### 3.1434 $$\int \frac{(b d+2 c d x)^m}{(a+b x+c x^2)^{5/2}} \, dx$$

Optimal. Leaf size=96 $-\frac{16 (b d+2 c d x)^{m+1} \, _2F_1\left (1,\frac{m-2}{2};\frac{m+3}{2};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{d (m+1) \left (b^2-4 a c\right ) \left (4 a-\frac{b^2}{c}+\frac{(b+2 c x)^2}{c}\right )^{3/2}}$

[Out]

(-16*(b*d + 2*c*d*x)^(1 + m)*Hypergeometric2F1[1, (-2 + m)/2, (3 + m)/2, (b + 2*c*x)^2/(b^2 - 4*a*c)])/((b^2 -
4*a*c)*d*(1 + m)*(4*a - b^2/c + (b + 2*c*x)^2/c)^(3/2))

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Rubi [A]  time = 0.116954, antiderivative size = 110, normalized size of antiderivative = 1.15, 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} $\frac{8 c \sqrt{1-\frac{(b+2 c x)^2}{b^2-4 a c}} (d (b+2 c x))^{m+1} \, _2F_1\left (\frac{5}{2},\frac{m+1}{2};\frac{m+3}{2};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{d (m+1) \left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(8*c*(d*(b + 2*c*x))^(1 + m)*Sqrt[1 - (b + 2*c*x)^2/(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*d*(1 + m)*Sqrt[a + b*x + c*x^2])

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 \frac{(b d+2 c d x)^m}{\left (a+b x+c x^2\right )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{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{\sqrt{4+\frac{(b d+2 c d x)^2}{\left (a-\frac{b^2}{4 c}\right ) c d^2}} \operatorname{Subst}\left (\int \frac{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 )}{4 \left (a-\frac{b^2}{4 c}\right )^2 c d \sqrt{a+b x+c x^2}}\\ &=\frac{8 c (d (b+2 c x))^{1+m} \sqrt{1-\frac{(b+2 c x)^2}{b^2-4 a c}} \, _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 )}{\left (b^2-4 a c\right )^2 d (1+m) \sqrt{a+b x+c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0493093, size = 111, normalized size = 1.16 $\frac{16 c (b+2 c x) \sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}} (d (b+2 c x))^m \, _2F_1\left (\frac{5}{2},\frac{m+1}{2};\frac{m+3}{2};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{(m+1) \left (b^2-4 a c\right )^2 \sqrt{a+x (b+c x)}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(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)])

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Maple [F]  time = 1.163, 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)

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{c x^{2} + b x + a}{\left (2 \, c d x + b d\right )}^{m}}{c^{3} x^{6} + 3 \, b c^{2} x^{5} + 3 \,{\left (b^{2} c + a c^{2}\right )} x^{4} + 3 \, a^{2} b x +{\left (b^{3} + 6 \, a b c\right )} x^{3} + a^{3} + 3 \,{\left (a b^{2} + a^{2} c\right )} x^{2}}, 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(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)

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

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

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