Integrand size = 20, antiderivative size = 163 \[ \int \frac {x^3 \left (a+b x^2\right )^p}{d+e x} \, dx=-\frac {d \left (a+b x^2\right )^{1+p}}{2 b e^2 (1+p)}-\frac {e x^5 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {5}{2},-p,1,\frac {7}{2},-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{5 d^2}+\frac {d^3 \left (a+b x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,\frac {e^2 \left (a+b x^2\right )}{b d^2+a e^2}\right )}{2 e^2 \left (b d^2+a e^2\right ) (1+p)} \]
-1/2*d*(b*x^2+a)^(p+1)/b/e^2/(p+1)-1/5*e*x^5*(b*x^2+a)^p*AppellF1(5/2,1,-p ,7/2,e^2*x^2/d^2,-b*x^2/a)/d^2/((1+b*x^2/a)^p)+1/2*d^3*(b*x^2+a)^(p+1)*hyp ergeom([1, p+1],[2+p],e^2*(b*x^2+a)/(a*e^2+b*d^2))/e^2/(a*e^2+b*d^2)/(p+1)
Time = 0.33 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.60 \[ \int \frac {x^3 \left (a+b x^2\right )^p}{d+e x} \, dx=\frac {\left (a+b x^2\right )^p \left (-\frac {3 d^3 \left (\frac {e \left (-\sqrt {-\frac {a}{b}}+x\right )}{d+e x}\right )^{-p} \left (\frac {e \left (\sqrt {-\frac {a}{b}}+x\right )}{d+e x}\right )^{-p} \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,\frac {d-\sqrt {-\frac {a}{b}} e}{d+e x},\frac {d+\sqrt {-\frac {a}{b}} e}{d+e x}\right )}{p}+\frac {e \left (1+\frac {b x^2}{a}\right )^{-p} \left (6 b d^2 (1+p) x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )+e \left (-3 d \left (b x^2 \left (1+\frac {b x^2}{a}\right )^p+a \left (-1+\left (1+\frac {b x^2}{a}\right )^p\right )\right )+2 b e (1+p) x^3 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},-\frac {b x^2}{a}\right )\right )\right )}{b (1+p)}\right )}{6 e^4} \]
((a + b*x^2)^p*((-3*d^3*AppellF1[-2*p, -p, -p, 1 - 2*p, (d - Sqrt[-(a/b)]* e)/(d + e*x), (d + Sqrt[-(a/b)]*e)/(d + e*x)])/(p*((e*(-Sqrt[-(a/b)] + x)) /(d + e*x))^p*((e*(Sqrt[-(a/b)] + x))/(d + e*x))^p) + (e*(6*b*d^2*(1 + p)* x*Hypergeometric2F1[1/2, -p, 3/2, -((b*x^2)/a)] + e*(-3*d*(b*x^2*(1 + (b*x ^2)/a)^p + a*(-1 + (1 + (b*x^2)/a)^p)) + 2*b*e*(1 + p)*x^3*Hypergeometric2 F1[3/2, -p, 5/2, -((b*x^2)/a)])))/(b*(1 + p)*(1 + (b*x^2)/a)^p)))/(6*e^4)
Time = 0.30 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {621, 354, 90, 78, 395, 394}
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.
