Integrand size = 19, antiderivative size = 297 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{(c x)^{7/2}} \, dx=-\frac {12 b \sqrt {a+b x^2}}{5 c^3 \sqrt {c x}}+\frac {24 b^{3/2} \sqrt {c x} \sqrt {a+b x^2}}{5 c^4 \left (\sqrt {a}+\sqrt {b} x\right )}-\frac {2 \left (a+b x^2\right )^{3/2}}{5 c (c x)^{5/2}}-\frac {24 \sqrt [4]{a} b^{5/4} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )|\frac {1}{2}\right )}{5 c^{7/2} \sqrt {a+b x^2}}+\frac {12 \sqrt [4]{a} b^{5/4} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right ),\frac {1}{2}\right )}{5 c^{7/2} \sqrt {a+b x^2}} \]
-2/5*(b*x^2+a)^(3/2)/c/(c*x)^(5/2)-12/5*b*(b*x^2+a)^(1/2)/c^3/(c*x)^(1/2)+ 24/5*b^(3/2)*(c*x)^(1/2)*(b*x^2+a)^(1/2)/c^4/(a^(1/2)+x*b^(1/2))-24/5*a^(1 /4)*b^(5/4)*(cos(2*arctan(b^(1/4)*(c*x)^(1/2)/a^(1/4)/c^(1/2)))^2)^(1/2)/c os(2*arctan(b^(1/4)*(c*x)^(1/2)/a^(1/4)/c^(1/2)))*EllipticE(sin(2*arctan(b ^(1/4)*(c*x)^(1/2)/a^(1/4)/c^(1/2))),1/2*2^(1/2))*(a^(1/2)+x*b^(1/2))*((b* x^2+a)/(a^(1/2)+x*b^(1/2))^2)^(1/2)/c^(7/2)/(b*x^2+a)^(1/2)+12/5*a^(1/4)*b ^(5/4)*(cos(2*arctan(b^(1/4)*(c*x)^(1/2)/a^(1/4)/c^(1/2)))^2)^(1/2)/cos(2* arctan(b^(1/4)*(c*x)^(1/2)/a^(1/4)/c^(1/2)))*EllipticF(sin(2*arctan(b^(1/4 )*(c*x)^(1/2)/a^(1/4)/c^(1/2))),1/2*2^(1/2))*(a^(1/2)+x*b^(1/2))*((b*x^2+a )/(a^(1/2)+x*b^(1/2))^2)^(1/2)/c^(7/2)/(b*x^2+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.02 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.19 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{(c x)^{7/2}} \, dx=-\frac {2 a x \sqrt {a+b x^2} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {5}{4},-\frac {1}{4},-\frac {b x^2}{a}\right )}{5 (c x)^{7/2} \sqrt {1+\frac {b x^2}{a}}} \]
(-2*a*x*Sqrt[a + b*x^2]*Hypergeometric2F1[-3/2, -5/4, -1/4, -((b*x^2)/a)]) /(5*(c*x)^(7/2)*Sqrt[1 + (b*x^2)/a])
Time = 0.38 (sec) , antiderivative size = 337, normalized size of antiderivative = 1.13, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {247, 247, 266, 834, 27, 761, 1510}
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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^{3/2}}{(c x)^{7/2}} \, dx\) |
\(\Big \downarrow \) 247 |
\(\displaystyle \frac {6 b \int \frac {\sqrt {b x^2+a}}{(c x)^{3/2}}dx}{5 c^2}-\frac {2 \left (a+b x^2\right )^{3/2}}{5 c (c x)^{5/2}}\) |
\(\Big \downarrow \) 247 |
\(\displaystyle \frac {6 b \left (\frac {2 b \int \frac {\sqrt {c x}}{\sqrt {b x^2+a}}dx}{c^2}-\frac {2 \sqrt {a+b x^2}}{c \sqrt {c x}}\right )}{5 c^2}-\frac {2 \left (a+b x^2\right )^{3/2}}{5 c (c x)^{5/2}}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {6 b \left (\frac {4 b \int \frac {c x}{\sqrt {b x^2+a}}d\sqrt {c x}}{c^3}-\frac {2 \sqrt {a+b x^2}}{c \sqrt {c x}}\right )}{5 c^2}-\frac {2 \left (a+b x^2\right )^{3/2}}{5 c (c x)^{5/2}}\) |
\(\Big \downarrow \) 834 |
