3.2166 \(\int \frac{(A+B x) (a c+b c x)^m}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=112 \[ -\frac{c^2 (a+b x) (A b-a B) (a c+b c x)^{m-2}}{b^2 (2-m) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{B c (a+b x) (a c+b c x)^{m-1}}{b^2 (1-m) \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0850823, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {770, 21, 43} \[ -\frac{c^2 (a+b x) (A b-a B) (a c+b c x)^{m-2}}{b^2 (2-m) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{B c (a+b x) (a c+b c x)^{m-1}}{b^2 (1-m) \sqrt{a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a*c + b*c*x)^m)/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

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

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{(A+B x) (a c+b c x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx &=\frac{\left (b^2 \left (a b+b^2 x\right )\right ) \int \frac{(A+B x) (a c+b c x)^m}{\left (a b+b^2 x\right )^3} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{\left (c^3 \left (a b+b^2 x\right )\right ) \int (A+B x) (a c+b c x)^{-3+m} \, dx}{b \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{\left (c^3 \left (a b+b^2 x\right )\right ) \int \left (\frac{(A b-a B) (a c+b c x)^{-3+m}}{b}+\frac{B (a c+b c x)^{-2+m}}{b c}\right ) \, dx}{b \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{(A b-a B) c^2 (a+b x) (a c+b c x)^{-2+m}}{b^2 (2-m) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{B c (a+b x) (a c+b c x)^{-1+m}}{b^2 (1-m) \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0624824, size = 55, normalized size = 0.49 \[ \frac{c (c (a+b x))^{m-1} (-a B+A b (m-1)+b B (m-2) x)}{b^2 (m-2) (m-1) \sqrt{(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a*c + b*c*x)^m)/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

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

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Maple [A]  time = 0.002, size = 62, normalized size = 0.6 \begin{align*}{\frac{ \left ( bcx+ac \right ) ^{m} \left ( Bbmx+Abm-2\,bBx-Ab-aB \right ) \left ( bx+a \right ) }{{b}^{2} \left ({m}^{2}-3\,m+2 \right ) } \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(b*c*x+a*c)^m/(b^2*x^2+2*a*b*x+a^2)^(3/2),x)

[Out]

(b*c*x+a*c)^m*(B*b*m*x+A*b*m-2*B*b*x-A*b-B*a)*(b*x+a)/((b*x+a)^2)^(3/2)/b^2/(m^2-3*m+2)

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Maxima [A]  time = 1.05959, size = 158, normalized size = 1.41 \begin{align*} \frac{{\left (b c^{m}{\left (m - 2\right )} x - a c^{m}\right )}{\left (b x + a\right )}^{m} B}{{\left (m^{2} - 3 \, m + 2\right )} b^{4} x^{2} + 2 \,{\left (m^{2} - 3 \, m + 2\right )} a b^{3} x +{\left (m^{2} - 3 \, m + 2\right )} a^{2} b^{2}} + \frac{{\left (b x + a\right )}^{m} A c^{m}}{b^{3}{\left (m - 2\right )} x^{2} + 2 \, a b^{2}{\left (m - 2\right )} x + a^{2} b{\left (m - 2\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b*c*x+a*c)^m/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")

[Out]

(b*c^m*(m - 2)*x - a*c^m)*(b*x + a)^m*B/((m^2 - 3*m + 2)*b^4*x^2 + 2*(m^2 - 3*m + 2)*a*b^3*x + (m^2 - 3*m + 2)
*a^2*b^2) + (b*x + a)^m*A*c^m/(b^3*(m - 2)*x^2 + 2*a*b^2*(m - 2)*x + a^2*b*(m - 2))

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Fricas [A]  time = 1.5747, size = 231, normalized size = 2.06 \begin{align*} \frac{{\left (A b m - B a - A b +{\left (B b m - 2 \, B b\right )} x\right )}{\left (b c x + a c\right )}^{m}}{a^{2} b^{2} m^{2} - 3 \, a^{2} b^{2} m + 2 \, a^{2} b^{2} +{\left (b^{4} m^{2} - 3 \, b^{4} m + 2 \, b^{4}\right )} x^{2} + 2 \,{\left (a b^{3} m^{2} - 3 \, a b^{3} m + 2 \, a b^{3}\right )} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b*c*x+a*c)^m/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")

[Out]

(A*b*m - B*a - A*b + (B*b*m - 2*B*b)*x)*(b*c*x + a*c)^m/(a^2*b^2*m^2 - 3*a^2*b^2*m + 2*a^2*b^2 + (b^4*m^2 - 3*
b^4*m + 2*b^4)*x^2 + 2*(a*b^3*m^2 - 3*a*b^3*m + 2*a*b^3)*x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c \left (a + b x\right )\right )^{m} \left (A + B x\right )}{\left (\left (a + b x\right )^{2}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b*c*x+a*c)**m/(b**2*x**2+2*a*b*x+a**2)**(3/2),x)

[Out]

Integral((c*(a + b*x))**m*(A + B*x)/((a + b*x)**2)**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x + A\right )}{\left (b c x + a c\right )}^{m}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b*c*x+a*c)^m/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")

[Out]

integrate((B*x + A)*(b*c*x + a*c)^m/(b^2*x^2 + 2*a*b*x + a^2)^(3/2), x)