Integrand size = 20, antiderivative size = 168 \[ \int \frac {(d+e x)^3 \left (a+b x^2\right )^p}{x^3} \, dx=-\frac {d^3 \left (a+b x^2\right )^{1+p}}{2 a x^2}-\frac {3 d^2 e \left (a+b x^2\right )^{1+p}}{a x}+\frac {e \left (a e^2+3 b d^2 (1+2 p)\right ) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{a}-\frac {d \left (3 a e^2+b d^2 p\right ) \left (a+b x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b x^2}{a}\right )}{2 a^2 (1+p)} \]
-1/2*d^3*(b*x^2+a)^(p+1)/a/x^2-3*d^2*e*(b*x^2+a)^(p+1)/a/x+e*(a*e^2+3*b*d^ 2*(1+2*p))*x*(b*x^2+a)^p*hypergeom([1/2, -p],[3/2],-b*x^2/a)/a/((1+b*x^2/a )^p)-1/2*d*(b*d^2*p+3*a*e^2)*(b*x^2+a)^(p+1)*hypergeom([1, p+1],[2+p],1+b* x^2/a)/a^2/(p+1)
Time = 0.25 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.04 \[ \int \frac {(d+e x)^3 \left (a+b x^2\right )^p}{x^3} \, dx=-\frac {\left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (6 a^2 d^2 e (1+p) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b x^2}{a}\right )+x \left (-2 a^2 e^3 (1+p) x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )+d \left (a+b x^2\right ) \left (1+\frac {b x^2}{a}\right )^p \left (3 a e^2 \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b x^2}{a}\right )-b d^2 \operatorname {Hypergeometric2F1}\left (2,1+p,2+p,1+\frac {b x^2}{a}\right )\right )\right )\right )}{2 a^2 (1+p) x} \]
-1/2*((a + b*x^2)^p*(6*a^2*d^2*e*(1 + p)*Hypergeometric2F1[-1/2, -p, 1/2, -((b*x^2)/a)] + x*(-2*a^2*e^3*(1 + p)*x*Hypergeometric2F1[1/2, -p, 3/2, -( (b*x^2)/a)] + d*(a + b*x^2)*(1 + (b*x^2)/a)^p*(3*a*e^2*Hypergeometric2F1[1 , 1 + p, 2 + p, 1 + (b*x^2)/a] - b*d^2*Hypergeometric2F1[2, 1 + p, 2 + p, 1 + (b*x^2)/a]))))/(a^2*(1 + p)*x*(1 + (b*x^2)/a)^p)
Time = 0.29 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.01, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {543, 354, 27, 87, 75, 359, 238, 237}
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.
\(\displaystyle \int \frac {(d+e x)^3 \left (a+b x^2\right )^p}{x^3} \, dx\) |
\(\Big \downarrow \) 543 |
\(\displaystyle \int \frac {\left (b x^2+a\right )^p \left (d^3+3 e^2 x^2 d\right )}{x^3}dx+\int \frac {\left (b x^2+a\right )^p \left (x^2 e^3+3 d^2 e\right )}{x^2}dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \int \frac {\left (b x^2+a\right )^p \left (x^2 e^3+3 d^2 e\right )}{x^2}dx+\frac {1}{2} \int \frac {d \left (b x^2+a\right )^p \left (d^2+3 e^2 x^2\right )}{x^4}dx^2\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {\left (b x^2+a\right )^p \left (x^2 e^3+3 d^2 e\right )}{x^2}dx+\frac {1}{2} d \int \frac {\left (b x^2+a\right )^p \left (d^2+3 e^2 x^2\right )}{x^4}dx^2\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \int \frac {\left (b x^2+a\right )^p \left (x^2 e^3+3 d^2 e\right )}{x^2}dx+\frac {1}{2} d \left (\frac {\left (3 a e^2+b d^2 p\right ) \int \frac {\left (b x^2+a\right )^p}{x^2}dx^2}{a}-\frac {d^2 \left (a+b x^2\right )^{p+1}}{a x^2}\right )\) |
\(\Big \downarrow \) 75 |
\(\displaystyle \int \frac {\left (b x^2+a\right )^p \left (x^2 e^3+3 d^2 e\right )}{x^2}dx+\frac {1}{2} d \left (-\frac {\left (a+b x^2\right )^{p+1} \left (3 a e^2+b d^2 p\right ) \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {b x^2}{a}+1\right )}{a^2 (p+1)}-\frac {d^2 \left (a+b x^2\right )^{p+1}}{a x^2}\right )\) |
\(\Big \downarrow \) 359 |
\(\displaystyle \frac {e \left (a e^2+3 b d^2 (2 p+1)\right ) \int \left (b x^2+a\right )^pdx}{a}+\frac {1}{2} d \left (-\frac {\left (a+b x^2\right )^{p+1} \left (3 a e^2+b d^2 p\right ) \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {b x^2}{a}+1\right )}{a^2 (p+1)}-\frac {d^2 \left (a+b x^2\right )^{p+1}}{a x^2}\right )-\frac {3 d^2 e \left (a+b x^2\right )^{p+1}}{a x}\) |
\(\Big \downarrow \) 238 |
\(\displaystyle \frac {e \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (a e^2+3 b d^2 (2 p+1)\right ) \int \left (\frac {b x^2}{a}+1\right )^pdx}{a}+\frac {1}{2} d \left (-\frac {\left (a+b x^2\right )^{p+1} \left (3 a e^2+b d^2 p\right ) \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {b x^2}{a}+1\right )}{a^2 (p+1)}-\frac {d^2 \left (a+b x^2\right )^{p+1}}{a x^2}\right )-\frac {3 d^2 e \left (a+b x^2\right )^{p+1}}{a x}\) |
