3.7.7 \(\int \frac {(a+b x^2)^2 \sqrt {c+d x^2}}{x^2} \, dx\) [607]

Optimal. Leaf size=133 \[ -\frac {\left (b^2 c^2-8 a d (b c+a d)\right ) x \sqrt {c+d x^2}}{8 c d}-\frac {a^2 \left (c+d x^2\right )^{3/2}}{c x}+\frac {b^2 x \left (c+d x^2\right )^{3/2}}{4 d}-\frac {\left (b^2 c^2-8 a d (b c+a d)\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{8 d^{3/2}} \]

[Out]

-a^2*(d*x^2+c)^(3/2)/c/x+1/4*b^2*x*(d*x^2+c)^(3/2)/d-1/8*(b^2*c^2-8*a*d*(a*d+b*c))*arctanh(x*d^(1/2)/(d*x^2+c)
^(1/2))/d^(3/2)-1/8*(b^2*c^2-8*a*d*(a*d+b*c))*x*(d*x^2+c)^(1/2)/c/d

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Rubi [A]
time = 0.06, antiderivative size = 130, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {473, 396, 201, 223, 212} \begin {gather*} -\frac {a^2 \left (c+d x^2\right )^{3/2}}{c x}-\frac {\left (b^2 c^2-8 a d (a d+b c)\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{8 d^{3/2}}-\frac {1}{8} x \sqrt {c+d x^2} \left (\frac {b^2 c}{d}-\frac {8 a (a d+b c)}{c}\right )+\frac {b^2 x \left (c+d x^2\right )^{3/2}}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*(((b^2*c)/d - (8*a*(b*c + a*d))/c)*x*Sqrt[c + d*x^2]) - (a^2*(c + d*x^2)^(3/2))/(c*x) + (b^2*x*(c + d*x^2
)^(3/2))/(4*d) - ((b^2*c^2 - 8*a*d*(b*c + a*d))*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/(8*d^(3/2))

Rule 201

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 212

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

Rule 223

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 396

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 473

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 \sqrt {c+d x^2}}{x^2} \, dx &=-\frac {a^2 \left (c+d x^2\right )^{3/2}}{c x}+\frac {\int \left (2 a (b c+a d)+b^2 c x^2\right ) \sqrt {c+d x^2} \, dx}{c}\\ &=-\frac {a^2 \left (c+d x^2\right )^{3/2}}{c x}+\frac {b^2 x \left (c+d x^2\right )^{3/2}}{4 d}-\frac {1}{4} \left (\frac {b^2 c}{d}-\frac {8 a (b c+a d)}{c}\right ) \int \sqrt {c+d x^2} \, dx\\ &=-\frac {1}{8} \left (\frac {b^2 c}{d}-\frac {8 a (b c+a d)}{c}\right ) x \sqrt {c+d x^2}-\frac {a^2 \left (c+d x^2\right )^{3/2}}{c x}+\frac {b^2 x \left (c+d x^2\right )^{3/2}}{4 d}-\frac {1}{8} \left (\frac {b^2 c^2}{d}-8 a (b c+a d)\right ) \int \frac {1}{\sqrt {c+d x^2}} \, dx\\ &=-\frac {1}{8} \left (\frac {b^2 c}{d}-\frac {8 a (b c+a d)}{c}\right ) x \sqrt {c+d x^2}-\frac {a^2 \left (c+d x^2\right )^{3/2}}{c x}+\frac {b^2 x \left (c+d x^2\right )^{3/2}}{4 d}-\frac {1}{8} \left (\frac {b^2 c^2}{d}-8 a (b c+a d)\right ) \text {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )\\ &=-\frac {1}{8} \left (\frac {b^2 c}{d}-\frac {8 a (b c+a d)}{c}\right ) x \sqrt {c+d x^2}-\frac {a^2 \left (c+d x^2\right )^{3/2}}{c x}+\frac {b^2 x \left (c+d x^2\right )^{3/2}}{4 d}-\frac {\left (\frac {b^2 c^2}{d}-8 a (b c+a d)\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{8 \sqrt {d}}\\ \end {align*}

