Integrand size = 21, antiderivative size = 108 \[ \int \frac {\left (c+d x^2\right )^2}{\sqrt {a+b x^2}} \, dx=\frac {3 d (2 b c-a d) x \sqrt {a+b x^2}}{8 b^2}+\frac {d x \sqrt {a+b x^2} \left (c+d x^2\right )}{4 b}+\frac {\left (8 b^2 c^2-8 a b c d+3 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{5/2}} \]
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
Time = 0.09 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.83 \[ \int \frac {\left (c+d x^2\right )^2}{\sqrt {a+b x^2}} \, dx=\frac {d x \sqrt {a+b x^2} \left (8 b c-3 a d+2 b d x^2\right )}{8 b^2}+\frac {\left (-8 b^2 c^2+8 a b c d-3 a^2 d^2\right ) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{8 b^{5/2}} \]
(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))
Time = 0.22 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.07, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {318, 299, 224, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (c+d x^2\right )^2}{\sqrt {a+b x^2}} \, dx\) |
\(\Big \downarrow \) 318 |
\(\displaystyle \frac {\int \frac {3 d (2 b c-a d) x^2+c (4 b c-a d)}{\sqrt {b x^2+a}}dx}{4 b}+\frac {d x \sqrt {a+b x^2} \left (c+d x^2\right )}{4 b}\) |
\(\Big \downarrow \) 299 |
\(\displaystyle \frac {\frac {\left (3 a^2 d^2-8 a b c d+8 b^2 c^2\right ) \int \frac {1}{\sqrt {b x^2+a}}dx}{2 b}+\frac {3 d x \sqrt {a+b x^2} (2 b c-a d)}{2 b}}{4 b}+\frac {d x \sqrt {a+b x^2} \left (c+d x^2\right )}{4 b}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {\frac {\left (3 a^2 d^2-8 a b c d+8 b^2 c^2\right ) \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{2 b}+\frac {3 d x \sqrt {a+b x^2} (2 b c-a d)}{2 b}}{4 b}+\frac {d x \sqrt {a+b x^2} \left (c+d x^2\right )}{4 b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {\text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \left (3 a^2 d^2-8 a b c d+8 b^2 c^2\right )}{2 b^{3/2}}+\frac {3 d x \sqrt {a+b x^2} (2 b c-a d)}{2 b}}{4 b}+\frac {d x \sqrt {a+b x^2} \left (c+d x^2\right )}{4 b}\) |
(d*x*Sqrt[a + b*x^2]*(c + d*x^2))/(4*b) + ((3*d*(2*b*c - a*d)*x*Sqrt[a + b *x^2])/(2*b) + ((8*b^2*c^2 - 8*a*b*c*d + 3*a^2*d^2)*ArcTanh[(Sqrt[b]*x)/Sq rt[a + b*x^2]])/(2*b^(3/2)))/(4*b)
3.1.75.3.1 Defintions of rubi rules used
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] && (Gt Q[a, 0] || LtQ[b, 0])
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]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x *((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 *p + 3)) Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && NeQ[2*p + 3, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[d*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(b*(2*(p + q) + 1))), x] + S imp[1/(b*(2*(p + q) + 1)) Int[(a + b*x^2)^p*(c + d*x^2)^(q - 2)*Simp[c*(b *c*(2*(p + q) + 1) - a*d) + d*(b*c*(2*(p + 2*q - 1) + 1) - a*d*(2*(q - 1) + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && G tQ[q, 1] && NeQ[2*(p + q) + 1, 0] && !IGtQ[p, 1] && IntBinomialQ[a, b, c, d, 2, p, q, x]
Time = 2.35 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.72
method | result | size |
risch | \(-\frac {d x \left (-2 b d \,x^{2}+3 a d -8 b c \right ) \sqrt {b \,x^{2}+a}}{8 b^{2}}+\frac {\left (3 a^{2} d^{2}-8 a b c d +8 b^{2} c^{2}\right ) \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{8 b^{\frac {5}{2}}}\) | \(78\) |
pseudoelliptic | \(\frac {\frac {3 \left (a^{2} d^{2}-\frac {8}{3} a b c d +\frac {8}{3} b^{2} c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )}{8}-\frac {3 x \sqrt {b \,x^{2}+a}\, \left (\frac {2 \left (-d \,x^{2}-4 c \right ) b^{\frac {3}{2}}}{3}+a d \sqrt {b}\right ) d}{8}}{b^{\frac {5}{2}}}\) | \(82\) |
