3.8.36 \(\int \sqrt {f^x} (a+b x)^2 \, dx\) [736]

Optimal. Leaf size=56 \[ \frac {16 b^2 \sqrt {f^x}}{\log ^3(f)}-\frac {8 b \sqrt {f^x} (a+b x)}{\log ^2(f)}+\frac {2 \sqrt {f^x} (a+b x)^2}{\log (f)} \]

[Out]

16*b^2*(f^x)^(1/2)/ln(f)^3-8*b*(b*x+a)*(f^x)^(1/2)/ln(f)^2+2*(b*x+a)^2*(f^x)^(1/2)/ln(f)

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Rubi [A]
time = 0.03, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2207, 2225} \begin {gather*} -\frac {8 b \sqrt {f^x} (a+b x)}{\log ^2(f)}+\frac {2 \sqrt {f^x} (a+b x)^2}{\log (f)}+\frac {16 b^2 \sqrt {f^x}}{\log ^3(f)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[f^x]*(a + b*x)^2,x]

[Out]

(16*b^2*Sqrt[f^x])/Log[f]^3 - (8*b*Sqrt[f^x]*(a + b*x))/Log[f]^2 + (2*Sqrt[f^x]*(a + b*x)^2)/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]

Rubi steps

\begin {align*} \int \sqrt {f^x} (a+b x)^2 \, dx &=\frac {2 \sqrt {f^x} (a+b x)^2}{\log (f)}-\frac {(4 b) \int \sqrt {f^x} (a+b x) \, dx}{\log (f)}\\ &=-\frac {8 b \sqrt {f^x} (a+b x)}{\log ^2(f)}+\frac {2 \sqrt {f^x} (a+b x)^2}{\log (f)}+\frac {\left (8 b^2\right ) \int \sqrt {f^x} \, dx}{\log ^2(f)}\\ &=\frac {16 b^2 \sqrt {f^x}}{\log ^3(f)}-\frac {8 b \sqrt {f^x} (a+b x)}{\log ^2(f)}+\frac {2 \sqrt {f^x} (a+b x)^2}{\log (f)}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 41, normalized size = 0.73 \begin {gather*} \frac {2 \sqrt {f^x} \left (8 b^2-4 b (a+b x) \log (f)+(a+b x)^2 \log ^2(f)\right )}{\log ^3(f)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[f^x]*(a + b*x)^2,x]

[Out]

(2*Sqrt[f^x]*(8*b^2 - 4*b*(a + b*x)*Log[f] + (a + b*x)^2*Log[f]^2))/Log[f]^3

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Maple [A]
time = 0.08, size = 60, normalized size = 1.07

method	result	size
gosper	\(\frac {2 \left (b^{2} x^{2} \ln \left (f \right )^{2}+2 \ln \left (f \right )^{2} a b x +\ln \left (f \right )^{2} a^{2}-4 \ln \left (f \right ) b^{2} x -4 \ln \left (f \right ) b a +8 b^{2}\right ) \sqrt {f^{x}}}{\ln \left (f \right )^{3}}\)	\(60\)
risch	\(\frac {2 \left (b^{2} x^{2} \ln \left (f \right )^{2}+2 \ln \left (f \right )^{2} a b x +\ln \left (f \right )^{2} a^{2}-4 \ln \left (f \right ) b^{2} x -4 \ln \left (f \right ) b a +8 b^{2}\right ) \sqrt {f^{x}}}{\ln \left (f \right )^{3}}\)	\(60\)
meijerg	\(-\frac {8 b^{2} \sqrt {f^{x}}\, f^{-\frac {x}{2}} \left (2-\frac {\left (\frac {3 \ln \left (f \right )^{2} x^{2}}{4}-3 x \ln \left (f \right )+6\right ) {\mathrm e}^{\frac {x \ln \left (f \right )}{2}}}{3}\right )}{\ln \left (f \right )^{3}}+\frac {8 b a \sqrt {f^{x}}\, f^{-\frac {x}{2}} \left (1-\frac {\left (2-x \ln \left (f \right )\right ) {\mathrm e}^{\frac {x \ln \left (f \right )}{2}}}{2}\right )}{\ln \left (f \right )^{2}}-\frac {2 a^{2} \sqrt {f^{x}}\, f^{-\frac {x}{2}} \left (1-{\mathrm e}^{\frac {x \ln \left (f \right )}{2}}\right )}{\ln \left (f \right )}\)	\(111\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^2*(f^x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

2*(b^2*x^2*ln(f)^2+2*ln(f)^2*a*b*x+ln(f)^2*a^2-4*ln(f)*b^2*x-4*ln(f)*b*a+8*b^2)*(f^x)^(1/2)/ln(f)^3

