Integrand size = 20, antiderivative size = 321 \[ \int \frac {x^3 \left (a+b x^2\right )^p}{(c+d x)^2} \, dx=\frac {\left (a+b x^2\right )^{1+p}}{2 b d^2 (1+p)}+\frac {c^3 \left (a+b x^2\right )^{1+p}}{d^2 \left (b c^2+a d^2\right ) (c+d x)}+\frac {c \left (3 a d^2+b c^2 (3+2 p)\right ) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{d^3 \left (b c^2+a d^2\right )}-\frac {c \left (2 a d^2+b c^2 (3+2 p)\right ) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{d^3 \left (b c^2+a d^2\right )}-\frac {c^2 \left (3 a d^2+b c^2 (3+2 p)\right ) \left (a+b x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,\frac {d^2 \left (a+b x^2\right )}{b c^2+a d^2}\right )}{2 d^2 \left (b c^2+a d^2\right )^2 (1+p)} \] Output:
1/2*(b*x^2+a)^(p+1)/b/d^2/(p+1)+c^3*(b*x^2+a)^(p+1)/d^2/(a*d^2+b*c^2)/(d*x +c)+c*(3*a*d^2+b*c^2*(3+2*p))*x*(b*x^2+a)^p*AppellF1(1/2,1,-p,3/2,d^2*x^2/ c^2,-b*x^2/a)/d^3/(a*d^2+b*c^2)/((1+b*x^2/a)^p)-c*(2*a*d^2+b*c^2*(3+2*p))* x*(b*x^2+a)^p*hypergeom([1/2, -p],[3/2],-b*x^2/a)/d^3/(a*d^2+b*c^2)/((1+b* x^2/a)^p)-1/2*c^2*(3*a*d^2+b*c^2*(3+2*p))*(b*x^2+a)^(p+1)*hypergeom([1, p+ 1],[2+p],d^2*(b*x^2+a)/(a*d^2+b*c^2))/d^2/(a*d^2+b*c^2)^2/(p+1)
Time = 0.67 (sec) , antiderivative size = 343, normalized size of antiderivative = 1.07 \[ \int \frac {x^3 \left (a+b x^2\right )^p}{(c+d x)^2} \, dx=\frac {\left (a+b x^2\right )^p \left (-\frac {2 c^3 \left (\frac {d \left (-\sqrt {-\frac {a}{b}}+x\right )}{c+d x}\right )^{-p} \left (\frac {d \left (\sqrt {-\frac {a}{b}}+x\right )}{c+d x}\right )^{-p} \operatorname {AppellF1}\left (1-2 p,-p,-p,2-2 p,\frac {c-\sqrt {-\frac {a}{b}} d}{c+d x},\frac {c+\sqrt {-\frac {a}{b}} d}{c+d x}\right )}{(-1+2 p) (c+d x)}+\frac {3 c^2 \left (\frac {d \left (-\sqrt {-\frac {a}{b}}+x\right )}{c+d x}\right )^{-p} \left (\frac {d \left (\sqrt {-\frac {a}{b}}+x\right )}{c+d x}\right )^{-p} \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,\frac {c-\sqrt {-\frac {a}{b}} d}{c+d x},\frac {c+\sqrt {-\frac {a}{b}} d}{c+d x}\right )}{p}+d \left (\frac {d \left (a+b x^2-a \left (1+\frac {b x^2}{a}\right )^{-p}\right )}{b+b p}-4 c x \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )\right )\right )}{2 d^4} \] Input:
Integrate[(x^3*(a + b*x^2)^p)/(c + d*x)^2,x]
Output:
((a + b*x^2)^p*((-2*c^3*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)*((d*(-Sqrt[- (a/b)] + x))/(c + d*x))^p*((d*(Sqrt[-(a/b)] + x))/(c + d*x))^p*(c + d*x)) + (3*c^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*((d*(-Sqrt[-(a/b)] + x))/(c + d*x))^p*( (d*(Sqrt[-(a/b)] + x))/(c + d*x))^p) + d*((d*(a + b*x^2 - a/(1 + (b*x^2)/a )^p))/(b + b*p) - (4*c*x*Hypergeometric2F1[1/2, -p, 3/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^p)))/(2*d^4)
Time = 1.11 (sec) , antiderivative size = 309, normalized size of antiderivative = 0.96, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {603, 25, 2185, 27, 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.
