3.1.75 \(\int \frac {(c+d x^2)^2}{\sqrt {a+b x^2}} \, dx\) [75]

Optimal. Leaf size=108 \[ \frac {3 d (2 b c-a d) x \sqrt {a+b x^2}}{8 b^2}+\frac {d x \sqrt {a+b x^2} \left (c+d x^2\right )}{4 b}+\frac {\left (8 b^2 c^2-8 a b c d+3 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{5/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.04, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {427, 396, 223, 212} \begin {gather*} \frac {\left (3 a^2 d^2-8 a b c d+8 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{5/2}}+\frac {3 d x \sqrt {a+b x^2} (2 b c-a d)}{8 b^2}+\frac {d x \sqrt {a+b x^2} \left (c+d x^2\right )}{4 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x^2)^2/Sqrt[a + b*x^2],x]

[Out]

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

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 427

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[d*x*(a + b*x^n)^(p + 1)*((c
+ d*x^n)^(q - 1)/(b*(n*(p + q) + 1))), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntB
inomialQ[a, b, c, d, n, p, q, x]

Rubi steps

\begin {align*} \int \frac {\left (c+d x^2\right )^2}{\sqrt {a+b x^2}} \, dx &=\frac {d x \sqrt {a+b x^2} \left (c+d x^2\right )}{4 b}+\frac {\int \frac {c (4 b c-a d)+3 d (2 b c-a d) x^2}{\sqrt {a+b x^2}} \, dx}{4 b}\\ &=\frac {3 d (2 b c-a d) x \sqrt {a+b x^2}}{8 b^2}+\frac {d x \sqrt {a+b x^2} \left (c+d x^2\right )}{4 b}-\frac {(3 a d (2 b c-a d)-2 b c (4 b c-a d)) \int \frac {1}{\sqrt {a+b x^2}} \, dx}{8 b^2}\\ &=\frac {3 d (2 b c-a d) x \sqrt {a+b x^2}}{8 b^2}+\frac {d x \sqrt {a+b x^2} \left (c+d x^2\right )}{4 b}-\frac {(3 a d (2 b c-a d)-2 b c (4 b c-a d)) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )}{8 b^2}\\ &=\frac {3 d (2 b c-a d) x \sqrt {a+b x^2}}{8 b^2}+\frac {d x \sqrt {a+b x^2} \left (c+d x^2\right )}{4 b}+\frac {\left (8 b^2 c^2-8 a b c d+3 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{5/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.10, size = 90, normalized size = 0.83 \begin {gather*} \frac {d x \sqrt {a+b x^2} \left (8 b c-3 a d+2 b d x^2\right )}{8 b^2}+\frac {\left (-8 b^2 c^2+8 a b c d-3 a^2 d^2\right ) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{8 b^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d*x*Sqrt[a + b*x^2]*(8*b*c - 3*a*d + 2*b*d*x^2))/(8*b^2) + ((-8*b^2*c^2 + 8*a*b*c*d - 3*a^2*d^2)*Log[-(Sqrt[b
]*x) + Sqrt[a + b*x^2]])/(8*b^(5/2))

________________________________________________________________________________________

Maple [A]
time = 0.06, size = 133, normalized size = 1.23

method	result	size
risch	\(-\frac {d x \left (-2 b d \,x^{2}+3 a d -8 b c \right ) \sqrt {b \,x^{2}+a}}{8 b^{2}}+\frac {3 \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right ) a^{2} d^{2}}{8 b^{\frac {5}{2}}}-\frac {\ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right ) a c d}{b^{\frac {3}{2}}}+\frac {c^{2} \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{\sqrt {b}}\)	\(108\)
default	\(d^{2} \left (\frac {x^{3} \sqrt {b \,x^{2}+a}}{4 b}-\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}\right )+2 c d \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )+\frac {c^{2} \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{\sqrt {b}}\)	\(133\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.30, size = 109, normalized size = 1.01 \begin {gather*} \frac {\sqrt {b x^{2} + a} d^{2} x^{3}}{4 \, b} + \frac {\sqrt {b x^{2} + a} c d x}{b} - \frac {3 \, \sqrt {b x^{2} + a} a d^{2} x}{8 \, b^{2}} + \frac {c^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {b}} - \frac {a c d \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{b^{\frac {3}{2}}} + \frac {3 \, a^{2} d^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(b*x^2 + a)*d^2*x^3/b + sqrt(b*x^2 + a)*c*d*x/b - 3/8*sqrt(b*x^2 + a)*a*d^2*x/b^2 + c^2*arcsinh(b*x/sq
rt(a*b))/sqrt(b) - a*c*d*arcsinh(b*x/sqrt(a*b))/b^(3/2) + 3/8*a^2*d^2*arcsinh(b*x/sqrt(a*b))/b^(5/2)

