Integrand size = 17, antiderivative size = 107 \[ \int (A+B x) \left (a+c x^2\right )^{5/2} \, dx=\frac {5}{16} a^2 A x \sqrt {a+c x^2}+\frac {5}{24} a A x \left (a+c x^2\right )^{3/2}+\frac {1}{6} A x \left (a+c x^2\right )^{5/2}+\frac {B \left (a+c x^2\right )^{7/2}}{7 c}+\frac {5 a^3 A \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{16 \sqrt {c}} \]
5/24*a*A*x*(c*x^2+a)^(3/2)+1/6*A*x*(c*x^2+a)^(5/2)+1/7*B*(c*x^2+a)^(7/2)/c +5/16*a^3*A*arctanh(x*c^(1/2)/(c*x^2+a)^(1/2))/c^(1/2)+5/16*a^2*A*x*(c*x^2 +a)^(1/2)
Time = 0.29 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00 \[ \int (A+B x) \left (a+c x^2\right )^{5/2} \, dx=\frac {\sqrt {a+c x^2} \left (48 a^3 B+8 c^3 x^5 (7 A+6 B x)+3 a^2 c x (77 A+48 B x)+2 a c^2 x^3 (91 A+72 B x)\right )-105 a^3 A \sqrt {c} \log \left (-\sqrt {c} x+\sqrt {a+c x^2}\right )}{336 c} \]
(Sqrt[a + c*x^2]*(48*a^3*B + 8*c^3*x^5*(7*A + 6*B*x) + 3*a^2*c*x*(77*A + 4 8*B*x) + 2*a*c^2*x^3*(91*A + 72*B*x)) - 105*a^3*A*Sqrt[c]*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2]])/(336*c)
Time = 0.21 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {455, 211, 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.
\(\displaystyle \int \left (a+c x^2\right )^{5/2} (A+B x) \, dx\) |
\(\Big \downarrow \) 455 |
\(\displaystyle A \int \left (c x^2+a\right )^{5/2}dx+\frac {B \left (a+c x^2\right )^{7/2}}{7 c}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle A \left (\frac {5}{6} a \int \left (c x^2+a\right )^{3/2}dx+\frac {1}{6} x \left (a+c x^2\right )^{5/2}\right )+\frac {B \left (a+c x^2\right )^{7/2}}{7 c}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle A \left (\frac {5}{6} a \left (\frac {3}{4} a \int \sqrt {c x^2+a}dx+\frac {1}{4} x \left (a+c x^2\right )^{3/2}\right )+\frac {1}{6} x \left (a+c x^2\right )^{5/2}\right )+\frac {B \left (a+c x^2\right )^{7/2}}{7 c}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle A \left (\frac {5}{6} a \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{\sqrt {c x^2+a}}dx+\frac {1}{2} x \sqrt {a+c x^2}\right )+\frac {1}{4} x \left (a+c x^2\right )^{3/2}\right )+\frac {1}{6} x \left (a+c x^2\right )^{5/2}\right )+\frac {B \left (a+c x^2\right )^{7/2}}{7 c}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle A \left (\frac {5}{6} a \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{1-\frac {c x^2}{c x^2+a}}d\frac {x}{\sqrt {c x^2+a}}+\frac {1}{2} x \sqrt {a+c x^2}\right )+\frac {1}{4} x \left (a+c x^2\right )^{3/2}\right )+\frac {1}{6} x \left (a+c x^2\right )^{5/2}\right )+\frac {B \left (a+c x^2\right )^{7/2}}{7 c}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle A \left (\frac {5}{6} a \left (\frac {3}{4} a \left (\frac {a \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 \sqrt {c}}+\frac {1}{2} x \sqrt {a+c x^2}\right )+\frac {1}{4} x \left (a+c x^2\right )^{3/2}\right )+\frac {1}{6} x \left (a+c x^2\right )^{5/2}\right )+\frac {B \left (a+c x^2\right )^{7/2}}{7 c}\) |
(B*(a + c*x^2)^(7/2))/(7*c) + A*((x*(a + c*x^2)^(5/2))/6 + (5*a*((x*(a + c *x^2)^(3/2))/4 + (3*a*((x*Sqrt[a + c*x^2])/2 + (a*ArcTanh[(Sqrt[c]*x)/Sqrt [a + c*x^2]])/(2*Sqrt[c])))/4))/6)
3.4.43.3.1 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 = 0.08 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.80
method | result | size |
default | \(A \left (\frac {x \left (c \,x^{2}+a \right )^{\frac {5}{2}}}{6}+\frac {5 a \left (\frac {x \left (c \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {c \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+a}\right )}{2 \sqrt {c}}\right )}{4}\right )}{6}\right )+\frac {B \left (c \,x^{2}+a \right )^{\frac {7}{2}}}{7 c}\) | \(86\) |
risch | \(\frac {\left (48 B \,c^{3} x^{6}+56 A \,c^{3} x^{5}+144 a B \,c^{2} x^{4}+182 a A \,c^{2} x^{3}+144 a^{2} B c \,x^{2}+231 a^{2} A c x +48 B \,a^{3}\right ) \sqrt {c \,x^{2}+a}}{336 c}+\frac {5 A \,a^{3} \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+a}\right )}{16 \sqrt {c}}\) | \(104\) |