\(\displaystyle \int \frac {x^3 \left (a+b x^2\right )^p}{(c+d x)^2} \, dx\) |
\(\Big \downarrow \) 603 |
\(\displaystyle \frac {c^3 \left (a+b x^2\right )^{p+1}}{d^2 (c+d x) \left (a d^2+b c^2\right )}-\frac {\int -\frac {\left (b x^2+a\right )^p \left (\frac {a c^2}{d}-\left (\frac {2 b (p+1) c^2}{d^2}+a\right ) x c+\frac {\left (b c^2+a d^2\right ) x^2}{d}\right )}{c+d x}dx}{a d^2+b c^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {\left (b x^2+a\right )^p \left (\frac {a c^2}{d}-\left (\frac {2 b (p+1) c^2}{d^2}+a\right ) x c+\frac {\left (b c^2+a d^2\right ) x^2}{d}\right )}{c+d x}dx}{a d^2+b c^2}+\frac {c^3 \left (a+b x^2\right )^{p+1}}{d^2 (c+d x) \left (a d^2+b c^2\right )}\) |
\(\Big \downarrow \) 2185 |
\(\displaystyle \frac {\frac {\int \frac {2 b c (p+1) \left (a c d-\left (b (2 p+3) c^2+2 a d^2\right ) x\right ) \left (b x^2+a\right )^p}{c+d x}dx}{2 b d^2 (p+1)}+\frac {\left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{2 b d^2 (p+1)}}{a d^2+b c^2}+\frac {c^3 \left (a+b x^2\right )^{p+1}}{d^2 (c+d x) \left (a d^2+b c^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {c \int \frac {\left (a c d-\left (b (2 p+3) c^2+2 a d^2\right ) x\right ) \left (b x^2+a\right )^p}{c+d x}dx}{d^2}+\frac {\left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{2 b d^2 (p+1)}}{a d^2+b c^2}+\frac {c^3 \left (a+b x^2\right )^{p+1}}{d^2 (c+d x) \left (a d^2+b c^2\right )}\) |
\(\Big \downarrow \) 719 |
\(\displaystyle \frac {\frac {c \left (\frac {c \left (3 a d^2+b c^2 (2 p+3)\right ) \int \frac {\left (b x^2+a\right )^p}{c+d x}dx}{d}-\frac {\left (2 a d^2+b c^2 (2 p+3)\right ) \int \left (b x^2+a\right )^pdx}{d}\right )}{d^2}+\frac {\left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{2 b d^2 (p+1)}}{a d^2+b c^2}+\frac {c^3 \left (a+b x^2\right )^{p+1}}{d^2 (c+d x) \left (a d^2+b c^2\right )}\) |
\(\Big \downarrow \) 238 |
\(\displaystyle \frac {\frac {c \left (\frac {c \left (3 a d^2+b c^2 (2 p+3)\right ) \int \frac {\left (b x^2+a\right )^p}{c+d x}dx}{d}-\frac {\left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (2 a d^2+b c^2 (2 p+3)\right ) \int \left (\frac {b x^2}{a}+1\right )^pdx}{d}\right )}{d^2}+\frac {\left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{2 b d^2 (p+1)}}{a d^2+b c^2}+\frac {c^3 \left (a+b x^2\right )^{p+1}}{d^2 (c+d x) \left (a d^2+b c^2\right )}\) |
\(\Big \downarrow \) 237 |
\(\displaystyle \frac {\frac {c \left (\frac {c \left (3 a d^2+b c^2 (2 p+3)\right ) \int \frac {\left (b x^2+a\right )^p}{c+d x}dx}{d}-\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (2 a d^2+b c^2 (2 p+3)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{d}\right )}{d^2}+\frac {\left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{2 b d^2 (p+1)}}{a d^2+b c^2}+\frac {c^3 \left (a+b x^2\right )^{p+1}}{d^2 (c+d x) \left (a d^2+b c^2\right )}\) |
\(\Big \downarrow \) 504 |
\(\displaystyle \frac {\frac {c \left (\frac {c \left (3 a d^2+b c^2 (2 p+3)\right ) \left (c \int \frac {\left (b x^2+a\right )^p}{c^2-d^2 x^2}dx-d \int \frac {x \left (b x^2+a\right )^p}{c^2-d^2 x^2}dx\right )}{d}-\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (2 a d^2+b c^2 (2 p+3)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{d}\right )}{d^2}+\frac {\left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{2 b d^2 (p+1)}}{a d^2+b c^2}+\frac {c^3 \left (a+b x^2\right )^{p+1}}{d^2 (c+d x) \left (a d^2+b c^2\right )}\) |
