sqrt {c} d x \sqrt {1 + \frac {d x^{2}}{c}}}{2} - \frac {a^{2} \sqrt {c} d x}{\sqrt {1 + \frac {d x^{2}}{c}}} + \frac {3 a^{2} c \sqrt {d} \operatorname {asinh}{\left (\frac {\sqrt {d} x}{\sqrt {c}} \right )}}{2} + a b c^{\frac {3}{2}} x \sqrt {1 + \frac {d x^{2}}{c}} + \frac {a b c^{\frac {3}{2}} x}{4 \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {3 a b \sqrt {c} d x^{3}}{4 \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {3 a b c^{2} \operatorname {asinh}{\left (\frac {\sqrt {d} x}{\sqrt {c}} \right )}}{4 \sqrt {d}} + \frac {a b d^{2} x^{5}}{2 \sqrt {c} \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {b^{2} c^{\frac {5}{2}} x}{16 d \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {17 b^{2} c^{\frac {3}{2}} x^{3}}{48 \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {11 b^{2} \sqrt {c} d x^{5}}{24 \sqrt {1 + \frac {d x^{2}}{c}}} - \frac {b^{2} c^{3} \operatorname {asinh}{\left (\frac {\sqrt {d} x}{\sqrt {c}} \right )}}{16 d^{\frac {3}{2}}} + \frac {b^{2} d^{2} x^{7}}{6 \sqrt {c} \sqrt {1 + \frac {d x^{2}}{c}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**2*c**(3/2)/(x*sqrt(1 + d*x**2/c)) + a**2*sqrt(c)*d*x*sqrt(1 + d*x**2/c)/2 - a**2*sqrt(c)*d*x/sqrt(1 + d*x*
*2/c) + 3*a**2*c*sqrt(d)*asinh(sqrt(d)*x/sqrt(c))/2 + a*b*c**(3/2)*x*sqrt(1 + d*x**2/c) + a*b*c**(3/2)*x/(4*sq
rt(1 + d*x**2/c)) + 3*a*b*sqrt(c)*d*x**3/(4*sqrt(1 + d*x**2/c)) + 3*a*b*c**2*asinh(sqrt(d)*x/sqrt(c))/(4*sqrt(
d)) + a*b*d**2*x**5/(2*sqrt(c)*sqrt(1 + d*x**2/c)) + b**2*c**(5/2)*x/(16*d*sqrt(1 + d*x**2/c)) + 17*b**2*c**(3
/2)*x**3/(48*sqrt(1 + d*x**2/c)) + 11*b**2*sqrt(c)*d*x**5/(24*sqrt(1 + d*x**2/c)) - b**2*c**3*asinh(sqrt(d)*x/
sqrt(c))/(16*d**(3/2)) + b**2*d**2*x**7/(6*sqrt(c)*sqrt(1 + d*x**2/c))

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