\(\displaystyle \frac {6 b \left (\frac {4 b \left (\frac {\sqrt {a} c \int \frac {1}{\sqrt {b x^2+a}}d\sqrt {c x}}{\sqrt {b}}-\frac {\sqrt {a} c \int \frac {\sqrt {a} c-\sqrt {b} c x}{\sqrt {a} c \sqrt {b x^2+a}}d\sqrt {c x}}{\sqrt {b}}\right )}{c^3}-\frac {2 \sqrt {a+b x^2}}{c \sqrt {c x}}\right )}{5 c^2}-\frac {2 \left (a+b x^2\right )^{3/2}}{5 c (c x)^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {6 b \left (\frac {4 b \left (\frac {\sqrt {a} c \int \frac {1}{\sqrt {b x^2+a}}d\sqrt {c x}}{\sqrt {b}}-\frac {\int \frac {\sqrt {a} c-\sqrt {b} c x}{\sqrt {b x^2+a}}d\sqrt {c x}}{\sqrt {b}}\right )}{c^3}-\frac {2 \sqrt {a+b x^2}}{c \sqrt {c x}}\right )}{5 c^2}-\frac {2 \left (a+b x^2\right )^{3/2}}{5 c (c x)^{5/2}}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {6 b \left (\frac {4 b \left (\frac {\sqrt [4]{a} \sqrt {c} \left (\sqrt {a} c+\sqrt {b} c x\right ) \sqrt {\frac {a c^2+b c^2 x^2}{\left (\sqrt {a} c+\sqrt {b} c x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+b x^2}}-\frac {\int \frac {\sqrt {a} c-\sqrt {b} c x}{\sqrt {b x^2+a}}d\sqrt {c x}}{\sqrt {b}}\right )}{c^3}-\frac {2 \sqrt {a+b x^2}}{c \sqrt {c x}}\right )}{5 c^2}-\frac {2 \left (a+b x^2\right )^{3/2}}{5 c (c x)^{5/2}}\) |
\(\Big \downarrow \) 1510 |
\(\displaystyle \frac {6 b \left (\frac {4 b \left (\frac {\sqrt [4]{a} \sqrt {c} \left (\sqrt {a} c+\sqrt {b} c x\right ) \sqrt {\frac {a c^2+b c^2 x^2}{\left (\sqrt {a} c+\sqrt {b} c x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+b x^2}}-\frac {\frac {\sqrt [4]{a} \sqrt {c} \left (\sqrt {a} c+\sqrt {b} c x\right ) \sqrt {\frac {a c^2+b c^2 x^2}{\left (\sqrt {a} c+\sqrt {b} c x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )|\frac {1}{2}\right )}{\sqrt [4]{b} \sqrt {a+b x^2}}-\frac {c^2 \sqrt {c x} \sqrt {a+b x^2}}{\sqrt {a} c+\sqrt {b} c x}}{\sqrt {b}}\right )}{c^3}-\frac {2 \sqrt {a+b x^2}}{c \sqrt {c x}}\right )}{5 c^2}-\frac {2 \left (a+b x^2\right )^{3/2}}{5 c (c x)^{5/2}}\) |
(-2*(a + b*x^2)^(3/2))/(5*c*(c*x)^(5/2)) + (6*b*((-2*Sqrt[a + b*x^2])/(c*S qrt[c*x]) + (4*b*(-((-((c^2*Sqrt[c*x]*Sqrt[a + b*x^2])/(Sqrt[a]*c + Sqrt[b ]*c*x)) + (a^(1/4)*Sqrt[c]*(Sqrt[a]*c + Sqrt[b]*c*x)*Sqrt[(a*c^2 + b*c^2*x ^2)/(Sqrt[a]*c + Sqrt[b]*c*x)^2]*EllipticE[2*ArcTan[(b^(1/4)*Sqrt[c*x])/(a ^(1/4)*Sqrt[c])], 1/2])/(b^(1/4)*Sqrt[a + b*x^2]))/Sqrt[b]) + (a^(1/4)*Sqr t[c]*(Sqrt[a]*c + Sqrt[b]*c*x)*Sqrt[(a*c^2 + b*c^2*x^2)/(Sqrt[a]*c + Sqrt[ b]*c*x)^2]*EllipticF[2*ArcTan[(b^(1/4)*Sqrt[c*x])/(a^(1/4)*Sqrt[c])], 1/2] )/(2*b^(3/4)*Sqrt[a + b*x^2])))/c^3))/(5*c^2)
3.7.4.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^p/(c*(m + 1))), x] - Simp[2*b*(p/(c^2*(m + 1))) Int[ (c*x)^(m + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && LtQ[m, -1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, S imp[1/q Int[1/Sqrt[a + b*x^4], x], x] - Simp[1/q Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* (1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e }, x] && PosQ[c/a]
Time = 2.04 (sec) , antiderivative size = 216, normalized size of antiderivative = 0.73
method | result | size |
default | \(\frac {\frac {24 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, E\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) a b \,x^{2}}{5}-\frac {12 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) a b \,x^{2}}{5}-\frac {14 b^{2} x^{4}}{5}-\frac {16 a b \,x^{2}}{5}-\frac {2 a^{2}}{5}}{x^{2} \sqrt {b \,x^{2}+a}\, c^{3} \sqrt {c x}}\) | \(216\) |