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Maxima [A]
time = 0.28, size = 63, normalized size = 1.12 \begin {gather*} \frac {4 \, {\left (x \log \left (f\right ) - 2\right )} a b f^{\frac {1}{2} \, x}}{\log \left (f\right )^{2}} + \frac {2 \, a^{2} f^{\frac {1}{2} \, x}}{\log \left (f\right )} + \frac {2 \, {\left (x^{2} \log \left (f\right )^{2} - 4 \, x \log \left (f\right ) + 8\right )} b^{2} f^{\frac {1}{2} \, x}}{\log \left (f\right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*(x*log(f) - 2)*a*b*f^(1/2*x)/log(f)^2 + 2*a^2*f^(1/2*x)/log(f) + 2*(x^2*log(f)^2 - 4*x*log(f) + 8)*b^2*f^(1/
2*x)/log(f)^3

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

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Sympy [A]
time = 0.06, size = 94, normalized size = 1.68 \begin {gather*} \begin {cases} \frac {\left (2 a^{2} \log {\left (f \right )}^{2} + 4 a b x \log {\left (f \right )}^{2} - 8 a b \log {\left (f \right )} + 2 b^{2} x^{2} \log {\left (f \right )}^{2} - 8 b^{2} x \log {\left (f \right )} + 16 b^{2}\right ) \sqrt {f^{x}}}{\log {\left (f \right )}^{3}} & \text {for}\: \log {\left (f \right )}^{3} \neq 0 \\a^{2} x + a b x^{2} + \frac {b^{2} x^{3}}{3} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((2*a**2*log(f)**2 + 4*a*b*x*log(f)**2 - 8*a*b*log(f) + 2*b**2*x**2*log(f)**2 - 8*b**2*x*log(f) + 16
*b**2)*sqrt(f**x)/log(f)**3, Ne(log(f)**3, 0)), (a**2*x + a*b*x**2 + b**2*x**3/3, True))

