### 3.1429 $$\int (b d+2 c d x)^m (a+b x+c x^2)^{5/2} \, dx$$

Optimal. Leaf size=82 $-\frac{2 \left (a+b x+c x^2\right )^{7/2} (b d+2 c d x)^{m+1} \, _2F_1\left (1,\frac{m+8}{2};\frac{m+3}{2};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{d (m+1) \left (b^2-4 a c\right )}$

[Out]

(-2*(b*d + 2*c*d*x)^(1 + m)*(a + b*x + c*x^2)^(7/2)*Hypergeometric2F1[1, (8 + m)/2, (3 + m)/2, (b + 2*c*x)^2/(
b^2 - 4*a*c)])/((b^2 - 4*a*c)*d*(1 + m))

________________________________________________________________________________________

Rubi [A]  time = 0.119357, antiderivative size = 114, normalized size of antiderivative = 1.39, number of steps used = 3, number of rules used = 3, integrand size = 26, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.115, Rules used = {694, 365, 364} $\frac{\left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2} (d (b+2 c x))^{m+1} \, _2F_1\left (-\frac{5}{2},\frac{m+1}{2};\frac{m+3}{2};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{32 c^3 d (m+1) \sqrt{1-\frac{(b+2 c x)^2}{b^2-4 a c}}}$

Antiderivative was successfully veriﬁed.

[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 694

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]

Rule 365

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 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int (b d+2 c d x)^m \left (a+b x+c x^2\right )^{5/2} \, dx &=\frac{\operatorname{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 ) \operatorname{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]  time = 0.0841635, size = 115, normalized size = 1.4 $\frac{\left (b^2-4 a c\right )^2 (b+2 c x) \sqrt{a+x (b+c x)} (d (b+2 c x))^m \, _2F_1\left (-\frac{5}{2},\frac{m+1}{2};\frac{m+3}{2};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{64 c^3 (m+1) \sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}}}$

Antiderivative was successfully veriﬁed.

[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]  time = 1.167, size = 0, normalized size = 0. \begin{align*} \int \left ( 2\,cdx+bd \right ) ^{m} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{5}{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (c x^{2} + b x + a\right )}^{\frac{5}{2}}{\left (2 \, c d x + b d\right )}^{m}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x +{\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}\right )} \sqrt{c x^{2} + b x + a}{\left (2 \, c d x + b d\right )}^{m}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (c x^{2} + b x + a\right )}^{\frac{5}{2}}{\left (2 \, c d x + b d\right )}^{m}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)