3.1.3 \(\int F^{c (a+b x)} \sin ^2(d+e x) \, dx\) [3]

Optimal. Leaf size=128 \[ \frac {2 e^2 F^{c (a+b x)}}{b c \log (F) \left (4 e^2+b^2 c^2 \log ^2(F)\right )}-\frac {2 e F^{c (a+b x)} \cos (d+e x) \sin (d+e x)}{4 e^2+b^2 c^2 \log ^2(F)}+\frac {b c F^{c (a+b x)} \log (F) \sin ^2(d+e x)}{4 e^2+b^2 c^2 \log ^2(F)} \]

[Out]

2*e^2*F^(c*(b*x+a))/b/c/ln(F)/(4*e^2+b^2*c^2*ln(F)^2)-2*e*F^(c*(b*x+a))*cos(e*x+d)*sin(e*x+d)/(4*e^2+b^2*c^2*l
n(F)^2)+b*c*F^(c*(b*x+a))*ln(F)*sin(e*x+d)^2/(4*e^2+b^2*c^2*ln(F)^2)

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Rubi [A]
time = 0.03, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {4519, 2225} \begin {gather*} \frac {b c \log (F) \sin ^2(d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+4 e^2}-\frac {2 e \sin (d+e x) \cos (d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+4 e^2}+\frac {2 e^2 F^{c (a+b x)}}{b c \log (F) \left (b^2 c^2 \log ^2(F)+4 e^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[F^(c*(a + b*x))*Sin[d + e*x]^2,x]

[Out]

(2*e^2*F^(c*(a + b*x)))/(b*c*Log[F]*(4*e^2 + b^2*c^2*Log[F]^2)) - (2*e*F^(c*(a + b*x))*Cos[d + e*x]*Sin[d + e*
x])/(4*e^2 + b^2*c^2*Log[F]^2) + (b*c*F^(c*(a + b*x))*Log[F]*Sin[d + e*x]^2)/(4*e^2 + b^2*c^2*Log[F]^2)

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 4519

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sin[(d_.) + (e_.)*(x_)]^(n_), x_Symbol] :> Simp[b*c*Log[F]*F^(c*(a + b*x
))*(Sin[d + e*x]^n/(e^2*n^2 + b^2*c^2*Log[F]^2)), x] + (Dist[(n*(n - 1)*e^2)/(e^2*n^2 + b^2*c^2*Log[F]^2), Int
[F^(c*(a + b*x))*Sin[d + e*x]^(n - 2), x], x] - Simp[e*n*F^(c*(a + b*x))*Cos[d + e*x]*(Sin[d + e*x]^(n - 1)/(e
^2*n^2 + b^2*c^2*Log[F]^2)), x]) /; FreeQ[{F, a, b, c, d, e}, x] && NeQ[e^2*n^2 + b^2*c^2*Log[F]^2, 0] && GtQ[
n, 1]

Rubi steps

\begin {align*} \int F^{c (a+b x)} \sin ^2(d+e x) \, dx &=-\frac {2 e F^{c (a+b x)} \cos (d+e x) \sin (d+e x)}{4 e^2+b^2 c^2 \log ^2(F)}+\frac {b c F^{c (a+b x)} \log (F) \sin ^2(d+e x)}{4 e^2+b^2 c^2 \log ^2(F)}+\frac {\left (2 e^2\right ) \int F^{c (a+b x)} \, dx}{4 e^2+b^2 c^2 \log ^2(F)}\\ &=\frac {2 e^2 F^{c (a+b x)}}{b c \log (F) \left (4 e^2+b^2 c^2 \log ^2(F)\right )}-\frac {2 e F^{c (a+b x)} \cos (d+e x) \sin (d+e x)}{4 e^2+b^2 c^2 \log ^2(F)}+\frac {b c F^{c (a+b x)} \log (F) \sin ^2(d+e x)}{4 e^2+b^2 c^2 \log ^2(F)}\\ \end {align*}