________________________________________________________________________________________

Fricas [A]
time = 0.51, size = 192, normalized size = 1.78 \begin {gather*} \left [\frac {{\left (8 \, b^{2} c^{2} - 8 \, a b c d + 3 \, a^{2} d^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (2 \, b^{2} d^{2} x^{3} + {\left (8 \, b^{2} c d - 3 \, a b d^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{16 \, b^{3}}, -\frac {{\left (8 \, b^{2} c^{2} - 8 \, a b c d + 3 \, a^{2} d^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (2 \, b^{2} d^{2} x^{3} + {\left (8 \, b^{2} c d - 3 \, a b d^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{8 \, b^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*((8*b^2*c^2 - 8*a*b*c*d + 3*a^2*d^2)*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(2*b^2*
d^2*x^3 + (8*b^2*c*d - 3*a*b*d^2)*x)*sqrt(b*x^2 + a))/b^3, -1/8*((8*b^2*c^2 - 8*a*b*c*d + 3*a^2*d^2)*sqrt(-b)*
arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (2*b^2*d^2*x^3 + (8*b^2*c*d - 3*a*b*d^2)*x)*sqrt(b*x^2 + a))/b^3]

________________________________________________________________________________________

Sympy [A]
time = 4.42, size = 238, normalized size = 2.20 \begin {gather*} - \frac {3 a^{\frac {3}{2}} d^{2} x}{8 b^{2} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {\sqrt {a} c d x \sqrt {1 + \frac {b x^{2}}{a}}}{b} - \frac {\sqrt {a} d^{2} x^{3}}{8 b \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {3 a^{2} d^{2} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{8 b^{\frac {5}{2}}} - \frac {a c d \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{b^{\frac {3}{2}}} + c^{2} \left (\begin {cases} \frac {\sqrt {- \frac {a}{b}} \operatorname {asin}{\left (x \sqrt {- \frac {b}{a}} \right )}}{\sqrt {a}} & \text {for}\: a > 0 \wedge b < 0 \\\frac {\sqrt {\frac {a}{b}} \operatorname {asinh}{\left (x \sqrt {\frac {b}{a}} \right )}}{\sqrt {a}} & \text {for}\: a > 0 \wedge b > 0 \\\frac {\sqrt {- \frac {a}{b}} \operatorname {acosh}{\left (x \sqrt {- \frac {b}{a}} \right )}}{\sqrt {- a}} & \text {for}\: b > 0 \wedge a < 0 \end {cases}\right ) + \frac {d^{2} x^{5}}{4 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*a**(3/2)*d**2*x/(8*b**2*sqrt(1 + b*x**2/a)) + sqrt(a)*c*d*x*sqrt(1 + b*x**2/a)/b - sqrt(a)*d**2*x**3/(8*b*s
qrt(1 + b*x**2/a)) + 3*a**2*d**2*asinh(sqrt(b)*x/sqrt(a))/(8*b**(5/2)) - a*c*d*asinh(sqrt(b)*x/sqrt(a))/b**(3/
2) + c**2*Piecewise((sqrt(-a/b)*asin(x*sqrt(-b/a))/sqrt(a), (a > 0) & (b < 0)), (sqrt(a/b)*asinh(x*sqrt(b/a))/
sqrt(a), (a > 0) & (b > 0)), (sqrt(-a/b)*acosh(x*sqrt(-b/a))/sqrt(-a), (b > 0) & (a < 0))) + d**2*x**5/(4*sqrt
(a)*sqrt(1 + b*x**2/a))

________________________________________________________________________________________

Giac [A]
time = 1.08, size = 90, normalized size = 0.83 \begin {gather*} \frac {1}{8} \, \sqrt {b x^{2} + a} {\left (\frac {2 \, d^{2} x^{2}}{b} + \frac {8 \, b^{2} c d - 3 \, a b d^{2}}{b^{3}}\right )} x - \frac {{\left (8 \, b^{2} c^{2} - 8 \, a b c d + 3 \, a^{2} d^{2}\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{8 \, b^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*sqrt(b*x^2 + a)*(2*d^2*x^2/b + (8*b^2*c*d - 3*a*b*d^2)/b^3)*x - 1/8*(8*b^2*c^2 - 8*a*b*c*d + 3*a^2*d^2)*lo
g(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(5/2)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________