3.2165 \(\int (A+B x) (a c+b c x)^m (a^2+2 a b x+b^2 x^2)^{3/2} \, dx\)

Optimal. Leaf size=112 \[ \frac{\sqrt{a^2+2 a b x+b^2 x^2} (A b-a B) (a c+b c x)^{m+4}}{b^2 c^4 (m+4) (a+b x)}+\frac{B \sqrt{a^2+2 a b x+b^2 x^2} (a c+b c x)^{m+5}}{b^2 c^5 (m+5) (a+b x)} \]

[Out]

((A*b - a*B)*(a*c + b*c*x)^(4 + m)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(b^2*c^4*(4 + m)*(a + b*x)) + (B*(a*c + b*c*
x)^(5 + m)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(b^2*c^5*(5 + m)*(a + b*x))

________________________________________________________________________________________

Rubi [A]  time = 0.0759646, 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} \[ \frac{\sqrt{a^2+2 a b x+b^2 x^2} (A b-a B) (a c+b c x)^{m+4}}{b^2 c^4 (m+4) (a+b x)}+\frac{B \sqrt{a^2+2 a b x+b^2 x^2} (a c+b c x)^{m+5}}{b^2 c^5 (m+5) (a+b x)} \]

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)*(a*c + b*c*x)^(4 + m)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(b^2*c^4*(4 + m)*(a + b*x)) + (B*(a*c + b*c*
x)^(5 + m)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(b^2*c^5*(5 + m)*(a + b*x))

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 (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{\sqrt{a^2+2 a b x+b^2 x^2} \int \left (a b+b^2 x\right )^3 (A+B x) (a c+b c x)^m \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=\frac{\left (b \sqrt{a^2+2 a b x+b^2 x^2}\right ) \int (A+B x) (a c+b c x)^{3+m} \, dx}{c^3 \left (a b+b^2 x\right )}\\ &=\frac{\left (b \sqrt{a^2+2 a b x+b^2 x^2}\right ) \int \left (\frac{(A b-a B) (a c+b c x)^{3+m}}{b}+\frac{B (a c+b c x)^{4+m}}{b c}\right ) \, dx}{c^3 \left (a b+b^2 x\right )}\\ &=\frac{(A b-a B) (a c+b c x)^{4+m} \sqrt{a^2+2 a b x+b^2 x^2}}{b^2 c^4 (4+m) (a+b x)}+\frac{B (a c+b c x)^{5+m} \sqrt{a^2+2 a b x+b^2 x^2}}{b^2 c^5 (5+m) (a+b x)}\\ \end{align*}

Mathematica [A]  time = 0.0675136, size = 59, normalized size = 0.53 \[ \frac{(a+b x)^3 \sqrt{(a+b x)^2} (c (a+b x))^m (-a B+A b (m+5)+b B (m+4) x)}{b^2 (m+4) (m+5)} \]

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]

((a + b*x)^3*(c*(a + b*x))^m*Sqrt[(a + b*x)^2]*(-(a*B) + A*b*(5 + m) + b*B*(4 + m)*x))/(b^2*(4 + m)*(5 + m))

________________________________________________________________________________________

Maple [A]  time = 0.005, size = 62, normalized size = 0.6 \begin{align*}{\frac{ \left ( bcx+ac \right ) ^{m} \left ( Bbmx+Abm+4\,bBx+5\,Ab-aB \right ) \left ( bx+a \right ) }{{b}^{2} \left ({m}^{2}+9\,m+20 \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*x+a)^2)^(3/2)*(b*c*x+a*c)^m*(B*b*m*x+A*b*m+4*B*b*x+5*A*b-B*a)*(b*x+a)/b^2/(m^2+9*m+20)

