\(\int f^{a+b \sqrt {c+d x}} \, dx\) [299]

Optimal result

Integrand size = 15, antiderivative size = 64 \[ \int f^{a+b \sqrt {c+d x}} \, dx=-\frac {2 f^{a+b \sqrt {c+d x}}}{b^2 d \log ^2(f)}+\frac {2 f^{a+b \sqrt {c+d x}} \sqrt {c+d x}}{b d \log (f)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2238, 2207, 2225} \[ \int f^{a+b \sqrt {c+d x}} \, dx=\frac {2 \sqrt {c+d x} f^{a+b \sqrt {c+d x}}}{b d \log (f)}-\frac {2 f^{a+b \sqrt {c+d x}}}{b^2 d \log ^2(f)} \]

[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[((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] &&  !TrueQ[$UseGamma]

Rule 2225

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]

Rule 2238

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]

Rubi steps \begin{align*} \text {integral}& = \frac {2 \text {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 \text {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] (verified)

Time = 0.04 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.66 \[ \int f^{a+b \sqrt {c+d x}} \, dx=\frac {2 f^{a+b \sqrt {c+d x}} \left (-1+b \sqrt {c+d x} \log (f)\right )}{b^2 d \log ^2(f)} \]

[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]

\[\int f^{a +b \sqrt {d x +c}}d x\]

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.66 \[ \int f^{a+b \sqrt {c+d x}} \, dx=\frac {2 \, {\left (\sqrt {d x + c} b \log \left (f\right ) - 1\right )} e^{\left (\sqrt {d x + c} b \log \left (f\right ) + a \log \left (f\right )\right )}}{b^{2} d \log \left (f\right )^{2}} \]

[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] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.14 \[ \int f^{a+b \sqrt {c+d x}} \, dx=\begin {cases} x & \text {for}\: b = 0 \wedge d = 0 \wedge f = 1 \\f^{a} x & \text {for}\: b = 0 \\f^{a + b \sqrt {c}} x & \text {for}\: d = 0 \\x & \text {for}\: f = 1 \\\frac {2 f^{a + b \sqrt {c + d x}} \sqrt {c + d x}}{b d \log {\left (f \right )}} - \frac {2 f^{a + b \sqrt {c + d x}}}{b^{2} d \log {\left (f \right )}^{2}} & \text {otherwise} \end {cases} \]

[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)), (f**(a + b*sqrt(c))*x, Eq(d, 0)), (x, Eq(f,
1)), (2*f**(a + b*sqrt(c + d*x))*sqrt(c + d*x)/(b*d*log(f)) - 2*f**(a + b*sqrt(c + d*x))/(b**2*d*log(f)**2),
True))

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.67 \[ \int f^{a+b \sqrt {c+d x}} \, dx=\frac {2 \, {\left (\sqrt {d x + c} b f^{a} \log \left (f\right ) - f^{a}\right )} f^{\sqrt {d x + c} b}}{b^{2} d \log \left (f\right )^{2}} \]

[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)

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.35 (sec) , antiderivative size = 781, normalized size of antiderivative = 12.20 \[ \int f^{a+b \sqrt {c+d x}} \, dx=\text {Too large to display} \]

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

Mupad [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.59 \[ \int f^{a+b \sqrt {c+d x}} \, dx=\frac {f^{a+b\,\sqrt {c+d\,x}}\,\left (2\,b\,\ln \left (f\right )\,\sqrt {c+d\,x}-2\right )}{b^2\,d\,{\ln \left (f\right )}^2} \]

[In]

int(f^(a + b*(c + d*x)^(1/2)),x)

[Out]

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