Integrand size = 20, antiderivative size = 324 \[ \int \frac {x^2 \left (a+b x^2\right )^p}{(c+d x)^3} \, dx=-\frac {c^4 \left (a+b x^2\right )^{1+p}}{d \left (b c^2+a d^2\right ) \left (c^2-d^2 x^2\right )^2}-\frac {\left (a+b x^2\right )^{1+p}}{2 b d p \left (c^2-d^2 x^2\right )}+\frac {x^3 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{2},-p,3,\frac {5}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{3 c^3}+\frac {3 d^2 x^5 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {5}{2},-p,3,\frac {7}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{5 c^5}+\frac {\left (a^2 d^4+a b c^2 d^2 (2+5 p)+b^2 c^4 \left (1+3 p+2 p^2\right )\right ) \left (a+b x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (2,1+p,2+p,\frac {d^2 \left (a+b x^2\right )}{b c^2+a d^2}\right )}{2 d \left (b c^2+a d^2\right )^3 p (1+p)} \] Output:
-c^4*(b*x^2+a)^(p+1)/d/(a*d^2+b*c^2)/(-d^2*x^2+c^2)^2-1/2*(b*x^2+a)^(p+1)/ b/d/p/(-d^2*x^2+c^2)+1/3*x^3*(b*x^2+a)^p*AppellF1(3/2,3,-p,5/2,d^2*x^2/c^2 ,-b*x^2/a)/c^3/((1+b*x^2/a)^p)+3/5*d^2*x^5*(b*x^2+a)^p*AppellF1(5/2,3,-p,7 /2,d^2*x^2/c^2,-b*x^2/a)/c^5/((1+b*x^2/a)^p)+1/2*(a^2*d^4+a*b*c^2*d^2*(2+5 *p)+b^2*c^4*(2*p^2+3*p+1))*(b*x^2+a)^(p+1)*hypergeom([2, p+1],[2+p],d^2*(b *x^2+a)/(a*d^2+b*c^2))/d/(a*d^2+b*c^2)^3/p/(p+1)
Time = 0.25 (sec) , antiderivative size = 290, normalized size of antiderivative = 0.90 \[ \int \frac {x^2 \left (a+b x^2\right )^p}{(c+d x)^3} \, dx=\frac {\left (\frac {d \left (-\sqrt {-\frac {a}{b}}+x\right )}{c+d x}\right )^{-p} \left (\frac {d \left (\sqrt {-\frac {a}{b}}+x\right )}{c+d x}\right )^{-p} \left (a+b x^2\right )^p \left (-\frac {4 c \operatorname {AppellF1}\left (1-2 p,-p,-p,2-2 p,\frac {c-\sqrt {-\frac {a}{b}} d}{c+d x},\frac {c+\sqrt {-\frac {a}{b}} d}{c+d x}\right )}{(-1+2 p) (c+d x)}+\frac {c^2 \operatorname {AppellF1}\left (2-2 p,-p,-p,3-2 p,\frac {c-\sqrt {-\frac {a}{b}} d}{c+d x},\frac {c+\sqrt {-\frac {a}{b}} d}{c+d x}\right )}{(-1+p) (c+d x)^2}+\frac {\operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,\frac {c-\sqrt {-\frac {a}{b}} d}{c+d x},\frac {c+\sqrt {-\frac {a}{b}} d}{c+d x}\right )}{p}\right )}{2 d^3} \] Input:
Integrate[(x^2*(a + b*x^2)^p)/(c + d*x)^3,x]
Output:
((a + b*x^2)^p*((-4*c*AppellF1[1 - 2*p, -p, -p, 2 - 2*p, (c - Sqrt[-(a/b)] *d)/(c + d*x), (c + Sqrt[-(a/b)]*d)/(c + d*x)])/((-1 + 2*p)*(c + d*x)) + ( c^2*AppellF1[2 - 2*p, -p, -p, 3 - 2*p, (c - Sqrt[-(a/b)]*d)/(c + d*x), (c + Sqrt[-(a/b)]*d)/(c + d*x)])/((-1 + p)*(c + d*x)^2) + AppellF1[-2*p, -p, -p, 1 - 2*p, (c - Sqrt[-(a/b)]*d)/(c + d*x), (c + Sqrt[-(a/b)]*d)/(c + d*x )]/p))/(2*d^3*((d*(-Sqrt[-(a/b)] + x))/(c + d*x))^p*((d*(Sqrt[-(a/b)] + x) )/(c + d*x))^p)
Time = 1.02 (sec) , antiderivative size = 364, normalized size of antiderivative = 1.12, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.550, Rules used = {603, 27, 688, 719, 238, 237, 504, 334, 333, 353, 78}
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.
