3.655 \(\int x (a+b \sqrt {c+d x})^p \, dx\)

Optimal. Leaf size=145 \[ -\frac {2 a \left (a^2-b^2 c\right ) \left (a+b \sqrt {c+d x}\right )^{p+1}}{b^4 d^2 (p+1)}+\frac {2 \left (3 a^2-b^2 c\right ) \left (a+b \sqrt {c+d x}\right )^{p+2}}{b^4 d^2 (p+2)}-\frac {6 a \left (a+b \sqrt {c+d x}\right )^{p+3}}{b^4 d^2 (p+3)}+\frac {2 \left (a+b \sqrt {c+d x}\right )^{p+4}}{b^4 d^2 (p+4)} \]

[Out]

-2*a*(-b^2*c+a^2)*(a+b*(d*x+c)^(1/2))^(1+p)/b^4/d^2/(1+p)+2*(-b^2*c+3*a^2)*(a+b*(d*x+c)^(1/2))^(2+p)/b^4/d^2/(
2+p)-6*a*(a+b*(d*x+c)^(1/2))^(3+p)/b^4/d^2/(3+p)+2*(a+b*(d*x+c)^(1/2))^(4+p)/b^4/d^2/(4+p)

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Rubi [A]  time = 0.11, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {371, 1398, 772} \[ -\frac {2 a \left (a^2-b^2 c\right ) \left (a+b \sqrt {c+d x}\right )^{p+1}}{b^4 d^2 (p+1)}+\frac {2 \left (3 a^2-b^2 c\right ) \left (a+b \sqrt {c+d x}\right )^{p+2}}{b^4 d^2 (p+2)}-\frac {6 a \left (a+b \sqrt {c+d x}\right )^{p+3}}{b^4 d^2 (p+3)}+\frac {2 \left (a+b \sqrt {c+d x}\right )^{p+4}}{b^4 d^2 (p+4)} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*Sqrt[c + d*x])^p,x]

[Out]

(-2*a*(a^2 - b^2*c)*(a + b*Sqrt[c + d*x])^(1 + p))/(b^4*d^2*(1 + p)) + (2*(3*a^2 - b^2*c)*(a + b*Sqrt[c + d*x]
)^(2 + p))/(b^4*d^2*(2 + p)) - (6*a*(a + b*Sqrt[c + d*x])^(3 + p))/(b^4*d^2*(3 + p)) + (2*(a + b*Sqrt[c + d*x]
)^(4 + p))/(b^4*d^2*(4 + p))

Rule 371

Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coefficient[v, x, 0], d = Coefficient[v,
x, 1]}, Dist[1/d^(m + 1), Subst[Int[SimplifyIntegrand[(x - c)^m*(a + b*x^n)^p, x], x], x, v], x] /; NeQ[c, 0]
] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && IntegerQ[m]

Rule 772

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegr
and[(d + e*x)^m*(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IGtQ[p, 0]

Rule 1398

Int[((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> With[{g = Denominator[n]}, D
ist[g, Subst[Int[x^(g - 1)*(d + e*x^(g*n))^q*(a + c*x^(2*g*n))^p, x], x, x^(1/g)], x]] /; FreeQ[{a, c, d, e, p
, q}, x] && EqQ[n2, 2*n] && FractionQ[n]

Rubi steps

\begin {align*} \int x \left (a+b \sqrt {c+d x}\right )^p \, dx &=\frac {\operatorname {Subst}\left (\int \left (a+b \sqrt {x}\right )^p (-c+x) \, dx,x,c+d x\right )}{d^2}\\ &=\frac {2 \operatorname {Subst}\left (\int x (a+b x)^p \left (-c+x^2\right ) \, dx,x,\sqrt {c+d x}\right )}{d^2}\\ &=\frac {2 \operatorname {Subst}\left (\int \left (\frac {\left (-a^3+a b^2 c\right ) (a+b x)^p}{b^3}+\frac {\left (3 a^2-b^2 c\right ) (a+b x)^{1+p}}{b^3}-\frac {3 a (a+b x)^{2+p}}{b^3}+\frac {(a+b x)^{3+p}}{b^3}\right ) \, dx,x,\sqrt {c+d x}\right )}{d^2}\\ &=-\frac {2 a \left (a^2-b^2 c\right ) \left (a+b \sqrt {c+d x}\right )^{1+p}}{b^4 d^2 (1+p)}+\frac {2 \left (3 a^2-b^2 c\right ) \left (a+b \sqrt {c+d x}\right )^{2+p}}{b^4 d^2 (2+p)}-\frac {6 a \left (a+b \sqrt {c+d x}\right )^{3+p}}{b^4 d^2 (3+p)}+\frac {2 \left (a+b \sqrt {c+d x}\right )^{4+p}}{b^4 d^2 (4+p)}\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 128, normalized size = 0.88 \[ -\frac {2 \left (a+b \sqrt {c+d x}\right )^{p+1} \left (6 a^3-6 a^2 b (p+1) \sqrt {c+d x}+a b^2 \left (2 c \left (p^2+p-3\right )+3 d \left (p^2+3 p+2\right ) x\right )-b^3 \left (p^2+4 p+3\right ) \sqrt {c+d x} (d (p+2) x-2 c)\right )}{b^4 d^2 (p+1) (p+2) (p+3) (p+4)} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(a + b*Sqrt[c + d*x])^p,x]