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Mathematica [A]
time = 0.21, size = 86, normalized size = 0.67 \begin {gather*} \frac {F^{c (a+b x)} \left (4 e^2+b^2 c^2 \log ^2(F)-b^2 c^2 \cos (2 (d+e x)) \log ^2(F)-2 b c e \log (F) \sin (2 (d+e x))\right )}{8 b c e^2 \log (F)+2 b^3 c^3 \log ^3(F)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[F^(c*(a + b*x))*Sin[d + e*x]^2,x]

[Out]

(F^(c*(a + b*x))*(4*e^2 + b^2*c^2*Log[F]^2 - b^2*c^2*Cos[2*(d + e*x)]*Log[F]^2 - 2*b*c*e*Log[F]*Sin[2*(d + e*x
)]))/(8*b*c*e^2*Log[F] + 2*b^3*c^3*Log[F]^3)

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Maple [A]
time = 0.28, size = 106, normalized size = 0.83

method	result	size
risch	\(\frac {F^{c \left (b x +a \right )}}{2 b c \ln \left (F \right )}-\frac {\ln \left (F \right ) c b \,F^{c \left (b x +a \right )} \cos \left (2 e x +2 d \right )}{2 \left (4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}-\frac {e \,F^{c \left (b x +a \right )} \sin \left (2 e x +2 d \right )}{4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}\)	\(106\)
norman	\(\frac {-\frac {4 e \,{\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )}{4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {4 e \,{\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \left (\tan ^{3}\left (\frac {d}{2}+\frac {e x}{2}\right )\right )}{4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {2 e^{2} {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{b c \ln \left (F \right ) \left (4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}+\frac {2 e^{2} {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \left (\tan ^{4}\left (\frac {d}{2}+\frac {e x}{2}\right )\right )}{b c \ln \left (F \right ) \left (4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}+\frac {4 \left (e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \left (\tan ^{2}\left (\frac {d}{2}+\frac {e x}{2}\right )\right )}{b c \ln \left (F \right ) \left (4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}}{\left (1+\tan ^{2}\left (\frac {d}{2}+\frac {e x}{2}\right )\right )^{2}}\)	\(268\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(c*(b*x+a))*sin(e*x+d)^2,x,method=_RETURNVERBOSE)

[Out]

1/2*F^(c*(b*x+a))/b/c/ln(F)-1/2/(4*e^2+b^2*c^2*ln(F)^2)*ln(F)*c*b*F^(c*(b*x+a))*cos(2*e*x+2*d)-e*F^(c*(b*x+a))
/(4*e^2+b^2*c^2*ln(F)^2)*sin(2*e*x+2*d)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 348 vs. \(2 (128) = 256\).
time = 0.29, size = 348, normalized size = 2.72 \begin {gather*} -\frac {{\left (F^{a c} b^{2} c^{2} \cos \left (2 \, d\right ) \log \left (F\right )^{2} + 2 \, F^{a c} b c e \log \left (F\right ) \sin \left (2 \, d\right )\right )} F^{b c x} \cos \left (2 \, x e\right ) + {\left (F^{a c} b^{2} c^{2} \cos \left (2 \, d\right ) \log \left (F\right )^{2} - 2 \, F^{a c} b c e \log \left (F\right ) \sin \left (2 \, d\right )\right )} F^{b c x} \cos \left (2 \, x e + 4 \, d\right ) - {\left (F^{a c} b^{2} c^{2} \log \left (F\right )^{2} \sin \left (2 \, d\right ) - 2 \, F^{a c} b c \cos \left (2 \, d\right ) e \log \left (F\right )\right )} F^{b c x} \sin \left (2 \, x e\right ) + {\left (F^{a c} b^{2} c^{2} \log \left (F\right )^{2} \sin \left (2 \, d\right ) + 2 \, F^{a c} b c \cos \left (2 \, d\right ) e \log \left (F\right )\right )} F^{b c x} \sin \left (2 \, x e + 4 \, d\right ) - 2 \, {\left ({\left (F^{a c} b^{2} c^{2} \log \left (F\right )^{2} + 4 \, F^{a c} e^{2}\right )} \cos \left (2 \, d\right )^{2} + {\left (F^{a c} b^{2} c^{2} \log \left (F\right )^{2} + 4 \, F^{a c} e^{2}\right )} \sin \left (2 \, d\right )^{2}\right )} F^{b c x}}{4 \, {\left ({\left (b^{3} c^{3} \log \left (F\right )^{3} + 4 \, b c e^{2} \log \left (F\right )\right )} \cos \left (2 \, d\right )^{2} + {\left (b^{3} c^{3} \log \left (F\right )^{3} + 4 \, b c e^{2} \log \left (F\right )\right )} \sin \left (2 \, d\right )^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))*sin(e*x+d)^2,x, algorithm="maxima")