\(\Big \downarrow \) 237 |
\(\displaystyle \frac {1}{2} d \left (-\frac {\left (a+b x^2\right )^{p+1} \left (3 a e^2+b d^2 p\right ) \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {b x^2}{a}+1\right )}{a^2 (p+1)}-\frac {d^2 \left (a+b x^2\right )^{p+1}}{a x^2}\right )+\frac {e x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (a e^2+3 b d^2 (2 p+1)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{a}-\frac {3 d^2 e \left (a+b x^2\right )^{p+1}}{a x}\) |
(-3*d^2*e*(a + b*x^2)^(1 + p))/(a*x) + (e*(a*e^2 + 3*b*d^2*(1 + 2*p))*x*(a + b*x^2)^p*Hypergeometric2F1[1/2, -p, 3/2, -((b*x^2)/a)])/(a*(1 + (b*x^2) /a)^p) + (d*(-((d^2*(a + b*x^2)^(1 + p))/(a*x^2)) - ((3*a*e^2 + b*d^2*p)*( a + b*x^2)^(1 + p)*Hypergeometric2F1[1, 1 + p, 2 + p, 1 + (b*x^2)/a])/(a^2 *(1 + p))))/2
3.5.7.3.1 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[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x )^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))*Hypergeometric2F1[-m, n + 1, n + 2, 1 + d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] && !IntegerQ[n] && (IntegerQ[m] || GtQ[-d/(b*c), 0])
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[- p, 1/2, 1/2 + 1, (-b)*(x^2/a)], x] /; FreeQ[{a, b, p}, x] && !IntegerQ[2*p ] && GtQ[a, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^2) ^FracPart[p]/(1 + b*(x^2/a))^FracPart[p]) Int[(1 + b*(x^2/a))^p, x], x] / ; FreeQ[{a, b, p}, x] && !IntegerQ[2*p] && !GtQ[a, 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), x _Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)* (a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !ILtQ[p, -1]
Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> Module[{k}, Int[x^m*Sum[Binomial[n, 2*k]*c^(n - 2*k)*d^(2*k)*x^(2*k), {k, 0, n/2}]*(a + b*x^2)^p, x] + Int[x^(m + 1)*Sum[Binomial[n, 2*k + 1]*c^ (n - 2*k - 1)*d^(2*k + 1)*x^(2*k), {k, 0, (n - 1)/2}]*(a + b*x^2)^p, x]] /; FreeQ[{a, b, c, d, p}, x] && IGtQ[n, 1] && IntegerQ[m] && !IntegerQ[2*p] && !(EqQ[m, 1] && EqQ[b*c^2 + a*d^2, 0])
\[\int \frac {\left (e x +d \right )^{3} \left (b \,x^{2}+a \right )^{p}}{x^{3}}d x\]
\[ \int \frac {(d+e x)^3 \left (a+b x^2\right )^p}{x^3} \, dx=\int { \frac {{\left (e x + d\right )}^{3} {\left (b x^{2} + a\right )}^{p}}{x^{3}} \,d x } \]
Result contains complex when optimal does not.
Time = 8.17 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.88 \[ \int \frac {(d+e x)^3 \left (a+b x^2\right )^p}{x^3} \, dx=- \frac {3 a^{p} d^{2} e {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - p \\ \frac {1}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{x} + a^{p} e^{3} x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )} - \frac {b^{p} d^{3} x^{2 p - 2} \Gamma \left (1 - p\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, 1 - p \\ 2 - p \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{2 \Gamma \left (2 - p\right )} - \frac {3 b^{p} d e^{2} x^{2 p} \Gamma \left (- p\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, - p \\ 1 - p \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{2 \Gamma \left (1 - p\right )} \]
-3*a**p*d**2*e*hyper((-1/2, -p), (1/2,), b*x**2*exp_polar(I*pi)/a)/x + a** p*e**3*x*hyper((1/2, -p), (3/2,), b*x**2*exp_polar(I*pi)/a) - b**p*d**3*x* *(2*p - 2)*gamma(1 - p)*hyper((-p, 1 - p), (2 - p,), a*exp_polar(I*pi)/(b* x**2))/(2*gamma(2 - p)) - 3*b**p*d*e**2*x**(2*p)*gamma(-p)*hyper((-p, -p), (1 - p,), a*exp_polar(I*pi)/(b*x**2))/(2*gamma(1 - p))
\[ \int \frac {(d+e x)^3 \left (a+b x^2\right )^p}{x^3} \, dx=\int { \frac {{\left (e x + d\right )}^{3} {\left (b x^{2} + a\right )}^{p}}{x^{3}} \,d x } \]
\[ \int \frac {(d+e x)^3 \left (a+b x^2\right )^p}{x^3} \, dx=\int { \frac {{\left (e x + d\right )}^{3} {\left (b x^{2} + a\right )}^{p}}{x^{3}} \,d x } \]
Timed out. \[ \int \frac {(d+e x)^3 \left (a+b x^2\right )^p}{x^3} \, dx=\int \frac {{\left (b\,x^2+a\right )}^p\,{\left (d+e\,x\right )}^3}{x^3} \,d x \]