\(\Big \downarrow \) 334 |
\(\displaystyle \frac {\frac {c \left (\frac {c \left (3 a d^2+b c^2 (2 p+3)\right ) \left (c \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \int \frac {\left (\frac {b x^2}{a}+1\right )^p}{c^2-d^2 x^2}dx-d \int \frac {x \left (b x^2+a\right )^p}{c^2-d^2 x^2}dx\right )}{d}-\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (2 a d^2+b c^2 (2 p+3)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{d}\right )}{d^2}+\frac {\left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{2 b d^2 (p+1)}}{a d^2+b c^2}+\frac {c^3 \left (a+b x^2\right )^{p+1}}{d^2 (c+d x) \left (a d^2+b c^2\right )}\) |
\(\Big \downarrow \) 333 |
\(\displaystyle \frac {\frac {c \left (\frac {c \left (3 a d^2+b c^2 (2 p+3)\right ) \left (\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c}-d \int \frac {x \left (b x^2+a\right )^p}{c^2-d^2 x^2}dx\right )}{d}-\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (2 a d^2+b c^2 (2 p+3)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{d}\right )}{d^2}+\frac {\left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{2 b d^2 (p+1)}}{a d^2+b c^2}+\frac {c^3 \left (a+b x^2\right )^{p+1}}{d^2 (c+d x) \left (a d^2+b c^2\right )}\) |
\(\Big \downarrow \) 353 |
\(\displaystyle \frac {\frac {c \left (\frac {c \left (3 a d^2+b c^2 (2 p+3)\right ) \left (\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c}-\frac {1}{2} d \int \frac {\left (b x^2+a\right )^p}{c^2-d^2 x^2}dx^2\right )}{d}-\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (2 a d^2+b c^2 (2 p+3)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{d}\right )}{d^2}+\frac {\left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{2 b d^2 (p+1)}}{a d^2+b c^2}+\frac {c^3 \left (a+b x^2\right )^{p+1}}{d^2 (c+d x) \left (a d^2+b c^2\right )}\) |
\(\Big \downarrow \) 78 |
\(\displaystyle \frac {\frac {c \left (\frac {c \left (3 a d^2+b c^2 (2 p+3)\right ) \left (\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c}-\frac {d \left (a+b x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {d^2 \left (b x^2+a\right )}{b c^2+a d^2}\right )}{2 (p+1) \left (a d^2+b c^2\right )}\right )}{d}-\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (2 a d^2+b c^2 (2 p+3)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{d}\right )}{d^2}+\frac {\left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{2 b d^2 (p+1)}}{a d^2+b c^2}+\frac {c^3 \left (a+b x^2\right )^{p+1}}{d^2 (c+d x) \left (a d^2+b c^2\right )}\) |
Input:
Int[(x^3*(a + b*x^2)^p)/(c + d*x)^2,x]
Output:
(c^3*(a + b*x^2)^(1 + p))/(d^2*(b*c^2 + a*d^2)*(c + d*x)) + (((b*c^2 + a*d ^2)*(a + b*x^2)^(1 + p))/(2*b*d^2*(1 + p)) + (c*(-(((2*a*d^2 + b*c^2*(3 + 2*p))*x*(a + b*x^2)^p*Hypergeometric2F1[1/2, -p, 3/2, -((b*x^2)/a)])/(d*(1 + (b*x^2)/a)^p)) + (c*(3*a*d^2 + b*c^2*(3 + 2*p))*((x*(a + b*x^2)^p*Appel lF1[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))/d^2)/(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[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[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : > With[{q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[f*(d + e*x) ^(m + q - 1)*((a + b*x^2)^(p + 1)/(b*e^(q - 1)*(m + q + 2*p + 1))), x] + Si mp[1/(b*e^q*(m + q + 2*p + 1)) Int[(d + e*x)^m*(a + b*x^2)^p*ExpandToSum[ b*e^q*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x )^(q - 2)*(a*e^2*(m + q - 1) - b*d^2*(m + q + 2*p + 1) - 2*b*d*e*(m + q + p )*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, d , e, m, p}, x] && PolyQ[Pq, x] && NeQ[b*d^2 + a*e^2, 0] && !(EqQ[d, 0] && True) && !(IGtQ[m, 0] && RationalQ[a, b, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))