[Out]

(-2*(a + b*Sqrt[c + d*x])^(1 + p)*(6*a^3 - 6*a^2*b*(1 + p)*Sqrt[c + d*x] - b^3*(3 + 4*p + p^2)*Sqrt[c + d*x]*(
-2*c + d*(2 + p)*x) + a*b^2*(2*c*(-3 + p + p^2) + 3*d*(2 + 3*p + p^2)*x)))/(b^4*d^2*(1 + p)*(2 + p)*(3 + p)*(4
+ p))

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fricas [B]  time = 0.54, size = 294, normalized size = 2.03 \[ -\frac {2 \, {\left (6 \, b^{4} c^{2} - 12 \, a^{2} b^{2} c + 6 \, a^{4} + 2 \, {\left (b^{4} c^{2} + a^{2} b^{2} c\right )} p^{2} - {\left (b^{4} d^{2} p^{3} + 6 \, b^{4} d^{2} p^{2} + 11 \, b^{4} d^{2} p + 6 \, b^{4} d^{2}\right )} x^{2} + 4 \, {\left (2 \, b^{4} c^{2} - a^{2} b^{2} c\right )} p - {\left (b^{4} c d p^{3} + {\left (4 \, b^{4} c - 3 \, a^{2} b^{2}\right )} d p^{2} + 3 \, {\left (b^{4} c - a^{2} b^{2}\right )} d p\right )} x + {\left (4 \, a b^{3} c p^{2} + 2 \, {\left (5 \, a b^{3} c - 3 \, a^{3} b\right )} p - {\left (a b^{3} d p^{3} + 3 \, a b^{3} d p^{2} + 2 \, a b^{3} d p\right )} x\right )} \sqrt {d x + c}\right )} {\left (\sqrt {d x + c} b + a\right )}^{p}}{b^{4} d^{2} p^{4} + 10 \, b^{4} d^{2} p^{3} + 35 \, b^{4} d^{2} p^{2} + 50 \, b^{4} d^{2} p + 24 \, b^{4} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*(d*x+c)^(1/2))^p,x, algorithm="fricas")

[Out]

-2*(6*b^4*c^2 - 12*a^2*b^2*c + 6*a^4 + 2*(b^4*c^2 + a^2*b^2*c)*p^2 - (b^4*d^2*p^3 + 6*b^4*d^2*p^2 + 11*b^4*d^2
*p + 6*b^4*d^2)*x^2 + 4*(2*b^4*c^2 - a^2*b^2*c)*p - (b^4*c*d*p^3 + (4*b^4*c - 3*a^2*b^2)*d*p^2 + 3*(b^4*c - a^
2*b^2)*d*p)*x + (4*a*b^3*c*p^2 + 2*(5*a*b^3*c - 3*a^3*b)*p - (a*b^3*d*p^3 + 3*a*b^3*d*p^2 + 2*a*b^3*d*p)*x)*sq
rt(d*x + c))*(sqrt(d*x + c)*b + a)^p/(b^4*d^2*p^4 + 10*b^4*d^2*p^3 + 35*b^4*d^2*p^2 + 50*b^4*d^2*p + 24*b^4*d^
2)

