Integrand size = 26, antiderivative size = 96 \[ \int \frac {(b d+2 c d x)^m}{\left (a+b x+c x^2\right )^{5/2}} \, dx=-\frac {16 (b d+2 c d x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (-2+m),\frac {3+m}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{\left (b^2-4 a c\right ) d (1+m) \left (4 a-\frac {b^2}{c}+\frac {(b+2 c x)^2}{c}\right )^{3/2}} \]
-16*(2*c*d*x+b*d)^(1+m)*hypergeom([1, -1+1/2*m],[3/2+1/2*m],(2*c*x+b)^2/(- 4*a*c+b^2))/(-4*a*c+b^2)/d/(1+m)/(4*a-b^2/c+(2*c*x+b)^2/c)^(3/2)
Time = 1.09 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.16 \[ \int \frac {(b d+2 c d x)^m}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\frac {16 c (b+2 c x) (d (b+2 c x))^m \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {1+m}{2},\frac {3+m}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{\left (b^2-4 a c\right )^2 (1+m) \sqrt {a+x (b+c x)}} \]
(16*c*(b + 2*c*x)*(d*(b + 2*c*x))^m*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a *c)]*Hypergeometric2F1[5/2, (1 + m)/2, (3 + m)/2, (b + 2*c*x)^2/(b^2 - 4*a *c)])/((b^2 - 4*a*c)^2*(1 + m)*Sqrt[a + x*(b + c*x)])
Time = 0.25 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.49, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1118, 27, 279, 278}
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.
\(\displaystyle \int \frac {(b d+2 c d x)^m}{\left (a+b x+c x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 1118 |
\(\displaystyle \frac {\int \frac {32 (b d+2 c x d)^m}{\left (-\frac {b^2}{c}+\frac {(b d+2 c x d)^2}{c d^2}+4 a\right )^{5/2}}d(b d+2 c x d)}{2 c d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {16 \int \frac {(b d+2 c x d)^m}{\left (-\frac {b^2}{c}+\frac {(b d+2 c x d)^2}{c d^2}+4 a\right )^{5/2}}d(b d+2 c x d)}{c d}\) |
\(\Big \downarrow \) 279 |
\(\displaystyle \frac {16 c \sqrt {1-\frac {(b d+2 c d x)^2}{d^2 \left (b^2-4 a c\right )}} \int \frac {(b d+2 c x d)^m}{\left (1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}\right )^{5/2}}d(b d+2 c x d)}{d \left (b^2-4 a c\right )^2 \sqrt {4 a-\frac {b^2}{c}+\frac {(b d+2 c d x)^2}{c d^2}}}\) |
\(\Big \downarrow \) 278 |
\(\displaystyle \frac {16 c \sqrt {1-\frac {(b d+2 c d x)^2}{d^2 \left (b^2-4 a c\right )}} (b d+2 c d x)^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {m+1}{2},\frac {m+3}{2},\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}\right )}{d (m+1) \left (b^2-4 a c\right )^2 \sqrt {4 a-\frac {b^2}{c}+\frac {(b d+2 c d x)^2}{c d^2}}}\) |
(16*c*(b*d + 2*c*d*x)^(1 + m)*Sqrt[1 - (b*d + 2*c*d*x)^2/((b^2 - 4*a*c)*d^ 2)]*Hypergeometric2F1[5/2, (1 + m)/2, (3 + m)/2, (b*d + 2*c*d*x)^2/((b^2 - 4*a*c)*d^2)])/((b^2 - 4*a*c)^2*d*(1 + m)*Sqrt[4*a - b^2/c + (b*d + 2*c*d* x)^2/(c*d^2)])
3.15.34.3.1 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[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( -b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && (ILtQ[p, 0 ] || GtQ[a, 0])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntP art[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p]) Int[(c*x)^m* (1 + b*(x^2/a))^p, x], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && !(ILtQ[p, 0] || GtQ[a, 0])
Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S ymbol] :> Simp[1/e Subst[Int[x^m*(a - b^2/(4*c) + (c*x^2)/e^2)^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[2*c*d - b*e, 0]
\[\int \frac {\left (2 c d x +b d \right )^{m}}{\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}}}d x\]
\[ \int \frac {(b d+2 c d x)^m}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\int { \frac {{\left (2 \, c d x + b d\right )}^{m}}{{\left (c x^{2} + b x + a\right )}^{\frac {5}{2}}} \,d x } \]
integral(sqrt(c*x^2 + b*x + a)*(2*c*d*x + b*d)^m/(c^3*x^6 + 3*b*c^2*x^5 + 3*(b^2*c + a*c^2)*x^4 + 3*a^2*b*x + (b^3 + 6*a*b*c)*x^3 + a^3 + 3*(a*b^2 + a^2*c)*x^2), x)
\[ \int \frac {(b d+2 c d x)^m}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\int \frac {\left (d \left (b + 2 c x\right )\right )^{m}}{\left (a + b x + c x^{2}\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {(b d+2 c d x)^m}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\int { \frac {{\left (2 \, c d x + b d\right )}^{m}}{{\left (c x^{2} + b x + a\right )}^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {(b d+2 c d x)^m}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\int { \frac {{\left (2 \, c d x + b d\right )}^{m}}{{\left (c x^{2} + b x + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {(b d+2 c d x)^m}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\int \frac {{\left (b\,d+2\,c\,d\,x\right )}^m}{{\left (c\,x^2+b\,x+a\right )}^{5/2}} \,d x \]