\[\int \frac {x^{3} \left (b \,x^{2}+a \right )^{p}}{\left (d x +c \right )^{2}}d x\]
Input:
int(x^3*(b*x^2+a)^p/(d*x+c)^2,x)
Output:
int(x^3*(b*x^2+a)^p/(d*x+c)^2,x)
\[ \int \frac {x^3 \left (a+b x^2\right )^p}{(c+d x)^2} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p} x^{3}}{{\left (d x + c\right )}^{2}} \,d x } \] Input:
integrate(x^3*(b*x^2+a)^p/(d*x+c)^2,x, algorithm="fricas")
Output:
integral((b*x^2 + a)^p*x^3/(d^2*x^2 + 2*c*d*x + c^2), x)
Timed out. \[ \int \frac {x^3 \left (a+b x^2\right )^p}{(c+d x)^2} \, dx=\text {Timed out} \] Input:
integrate(x**3*(b*x**2+a)**p/(d*x+c)**2,x)
Output:
Timed out
\[ \int \frac {x^3 \left (a+b x^2\right )^p}{(c+d x)^2} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p} x^{3}}{{\left (d x + c\right )}^{2}} \,d x } \] Input:
integrate(x^3*(b*x^2+a)^p/(d*x+c)^2,x, algorithm="maxima")
Output:
integrate((b*x^2 + a)^p*x^3/(d*x + c)^2, x)
Exception generated. \[ \int \frac {x^3 \left (a+b x^2\right )^p}{(c+d x)^2} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(x^3*(b*x^2+a)^p/(d*x+c)^2,x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro unding error%%%{-1,[0,3,1,0]%%%} / %%%{1,[0,0,0,3]%%%} Error: Bad Argument Value
Timed out. \[ \int \frac {x^3 \left (a+b x^2\right )^p}{(c+d x)^2} \, dx=\int \frac {x^3\,{\left (b\,x^2+a\right )}^p}{{\left (c+d\,x\right )}^2} \,d x \] Input:
int((x^3*(a + b*x^2)^p)/(c + d*x)^2,x)
Output:
int((x^3*(a + b*x^2)^p)/(c + d*x)^2, x)
\[ \int \frac {x^3 \left (a+b x^2\right )^p}{(c+d x)^2} \, dx=\text {too large to display} \] Input:
int(x^3*(b*x^2+a)^p/(d*x+c)^2,x)
Output:
(4*(a + b*x**2)**p*a*c*p**2 + 8*(a + b*x**2)**p*a*c*p + 3*(a + b*x**2)**p* a*c - 8*(a + b*x**2)**p*a*d*p**2*x - 4*(a + b*x**2)**p*a*d*p*x + 3*(a + b* x**2)**p*a*d*x - 4*(a + b*x**2)**p*b*c*p**3*x**2 - 8*(a + b*x**2)**p*b*c*p **2*x**2 - 3*(a + b*x**2)**p*b*c*p*x**2 + 4*(a + b*x**2)**p*b*d*p**3*x**3 + 4*(a + b*x**2)**p*b*d*p**2*x**3 + (a + b*x**2)**p*b*d*p*x**3 + 48*int((a + b*x**2)**p/(4*a*c**2*p**2 + 4*a*c**2*p + a*c**2 + 8*a*c*d*p**2*x + 8*a* c*d*p*x + 2*a*c*d*x + 4*a*d**2*p**2*x**2 + 4*a*d**2*p*x**2 + a*d**2*x**2 + 4*b*c**2*p**2*x**2 + 4*b*c**2*p*x**2 + b*c**2*x**2 + 8*b*c*d*p**2*x**3 + 8*b*c*d*p*x**3 + 2*b*c*d*x**3 + 4*b*d**2*p**2*x**4 + 4*b*d**2*p*x**4 + b*d **2*x**4),x)*a**2*c**2*d*p**4 + 96*int((a + b*x**2)**p/(4*a*c**2*p**2 + 4* a*c**2*p + a*c**2 + 8*a*c*d*p**2*x + 8*a*c*d*p*x + 2*a*c*d*x + 4*a*d**2*p* *2*x**2 + 4*a*d**2*p*x**2 + a*d**2*x**2 + 4*b*c**2*p**2*x**2 + 4*b*c**2*p* x**2 + b*c**2*x**2 + 8*b*c*d*p**2*x**3 + 8*b*c*d*p*x**3 + 2*b*c*d*x**3 + 4 *b*d**2*p**2*x**4 + 4*b*d**2*p*x**4 + b*d**2*x**4),x)*a**2*c**2*d*p**3 + 6 0*int((a + b*x**2)**p/(4*a*c**2*p**2 + 4*a*c**2*p + a*c**2 + 8*a*c*d*p**2* x + 8*a*c*d*p*x + 2*a*c*d*x + 4*a*d**2*p**2*x**2 + 4*a*d**2*p*x**2 + a*d** 2*x**2 + 4*b*c**2*p**2*x**2 + 4*b*c**2*p*x**2 + b*c**2*x**2 + 8*b*c*d*p**2 *x**3 + 8*b*c*d*p*x**3 + 2*b*c*d*x**3 + 4*b*d**2*p**2*x**4 + 4*b*d**2*p*x* *4 + b*d**2*x**4),x)*a**2*c**2*d*p**2 + 12*int((a + b*x**2)**p/(4*a*c**2*p **2 + 4*a*c**2*p + a*c**2 + 8*a*c*d*p**2*x + 8*a*c*d*p*x + 2*a*c*d*x + ...