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giac [B]  time = 0.44, size = 806, normalized size = 5.56 \[ -\frac {2 \, {\left ({\left (\sqrt {d x + c} b + a\right )}^{2} {\left (\sqrt {d x + c} b + a\right )}^{p} b^{2} c p^{3} - {\left (\sqrt {d x + c} b + a\right )} {\left (\sqrt {d x + c} b + a\right )}^{p} a b^{2} c p^{3} + 8 \, {\left (\sqrt {d x + c} b + a\right )}^{2} {\left (\sqrt {d x + c} b + a\right )}^{p} b^{2} c p^{2} - 9 \, {\left (\sqrt {d x + c} b + a\right )} {\left (\sqrt {d x + c} b + a\right )}^{p} a b^{2} c p^{2} - {\left (\sqrt {d x + c} b + a\right )}^{4} {\left (\sqrt {d x + c} b + a\right )}^{p} p^{3} + 3 \, {\left (\sqrt {d x + c} b + a\right )}^{3} {\left (\sqrt {d x + c} b + a\right )}^{p} a p^{3} - 3 \, {\left (\sqrt {d x + c} b + a\right )}^{2} {\left (\sqrt {d x + c} b + a\right )}^{p} a^{2} p^{3} + {\left (\sqrt {d x + c} b + a\right )} {\left (\sqrt {d x + c} b + a\right )}^{p} a^{3} p^{3} + 19 \, {\left (\sqrt {d x + c} b + a\right )}^{2} {\left (\sqrt {d x + c} b + a\right )}^{p} b^{2} c p - 26 \, {\left (\sqrt {d x + c} b + a\right )} {\left (\sqrt {d x + c} b + a\right )}^{p} a b^{2} c p - 6 \, {\left (\sqrt {d x + c} b + a\right )}^{4} {\left (\sqrt {d x + c} b + a\right )}^{p} p^{2} + 21 \, {\left (\sqrt {d x + c} b + a\right )}^{3} {\left (\sqrt {d x + c} b + a\right )}^{p} a p^{2} - 24 \, {\left (\sqrt {d x + c} b + a\right )}^{2} {\left (\sqrt {d x + c} b + a\right )}^{p} a^{2} p^{2} + 9 \, {\left (\sqrt {d x + c} b + a\right )} {\left (\sqrt {d x + c} b + a\right )}^{p} a^{3} p^{2} + 12 \, {\left (\sqrt {d x + c} b + a\right )}^{2} {\left (\sqrt {d x + c} b + a\right )}^{p} b^{2} c - 24 \, {\left (\sqrt {d x + c} b + a\right )} {\left (\sqrt {d x + c} b + a\right )}^{p} a b^{2} c - 11 \, {\left (\sqrt {d x + c} b + a\right )}^{4} {\left (\sqrt {d x + c} b + a\right )}^{p} p + 42 \, {\left (\sqrt {d x + c} b + a\right )}^{3} {\left (\sqrt {d x + c} b + a\right )}^{p} a p - 57 \, {\left (\sqrt {d x + c} b + a\right )}^{2} {\left (\sqrt {d x + c} b + a\right )}^{p} a^{2} p + 26 \, {\left (\sqrt {d x + c} b + a\right )} {\left (\sqrt {d x + c} b + a\right )}^{p} a^{3} p - 6 \, {\left (\sqrt {d x + c} b + a\right )}^{4} {\left (\sqrt {d x + c} b + a\right )}^{p} + 24 \, {\left (\sqrt {d x + c} b + a\right )}^{3} {\left (\sqrt {d x + c} b + a\right )}^{p} a - 36 \, {\left (\sqrt {d x + c} b + a\right )}^{2} {\left (\sqrt {d x + c} b + a\right )}^{p} a^{2} + 24 \, {\left (\sqrt {d x + c} b + a\right )} {\left (\sqrt {d x + c} b + a\right )}^{p} a^{3}\right )}}{{\left (b^{2} p^{4} + 10 \, b^{2} p^{3} + 35 \, b^{2} p^{2} + 50 \, b^{2} p + 24 \, b^{2}\right )} b^{2} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*(d*x+c)^(1/2))^p,x, algorithm="giac")

[Out]