\(\displaystyle \int \frac {x^2 \left (a+b x^2\right )^p}{(c+d x)^3} \, dx\) |
\(\Big \downarrow \) 603 |
\(\displaystyle -\frac {\int \frac {2 \left (a c d-\left (b (p+1) c^2+a d^2\right ) x\right ) \left (b x^2+a\right )^p}{d (c+d x)^2}dx}{2 \left (a d^2+b c^2\right )}-\frac {c^2 \left (a+b x^2\right )^{p+1}}{2 d (c+d x)^2 \left (a d^2+b c^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \frac {\left (a c d-\left (b (p+1) c^2+a d^2\right ) x\right ) \left (b x^2+a\right )^p}{(c+d x)^2}dx}{d \left (a d^2+b c^2\right )}-\frac {c^2 \left (a+b x^2\right )^{p+1}}{2 d (c+d x)^2 \left (a d^2+b c^2\right )}\) |
\(\Big \downarrow \) 688 |
\(\displaystyle -\frac {-\frac {\int \frac {\left (a d \left (b p c^2+a d^2\right )-b c (2 p+1) \left (b (p+1) c^2+2 a d^2\right ) x\right ) \left (b x^2+a\right )^p}{c+d x}dx}{a d^2+b c^2}-\frac {c \left (a+b x^2\right )^{p+1} \left (2 a d^2+b c^2 (p+1)\right )}{(c+d x) \left (a d^2+b c^2\right )}}{d \left (a d^2+b c^2\right )}-\frac {c^2 \left (a+b x^2\right )^{p+1}}{2 d (c+d x)^2 \left (a d^2+b c^2\right )}\) |
\(\Big \downarrow \) 719 |
\(\displaystyle -\frac {-\frac {\frac {\left (a^2 d^4+a b c^2 d^2 (5 p+2)+b^2 c^4 \left (2 p^2+3 p+1\right )\right ) \int \frac {\left (b x^2+a\right )^p}{c+d x}dx}{d}-\frac {b c (2 p+1) \left (2 a d^2+b c^2 (p+1)\right ) \int \left (b x^2+a\right )^pdx}{d}}{a d^2+b c^2}-\frac {c \left (a+b x^2\right )^{p+1} \left (2 a d^2+b c^2 (p+1)\right )}{(c+d x) \left (a d^2+b c^2\right )}}{d \left (a d^2+b c^2\right )}-\frac {c^2 \left (a+b x^2\right )^{p+1}}{2 d (c+d x)^2 \left (a d^2+b c^2\right )}\) |
\(\Big \downarrow \) 238 |
\(\displaystyle -\frac {-\frac {\frac {\left (a^2 d^4+a b c^2 d^2 (5 p+2)+b^2 c^4 \left (2 p^2+3 p+1\right )\right ) \int \frac {\left (b x^2+a\right )^p}{c+d x}dx}{d}-\frac {b c (2 p+1) \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (2 a d^2+b c^2 (p+1)\right ) \int \left (\frac {b x^2}{a}+1\right )^pdx}{d}}{a d^2+b c^2}-\frac {c \left (a+b x^2\right )^{p+1} \left (2 a d^2+b c^2 (p+1)\right )}{(c+d x) \left (a d^2+b c^2\right )}}{d \left (a d^2+b c^2\right )}-\frac {c^2 \left (a+b x^2\right )^{p+1}}{2 d (c+d x)^2 \left (a d^2+b c^2\right )}\) |
\(\Big \downarrow \) 237 |
\(\displaystyle -\frac {-\frac {\frac {\left (a^2 d^4+a b c^2 d^2 (5 p+2)+b^2 c^4 \left (2 p^2+3 p+1\right )\right ) \int \frac {\left (b x^2+a\right )^p}{c+d x}dx}{d}-\frac {b c (2 p+1) x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (2 a d^2+b c^2 (p+1)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{d}}{a d^2+b c^2}-\frac {c \left (a+b x^2\right )^{p+1} \left (2 a d^2+b c^2 (p+1)\right )}{(c+d x) \left (a d^2+b c^2\right )}}{d \left (a d^2+b c^2\right )}-\frac {c^2 \left (a+b x^2\right )^{p+1}}{2 d (c+d x)^2 \left (a d^2+b c^2\right )}\) |
\(\Big \downarrow \) 504 |
