∫ fa+b√ __

3.299 $\int f^{a+b \sqrt{c+d x}} \, dx$

Optimal. Leaf size=64 \[ \frac{2 \sqrt{c+d x} f^{a+b \sqrt{c+d x}}}{b d \log (f)}-\frac{2 f^{a+b \sqrt{c+d x}}}{b^2 d \log ^2(f)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A] time = 0.0324213, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 15, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.2, Rules used = {2207, 2176, 2194} \[ \frac{2 \sqrt{c+d x} f^{a+b \sqrt{c+d x}}}{b d \log (f)}-\frac{2 f^{a+b \sqrt{c+d x}}}{b^2 d \log ^2(f)} \]

Antiderivative was successfully veriﬁed.

[In]

Int[f^(a + b*Sqrt[c + d*x]),x]

[Out]

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

Rule 2207

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_)), x_Symbol] :> With[{k = Denominator[n]}, Dist[k/d, Subst[In

t[x^(k - 1)*F^(a + b*x^(k*n)), x], x, (c + d*x)^(1/k)], x]] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2/n] &&

!IntegerQ[n]

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m

*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x

)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] && !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre

eQ[{F, a, b, c, n}, x]

Rubi steps

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

Mathematica [A] time = 0.0340865, size = 42, normalized size = 0.66 \[ \frac{2 f^{a+b \sqrt{c+d x}} \left (b \log (f) \sqrt{c+d x}-1\right )}{b^2 d \log ^2(f)} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F] time = 0.005, size = 0, normalized size = 0. \begin{align*} \int{f}^{a+b\sqrt{dx+c}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A] time = 1.01327, size = 58, normalized size = 0.91 \begin{align*} \frac{2 \,{\left (\sqrt{d x + c} b f^{a} \log \left (f\right ) - f^{a}\right )} f^{\sqrt{d x + c} b}}{b^{2} d \log \left (f\right )^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(sqrt(d*x + c)*b*f^a*log(f) - f^a)*f^(sqrt(d*x + c)*b)/(b^2*d*log(f)^2)

________________________________________________________________________________________

Fricas [A] time = 1.52162, size = 117, normalized size = 1.83 \begin{align*} \frac{2 \,{\left (\sqrt{d x + c} b \log \left (f\right ) - 1\right )} e^{\left (\sqrt{d x + c} b \log \left (f\right ) + a \log \left (f\right )\right )}}{b^{2} d \log \left (f\right )^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(sqrt(d*x + c)*b*log(f) - 1)*e^(sqrt(d*x + c)*b*log(f) + a*log(f))/(b^2*d*log(f)^2)

________________________________________________________________________________________

Sympy [A] time = 0.75721, size = 76, normalized size = 1.19 \begin{align*} \begin{cases} x & \text{for}\: b = 0 \wedge d = 0 \wedge f = 1 \\f^{a} x & \text{for}\: b = 0 \\x & \text{for}\: f = 1 \\f^{a + b \sqrt{c}} x & \text{for}\: d = 0 \\\frac{2 f^{a} f^{b \sqrt{c + d x}} \sqrt{c + d x}}{b d \log{\left (f \right )}} - \frac{2 f^{a} f^{b \sqrt{c + d x}}}{b^{2} d \log{\left (f \right )}^{2}} & \text{otherwise} \end{cases} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x, Eq(b, 0) & Eq(d, 0) & Eq(f, 1)), (f**a*x, Eq(b, 0)), (x, Eq(f, 1)), (f**(a + b*sqrt(c))*x, Eq(d,

0)), (2*f**a*f**(b*sqrt(c + d*x))*sqrt(c + d*x)/(b*d*log(f)) - 2*f**a*f**(b*sqrt(c + d*x))/(b**2*d*log(f)**2)

, True))

________________________________________________________________________________________

Giac [B] time = 2.13246, size = 1436, normalized size = 22.44 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(2*((pi*b^2*log(abs(f))*sgn(f) - pi*b^2*log(abs(f)))*(pi*sqrt(d*x + c)*b*sgn(f) - pi*sqrt(d*x + c)*b)/((pi^

2*b^2*sgn(f) - pi^2*b^2 + 2*b^2*log(abs(f))^2)^2 + 4*(pi*b^2*log(abs(f))*sgn(f) - pi*b^2*log(abs(f)))^2) + (pi

^2*b^2*sgn(f) - pi^2*b^2 + 2*b^2*log(abs(f))^2)*(sqrt(d*x + c)*b*log(abs(f)) - 1)/((pi^2*b^2*sgn(f) - pi^2*b^2