A*(1/6*x*(c*x^2+a)^(5/2)+5/6*a*(1/4*x*(c*x^2+a)^(3/2)+3/4*a*(1/2*x*(c*x^2+ a)^(1/2)+1/2*a/c^(1/2)*ln(x*c^(1/2)+(c*x^2+a)^(1/2)))))+1/7*B*(c*x^2+a)^(7 /2)/c
Time = 0.40 (sec) , antiderivative size = 224, normalized size of antiderivative = 2.09 \[ \int (A+B x) \left (a+c x^2\right )^{5/2} \, dx=\left [\frac {105 \, A a^{3} \sqrt {c} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 2 \, {\left (48 \, B c^{3} x^{6} + 56 \, A c^{3} x^{5} + 144 \, B a c^{2} x^{4} + 182 \, A a c^{2} x^{3} + 144 \, B a^{2} c x^{2} + 231 \, A a^{2} c x + 48 \, B a^{3}\right )} \sqrt {c x^{2} + a}}{672 \, c}, -\frac {105 \, A a^{3} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - {\left (48 \, B c^{3} x^{6} + 56 \, A c^{3} x^{5} + 144 \, B a c^{2} x^{4} + 182 \, A a c^{2} x^{3} + 144 \, B a^{2} c x^{2} + 231 \, A a^{2} c x + 48 \, B a^{3}\right )} \sqrt {c x^{2} + a}}{336 \, c}\right ] \]
[1/672*(105*A*a^3*sqrt(c)*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 2*(48*B*c^3*x^6 + 56*A*c^3*x^5 + 144*B*a*c^2*x^4 + 182*A*a*c^2*x^3 + 144 *B*a^2*c*x^2 + 231*A*a^2*c*x + 48*B*a^3)*sqrt(c*x^2 + a))/c, -1/336*(105*A *a^3*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - (48*B*c^3*x^6 + 56*A*c^ 3*x^5 + 144*B*a*c^2*x^4 + 182*A*a*c^2*x^3 + 144*B*a^2*c*x^2 + 231*A*a^2*c* x + 48*B*a^3)*sqrt(c*x^2 + a))/c]
Time = 0.49 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.40 \[ \int (A+B x) \left (a+c x^2\right )^{5/2} \, dx=\begin {cases} \frac {5 A a^{3} \left (\begin {cases} \frac {\log {\left (2 \sqrt {c} \sqrt {a + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {c x^{2}}} & \text {otherwise} \end {cases}\right )}{16} + \sqrt {a + c x^{2}} \cdot \left (\frac {11 A a^{2} x}{16} + \frac {13 A a c x^{3}}{24} + \frac {A c^{2} x^{5}}{6} + \frac {B a^{3}}{7 c} + \frac {3 B a^{2} x^{2}}{7} + \frac {3 B a c x^{4}}{7} + \frac {B c^{2} x^{6}}{7}\right ) & \text {for}\: c \neq 0 \\a^{\frac {5}{2}} \left (A x + \frac {B x^{2}}{2}\right ) & \text {otherwise} \end {cases} \]
Piecewise((5*A*a**3*Piecewise((log(2*sqrt(c)*sqrt(a + c*x**2) + 2*c*x)/sqr t(c), Ne(a, 0)), (x*log(x)/sqrt(c*x**2), True))/16 + sqrt(a + c*x**2)*(11* A*a**2*x/16 + 13*A*a*c*x**3/24 + A*c**2*x**5/6 + B*a**3/(7*c) + 3*B*a**2*x **2/7 + 3*B*a*c*x**4/7 + B*c**2*x**6/7), Ne(c, 0)), (a**(5/2)*(A*x + B*x** 2/2), True))
Time = 0.24 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.72 \[ \int (A+B x) \left (a+c x^2\right )^{5/2} \, dx=\frac {1}{6} \, {\left (c x^{2} + a\right )}^{\frac {5}{2}} A x + \frac {5}{24} \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} A a x + \frac {5}{16} \, \sqrt {c x^{2} + a} A a^{2} x + \frac {5 \, A a^{3} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{16 \, \sqrt {c}} + \frac {{\left (c x^{2} + a\right )}^{\frac {7}{2}} B}{7 \, c} \]
1/6*(c*x^2 + a)^(5/2)*A*x + 5/24*(c*x^2 + a)^(3/2)*A*a*x + 5/16*sqrt(c*x^2 + a)*A*a^2*x + 5/16*A*a^3*arcsinh(c*x/sqrt(a*c))/sqrt(c) + 1/7*(c*x^2 + a )^(7/2)*B/c
Time = 0.30 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.94 \[ \int (A+B x) \left (a+c x^2\right )^{5/2} \, dx=-\frac {5 \, A a^{3} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right )}{16 \, \sqrt {c}} + \frac {1}{336} \, {\left (\frac {48 \, B a^{3}}{c} + {\left (231 \, A a^{2} + 2 \, {\left (72 \, B a^{2} + {\left (91 \, A a c + 4 \, {\left (18 \, B a c + {\left (6 \, B c^{2} x + 7 \, A c^{2}\right )} x\right )} x\right )} x\right )} x\right )} x\right )} \sqrt {c x^{2} + a} \]
-5/16*A*a^3*log(abs(-sqrt(c)*x + sqrt(c*x^2 + a)))/sqrt(c) + 1/336*(48*B*a ^3/c + (231*A*a^2 + 2*(72*B*a^2 + (91*A*a*c + 4*(18*B*a*c + (6*B*c^2*x + 7 *A*c^2)*x)*x)*x)*x)*x)*sqrt(c*x^2 + a)
Time = 10.02 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.50 \[ \int (A+B x) \left (a+c x^2\right )^{5/2} \, dx=\frac {B\,{\left (c\,x^2+a\right )}^{7/2}}{7\,c}+\frac {A\,x\,{\left (c\,x^2+a\right )}^{5/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{2},\frac {1}{2};\ \frac {3}{2};\ -\frac {c\,x^2}{a}\right )}{{\left (\frac {c\,x^2}{a}+1\right )}^{5/2}} \]