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

Optimal. Leaf size=62 \[ \frac {2 \left (a+b \sqrt {c+d x}\right )^{p+2}}{b^2 d (p+2)}-\frac {2 a \left (a+b \sqrt {c+d x}\right )^{p+1}}{b^2 d (p+1)} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {247, 190, 43} \[ \frac {2 \left (a+b \sqrt {c+d x}\right )^{p+2}}{b^2 d (p+2)}-\frac {2 a \left (a+b \sqrt {c+d x}\right )^{p+1}}{b^2 d (p+1)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sqrt[c + d*x])^p,x]

[Out]

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

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 190

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(1/n - 1)*(a + b*x)^p, x], x, x^n], x] /
; FreeQ[{a, b, p}, x] && FractionQ[n] && IntegerQ[1/n]

Rule 247

Int[((a_.) + (b_.)*(v_)^(n_))^(p_), x_Symbol] :> Dist[1/Coefficient[v, x, 1], Subst[Int[(a + b*x^n)^p, x], x,
v], x] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && NeQ[v, x]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 53, normalized size = 0.85 \[ \frac {2 \left (a+b \sqrt {c+d x}\right )^{p+1} \left (b (p+1) \sqrt {c+d x}-a\right )}{b^2 d (p+1) (p+2)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sqrt[c + d*x])^p,x]

[Out]

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

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fricas [A]  time = 0.53, size = 81, normalized size = 1.31 \[ \frac {2 \, {\left (b^{2} c p + \sqrt {d x + c} a b p + b^{2} c - a^{2} + {\left (b^{2} d p + b^{2} d\right )} x\right )} {\left (\sqrt {d x + c} b + a\right )}^{p}}{b^{2} d p^{2} + 3 \, b^{2} d p + 2 \, b^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(b^2*c*p + sqrt(d*x + c)*a*b*p + b^2*c - a^2 + (b^2*d*p + b^2*d)*x)*(sqrt(d*x + c)*b + a)^p/(b^2*d*p^2 + 3*b
^2*d*p + 2*b^2*d)

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giac [B]  time = 0.44, size = 129, normalized size = 2.08 \[ \frac {2 \, {\left ({\left (\sqrt {d x + c} b + a\right )}^{2} {\left (\sqrt {d x + c} b + a\right )}^{p} p - {\left (\sqrt {d x + c} b + a\right )} {\left (\sqrt {d x + c} b + a\right )}^{p} a p + {\left (\sqrt {d x + c} b + a\right )}^{2} {\left (\sqrt {d x + c} b + a\right )}^{p} - 2 \, {\left (\sqrt {d x + c} b + a\right )} {\left (\sqrt {d x + c} b + a\right )}^{p} a\right )}}{{\left (p^{2} + 3 \, p + 2\right )} b^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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*p - (sqrt(d*x + c)*b + a)*(sqrt(d*x + c)*b + a)^p*a*p + (sq
rt(d*x + c)*b + a)^2*(sqrt(d*x + c)*b + a)^p - 2*(sqrt(d*x + c)*b + a)*(sqrt(d*x + c)*b + a)^p*a)/((p^2 + 3*p
+ 2)*b^2*d)

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[ \int \left (a +\sqrt {d x +c}\, b \right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.93, size = 60, normalized size = 0.97 \[ \frac {2 \, {\left ({\left (d x + c\right )} b^{2} {\left (p + 1\right )} + \sqrt {d x + c} a b p - a^{2}\right )} {\left (\sqrt {d x + c} b + a\right )}^{p}}{{\left (p^{2} + 3 \, p + 2\right )} b^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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/((p^2 + 3*p + 2)*b^2*d)

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mupad [B]  time = 3.58, size = 146, normalized size = 2.35 \[ \left \{\begin {array}{cl} -\frac {2\,a\,\ln \left (a+b\,\sqrt {c+d\,x}\right )-2\,b\,\sqrt {c+d\,x}}{b^2\,d} & \text {\ if\ \ }p=-1\\ \frac {2\,\left (\ln \left (a+b\,\sqrt {c+d\,x}\right )+\frac {a}{a+b\,\sqrt {c+d\,x}}\right )}{b^2\,d} & \text {\ if\ \ }p=-2\\ \frac {4\,{\left (a+b\,\sqrt {c+d\,x}\right )}^{p+2}}{b^2\,d\,\left (2\,p+4\right )}-\frac {4\,a\,{\left (a+b\,\sqrt {c+d\,x}\right )}^{p+1}}{b^2\,d\,\left (2\,p+2\right )} & \text {\ if\ \ }p\neq -1\wedge p\neq -2 \end {array}\right . \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*(c + d*x)^(1/2))^p,x)

[Out]

piecewise(p == -1, -(2*a*log(a + b*(c + d*x)^(1/2)) - 2*b*(c + d*x)^(1/2))/(b^2*d), p == -2, (2*(log(a + b*(c
+ d*x)^(1/2)) + a/(a + b*(c + d*x)^(1/2))))/(b^2*d), p ~= -1 & p ~= -2, (4*(a + b*(c + d*x)^(1/2))^(p + 2))/(b
^2*d*(2*p + 4)) - (4*a*(a + b*(c + d*x)^(1/2))^(p + 1))/(b^2*d*(2*p + 2)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*sqrt(c + d*x))**p, x)

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