[Out]

-1/4*((F^(a*c)*b^2*c^2*cos(2*d)*log(F)^2 + 2*F^(a*c)*b*c*e*log(F)*sin(2*d))*F^(b*c*x)*cos(2*x*e) + (F^(a*c)*b^
2*c^2*cos(2*d)*log(F)^2 - 2*F^(a*c)*b*c*e*log(F)*sin(2*d))*F^(b*c*x)*cos(2*x*e + 4*d) - (F^(a*c)*b^2*c^2*log(F
)^2*sin(2*d) - 2*F^(a*c)*b*c*cos(2*d)*e*log(F))*F^(b*c*x)*sin(2*x*e) + (F^(a*c)*b^2*c^2*log(F)^2*sin(2*d) + 2*
F^(a*c)*b*c*cos(2*d)*e*log(F))*F^(b*c*x)*sin(2*x*e + 4*d) - 2*((F^(a*c)*b^2*c^2*log(F)^2 + 4*F^(a*c)*e^2)*cos(
2*d)^2 + (F^(a*c)*b^2*c^2*log(F)^2 + 4*F^(a*c)*e^2)*sin(2*d)^2)*F^(b*c*x))/((b^3*c^3*log(F)^3 + 4*b*c*e^2*log(
F))*cos(2*d)^2 + (b^3*c^3*log(F)^3 + 4*b*c*e^2*log(F))*sin(2*d)^2)

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Fricas [A]
time = 2.77, size = 91, normalized size = 0.71 \begin {gather*} -\frac {{\left (2 \, b c \cos \left (x e + d\right ) e \log \left (F\right ) \sin \left (x e + d\right ) + {\left (b^{2} c^{2} \cos \left (x e + d\right )^{2} - b^{2} c^{2}\right )} \log \left (F\right )^{2} - 2 \, e^{2}\right )} F^{b c x + a c}}{b^{3} c^{3} \log \left (F\right )^{3} + 4 \, b c e^{2} \log \left (F\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))*sin(e*x+d)^2,x, algorithm="fricas")

[Out]

-(2*b*c*cos(x*e + d)*e*log(F)*sin(x*e + d) + (b^2*c^2*cos(x*e + d)^2 - b^2*c^2)*log(F)^2 - 2*e^2)*F^(b*c*x + a
*c)/(b^3*c^3*log(F)^3 + 4*b*c*e^2*log(F))