default | \(\frac {c^{2} \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{\sqrt {b}}+d^{2} \left (\frac {x^{3} \sqrt {b \,x^{2}+a}}{4 b}-\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}\right )+2 c d \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )\) | \(133\) |
-1/8*d*x*(-2*b*d*x^2+3*a*d-8*b*c)*(b*x^2+a)^(1/2)/b^2+1/8*(3*a^2*d^2-8*a*b *c*d+8*b^2*c^2)/b^(5/2)*ln(x*b^(1/2)+(b*x^2+a)^(1/2))
Time = 0.28 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.78 \[ \int \frac {\left (c+d x^2\right )^2}{\sqrt {a+b x^2}} \, dx=\left [\frac {{\left (8 \, b^{2} c^{2} - 8 \, a b c d + 3 \, a^{2} d^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (2 \, b^{2} d^{2} x^{3} + {\left (8 \, b^{2} c d - 3 \, a b d^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{16 \, b^{3}}, -\frac {{\left (8 \, b^{2} c^{2} - 8 \, a b c d + 3 \, a^{2} d^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (2 \, b^{2} d^{2} x^{3} + {\left (8 \, b^{2} c d - 3 \, a b d^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{8 \, b^{3}}\right ] \]
[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)*s qrt(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)*ar ctan(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]
Time = 0.33 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.24 \[ \int \frac {\left (c+d x^2\right )^2}{\sqrt {a+b x^2}} \, dx=\begin {cases} \sqrt {a + b x^{2}} \left (\frac {d^{2} x^{3}}{4 b} + \frac {x \left (- \frac {3 a d^{2}}{4 b} + 2 c d\right )}{2 b}\right ) + \left (- \frac {a \left (- \frac {3 a d^{2}}{4 b} + 2 c d\right )}{2 b} + c^{2}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: b \neq 0 \\\frac {c^{2} x + \frac {2 c d x^{3}}{3} + \frac {d^{2} x^{5}}{5}}{\sqrt {a}} & \text {otherwise} \end {cases} \]
Piecewise((sqrt(a + b*x**2)*(d**2*x**3/(4*b) + x*(-3*a*d**2/(4*b) + 2*c*d) /(2*b)) + (-a*(-3*a*d**2/(4*b) + 2*c*d)/(2*b) + c**2)*Piecewise((log(2*sqr t(b)*sqrt(a + b*x**2) + 2*b*x)/sqrt(b), Ne(a, 0)), (x*log(x)/sqrt(b*x**2), True)), Ne(b, 0)), ((c**2*x + 2*c*d*x**3/3 + d**2*x**5/5)/sqrt(a), True))
Time = 0.21 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.01 \[ \int \frac {\left (c+d x^2\right )^2}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {b x^{2} + a} d^{2} x^{3}}{4 \, b} + \frac {\sqrt {b x^{2} + a} c d x}{b} - \frac {3 \, \sqrt {b x^{2} + a} a d^{2} x}{8 \, b^{2}} + \frac {c^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {b}} - \frac {a c d \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{b^{\frac {3}{2}}} + \frac {3 \, a^{2} d^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {5}{2}}} \]
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/sqrt(a*b))/sqrt(b) - a*c*d*arcsinh(b*x/s qrt(a*b))/b^(3/2) + 3/8*a^2*d^2*arcsinh(b*x/sqrt(a*b))/b^(5/2)
Time = 0.29 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.83 \[ \int \frac {\left (c+d x^2\right )^2}{\sqrt {a+b x^2}} \, dx=\frac {1}{8} \, \sqrt {b x^{2} + a} {\left (\frac {2 \, d^{2} x^{2}}{b} + \frac {8 \, b^{2} c d - 3 \, a b d^{2}}{b^{3}}\right )} x - \frac {{\left (8 \, b^{2} c^{2} - 8 \, a b c d + 3 \, a^{2} d^{2}\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{8 \, b^{\frac {5}{2}}} \]
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)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b ^(5/2)
Timed out. \[ \int \frac {\left (c+d x^2\right )^2}{\sqrt {a+b x^2}} \, dx=\int \frac {{\left (d\,x^2+c\right )}^2}{\sqrt {b\,x^2+a}} \,d x \]