________________________________________________________________________________________

Maxima [A]  time = 1.05747, size = 243, normalized size = 2.17 \begin{align*} \frac{{\left (b^{4} c^{m} x^{4} + 4 \, a b^{3} c^{m} x^{3} + 6 \, a^{2} b^{2} c^{m} x^{2} + 4 \, a^{3} b c^{m} x + a^{4} c^{m}\right )}{\left (b x + a\right )}^{m} A}{b{\left (m + 4\right )}} + \frac{{\left (b^{5} c^{m}{\left (m + 4\right )} x^{5} + a b^{4} c^{m}{\left (4 \, m + 15\right )} x^{4} + 2 \, a^{2} b^{3} c^{m}{\left (3 \, m + 10\right )} x^{3} + 2 \, a^{3} b^{2} c^{m}{\left (2 \, m + 5\right )} x^{2} + a^{4} b c^{m} m x - a^{5} c^{m}\right )}{\left (b x + a\right )}^{m} B}{{\left (m^{2} + 9 \, m + 20\right )} b^{2}} \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^4*c^m*x^4 + 4*a*b^3*c^m*x^3 + 6*a^2*b^2*c^m*x^2 + 4*a^3*b*c^m*x + a^4*c^m)*(b*x + a)^m*A/(b*(m + 4)) + (b^5
*c^m*(m + 4)*x^5 + a*b^4*c^m*(4*m + 15)*x^4 + 2*a^2*b^3*c^m*(3*m + 10)*x^3 + 2*a^3*b^2*c^m*(2*m + 5)*x^2 + a^4
*b*c^m*m*x - a^5*c^m)*(b*x + a)^m*B/((m^2 + 9*m + 20)*b^2)

________________________________________________________________________________________

Fricas [B]  time = 1.61096, size = 458, normalized size = 4.09 \begin{align*} \frac{{\left (A a^{4} b m - B a^{5} + 5 \, A a^{4} b +{\left (B b^{5} m + 4 \, B b^{5}\right )} x^{5} +{\left (15 \, B a b^{4} + 5 \, A b^{5} +{\left (4 \, B a b^{4} + A b^{5}\right )} m\right )} x^{4} + 2 \,{\left (10 \, B a^{2} b^{3} + 10 \, A a b^{4} +{\left (3 \, B a^{2} b^{3} + 2 \, A a b^{4}\right )} m\right )} x^{3} + 2 \,{\left (5 \, B a^{3} b^{2} + 15 \, A a^{2} b^{3} +{\left (2 \, B a^{3} b^{2} + 3 \, A a^{2} b^{3}\right )} m\right )} x^{2} +{\left (20 \, A a^{3} b^{2} +{\left (B a^{4} b + 4 \, A a^{3} b^{2}\right )} m\right )} x\right )}{\left (b c x + a c\right )}^{m}}{b^{2} m^{2} + 9 \, b^{2} m + 20 \, b^{2}} \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*a^4*b*m - B*a^5 + 5*A*a^4*b + (B*b^5*m + 4*B*b^5)*x^5 + (15*B*a*b^4 + 5*A*b^5 + (4*B*a*b^4 + A*b^5)*m)*x^4
+ 2*(10*B*a^2*b^3 + 10*A*a*b^4 + (3*B*a^2*b^3 + 2*A*a*b^4)*m)*x^3 + 2*(5*B*a^3*b^2 + 15*A*a^2*b^3 + (2*B*a^3*b
^2 + 3*A*a^2*b^3)*m)*x^2 + (20*A*a^3*b^2 + (B*a^4*b + 4*A*a^3*b^2)*m)*x)*(b*c*x + a*c)^m/(b^2*m^2 + 9*b^2*m +
20*b^2)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \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)