-2*((sqrt(d*x + c)*b + a)^2*(sqrt(d*x + c)*b + a)^p*b^2*c*p^3 - (sqrt(d*x + c)*b + a)*(sqrt(d*x + c)*b + a)^p*
a*b^2*c*p^3 + 8*(sqrt(d*x + c)*b + a)^2*(sqrt(d*x + c)*b + a)^p*b^2*c*p^2 - 9*(sqrt(d*x + c)*b + a)*(sqrt(d*x
+ c)*b + a)^p*a*b^2*c*p^2 - (sqrt(d*x + c)*b + a)^4*(sqrt(d*x + c)*b + a)^p*p^3 + 3*(sqrt(d*x + c)*b + a)^3*(s
qrt(d*x + c)*b + a)^p*a*p^3 - 3*(sqrt(d*x + c)*b + a)^2*(sqrt(d*x + c)*b + a)^p*a^2*p^3 + (sqrt(d*x + c)*b + a
)*(sqrt(d*x + c)*b + a)^p*a^3*p^3 + 19*(sqrt(d*x + c)*b + a)^2*(sqrt(d*x + c)*b + a)^p*b^2*c*p - 26*(sqrt(d*x
+ c)*b + a)*(sqrt(d*x + c)*b + a)^p*a*b^2*c*p - 6*(sqrt(d*x + c)*b + a)^4*(sqrt(d*x + c)*b + a)^p*p^2 + 21*(sq
rt(d*x + c)*b + a)^3*(sqrt(d*x + c)*b + a)^p*a*p^2 - 24*(sqrt(d*x + c)*b + a)^2*(sqrt(d*x + c)*b + a)^p*a^2*p^
2 + 9*(sqrt(d*x + c)*b + a)*(sqrt(d*x + c)*b + a)^p*a^3*p^2 + 12*(sqrt(d*x + c)*b + a)^2*(sqrt(d*x + c)*b + a)
^p*b^2*c - 24*(sqrt(d*x + c)*b + a)*(sqrt(d*x + c)*b + a)^p*a*b^2*c - 11*(sqrt(d*x + c)*b + a)^4*(sqrt(d*x + c
)*b + a)^p*p + 42*(sqrt(d*x + c)*b + a)^3*(sqrt(d*x + c)*b + a)^p*a*p - 57*(sqrt(d*x + c)*b + a)^2*(sqrt(d*x +
c)*b + a)^p*a^2*p + 26*(sqrt(d*x + c)*b + a)*(sqrt(d*x + c)*b + a)^p*a^3*p - 6*(sqrt(d*x + c)*b + a)^4*(sqrt(
d*x + c)*b + a)^p + 24*(sqrt(d*x + c)*b + a)^3*(sqrt(d*x + c)*b + a)^p*a - 36*(sqrt(d*x + c)*b + a)^2*(sqrt(d*
x + c)*b + a)^p*a^2 + 24*(sqrt(d*x + c)*b + a)*(sqrt(d*x + c)*b + a)^p*a^3)/((b^2*p^4 + 10*b^2*p^3 + 35*b^2*p^
2 + 50*b^2*p + 24*b^2)*b^2*d^2)

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maple [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \left (a +\sqrt {d x +c}\, b \right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a+(d*x+c)^(1/2)*b)^p,x)

[Out]

int(x*(a+(d*x+c)^(1/2)*b)^p,x)

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maxima [A]  time = 0.99, size = 187, normalized size = 1.29 \[ -\frac {2 \, {\left (\frac {{\left ({\left (d x + c\right )} b^{2} {\left (p + 1\right )} + \sqrt {d x + c} a b p - a^{2}\right )} {\left (\sqrt {d x + c} b + a\right )}^{p} c}{{\left (p^{2} + 3 \, p + 2\right )} b^{2}} - \frac {{\left ({\left (p^{3} + 6 \, p^{2} + 11 \, p + 6\right )} {\left (d x + c\right )}^{2} b^{4} + {\left (p^{3} + 3 \, p^{2} + 2 \, p\right )} {\left (d x + c\right )}^{\frac {3}{2}} a b^{3} - 3 \, {\left (p^{2} + p\right )} {\left (d x + c\right )} a^{2} b^{2} + 6 \, \sqrt {d x + c} a^{3} b p - 6 \, a^{4}\right )} {\left (\sqrt {d x + c} b + a\right )}^{p}}{{\left (p^{4} + 10 \, p^{3} + 35 \, p^{2} + 50 \, p + 24\right )} b^{4}}\right )}}{d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*(d*x+c)^(1/2))^p,x, algorithm="maxima")

[Out]

-2*(((d*x + c)*b^2*(p + 1) + sqrt(d*x + c)*a*b*p - a^2)*(sqrt(d*x + c)*b + a)^p*c/((p^2 + 3*p + 2)*b^2) - ((p^
3 + 6*p^2 + 11*p + 6)*(d*x + c)^2*b^4 + (p^3 + 3*p^2 + 2*p)*(d*x + c)^(3/2)*a*b^3 - 3*(p^2 + p)*(d*x + c)*a^2*
b^2 + 6*sqrt(d*x + c)*a^3*b*p - 6*a^4)*(sqrt(d*x + c)*b + a)^p/((p^4 + 10*p^3 + 35*p^2 + 50*p + 24)*b^4))/d^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,{\left (a+b\,\sqrt {c+d\,x}\right )}^p \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b*(c + d*x)^(1/2))^p,x)

[Out]

int(x*(a + b*(c + d*x)^(1/2))^p, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \left (a + b \sqrt {c + d x}\right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*(d*x+c)**(1/2))**p,x)

[Out]

Integral(x*(a + b*sqrt(c + d*x))**p, x)

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