Integrand size = 17, antiderivative size = 87 \[ \int (A+B x) \left (a+b x^2\right )^{3/2} \, dx=\frac {3}{8} a A x \sqrt {a+b x^2}+\frac {1}{4} A x \left (a+b x^2\right )^{3/2}+\frac {B \left (a+b x^2\right )^{5/2}}{5 b}+\frac {3 a^2 A \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 \sqrt {b}} \]
1/4*A*x*(b*x^2+a)^(3/2)+1/5*B*(b*x^2+a)^(5/2)/b+3/8*a^2*A*arctanh(x*b^(1/2 )/(b*x^2+a)^(1/2))/b^(1/2)+3/8*a*A*x*(b*x^2+a)^(1/2)
Time = 0.23 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00 \[ \int (A+B x) \left (a+b x^2\right )^{3/2} \, dx=\frac {\sqrt {a+b x^2} \left (8 a^2 B+2 b^2 x^3 (5 A+4 B x)+a b x (25 A+16 B x)\right )-15 a^2 A \sqrt {b} \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{40 b} \]
(Sqrt[a + b*x^2]*(8*a^2*B + 2*b^2*x^3*(5*A + 4*B*x) + a*b*x*(25*A + 16*B*x )) - 15*a^2*A*Sqrt[b]*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(40*b)
Time = 0.18 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {455, 211, 211, 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.
\(\displaystyle \int \left (a+b x^2\right )^{3/2} (A+B x) \, dx\) |
\(\Big \downarrow \) 455 |
\(\displaystyle A \int \left (b x^2+a\right )^{3/2}dx+\frac {B \left (a+b x^2\right )^{5/2}}{5 b}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle A \left (\frac {3}{4} a \int \sqrt {b x^2+a}dx+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )+\frac {B \left (a+b x^2\right )^{5/2}}{5 b}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle A \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{\sqrt {b x^2+a}}dx+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )+\frac {B \left (a+b x^2\right )^{5/2}}{5 b}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle A \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )+\frac {B \left (a+b x^2\right )^{5/2}}{5 b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle A \left (\frac {3}{4} a \left (\frac {a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 \sqrt {b}}+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )+\frac {B \left (a+b x^2\right )^{5/2}}{5 b}\) |
(B*(a + b*x^2)^(5/2))/(5*b) + A*((x*(a + b*x^2)^(3/2))/4 + (3*a*((x*Sqrt[a + b*x^2])/2 + (a*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*Sqrt[b])))/4)
3.1.11.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
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[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && !LeQ[p, -1]
Time = 3.38 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.80
method | result | size |
default | \(A \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )+\frac {B \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{5 b}\) | \(70\) |
risch | \(\frac {\left (8 b^{2} B \,x^{4}+10 A \,b^{2} x^{3}+16 B a b \,x^{2}+25 a A b x +8 a^{2} B \right ) \sqrt {b \,x^{2}+a}}{40 b}+\frac {3 a^{2} A \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{8 \sqrt {b}}\) | \(80\) |
A*(1/4*x*(b*x^2+a)^(3/2)+3/4*a*(1/2*x*(b*x^2+a)^(1/2)+1/2*a/b^(1/2)*ln(x*b ^(1/2)+(b*x^2+a)^(1/2))))+1/5*B*(b*x^2+a)^(5/2)/b
Time = 0.32 (sec) , antiderivative size = 176, normalized size of antiderivative = 2.02 \[ \int (A+B x) \left (a+b x^2\right )^{3/2} \, dx=\left [\frac {15 \, A a^{2} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (8 \, B b^{2} x^{4} + 10 \, A b^{2} x^{3} + 16 \, B a b x^{2} + 25 \, A a b x + 8 \, B a^{2}\right )} \sqrt {b x^{2} + a}}{80 \, b}, -\frac {15 \, A a^{2} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (8 \, B b^{2} x^{4} + 10 \, A b^{2} x^{3} + 16 \, B a b x^{2} + 25 \, A a b x + 8 \, B a^{2}\right )} \sqrt {b x^{2} + a}}{40 \, b}\right ] \]
[1/80*(15*A*a^2*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(8*B*b^2*x^4 + 10*A*b^2*x^3 + 16*B*a*b*x^2 + 25*A*a*b*x + 8*B*a^2)*sqrt( b*x^2 + a))/b, -1/40*(15*A*a^2*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (8*B*b^2*x^4 + 10*A*b^2*x^3 + 16*B*a*b*x^2 + 25*A*a*b*x + 8*B*a^2)*sqrt (b*x^2 + a))/b]
Time = 0.40 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.37 \[ \int (A+B x) \left (a+b x^2\right )^{3/2} \, dx=\begin {cases} \frac {3 A a^{2} \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{8} + \sqrt {a + b x^{2}} \cdot \left (\frac {5 A a x}{8} + \frac {A b x^{3}}{4} + \frac {B a^{2}}{5 b} + \frac {2 B a x^{2}}{5} + \frac {B b x^{4}}{5}\right ) & \text {for}\: b \neq 0 \\a^{\frac {3}{2}} \left (A x + \frac {B x^{2}}{2}\right ) & \text {otherwise} \end {cases} \]
Piecewise((3*A*a**2*Piecewise((log(2*sqrt(b)*sqrt(a + b*x**2) + 2*b*x)/sqr t(b), Ne(a, 0)), (x*log(x)/sqrt(b*x**2), True))/8 + sqrt(a + b*x**2)*(5*A* a*x/8 + A*b*x**3/4 + B*a**2/(5*b) + 2*B*a*x**2/5 + B*b*x**4/5), Ne(b, 0)), (a**(3/2)*(A*x + B*x**2/2), True))
Time = 0.20 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.70 \[ \int (A+B x) \left (a+b x^2\right )^{3/2} \, dx=\frac {1}{4} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A x + \frac {3}{8} \, \sqrt {b x^{2} + a} A a x + \frac {3 \, A a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {b}} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B}{5 \, b} \]
1/4*(b*x^2 + a)^(3/2)*A*x + 3/8*sqrt(b*x^2 + a)*A*a*x + 3/8*A*a^2*arcsinh( b*x/sqrt(a*b))/sqrt(b) + 1/5*(b*x^2 + a)^(5/2)*B/b
Time = 0.29 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.87 \[ \int (A+B x) \left (a+b x^2\right )^{3/2} \, dx=-\frac {3 \, A a^{2} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{8 \, \sqrt {b}} + \frac {1}{40} \, \sqrt {b x^{2} + a} {\left (\frac {8 \, B a^{2}}{b} + {\left (25 \, A a + 2 \, {\left (8 \, B a + {\left (4 \, B b x + 5 \, A b\right )} x\right )} x\right )} x\right )} \]
-3/8*A*a^2*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/sqrt(b) + 1/40*sqrt(b*x^ 2 + a)*(8*B*a^2/b + (25*A*a + 2*(8*B*a + (4*B*b*x + 5*A*b)*x)*x)*x)
Time = 6.20 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.62 \[ \int (A+B x) \left (a+b x^2\right )^{3/2} \, dx=\frac {B\,{\left (b\,x^2+a\right )}^{5/2}}{5\,b}+\frac {A\,x\,{\left (b\,x^2+a\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},\frac {1}{2};\ \frac {3}{2};\ -\frac {b\,x^2}{a}\right )}{{\left (\frac {b\,x^2}{a}+1\right )}^{3/2}} \]