Integrand size = 24, antiderivative size = 59 \[ \int (d+e x)^m \left (d^2-e^2 x^2\right )^{7/2} \, dx=\frac {(d+e x)^m \left (d^2-e^2 x^2\right )^{9/2} \operatorname {Hypergeometric2F1}\left (1,9+m,\frac {11}{2}+m,\frac {d+e x}{2 d}\right )}{d e (9+2 m)} \]
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 1.39 (sec) , antiderivative size = 347, normalized size of antiderivative = 5.88 \[ \int (d+e x)^m \left (d^2-e^2 x^2\right )^{7/2} \, dx=\frac {(d+e x)^m \left (1+\frac {e x}{d}\right )^{-\frac {1}{2}-m} \left (-105 d^4 e^3 x^3 \sqrt {d-e x} \sqrt {d+e x} \operatorname {AppellF1}\left (3,-\frac {1}{2},-\frac {1}{2}-m,4,\frac {e x}{d},-\frac {e x}{d}\right )+63 d^2 e^5 x^5 \sqrt {d-e x} \sqrt {d+e x} \operatorname {AppellF1}\left (5,-\frac {1}{2},-\frac {1}{2}-m,6,\frac {e x}{d},-\frac {e x}{d}\right )-15 e^7 x^7 \sqrt {d-e x} \sqrt {d+e x} \operatorname {AppellF1}\left (7,-\frac {1}{2},-\frac {1}{2}-m,8,\frac {e x}{d},-\frac {e x}{d}\right )-35\ 2^{\frac {3}{2}+m} d^7 \sqrt {1-\frac {e x}{d}} \sqrt {d^2-e^2 x^2} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-\frac {1}{2}-m,\frac {5}{2},\frac {d-e x}{2 d}\right )+35\ 2^{\frac {3}{2}+m} d^6 e x \sqrt {1-\frac {e x}{d}} \sqrt {d^2-e^2 x^2} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-\frac {1}{2}-m,\frac {5}{2},\frac {d-e x}{2 d}\right )\right )}{105 e \sqrt {1-\frac {e x}{d}}} \]
((d + e*x)^m*(1 + (e*x)/d)^(-1/2 - m)*(-105*d^4*e^3*x^3*Sqrt[d - e*x]*Sqrt [d + e*x]*AppellF1[3, -1/2, -1/2 - m, 4, (e*x)/d, -((e*x)/d)] + 63*d^2*e^5 *x^5*Sqrt[d - e*x]*Sqrt[d + e*x]*AppellF1[5, -1/2, -1/2 - m, 6, (e*x)/d, - ((e*x)/d)] - 15*e^7*x^7*Sqrt[d - e*x]*Sqrt[d + e*x]*AppellF1[7, -1/2, -1/2 - m, 8, (e*x)/d, -((e*x)/d)] - 35*2^(3/2 + m)*d^7*Sqrt[1 - (e*x)/d]*Sqrt[ d^2 - e^2*x^2]*Hypergeometric2F1[3/2, -1/2 - m, 5/2, (d - e*x)/(2*d)] + 35 *2^(3/2 + m)*d^6*e*x*Sqrt[1 - (e*x)/d]*Sqrt[d^2 - e^2*x^2]*Hypergeometric2 F1[3/2, -1/2 - m, 5/2, (d - e*x)/(2*d)]))/(105*e*Sqrt[1 - (e*x)/d])
Time = 0.22 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.41, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {474, 473, 79}
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.
\(\displaystyle \int \left (d^2-e^2 x^2\right )^{7/2} (d+e x)^m \, dx\) |
\(\Big \downarrow \) 474 |
\(\displaystyle (d+e x)^m \left (\frac {e x}{d}+1\right )^{-m} \int \left (\frac {e x}{d}+1\right )^m \left (d^2-e^2 x^2\right )^{7/2}dx\) |
\(\Big \downarrow \) 473 |
\(\displaystyle \frac {\left (d^2-e^2 x^2\right )^{9/2} (d+e x)^m \left (\frac {e x}{d}+1\right )^{-m-\frac {9}{2}} \int \left (\frac {e x}{d}+1\right )^{m+\frac {7}{2}} \left (d^2-d e x\right )^{7/2}dx}{\left (d^2-d e x\right )^{9/2}}\) |
\(\Big \downarrow \) 79 |
\(\displaystyle -\frac {2^{m+\frac {9}{2}} \left (d^2-e^2 x^2\right )^{9/2} (d+e x)^m \left (\frac {e x}{d}+1\right )^{-m-\frac {9}{2}} \operatorname {Hypergeometric2F1}\left (\frac {9}{2},-m-\frac {7}{2},\frac {11}{2},\frac {d-e x}{2 d}\right )}{9 d e}\) |
-1/9*(2^(9/2 + m)*(d + e*x)^m*(1 + (e*x)/d)^(-9/2 - m)*(d^2 - e^2*x^2)^(9/ 2)*Hypergeometric2F1[9/2, -7/2 - m, 11/2, (d - e*x)/(2*d)])/(d*e)
3.10.51.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 , m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] && !IntegerQ[m] && !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] || !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ c^(n - 1)*((a + b*x^2)^(p + 1)/((1 + d*(x/c))^(p + 1)*(a/c + (b*x)/d)^(p + 1))) Int[(1 + d*(x/c))^(n + p)*(a/c + (b/d)*x)^p, x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c^2 + a*d^2, 0] && (IntegerQ[n] || GtQ[c, 0]) && !Gt Q[a, 0] && !(IntegerQ[n] && (IntegerQ[3*p] || IntegerQ[4*p]))
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ c^IntPart[n]*((c + d*x)^FracPart[n]/(1 + d*(x/c))^FracPart[n]) Int[(1 + d *(x/c))^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c^2 + a*d^2, 0] && !(IntegerQ[n] || GtQ[c, 0])
\[\int \left (e x +d \right )^{m} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {7}{2}}d x\]
\[ \int (d+e x)^m \left (d^2-e^2 x^2\right )^{7/2} \, dx=\int { {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}} {\left (e x + d\right )}^{m} \,d x } \]
integral(-(e^6*x^6 - 3*d^2*e^4*x^4 + 3*d^4*e^2*x^2 - d^6)*sqrt(-e^2*x^2 + d^2)*(e*x + d)^m, x)
\[ \int (d+e x)^m \left (d^2-e^2 x^2\right )^{7/2} \, dx=\int \left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {7}{2}} \left (d + e x\right )^{m}\, dx \]
\[ \int (d+e x)^m \left (d^2-e^2 x^2\right )^{7/2} \, dx=\int { {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}} {\left (e x + d\right )}^{m} \,d x } \]
\[ \int (d+e x)^m \left (d^2-e^2 x^2\right )^{7/2} \, dx=\int { {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}} {\left (e x + d\right )}^{m} \,d x } \]
Timed out. \[ \int (d+e x)^m \left (d^2-e^2 x^2\right )^{7/2} \, dx=\int {\left (d^2-e^2\,x^2\right )}^{7/2}\,{\left (d+e\,x\right )}^m \,d x \]