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Mathematica [A]
time = 0.16, size = 106, normalized size = 0.80 \begin {gather*} \frac {\sqrt {c+d x^2} \left (-8 a^2 d+b^2 c x^2+8 a b d x^2+2 b^2 d x^4\right )}{8 d x}+\frac {\left (b^2 c^2-8 a b c d-8 a^2 d^2\right ) \log \left (-\sqrt {d} x+\sqrt {c+d x^2}\right )}{8 d^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d*x^2]*(-8*a^2*d + b^2*c*x^2 + 8*a*b*d*x^2 + 2*b^2*d*x^4))/(8*d*x) + ((b^2*c^2 - 8*a*b*c*d - 8*a^2*d
^2)*Log[-(Sqrt[d]*x) + Sqrt[c + d*x^2]])/(8*d^(3/2))

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Maple [A]
time = 0.09, size = 165, normalized size = 1.24

method	result	size
risch	\(-\frac {\sqrt {d \,x^{2}+c}\, \left (-2 b^{2} d \,x^{4}-8 a b d \,x^{2}-b^{2} c \,x^{2}+8 a^{2} d \right )}{8 d x}+\sqrt {d}\, \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right ) a^{2}+\frac {\ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right ) a b c}{\sqrt {d}}-\frac {\ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right ) b^{2} c^{2}}{8 d^{\frac {3}{2}}}\)	\(125\)
default	\(b^{2} \left (\frac {x \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{4 d}-\frac {c \left (\frac {x \sqrt {d \,x^{2}+c}}{2}+\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 \sqrt {d}}\right )}{4 d}\right )+2 a b \left (\frac {x \sqrt {d \,x^{2}+c}}{2}+\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 \sqrt {d}}\right )+a^{2} \left (-\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}}}{c x}+\frac {2 d \left (\frac {x \sqrt {d \,x^{2}+c}}{2}+\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 \sqrt {d}}\right )}{c}\right )\)	\(165\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2*(d*x^2+c)^(1/2)/x^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.31, size = 120, normalized size = 0.90 \begin {gather*} \sqrt {d x^{2} + c} a b x + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} x}{4 \, d} - \frac {\sqrt {d x^{2} + c} b^{2} c x}{8 \, d} - \frac {b^{2} c^{2} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{8 \, d^{\frac {3}{2}}} + \frac {a b c \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{\sqrt {d}} + a^{2} \sqrt {d} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right ) - \frac {\sqrt {d x^{2} + c} 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)^(1/2)/x^2,x, algorithm="maxima")

[Out]

sqrt(d*x^2 + c)*a*b*x + 1/4*(d*x^2 + c)^(3/2)*b^2*x/d - 1/8*sqrt(d*x^2 + c)*b^2*c*x/d - 1/8*b^2*c^2*arcsinh(d*
x/sqrt(c*d))/d^(3/2) + a*b*c*arcsinh(d*x/sqrt(c*d))/sqrt(d) + a^2*sqrt(d)*arcsinh(d*x/sqrt(c*d)) - sqrt(d*x^2
+ c)*a^2/x

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Fricas [A]
time = 1.40, size = 215, normalized size = 1.62 \begin {gather*} \left [-\frac {{\left (b^{2} c^{2} - 8 \, a b c d - 8 \, a^{2} 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 (2 \, b^{2} d^{2} x^{4} - 8 \, a^{2} d^{2} + {\left (b^{2} c d + 8 \, a b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{16 \, d^{2} x}, \frac {{\left (b^{2} c^{2} - 8 \, a b c d - 8 \, a^{2} d^{2}\right )} \sqrt {-d} x \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) + {\left (2 \, b^{2} d^{2} x^{4} - 8 \, a^{2} d^{2} + {\left (b^{2} c d + 8 \, a b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{8 \, 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)^(1/2)/x^2,x, algorithm="fricas")

[Out]

