\(\int (b d+2 c d x)^m (a+b x+c x^2)^{5/2} \, dx\) [1429]

Optimal result

Integrand size = 26, antiderivative size = 82 \[ \int (b d+2 c d x)^m \left (a+b x+c x^2\right )^{5/2} \, dx=-\frac {2 (b d+2 c d x)^{1+m} \left (a+b x+c x^2\right )^{7/2} \operatorname {Hypergeometric2F1}\left (1,\frac {8+m}{2},\frac {3+m}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{\left (b^2-4 a c\right ) d (1+m)} \]

[Out]

-2*(2*c*d*x+b*d)^(1+m)*(c*x^2+b*x+a)^(7/2)*hypergeom([1, 4+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)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.39, 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} \[ \int (b d+2 c d x)^m \left (a+b x+c x^2\right )^{5/2} \, dx=\frac {\left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2} (d (b+2 c x))^{m+1} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {m+1}{2},\frac {m+3}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{32 c^3 d (m+1) \sqrt {1-\frac {(b+2 c x)^2}{b^2-4 a c}}} \]

[In]

Int[(b*d + 2*c*d*x)^m*(a + b*x + c*x^2)^(5/2),x]

[Out]

((b^2 - 4*a*c)^2*(d*(b + 2*c*x))^(1 + m)*Sqrt[a + b*x + c*x^2]*Hypergeometric2F1[-5/2, (1 + m)/2, (3 + m)/2, (
b + 2*c*x)^2/(b^2 - 4*a*c)])/(32*c^3*d*(1 + m)*Sqrt[1 - (b + 2*c*x)^2/(b^2 - 4*a*c)])

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 x^m \left (a-\frac {b^2}{4 c}+\frac {x^2}{4 c d^2}\right )^{5/2} \, dx,x,b d+2 c d x\right )}{2 c d} \\ & = \frac {\left (\left (a-\frac {b^2}{4 c}\right )^2 \sqrt {a+b x+c x^2}\right ) \text {Subst}\left (\int x^m \left (1+\frac {x^2}{4 \left (a-\frac {b^2}{4 c}\right ) c d^2}\right )^{5/2} \, dx,x,b d+2 c d x\right )}{c d \sqrt {4+\frac {(b d+2 c d x)^2}{\left (a-\frac {b^2}{4 c}\right ) c d^2}}} \\ & = \frac {\left (b^2-4 a c\right )^2 (d (b+2 c x))^{1+m} \sqrt {a+b x+c x^2} \, _2F_1\left (-\frac {5}{2},\frac {1+m}{2};\frac {3+m}{2};\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{32 c^3 d (1+m) \sqrt {1-\frac {(b+2 c x)^2}{b^2-4 a c}}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.59 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.40 \[ \int (b d+2 c d x)^m \left (a+b x+c x^2\right )^{5/2} \, dx=\frac {\left (b^2-4 a c\right )^2 (b+2 c x) (d (b+2 c x))^m \sqrt {a+x (b+c x)} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {1+m}{2},\frac {3+m}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{64 c^3 (1+m) \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}}} \]

[In]

Integrate[(b*d + 2*c*d*x)^m*(a + b*x + c*x^2)^(5/2),x]

[Out]

((b^2 - 4*a*c)^2*(b + 2*c*x)*(d*(b + 2*c*x))^m*Sqrt[a + x*(b + c*x)]*Hypergeometric2F1[-5/2, (1 + m)/2, (3 + m
)/2, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(64*c^3*(1 + m)*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])

Maple [F]

\[\int \left (2 c d x +b d \right )^{m} \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}}d x\]

[In]

int((2*c*d*x+b*d)^m*(c*x^2+b*x+a)^(5/2),x)

[Out]

int((2*c*d*x+b*d)^m*(c*x^2+b*x+a)^(5/2),x)

Fricas [F]

\[ \int (b d+2 c d x)^m \left (a+b x+c x^2\right )^{5/2} \, dx=\int { {\left (c x^{2} + b x + a\right )}^{\frac {5}{2}} {\left (2 \, c d x + b d\right )}^{m} \,d x } \]

[In]

integrate((2*c*d*x+b*d)^m*(c*x^2+b*x+a)^(5/2),x, algorithm="fricas")

[Out]

integral((c^2*x^4 + 2*b*c*x^3 + 2*a*b*x + (b^2 + 2*a*c)*x^2 + a^2)*sqrt(c*x^2 + b*x + a)*(2*c*d*x + b*d)^m, x)

Sympy [F]

\[ \int (b d+2 c d x)^m \left (a+b x+c x^2\right )^{5/2} \, dx=\int \left (d \left (b + 2 c x\right )\right )^{m} \left (a + b x + c x^{2}\right )^{\frac {5}{2}}\, dx \]

[In]

integrate((2*c*d*x+b*d)**m*(c*x**2+b*x+a)**(5/2),x)

[Out]

Integral((d*(b + 2*c*x))**m*(a + b*x + c*x**2)**(5/2), x)

Maxima [F]

\[ \int (b d+2 c d x)^m \left (a+b x+c x^2\right )^{5/2} \, dx=\int { {\left (c x^{2} + b x + a\right )}^{\frac {5}{2}} {\left (2 \, c d x + b d\right )}^{m} \,d x } \]

[In]

integrate((2*c*d*x+b*d)^m*(c*x^2+b*x+a)^(5/2),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x + a)^(5/2)*(2*c*d*x + b*d)^m, x)

Giac [F]

\[ \int (b d+2 c d x)^m \left (a+b x+c x^2\right )^{5/2} \, dx=\int { {\left (c x^{2} + b x + a\right )}^{\frac {5}{2}} {\left (2 \, c d x + b d\right )}^{m} \,d x } \]

[In]

integrate((2*c*d*x+b*d)^m*(c*x^2+b*x+a)^(5/2),x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x + a)^(5/2)*(2*c*d*x + b*d)^m, x)

Mupad [F(-1)]

Timed out. \[ \int (b d+2 c d x)^m \left (a+b x+c x^2\right )^{5/2} \, dx=\int {\left (b\,d+2\,c\,d\,x\right )}^m\,{\left (c\,x^2+b\,x+a\right )}^{5/2} \,d x \]

[In]

int((b*d + 2*c*d*x)^m*(a + b*x + c*x^2)^(5/2),x)

[Out]

int((b*d + 2*c*d*x)^m*(a + b*x + c*x^2)^(5/2), x)