\(\displaystyle \int \frac {x^3 \left (a+b x^2\right )^p}{d+e x} \, dx\) |
\(\Big \downarrow \) 621 |
\(\displaystyle d \int \frac {x^3 \left (b x^2+a\right )^p}{d^2-e^2 x^2}dx-e \int \frac {x^4 \left (b x^2+a\right )^p}{d^2-e^2 x^2}dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} d \int \frac {x^2 \left (b x^2+a\right )^p}{d^2-e^2 x^2}dx^2-e \int \frac {x^4 \left (b x^2+a\right )^p}{d^2-e^2 x^2}dx\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {1}{2} d \left (\frac {d^2 \int \frac {\left (b x^2+a\right )^p}{d^2-e^2 x^2}dx^2}{e^2}-\frac {\left (a+b x^2\right )^{p+1}}{b e^2 (p+1)}\right )-e \int \frac {x^4 \left (b x^2+a\right )^p}{d^2-e^2 x^2}dx\) |
\(\Big \downarrow \) 78 |
\(\displaystyle \frac {1}{2} d \left (\frac {d^2 \left (a+b x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{e^2 (p+1) \left (a e^2+b d^2\right )}-\frac {\left (a+b x^2\right )^{p+1}}{b e^2 (p+1)}\right )-e \int \frac {x^4 \left (b x^2+a\right )^p}{d^2-e^2 x^2}dx\) |
\(\Big \downarrow \) 395 |
\(\displaystyle \frac {1}{2} d \left (\frac {d^2 \left (a+b x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{e^2 (p+1) \left (a e^2+b d^2\right )}-\frac {\left (a+b x^2\right )^{p+1}}{b e^2 (p+1)}\right )-e \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \int \frac {x^4 \left (\frac {b x^2}{a}+1\right )^p}{d^2-e^2 x^2}dx\) |
\(\Big \downarrow \) 394 |
\(\displaystyle \frac {1}{2} d \left (\frac {d^2 \left (a+b x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{e^2 (p+1) \left (a e^2+b d^2\right )}-\frac {\left (a+b x^2\right )^{p+1}}{b e^2 (p+1)}\right )-\frac {e x^5 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {5}{2},-p,1,\frac {7}{2},-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{5 d^2}\) |
-1/5*(e*x^5*(a + b*x^2)^p*AppellF1[5/2, -p, 1, 7/2, -((b*x^2)/a), (e^2*x^2 )/d^2])/(d^2*(1 + (b*x^2)/a)^p) + (d*(-((a + b*x^2)^(1 + p)/(b*e^2*(1 + p) )) + (d^2*(a + b*x^2)^(1 + p)*Hypergeometric2F1[1, 1 + p, 2 + p, (e^2*(a + b*x^2))/(b*d^2 + a*e^2)])/(e^2*(b*d^2 + a*e^2)*(1 + p))))/2
3.5.9.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b *c - a*d)^n*((a + b*x)^(m + 1)/(b^(n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m}, x] && !IntegerQ[m] && IntegerQ[n]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/2 , -p, -q, 1 + (m + 1)/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; FreeQ[{a, b, c, d, e, m, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, 1] && (Int egerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^ FracPart[p]) Int[(e*x)^m*(1 + b*(x^2/a))^p*(c + d*x^2)^q, x], x] /; FreeQ [{a, b, c, d, e, m, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, 1] && !(IntegerQ[p] || GtQ[a, 0])
Int[((x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_))/((c_) + (d_.)*(x_)), x_Symbol] :> Simp[c Int[x^m*((a + b*x^2)^p/(c^2 - d^2*x^2)), x], x] - Simp[d Int[ x^(m + 1)*((a + b*x^2)^p/(c^2 - d^2*x^2)), x], x] /; FreeQ[{a, b, c, d, m, p}, x]
\[\int \frac {x^{3} \left (b \,x^{2}+a \right )^{p}}{e x +d}d x\]
\[ \int \frac {x^3 \left (a+b x^2\right )^p}{d+e x} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p} x^{3}}{e x + d} \,d x } \]
Timed out. \[ \int \frac {x^3 \left (a+b x^2\right )^p}{d+e x} \, dx=\text {Timed out} \]
\[ \int \frac {x^3 \left (a+b x^2\right )^p}{d+e x} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p} x^{3}}{e x + d} \,d x } \]
\[ \int \frac {x^3 \left (a+b x^2\right )^p}{d+e x} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p} x^{3}}{e x + d} \,d x } \]
Timed out. \[ \int \frac {x^3 \left (a+b x^2\right )^p}{d+e x} \, dx=\int \frac {x^3\,{\left (b\,x^2+a\right )}^p}{d+e\,x} \,d x \]