\(\int \frac {(b d+2 c d x)^m}{\sqrt {a+b x+c x^2}} \, dx\) [1432]

Optimal result

Integrand size = 26, antiderivative size = 96 \[ \int \frac {(b d+2 c d x)^m}{\sqrt {a+b x+c x^2}} \, dx=-\frac {(b d+2 c d x)^{1+m} \sqrt {4 a-\frac {b^2}{c}+\frac {(b+2 c x)^2}{c}} \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {3+m}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{\left (b^2-4 a c\right ) d (1+m)} \]

[Out]

-(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-b^2/c+(2*c*x+b)^2/c)^(1
/2)/(-4*a*c+b^2)/d/(1+m)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.08, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {708, 372, 371} \[ \int \frac {(b d+2 c d x)^m}{\sqrt {a+b x+c x^2}} \, dx=\frac {\sqrt {1-\frac {(b+2 c x)^2}{b^2-4 a c}} (d (b+2 c x))^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{2 c d (m+1) \sqrt {a+b x+c x^2}} \]

[In]

Int[(b*d + 2*c*d*x)^m/Sqrt[a + b*x + c*x^2],x]

[Out]

((d*(b + 2*c*x))^(1 + m)*Sqrt[1 - (b + 2*c*x)^2/(b^2 - 4*a*c)]*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, (b
+ 2*c*x)^2/(b^2 - 4*a*c)])/(2*c*d*(1 + m)*Sqrt[a + b*x + c*x^2])

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/
(1 + b*(x^n/a))^FracPart[p]), Int[(c*x)^m*(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 708

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[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] && NeQ[b^2 - 4*a*c, 0]
&& EqQ[2*c*d - b*e, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^m}{\sqrt {a-\frac {b^2}{4 c}+\frac {x^2}{4 c d^2}}} \, dx,x,b d+2 c d x\right )}{2 c d} \\ & = \frac {\sqrt {4+\frac {(b d+2 c d x)^2}{\left (a-\frac {b^2}{4 c}\right ) c d^2}} \text {Subst}\left (\int \frac {x^m}{\sqrt {1+\frac {x^2}{4 \left (a-\frac {b^2}{4 c}\right ) c d^2}}} \, dx,x,b d+2 c d x\right )}{4 c d \sqrt {a+b x+c x^2}} \\ & = \frac {(d (b+2 c x))^{1+m} \sqrt {1-\frac {(b+2 c x)^2}{b^2-4 a c}} \, _2F_1\left (\frac {1}{2},\frac {1+m}{2};\frac {3+m}{2};\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{2 c d (1+m) \sqrt {a+b x+c x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.06 \[ \int \frac {(b d+2 c d x)^m}{\sqrt {a+b x+c x^2}} \, dx=\frac {(b+2 c x) (d (b+2 c x))^m \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{c (1+m) \sqrt {a+x (b+c x)}} \]

[In]

Integrate[(b*d + 2*c*d*x)^m/Sqrt[a + b*x + c*x^2],x]

[Out]

((b + 2*c*x)*(d*(b + 2*c*x))^m*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)]*Hypergeometric2F1[1/2, (1 + m)/2, (3
+ m)/2, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(c*(1 + m)*Sqrt[a + x*(b + c*x)])

Maple [F]

\[\int \frac {\left (2 c d x +b d \right )^{m}}{\sqrt {c \,x^{2}+b x +a}}d x\]

[In]

int((2*c*d*x+b*d)^m/(c*x^2+b*x+a)^(1/2),x)

[Out]

int((2*c*d*x+b*d)^m/(c*x^2+b*x+a)^(1/2),x)

Fricas [F]

\[ \int \frac {(b d+2 c d x)^m}{\sqrt {a+b x+c x^2}} \, dx=\int { \frac {{\left (2 \, c d x + b d\right )}^{m}}{\sqrt {c x^{2} + b x + a}} \,d x } \]

[In]

integrate((2*c*d*x+b*d)^m/(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")

[Out]

integral((2*c*d*x + b*d)^m/sqrt(c*x^2 + b*x + a), x)

Sympy [F]

\[ \int \frac {(b d+2 c d x)^m}{\sqrt {a+b x+c x^2}} \, dx=\int \frac {\left (d \left (b + 2 c x\right )\right )^{m}}{\sqrt {a + b x + c x^{2}}}\, dx \]

[In]

integrate((2*c*d*x+b*d)**m/(c*x**2+b*x+a)**(1/2),x)

[Out]

Integral((d*(b + 2*c*x))**m/sqrt(a + b*x + c*x**2), x)

Maxima [F]

\[ \int \frac {(b d+2 c d x)^m}{\sqrt {a+b x+c x^2}} \, dx=\int { \frac {{\left (2 \, c d x + b d\right )}^{m}}{\sqrt {c x^{2} + b x + a}} \,d x } \]

[In]

integrate((2*c*d*x+b*d)^m/(c*x^2+b*x+a)^(1/2),x, algorithm="maxima")

[Out]

integrate((2*c*d*x + b*d)^m/sqrt(c*x^2 + b*x + a), x)

Giac [F]

\[ \int \frac {(b d+2 c d x)^m}{\sqrt {a+b x+c x^2}} \, dx=\int { \frac {{\left (2 \, c d x + b d\right )}^{m}}{\sqrt {c x^{2} + b x + a}} \,d x } \]

[In]

integrate((2*c*d*x+b*d)^m/(c*x^2+b*x+a)^(1/2),x, algorithm="giac")

[Out]

integrate((2*c*d*x + b*d)^m/sqrt(c*x^2 + b*x + a), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(b d+2 c d x)^m}{\sqrt {a+b x+c x^2}} \, dx=\int \frac {{\left (b\,d+2\,c\,d\,x\right )}^m}{\sqrt {c\,x^2+b\,x+a}} \,d x \]

[In]

int((b*d + 2*c*d*x)^m/(a + b*x + c*x^2)^(1/2),x)

[Out]

int((b*d + 2*c*d*x)^m/(a + b*x + c*x^2)^(1/2), x)