\(\displaystyle -\frac {-\frac {\frac {\left (a^2 d^4+a b c^2 d^2 (5 p+2)+b^2 c^4 \left (2 p^2+3 p+1\right )\right ) \left (c \int \frac {\left (b x^2+a\right )^p}{c^2-d^2 x^2}dx-d \int \frac {x \left (b x^2+a\right )^p}{c^2-d^2 x^2}dx\right )}{d}-\frac {b c (2 p+1) x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (2 a d^2+b c^2 (p+1)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{d}}{a d^2+b c^2}-\frac {c \left (a+b x^2\right )^{p+1} \left (2 a d^2+b c^2 (p+1)\right )}{(c+d x) \left (a d^2+b c^2\right )}}{d \left (a d^2+b c^2\right )}-\frac {c^2 \left (a+b x^2\right )^{p+1}}{2 d (c+d x)^2 \left (a d^2+b c^2\right )}\) |
\(\Big \downarrow \) 334 |
\(\displaystyle -\frac {-\frac {\frac {\left (a^2 d^4+a b c^2 d^2 (5 p+2)+b^2 c^4 \left (2 p^2+3 p+1\right )\right ) \left (c \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \int \frac {\left (\frac {b x^2}{a}+1\right )^p}{c^2-d^2 x^2}dx-d \int \frac {x \left (b x^2+a\right )^p}{c^2-d^2 x^2}dx\right )}{d}-\frac {b c (2 p+1) x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (2 a d^2+b c^2 (p+1)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{d}}{a d^2+b c^2}-\frac {c \left (a+b x^2\right )^{p+1} \left (2 a d^2+b c^2 (p+1)\right )}{(c+d x) \left (a d^2+b c^2\right )}}{d \left (a d^2+b c^2\right )}-\frac {c^2 \left (a+b x^2\right )^{p+1}}{2 d (c+d x)^2 \left (a d^2+b c^2\right )}\) |
\(\Big \downarrow \) 333 |
\(\displaystyle -\frac {-\frac {\frac {\left (a^2 d^4+a b c^2 d^2 (5 p+2)+b^2 c^4 \left (2 p^2+3 p+1\right )\right ) \left (\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c}-d \int \frac {x \left (b x^2+a\right )^p}{c^2-d^2 x^2}dx\right )}{d}-\frac {b c (2 p+1) x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (2 a d^2+b c^2 (p+1)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{d}}{a d^2+b c^2}-\frac {c \left (a+b x^2\right )^{p+1} \left (2 a d^2+b c^2 (p+1)\right )}{(c+d x) \left (a d^2+b c^2\right )}}{d \left (a d^2+b c^2\right )}-\frac {c^2 \left (a+b x^2\right )^{p+1}}{2 d (c+d x)^2 \left (a d^2+b c^2\right )}\) |
\(\Big \downarrow \) 353 |
\(\displaystyle -\frac {-\frac {\frac {\left (a^2 d^4+a b c^2 d^2 (5 p+2)+b^2 c^4 \left (2 p^2+3 p+1\right )\right ) \left (\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c}-\frac {1}{2} d \int \frac {\left (b x^2+a\right )^p}{c^2-d^2 x^2}dx^2\right )}{d}-\frac {b c (2 p+1) x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (2 a d^2+b c^2 (p+1)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{d}}{a d^2+b c^2}-\frac {c \left (a+b x^2\right )^{p+1} \left (2 a d^2+b c^2 (p+1)\right )}{(c+d x) \left (a d^2+b c^2\right )}}{d \left (a d^2+b c^2\right )}-\frac {c^2 \left (a+b x^2\right )^{p+1}}{2 d (c+d x)^2 \left (a d^2+b c^2\right )}\) |
\(\Big \downarrow \) 78 |
\(\displaystyle -\frac {-\frac {\frac {\left (a^2 d^4+a b c^2 d^2 (5 p+2)+b^2 c^4 \left (2 p^2+3 p+1\right )\right ) \left (\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c}-\frac {d \left (a+b x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {d^2 \left (b x^2+a\right )}{b c^2+a d^2}\right )}{2 (p+1) \left (a d^2+b c^2\right )}\right )}{d}-\frac {b c (2 p+1) x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (2 a d^2+b c^2 (p+1)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{d}}{a d^2+b c^2}-\frac {c \left (a+b x^2\right )^{p+1} \left (2 a d^2+b c^2 (p+1)\right )}{(c+d x) \left (a d^2+b c^2\right )}}{d \left (a d^2+b c^2\right )}-\frac {c^2 \left (a+b x^2\right )^{p+1}}{2 d (c+d x)^2 \left (a d^2+b c^2\right )}\) |