+ 2*b^2*log(abs(f))^2)^2 + 4*(pi*b^2*log(abs(f))*sgn(f) - pi*b^2*log(abs(f)))^2))*cos(-1/2*pi*sqrt(d*x + c)*b

*sgn(f) - 1/2*pi*a*sgn(f) + 1/2*pi*sqrt(d*x + c)*b + 1/2*pi*a) + ((pi^2*b^2*sgn(f) - pi^2*b^2 + 2*b^2*log(abs(

f))^2)*(pi*sqrt(d*x + c)*b*sgn(f) - pi*sqrt(d*x + c)*b)/((pi^2*b^2*sgn(f) - pi^2*b^2 + 2*b^2*log(abs(f))^2)^2

+ 4*(pi*b^2*log(abs(f))*sgn(f) - pi*b^2*log(abs(f)))^2) - 4*(pi*b^2*log(abs(f))*sgn(f) - pi*b^2*log(abs(f)))*(

sqrt(d*x + c)*b*log(abs(f)) - 1)/((pi^2*b^2*sgn(f) - pi^2*b^2 + 2*b^2*log(abs(f))^2)^2 + 4*(pi*b^2*log(abs(f))

*sgn(f) - pi*b^2*log(abs(f)))^2))*sin(-1/2*pi*sqrt(d*x + c)*b*sgn(f) - 1/2*pi*a*sgn(f) + 1/2*pi*sqrt(d*x + c)*

b + 1/2*pi*a))*e^(sqrt(d*x + c)*b*log(abs(f)) + a*log(abs(f))) - ((2*sqrt(d*x + c)*b*i*log(abs(f)) - pi*sqrt(d

*x + c)*b*sgn(f) + pi*sqrt(d*x + c)*b - 2*i)*e^(1/2*(pi*sqrt(d*x + c)*b*(sgn(f) - 1) + pi*a*(sgn(f) - 1))*i)/(

2*pi*b^2*i*log(abs(f))*sgn(f) - 2*pi*b^2*i*log(abs(f)) + pi^2*b^2*sgn(f) - pi^2*b^2 + 2*b^2*log(abs(f))^2) + (

2*sqrt(d*x + c)*b*i*log(abs(f)) + pi*sqrt(d*x + c)*b*sgn(f) - pi*sqrt(d*x + c)*b - 2*i)*e^(-1/2*(pi*sqrt(d*x +

c)*b*(sgn(f) - 1) + pi*a*(sgn(f) - 1))*i)/(2*pi*b^2*i*log(abs(f))*sgn(f) - 2*pi*b^2*i*log(abs(f)) - pi^2*b^2*

sgn(f) + pi^2*b^2 - 2*b^2*log(abs(f))^2))*e^(sqrt(d*x + c)*b*log(abs(f)) + a*log(abs(f)))/i - 2*(2*pi*i*abs(f)

^a*cos(1/2*pi*a*sgn(f) - 1/2*pi*a)*log(abs(f))*sgn(f) - pi^2*i*abs(f)^a*sgn(f)*sin(1/2*pi*a*sgn(f) - 1/2*pi*a)

- 2*pi*i*abs(f)^a*cos(1/2*pi*a*sgn(f) - 1/2*pi*a)*log(abs(f)) + pi^2*abs(f)^a*cos(1/2*pi*a*sgn(f) - 1/2*pi*a)

*sgn(f) + pi^2*i*abs(f)^a*sin(1/2*pi*a*sgn(f) - 1/2*pi*a) - 2*i*abs(f)^a*log(abs(f))^2*sin(1/2*pi*a*sgn(f) - 1

/2*pi*a) + 2*pi*abs(f)^a*log(abs(f))*sgn(f)*sin(1/2*pi*a*sgn(f) - 1/2*pi*a) - pi^2*abs(f)^a*cos(1/2*pi*a*sgn(f

) - 1/2*pi*a) + 2*abs(f)^a*cos(1/2*pi*a*sgn(f) - 1/2*pi*a)*log(abs(f))^2 - 2*pi*abs(f)^a*log(abs(f))*sin(1/2*p

i*a*sgn(f) - 1/2*pi*a))/(pi^4*b^2*sgn(f) + 2*pi^2*b^2*log(abs(f))^2*sgn(f) - pi^4*b^2 - 2*pi^2*b^2*log(abs(f))

^2 - 2*b^2*log(abs(f))^4))/d

[next] [prev] [prev-tail] [front] [up]

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