Integrand size = 29, antiderivative size = 213 \[ \int \frac {(g x)^m (d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {4 (g x)^{1+m} (d+e x)}{5 g \left (d^2-e^2 x^2\right )^{5/2}}+\frac {(1-4 m) (g x)^{1+m} \sqrt {1-\frac {e^2 x^2}{d^2}} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {1+m}{2},\frac {3+m}{2},\frac {e^2 x^2}{d^2}\right )}{5 d^3 g (1+m) \sqrt {d^2-e^2 x^2}}+\frac {e (7-4 m) (g x)^{2+m} \sqrt {1-\frac {e^2 x^2}{d^2}} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {2+m}{2},\frac {4+m}{2},\frac {e^2 x^2}{d^2}\right )}{5 d^4 g^2 (2+m) \sqrt {d^2-e^2 x^2}} \]
4/5*(g*x)^(1+m)*(e*x+d)/g/(-e^2*x^2+d^2)^(5/2)+1/5*(1-4*m)*(g*x)^(1+m)*hyp ergeom([5/2, 1/2+1/2*m],[3/2+1/2*m],e^2*x^2/d^2)*(1-e^2*x^2/d^2)^(1/2)/d^3 /g/(1+m)/(-e^2*x^2+d^2)^(1/2)+1/5*e*(7-4*m)*(g*x)^(2+m)*hypergeom([5/2, 1+ 1/2*m],[2+1/2*m],e^2*x^2/d^2)*(1-e^2*x^2/d^2)^(1/2)/d^4/g^2/(2+m)/(-e^2*x^ 2+d^2)^(1/2)
Time = 1.10 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.93 \[ \int \frac {(g x)^m (d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {x (g x)^m \sqrt {1-\frac {e^2 x^2}{d^2}} \left (\frac {d^3 \operatorname {Hypergeometric2F1}\left (\frac {7}{2},\frac {1+m}{2},\frac {3+m}{2},\frac {e^2 x^2}{d^2}\right )}{1+m}+e x \left (\frac {3 d^2 \operatorname {Hypergeometric2F1}\left (\frac {7}{2},\frac {2+m}{2},\frac {4+m}{2},\frac {e^2 x^2}{d^2}\right )}{2+m}+e x \left (\frac {3 d \operatorname {Hypergeometric2F1}\left (\frac {7}{2},\frac {3+m}{2},\frac {5+m}{2},\frac {e^2 x^2}{d^2}\right )}{3+m}+\frac {e x \operatorname {Hypergeometric2F1}\left (\frac {7}{2},\frac {4+m}{2},\frac {6+m}{2},\frac {e^2 x^2}{d^2}\right )}{4+m}\right )\right )\right )}{d^6 \sqrt {d^2-e^2 x^2}} \]
(x*(g*x)^m*Sqrt[1 - (e^2*x^2)/d^2]*((d^3*Hypergeometric2F1[7/2, (1 + m)/2, (3 + m)/2, (e^2*x^2)/d^2])/(1 + m) + e*x*((3*d^2*Hypergeometric2F1[7/2, ( 2 + m)/2, (4 + m)/2, (e^2*x^2)/d^2])/(2 + m) + e*x*((3*d*Hypergeometric2F1 [7/2, (3 + m)/2, (5 + m)/2, (e^2*x^2)/d^2])/(3 + m) + (e*x*Hypergeometric2 F1[7/2, (4 + m)/2, (6 + m)/2, (e^2*x^2)/d^2])/(4 + m)))))/(d^6*Sqrt[d^2 - e^2*x^2])
Time = 0.33 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {558, 25, 27, 557, 279, 278}
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.