risch | \(-\frac {2 \sqrt {b \,x^{2}+a}\, \left (7 b \,x^{2}+a \right )}{5 x^{2} c^{3} \sqrt {c x}}+\frac {12 b \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, E\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right ) \sqrt {c x \left (b \,x^{2}+a \right )}}{5 \sqrt {b c \,x^{3}+a c x}\, c^{3} \sqrt {c x}\, \sqrt {b \,x^{2}+a}}\) | \(219\) |
elliptic | \(\frac {\sqrt {c x \left (b \,x^{2}+a \right )}\, \left (-\frac {2 a \sqrt {b c \,x^{3}+a c x}}{5 c^{4} x^{3}}-\frac {14 \left (c b \,x^{2}+a c \right ) b}{5 c^{4} \sqrt {x \left (c b \,x^{2}+a c \right )}}+\frac {12 b \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, E\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right )}{5 c^{3} \sqrt {b c \,x^{3}+a c x}}\right )}{\sqrt {c x}\, \sqrt {b \,x^{2}+a}}\) | \(242\) |
2/5/x^2*(12*((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*2^(1/2)*((-b*x+(-a*b)^ (1/2))/(-a*b)^(1/2))^(1/2)*(-x/(-a*b)^(1/2)*b)^(1/2)*EllipticE(((b*x+(-a*b )^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*a*b*x^2-6*((b*x+(-a*b)^(1/2))/(- a*b)^(1/2))^(1/2)*2^(1/2)*((-b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*(-x/(-a *b)^(1/2)*b)^(1/2)*EllipticF(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2 ^(1/2))*a*b*x^2-7*b^2*x^4-8*a*b*x^2-a^2)/(b*x^2+a)^(1/2)/c^3/(c*x)^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.20 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{(c x)^{7/2}} \, dx=-\frac {2 \, {\left (12 \, \sqrt {b c} b x^{3} {\rm weierstrassZeta}\left (-\frac {4 \, a}{b}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right )\right ) + {\left (7 \, b x^{2} + a\right )} \sqrt {b x^{2} + a} \sqrt {c x}\right )}}{5 \, c^{4} x^{3}} \]
-2/5*(12*sqrt(b*c)*b*x^3*weierstrassZeta(-4*a/b, 0, weierstrassPInverse(-4 *a/b, 0, x)) + (7*b*x^2 + a)*sqrt(b*x^2 + a)*sqrt(c*x))/(c^4*x^3)
Result contains complex when optimal does not.
Time = 4.72 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.18 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{(c x)^{7/2}} \, dx=\frac {a^{\frac {3}{2}} \Gamma \left (- \frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, - \frac {5}{4} \\ - \frac {1}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 c^{\frac {7}{2}} x^{\frac {5}{2}} \Gamma \left (- \frac {1}{4}\right )} \]
a**(3/2)*gamma(-5/4)*hyper((-3/2, -5/4), (-1/4,), b*x**2*exp_polar(I*pi)/a )/(2*c**(7/2)*x**(5/2)*gamma(-1/4))
\[ \int \frac {\left (a+b x^2\right )^{3/2}}{(c x)^{7/2}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}}}{\left (c x\right )^{\frac {7}{2}}} \,d x } \]
\[ \int \frac {\left (a+b x^2\right )^{3/2}}{(c x)^{7/2}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}}}{\left (c x\right )^{\frac {7}{2}}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b x^2\right )^{3/2}}{(c x)^{7/2}} \, dx=\int \frac {{\left (b\,x^2+a\right )}^{3/2}}{{\left (c\,x\right )}^{7/2}} \,d x \]