Input:
Int[(x^2*(a + b*x^2)^p)/(c + d*x)^3,x]
Output:
-1/2*(c^2*(a + b*x^2)^(1 + p))/(d*(b*c^2 + a*d^2)*(c + d*x)^2) - (-((c*(2* a*d^2 + b*c^2*(1 + p))*(a + b*x^2)^(1 + p))/((b*c^2 + a*d^2)*(c + d*x))) - (-((b*c*(1 + 2*p)*(2*a*d^2 + b*c^2*(1 + p))*x*(a + b*x^2)^p*Hypergeometri c2F1[1/2, -p, 3/2, -((b*x^2)/a)])/(d*(1 + (b*x^2)/a)^p)) + ((a^2*d^4 + a*b *c^2*d^2*(2 + 5*p) + b^2*c^4*(1 + 3*p + 2*p^2))*((x*(a + b*x^2)^p*AppellF1 [1/2, -p, 1, 3/2, -((b*x^2)/a), (d^2*x^2)/c^2])/(c*(1 + (b*x^2)/a)^p) - (d *(a + b*x^2)^(1 + p)*Hypergeometric2F1[1, 1 + p, 2 + p, (d^2*(a + b*x^2))/ (b*c^2 + a*d^2)])/(2*(b*c^2 + a*d^2)*(1 + p))))/d)/(b*c^2 + a*d^2))/(d*(b* c^2 + a*d^2))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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_)^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[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[a^p*c^q*x*AppellF1[1/2, -p, -q, 3/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; F reeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[a^IntPart[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p]) Int[ (1 + b*(x^2/a))^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && !(IntegerQ[p] || GtQ[a, 0])
Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[1/2 Subst[Int[(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]
Int[((a_) + (b_.)*(x_)^2)^(p_)/((c_) + (d_.)*(x_)), x_Symbol] :> Simp[c I nt[(a + b*x^2)^p/(c^2 - d^2*x^2), x], x] - Simp[d Int[x*((a + b*x^2)^p/(c ^2 - d^2*x^2)), x], x] /; FreeQ[{a, b, c, d, p}, x]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol ] :> With[{Qx = PolynomialQuotient[x^m, c + d*x, x], R = PolynomialRemainde r[x^m, c + d*x, x]}, Simp[d*R*(c + d*x)^(n + 1)*((a + b*x^2)^(p + 1)/((n + 1)*(b*c^2 + a*d^2))), x] + Simp[1/((n + 1)*(b*c^2 + a*d^2)) Int[(c + d*x) ^(n + 1)*(a + b*x^2)^p*ExpandToSum[(n + 1)*(b*c^2 + a*d^2)*Qx + b*c*R*(n + 1) - b*d*R*(n + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d, p}, x] && IGt Q[m, 1] && LtQ[n, -1] && NeQ[b*c^2 + a*d^2, 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p _.), x_Symbol] :> Simp[(e*f - d*g)*(d + e*x)^(m + 1)*((a + c*x^2)^(p + 1)/( (m + 1)*(c*d^2 + a*e^2))), x] + Simp[1/((m + 1)*(c*d^2 + a*e^2)) Int[(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p _.), x_Symbol] :> Simp[g/e Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Simp[(e*f - d*g)/e Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && !IGtQ[m, 0]
\[\int \frac {x^{2} \left (b \,x^{2}+a \right )^{p}}{\left (d x +c \right )^{3}}d x\]
Input:
int(x^2*(b*x^2+a)^p/(d*x+c)^3,x)
Output:
int(x^2*(b*x^2+a)^p/(d*x+c)^3,x)