| \(\displaystyle \int \frac {(d+e x)^3 (g x)^m}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx\) |
| \(\Big \downarrow \) 558 |
| \(\displaystyle \frac {4 (d+e x) (g x)^{m+1}}{5 g \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int -\frac {d^2 (g x)^m (d (1-4 m)+e (7-4 m) x)}{\left (d^2-e^2 x^2\right )^{5/2}}dx}{5 d^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {d^2 (g x)^m (d (1-4 m)+e (7-4 m) x)}{\left (d^2-e^2 x^2\right )^{5/2}}dx}{5 d^2}+\frac {4 (d+e x) (g x)^{m+1}}{5 g \left (d^2-e^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int \frac {(g x)^m (d (1-4 m)+e (7-4 m) x)}{\left (d^2-e^2 x^2\right )^{5/2}}dx+\frac {4 (d+e x) (g x)^{m+1}}{5 g \left (d^2-e^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 557 |
| \(\displaystyle \frac {1}{5} \left (d (1-4 m) \int \frac {(g x)^m}{\left (d^2-e^2 x^2\right )^{5/2}}dx+\frac {e (7-4 m) \int \frac {(g x)^{m+1}}{\left (d^2-e^2 x^2\right )^{5/2}}dx}{g}\right )+\frac {4 (d+e x) (g x)^{m+1}}{5 g \left (d^2-e^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 279 |
| \(\displaystyle \frac {1}{5} \left (\frac {e (7-4 m) \sqrt {1-\frac {e^2 x^2}{d^2}} \int \frac {(g x)^{m+1}}{\left (1-\frac {e^2 x^2}{d^2}\right )^{5/2}}dx}{d^4 g \sqrt {d^2-e^2 x^2}}+\frac {(1-4 m) \sqrt {1-\frac {e^2 x^2}{d^2}} \int \frac {(g x)^m}{\left (1-\frac {e^2 x^2}{d^2}\right )^{5/2}}dx}{d^3 \sqrt {d^2-e^2 x^2}}\right )+\frac {4 (d+e x) (g x)^{m+1}}{5 g \left (d^2-e^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 278 |
| \(\displaystyle \frac {4 (d+e x) (g x)^{m+1}}{5 g \left (d^2-e^2 x^2\right )^{5/2}}+\frac {1}{5} \left (\frac {e (7-4 m) \sqrt {1-\frac {e^2 x^2}{d^2}} (g x)^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {m+2}{2},\frac {m+4}{2},\frac {e^2 x^2}{d^2}\right )}{d^4 g^2 (m+2) \sqrt {d^2-e^2 x^2}}+\frac {(1-4 m) \sqrt {1-\frac {e^2 x^2}{d^2}} (g x)^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {m+1}{2},\frac {m+3}{2},\frac {e^2 x^2}{d^2}\right )}{d^3 g (m+1) \sqrt {d^2-e^2 x^2}}\right )\) |
(4*(g*x)^(1 + m)*(d + e*x))/(5*g*(d^2 - e^2*x^2)^(5/2)) + (((1 - 4*m)*(g*x )^(1 + m)*Sqrt[1 - (e^2*x^2)/d^2]*Hypergeometric2F1[5/2, (1 + m)/2, (3 + m )/2, (e^2*x^2)/d^2])/(d^3*g*(1 + m)*Sqrt[d^2 - e^2*x^2]) + (e*(7 - 4*m)*(g *x)^(2 + m)*Sqrt[1 - (e^2*x^2)/d^2]*Hypergeometric2F1[5/2, (2 + m)/2, (4 + m)/2, (e^2*x^2)/d^2])/(d^4*g^2*(2 + m)*Sqrt[d^2 - e^2*x^2]))/5
3.3.33.3.1 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[a^p*(( c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( -b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && (ILtQ[p, 0 ] || GtQ[a, 0])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntP art[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p]) Int[(c*x)^m* (1 + b*(x^2/a))^p, x], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && !(ILtQ[p, 0] || GtQ[a, 0])
Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Sym bol] :> Simp[c Int[(e*x)^m*(a + b*x^2)^p, x], x] + Simp[d/e Int[(e*x)^( m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x]
Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Qx = PolynomialQuotient[(c + d*x)^n, a + b*x^2, x], f = Coeff[PolynomialRemainder[(c + d*x)^n, a + b*x^2, x], x, 0], g = Coeff[Pol ynomialRemainder[(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(-(e*x)^(m + 1))* (f + g*x)*((a + b*x^2)^(p + 1)/(2*a*e*(p + 1))), x] + Simp[1/(2*a*(p + 1)) Int[(e*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Qx + f*(m + 2*p + 3) + g*(m + 2*p + 4)*x, x], x], x]] /; FreeQ[{a, b, c, d, e, m}, x] && IGt Q[n, 1] && !IntegerQ[m] && LtQ[p, -1]
\[\int \frac {\left (g x \right )^{m} \left (e x +d \right )^{3}}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}}}d x\]
\[ \int \frac {(g x)^m (d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int { \frac {{\left (e x + d\right )}^{3} \left (g x\right )^{m}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}}} \,d x } \]
integral(sqrt(-e^2*x^2 + d^2)*(g*x)^m/(e^5*x^5 - 3*d*e^4*x^4 + 2*d^2*e^3*x ^3 + 2*d^3*e^2*x^2 - 3*d^4*e*x + d^5), x)
\[ \int \frac {(g x)^m (d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int \frac {\left (g x\right )^{m} \left (d + e x\right )^{3}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {7}{2}}}\, dx \]
\[ \int \frac {(g x)^m (d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int { \frac {{\left (e x + d\right )}^{3} \left (g x\right )^{m}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}}} \,d x } \]
\[ \int \frac {(g x)^m (d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int { \frac {{\left (e x + d\right )}^{3} \left (g x\right )^{m}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}}} \,d x } \]
Timed out. \[ \int \frac {(g x)^m (d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int \frac {{\left (g\,x\right )}^m\,{\left (d+e\,x\right )}^3}{{\left (d^2-e^2\,x^2\right )}^{7/2}} \,d x \]