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

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

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 112, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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}} \end {gather*}

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 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])

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]

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 = 0.06, size = 55, normalized size = 0.49 \begin {gather*} \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}} \end {gather*}

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])

________________________________________________________________________________________

IntegrateAlgebraic [F]  time = 2.10, size = 0, normalized size = 0.00 \begin {gather*} \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 \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.44, size = 113, normalized size = 1.01 \begin {gather*} \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 {gather*}

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)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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)

________________________________________________________________________________________

maple [A]  time = 0.05, size = 62, normalized size = 0.55 \begin {gather*} \frac {\left (B b m x +A b m -2 B b x -A b -B a \right ) \left (b x +a \right ) \left (b c x +a c \right )^{m}}{\left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} \left (m^{2}-3 m +2\right ) b^{2}} \end {gather*}

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)

________________________________________________________________________________________

maxima [A]  time = 0.75, size = 117, normalized size = 1.04 \begin {gather*} \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 {gather*}

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))

________________________________________________________________________________________

mupad [B]  time = 2.30, size = 105, normalized size = 0.94 \begin {gather*} -\frac {{\left (a\,c+b\,c\,x\right )}^m\,\left (\frac {A\,b+B\,a-A\,b\,m}{b^3\,\left (m^2-3\,m+2\right )}-\frac {B\,x\,\left (m-2\right )}{b^2\,\left (m^2-3\,m+2\right )}\right )}{x\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}+\frac {a\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((a*c + b*c*x)^m*((A*b + B*a - A*b*m)/(b^3*(m^2 - 3*m + 2)) - (B*x*(m - 2))/(b^2*(m^2 - 3*m + 2))))/(x*(a^2 +
b^2*x^2 + 2*a*b*x)^(1/2) + (a*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/b)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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)

________________________________________________________________________________________