________________________________________________________________________________________

Giac [B]  time = 1.1781, size = 736, normalized size = 6.57 \begin{align*} \frac{{\left (b c x + a c\right )}^{m} B b^{5} m x^{5} \mathrm{sgn}\left (b x + a\right ) + 4 \,{\left (b c x + a c\right )}^{m} B a b^{4} m x^{4} \mathrm{sgn}\left (b x + a\right ) +{\left (b c x + a c\right )}^{m} A b^{5} m x^{4} \mathrm{sgn}\left (b x + a\right ) + 4 \,{\left (b c x + a c\right )}^{m} B b^{5} x^{5} \mathrm{sgn}\left (b x + a\right ) + 6 \,{\left (b c x + a c\right )}^{m} B a^{2} b^{3} m x^{3} \mathrm{sgn}\left (b x + a\right ) + 4 \,{\left (b c x + a c\right )}^{m} A a b^{4} m x^{3} \mathrm{sgn}\left (b x + a\right ) + 15 \,{\left (b c x + a c\right )}^{m} B a b^{4} x^{4} \mathrm{sgn}\left (b x + a\right ) + 5 \,{\left (b c x + a c\right )}^{m} A b^{5} x^{4} \mathrm{sgn}\left (b x + a\right ) + 4 \,{\left (b c x + a c\right )}^{m} B a^{3} b^{2} m x^{2} \mathrm{sgn}\left (b x + a\right ) + 6 \,{\left (b c x + a c\right )}^{m} A a^{2} b^{3} m x^{2} \mathrm{sgn}\left (b x + a\right ) + 20 \,{\left (b c x + a c\right )}^{m} B a^{2} b^{3} x^{3} \mathrm{sgn}\left (b x + a\right ) + 20 \,{\left (b c x + a c\right )}^{m} A a b^{4} x^{3} \mathrm{sgn}\left (b x + a\right ) +{\left (b c x + a c\right )}^{m} B a^{4} b m x \mathrm{sgn}\left (b x + a\right ) + 4 \,{\left (b c x + a c\right )}^{m} A a^{3} b^{2} m x \mathrm{sgn}\left (b x + a\right ) + 10 \,{\left (b c x + a c\right )}^{m} B a^{3} b^{2} x^{2} \mathrm{sgn}\left (b x + a\right ) + 30 \,{\left (b c x + a c\right )}^{m} A a^{2} b^{3} x^{2} \mathrm{sgn}\left (b x + a\right ) +{\left (b c x + a c\right )}^{m} A a^{4} b m \mathrm{sgn}\left (b x + a\right ) + 20 \,{\left (b c x + a c\right )}^{m} A a^{3} b^{2} x \mathrm{sgn}\left (b x + a\right ) -{\left (b c x + a c\right )}^{m} B a^{5} \mathrm{sgn}\left (b x + a\right ) + 5 \,{\left (b c x + a c\right )}^{m} A a^{4} b \mathrm{sgn}\left (b x + a\right )}{b^{2} m^{2} + 9 \, b^{2} m + 20 \, b^{2}} \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]

((b*c*x + a*c)^m*B*b^5*m*x^5*sgn(b*x + a) + 4*(b*c*x + a*c)^m*B*a*b^4*m*x^4*sgn(b*x + a) + (b*c*x + a*c)^m*A*b
^5*m*x^4*sgn(b*x + a) + 4*(b*c*x + a*c)^m*B*b^5*x^5*sgn(b*x + a) + 6*(b*c*x + a*c)^m*B*a^2*b^3*m*x^3*sgn(b*x +
a) + 4*(b*c*x + a*c)^m*A*a*b^4*m*x^3*sgn(b*x + a) + 15*(b*c*x + a*c)^m*B*a*b^4*x^4*sgn(b*x + a) + 5*(b*c*x +
a*c)^m*A*b^5*x^4*sgn(b*x + a) + 4*(b*c*x + a*c)^m*B*a^3*b^2*m*x^2*sgn(b*x + a) + 6*(b*c*x + a*c)^m*A*a^2*b^3*m
*x^2*sgn(b*x + a) + 20*(b*c*x + a*c)^m*B*a^2*b^3*x^3*sgn(b*x + a) + 20*(b*c*x + a*c)^m*A*a*b^4*x^3*sgn(b*x + a
) + (b*c*x + a*c)^m*B*a^4*b*m*x*sgn(b*x + a) + 4*(b*c*x + a*c)^m*A*a^3*b^2*m*x*sgn(b*x + a) + 10*(b*c*x + a*c)
^m*B*a^3*b^2*x^2*sgn(b*x + a) + 30*(b*c*x + a*c)^m*A*a^2*b^3*x^2*sgn(b*x + a) + (b*c*x + a*c)^m*A*a^4*b*m*sgn(
b*x + a) + 20*(b*c*x + a*c)^m*A*a^3*b^2*x*sgn(b*x + a) - (b*c*x + a*c)^m*B*a^5*sgn(b*x + a) + 5*(b*c*x + a*c)^
m*A*a^4*b*sgn(b*x + a))/(b^2*m^2 + 9*b^2*m + 20*b^2)