\[ \int \frac {x^2 \left (a+b x^2\right )^p}{(c+d x)^3} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p} x^{2}}{{\left (d x + c\right )}^{3}} \,d x } \] Input:
integrate(x^2*(b*x^2+a)^p/(d*x+c)^3,x, algorithm="fricas")
Output:
integral((b*x^2 + a)^p*x^2/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3), x)
\[ \int \frac {x^2 \left (a+b x^2\right )^p}{(c+d x)^3} \, dx=\int \frac {x^{2} \left (a + b x^{2}\right )^{p}}{\left (c + d x\right )^{3}}\, dx \] Input:
integrate(x**2*(b*x**2+a)**p/(d*x+c)**3,x)
Output:
Integral(x**2*(a + b*x**2)**p/(c + d*x)**3, x)
\[ \int \frac {x^2 \left (a+b x^2\right )^p}{(c+d x)^3} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p} x^{2}}{{\left (d x + c\right )}^{3}} \,d x } \] Input:
integrate(x^2*(b*x^2+a)^p/(d*x+c)^3,x, algorithm="maxima")
Output:
integrate((b*x^2 + a)^p*x^2/(d*x + c)^3, x)
\[ \int \frac {x^2 \left (a+b x^2\right )^p}{(c+d x)^3} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p} x^{2}}{{\left (d x + c\right )}^{3}} \,d x } \] Input:
integrate(x^2*(b*x^2+a)^p/(d*x+c)^3,x, algorithm="giac")
Output:
integrate((b*x^2 + a)^p*x^2/(d*x + c)^3, x)
Timed out. \[ \int \frac {x^2 \left (a+b x^2\right )^p}{(c+d x)^3} \, dx=\int \frac {x^2\,{\left (b\,x^2+a\right )}^p}{{\left (c+d\,x\right )}^3} \,d x \] Input:
int((x^2*(a + b*x^2)^p)/(c + d*x)^3,x)
Output:
int((x^2*(a + b*x^2)^p)/(c + d*x)^3, x)
\[ \int \frac {x^2 \left (a+b x^2\right )^p}{(c+d x)^3} \, dx=\text {too large to display} \] Input:
int(x^2*(b*x^2+a)^p/(d*x+c)^3,x)
Output:
( - 3*(a + b*x**2)**p*a*d - 2*(a + b*x**2)**p*b*c*p*x - 2*(a + b*x**2)**p* b*c*x + 2*(a + b*x**2)**p*b*d*p*x**2 - (a + b*x**2)**p*b*d*x**2 - 12*int(( a + b*x**2)**p/(2*a*c**3*p - a*c**3 + 6*a*c**2*d*p*x - 3*a*c**2*d*x + 6*a* c*d**2*p*x**2 - 3*a*c*d**2*x**2 + 2*a*d**3*p*x**3 - a*d**3*x**3 + 2*b*c**3 *p*x**2 - b*c**3*x**2 + 6*b*c**2*d*p*x**3 - 3*b*c**2*d*x**3 + 6*b*c*d**2*p *x**4 - 3*b*c*d**2*x**4 + 2*b*d**3*p*x**5 - b*d**3*x**5),x)*a**2*c**2*d**2 *p + 6*int((a + b*x**2)**p/(2*a*c**3*p - a*c**3 + 6*a*c**2*d*p*x - 3*a*c** 2*d*x + 6*a*c*d**2*p*x**2 - 3*a*c*d**2*x**2 + 2*a*d**3*p*x**3 - a*d**3*x** 3 + 2*b*c**3*p*x**2 - b*c**3*x**2 + 6*b*c**2*d*p*x**3 - 3*b*c**2*d*x**3 + 6*b*c*d**2*p*x**4 - 3*b*c*d**2*x**4 + 2*b*d**3*p*x**5 - b*d**3*x**5),x)*a* *2*c**2*d**2 - 24*int((a + b*x**2)**p/(2*a*c**3*p - a*c**3 + 6*a*c**2*d*p* x - 3*a*c**2*d*x + 6*a*c*d**2*p*x**2 - 3*a*c*d**2*x**2 + 2*a*d**3*p*x**3 - a*d**3*x**3 + 2*b*c**3*p*x**2 - b*c**3*x**2 + 6*b*c**2*d*p*x**3 - 3*b*c** 2*d*x**3 + 6*b*c*d**2*p*x**4 - 3*b*c*d**2*x**4 + 2*b*d**3*p*x**5 - b*d**3* x**5),x)*a**2*c*d**3*p*x + 12*int((a + b*x**2)**p/(2*a*c**3*p - a*c**3 + 6 *a*c**2*d*p*x - 3*a*c**2*d*x + 6*a*c*d**2*p*x**2 - 3*a*c*d**2*x**2 + 2*a*d **3*p*x**3 - a*d**3*x**3 + 2*b*c**3*p*x**2 - b*c**3*x**2 + 6*b*c**2*d*p*x* *3 - 3*b*c**2*d*x**3 + 6*b*c*d**2*p*x**4 - 3*b*c*d**2*x**4 + 2*b*d**3*p*x* *5 - b*d**3*x**5),x)*a**2*c*d**3*x - 12*int((a + b*x**2)**p/(2*a*c**3*p - a*c**3 + 6*a*c**2*d*p*x - 3*a*c**2*d*x + 6*a*c*d**2*p*x**2 - 3*a*c*d**2...