[-1/16*((b^2*c^2 - 8*a*b*c*d - 8*a^2*d^2)*sqrt(d)*x*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) - 2*(2*b^2
*d^2*x^4 - 8*a^2*d^2 + (b^2*c*d + 8*a*b*d^2)*x^2)*sqrt(d*x^2 + c))/(d^2*x), 1/8*((b^2*c^2 - 8*a*b*c*d - 8*a^2*
d^2)*sqrt(-d)*x*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) + (2*b^2*d^2*x^4 - 8*a^2*d^2 + (b^2*c*d + 8*a*b*d^2)*x^2)*s
qrt(d*x^2 + c))/(d^2*x)]

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Sympy [A]
time = 4.17, size = 219, normalized size = 1.65 \begin {gather*} - \frac {a^{2} \sqrt {c}}{x \sqrt {1 + \frac {d x^{2}}{c}}} + a^{2} \sqrt {d} \operatorname {asinh}{\left (\frac {\sqrt {d} x}{\sqrt {c}} \right )} - \frac {a^{2} d x}{\sqrt {c} \sqrt {1 + \frac {d x^{2}}{c}}} + a b \sqrt {c} x \sqrt {1 + \frac {d x^{2}}{c}} + \frac {a b c \operatorname {asinh}{\left (\frac {\sqrt {d} x}{\sqrt {c}} \right )}}{\sqrt {d}} + \frac {b^{2} c^{\frac {3}{2}} x}{8 d \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {3 b^{2} \sqrt {c} x^{3}}{8 \sqrt {1 + \frac {d x^{2}}{c}}} - \frac {b^{2} c^{2} \operatorname {asinh}{\left (\frac {\sqrt {d} x}{\sqrt {c}} \right )}}{8 d^{\frac {3}{2}}} + \frac {b^{2} d x^{5}}{4 \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)**(1/2)/x**2,x)

[Out]

-a**2*sqrt(c)/(x*sqrt(1 + d*x**2/c)) + a**2*sqrt(d)*asinh(sqrt(d)*x/sqrt(c)) - a**2*d*x/(sqrt(c)*sqrt(1 + d*x*
*2/c)) + a*b*sqrt(c)*x*sqrt(1 + d*x**2/c) + a*b*c*asinh(sqrt(d)*x/sqrt(c))/sqrt(d) + b**2*c**(3/2)*x/(8*d*sqrt
(1 + d*x**2/c)) + 3*b**2*sqrt(c)*x**3/(8*sqrt(1 + d*x**2/c)) - b**2*c**2*asinh(sqrt(d)*x/sqrt(c))/(8*d**(3/2))
+ b**2*d*x**5/(4*sqrt(c)*sqrt(1 + d*x**2/c))

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Giac [A]
time = 1.25, size = 126, normalized size = 0.95 \begin {gather*} \frac {2 \, a^{2} c \sqrt {d}}{{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c} + \frac {1}{8} \, {\left (2 \, b^{2} x^{2} + \frac {b^{2} c d + 8 \, a b d^{2}}{d^{2}}\right )} \sqrt {d x^{2} + c} x + \frac {{\left (b^{2} c^{2} \sqrt {d} - 8 \, a b c d^{\frac {3}{2}} - 8 \, a^{2} d^{\frac {5}{2}}\right )} \log \left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2}\right )}{16 \, 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)^(1/2)/x^2,x, algorithm="giac")

[Out]

2*a^2*c*sqrt(d)/((sqrt(d)*x - sqrt(d*x^2 + c))^2 - c) + 1/8*(2*b^2*x^2 + (b^2*c*d + 8*a*b*d^2)/d^2)*sqrt(d*x^2
+ c)*x + 1/16*(b^2*c^2*sqrt(d) - 8*a*b*c*d^(3/2) - 8*a^2*d^(5/2))*log((sqrt(d)*x - sqrt(d*x^2 + c))^2)/d^2

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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\,\sqrt {d\,x^2+c}}{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)^(1/2))/x^2,x)

[Out]

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

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