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Giac [C] Result contains complex when optimal does not.
time = 5.52, size = 1392, normalized size = 24.86 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*((2*(pi*b^2*x^2*log(abs(f))*sgn(f) - pi*b^2*x^2*log(abs(f)) + 2*pi*a*b*x*log(abs(f))*sgn(f) - 2*pi*a*b*x*lo
g(abs(f)) - 2*pi*b^2*x*sgn(f) + pi*a^2*log(abs(f))*sgn(f) + 2*pi*b^2*x - pi*a^2*log(abs(f)) - 2*pi*a*b*sgn(f)
+ 2*pi*a*b)*(pi^3*sgn(f) - 3*pi*log(abs(f))^2*sgn(f) - pi^3 + 3*pi*log(abs(f))^2)/((pi^3*sgn(f) - 3*pi*log(abs
(f))^2*sgn(f) - pi^3 + 3*pi*log(abs(f))^2)^2 + (3*pi^2*log(abs(f))*sgn(f) - 3*pi^2*log(abs(f)) + 2*log(abs(f))
^3)^2) - (pi^2*b^2*x^2*sgn(f) - pi^2*b^2*x^2 + 2*b^2*x^2*log(abs(f))^2 + 2*pi^2*a*b*x*sgn(f) - 2*pi^2*a*b*x +
4*a*b*x*log(abs(f))^2 + pi^2*a^2*sgn(f) - pi^2*a^2 - 8*b^2*x*log(abs(f)) + 2*a^2*log(abs(f))^2 - 8*a*b*log(abs
(f)) + 16*b^2)*(3*pi^2*log(abs(f))*sgn(f) - 3*pi^2*log(abs(f)) + 2*log(abs(f))^3)/((pi^3*sgn(f) - 3*pi*log(abs
(f))^2*sgn(f) - pi^3 + 3*pi*log(abs(f))^2)^2 + (3*pi^2*log(abs(f))*sgn(f) - 3*pi^2*log(abs(f)) + 2*log(abs(f))
^3)^2))*cos(-1/4*pi*x*sgn(f) + 1/4*pi*x) - ((pi^2*b^2*x^2*sgn(f) - pi^2*b^2*x^2 + 2*b^2*x^2*log(abs(f))^2 + 2*
pi^2*a*b*x*sgn(f) - 2*pi^2*a*b*x + 4*a*b*x*log(abs(f))^2 + pi^2*a^2*sgn(f) - pi^2*a^2 - 8*b^2*x*log(abs(f)) +
2*a^2*log(abs(f))^2 - 8*a*b*log(abs(f)) + 16*b^2)*(pi^3*sgn(f) - 3*pi*log(abs(f))^2*sgn(f) - pi^3 + 3*pi*log(a
bs(f))^2)/((pi^3*sgn(f) - 3*pi*log(abs(f))^2*sgn(f) - pi^3 + 3*pi*log(abs(f))^2)^2 + (3*pi^2*log(abs(f))*sgn(f
) - 3*pi^2*log(abs(f)) + 2*log(abs(f))^3)^2) + 2*(pi*b^2*x^2*log(abs(f))*sgn(f) - pi*b^2*x^2*log(abs(f)) + 2*p
i*a*b*x*log(abs(f))*sgn(f) - 2*pi*a*b*x*log(abs(f)) - 2*pi*b^2*x*sgn(f) + pi*a^2*log(abs(f))*sgn(f) + 2*pi*b^2
*x - pi*a^2*log(abs(f)) - 2*pi*a*b*sgn(f) + 2*pi*a*b)*(3*pi^2*log(abs(f))*sgn(f) - 3*pi^2*log(abs(f)) + 2*log(
abs(f))^3)/((pi^3*sgn(f) - 3*pi*log(abs(f))^2*sgn(f) - pi^3 + 3*pi*log(abs(f))^2)^2 + (3*pi^2*log(abs(f))*sgn(
f) - 3*pi^2*log(abs(f)) + 2*log(abs(f))^3)^2))*sin(-1/4*pi*x*sgn(f) + 1/4*pi*x))*abs(f)^(1/2*x) - 4*I*abs(f)^(
1/2*x)*((-I*pi^2*b^2*x^2*sgn(f) + 2*pi*b^2*x^2*log(abs(f))*sgn(f) + I*pi^2*b^2*x^2 - 2*pi*b^2*x^2*log(abs(f))
- 2*I*b^2*x^2*log(abs(f))^2 - 2*I*pi^2*a*b*x*sgn(f) + 4*pi*a*b*x*log(abs(f))*sgn(f) + 2*I*pi^2*a*b*x - 4*pi*a*
b*x*log(abs(f)) - 4*I*a*b*x*log(abs(f))^2 - I*pi^2*a^2*sgn(f) - 4*pi*b^2*x*sgn(f) + 2*pi*a^2*log(abs(f))*sgn(f
) + I*pi^2*a^2 + 4*pi*b^2*x - 2*pi*a^2*log(abs(f)) + 8*I*b^2*x*log(abs(f)) - 2*I*a^2*log(abs(f))^2 - 4*pi*a*b*
sgn(f) + 4*pi*a*b + 8*I*a*b*log(abs(f)) - 16*I*b^2)*e^(1/4*I*pi*x*sgn(f) - 1/4*I*pi*x)/(-4*I*pi^3*sgn(f) + 12*
pi^2*log(abs(f))*sgn(f) + 12*I*pi*log(abs(f))^2*sgn(f) + 4*I*pi^3 - 12*pi^2*log(abs(f)) - 12*I*pi*log(abs(f))^
2 + 8*log(abs(f))^3) - (-I*pi^2*b^2*x^2*sgn(f) - 2*pi*b^2*x^2*log(abs(f))*sgn(f) + I*pi^2*b^2*x^2 + 2*pi*b^2*x
^2*log(abs(f)) - 2*I*b^2*x^2*log(abs(f))^2 - 2*I*pi^2*a*b*x*sgn(f) - 4*pi*a*b*x*log(abs(f))*sgn(f) + 2*I*pi^2*
a*b*x + 4*pi*a*b*x*log(abs(f)) - 4*I*a*b*x*log(abs(f))^2 - I*pi^2*a^2*sgn(f) + 4*pi*b^2*x*sgn(f) - 2*pi*a^2*lo
g(abs(f))*sgn(f) + I*pi^2*a^2 - 4*pi*b^2*x + 2*pi*a^2*log(abs(f)) + 8*I*b^2*x*log(abs(f)) - 2*I*a^2*log(abs(f)
)^2 + 4*pi*a*b*sgn(f) - 4*pi*a*b + 8*I*a*b*log(abs(f)) - 16*I*b^2)*e^(-1/4*I*pi*x*sgn(f) + 1/4*I*pi*x)/(4*I*pi
^3*sgn(f) + 12*pi^2*log(abs(f))*sgn(f) - 12*I*pi*log(abs(f))^2*sgn(f) - 4*I*pi^3 - 12*pi^2*log(abs(f)) + 12*I*
pi*log(abs(f))^2 + 8*log(abs(f))^3))

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Mupad [B]
time = 3.67, size = 62, normalized size = 1.11 \begin {gather*} \sqrt {f^x}\,\left (\frac {2\,a^2\,{\ln \left (f\right )}^2-8\,a\,b\,\ln \left (f\right )+16\,b^2}{{\ln \left (f\right )}^3}+\frac {2\,b^2\,x^2}{\ln \left (f\right )}-\frac {4\,b\,x\,\left (2\,b-a\,\ln \left (f\right )\right )}{{\ln \left (f\right )}^2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f^x)^(1/2)*(a + b*x)^2,x)

[Out]

(f^x)^(1/2)*((2*a^2*log(f)^2 + 16*b^2 - 8*a*b*log(f))/log(f)^3 + (2*b^2*x^2)/log(f) - (4*b*x*(2*b - a*log(f)))
/log(f)^2)

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