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Sympy [C] Result contains complex when optimal does not.
time = 15.47, size = 1117, normalized size = 8.73 \begin {gather*} \begin {cases} \frac {x \sin ^{2}{\left (d + e x \right )}}{2} + \frac {x \cos ^{2}{\left (d + e x \right )}}{2} - \frac {\sin {\left (d + e x \right )} \cos {\left (d + e x \right )}}{2 e} & \text {for}\: F = 1 \\\frac {b^{2} c^{2} \left (e^{- \frac {2 i e}{b c}}\right )^{a c} \left (e^{- \frac {2 i e}{b c}}\right )^{b c x} \log {\left (e^{- \frac {2 i e}{b c}} \right )}^{2} \sin ^{2}{\left (d + e x \right )}}{b^{3} c^{3} \log {\left (e^{- \frac {2 i e}{b c}} \right )}^{3} + 4 b c e^{2} \log {\left (e^{- \frac {2 i e}{b c}} \right )}} - \frac {2 b c e \left (e^{- \frac {2 i e}{b c}}\right )^{a c} \left (e^{- \frac {2 i e}{b c}}\right )^{b c x} \log {\left (e^{- \frac {2 i e}{b c}} \right )} \sin {\left (d + e x \right )} \cos {\left (d + e x \right )}}{b^{3} c^{3} \log {\left (e^{- \frac {2 i e}{b c}} \right )}^{3} + 4 b c e^{2} \log {\left (e^{- \frac {2 i e}{b c}} \right )}} + \frac {2 e^{2} \left (e^{- \frac {2 i e}{b c}}\right )^{a c} \left (e^{- \frac {2 i e}{b c}}\right )^{b c x} \sin ^{2}{\left (d + e x \right )}}{b^{3} c^{3} \log {\left (e^{- \frac {2 i e}{b c}} \right )}^{3} + 4 b c e^{2} \log {\left (e^{- \frac {2 i e}{b c}} \right )}} + \frac {2 e^{2} \left (e^{- \frac {2 i e}{b c}}\right )^{a c} \left (e^{- \frac {2 i e}{b c}}\right )^{b c x} \cos ^{2}{\left (d + e x \right )}}{b^{3} c^{3} \log {\left (e^{- \frac {2 i e}{b c}} \right )}^{3} + 4 b c e^{2} \log {\left (e^{- \frac {2 i e}{b c}} \right )}} & \text {for}\: F = e^{- \frac {2 i e}{b c}} \\\frac {b^{2} c^{2} \left (e^{\frac {2 i e}{b c}}\right )^{a c} \left (e^{\frac {2 i e}{b c}}\right )^{b c x} \log {\left (e^{\frac {2 i e}{b c}} \right )}^{2} \sin ^{2}{\left (d + e x \right )}}{b^{3} c^{3} \log {\left (e^{\frac {2 i e}{b c}} \right )}^{3} + 4 b c e^{2} \log {\left (e^{\frac {2 i e}{b c}} \right )}} - \frac {2 b c e \left (e^{\frac {2 i e}{b c}}\right )^{a c} \left (e^{\frac {2 i e}{b c}}\right )^{b c x} \log {\left (e^{\frac {2 i e}{b c}} \right )} \sin {\left (d + e x \right )} \cos {\left (d + e x \right )}}{b^{3} c^{3} \log {\left (e^{\frac {2 i e}{b c}} \right )}^{3} + 4 b c e^{2} \log {\left (e^{\frac {2 i e}{b c}} \right )}} + \frac {2 e^{2} \left (e^{\frac {2 i e}{b c}}\right )^{a c} \left (e^{\frac {2 i e}{b c}}\right )^{b c x} \sin ^{2}{\left (d + e x \right )}}{b^{3} c^{3} \log {\left (e^{\frac {2 i e}{b c}} \right )}^{3} + 4 b c e^{2} \log {\left (e^{\frac {2 i e}{b c}} \right )}} + \frac {2 e^{2} \left (e^{\frac {2 i e}{b c}}\right )^{a c} \left (e^{\frac {2 i e}{b c}}\right )^{b c x} \cos ^{2}{\left (d + e x \right )}}{b^{3} c^{3} \log {\left (e^{\frac {2 i e}{b c}} \right )}^{3} + 4 b c e^{2} \log {\left (e^{\frac {2 i e}{b c}} \right )}} & \text {for}\: F = e^{\frac {2 i e}{b c}} \\F^{a c} \left (\frac {x \sin ^{2}{\left (d + e x \right )}}{2} + \frac {x \cos ^{2}{\left (d + e x \right )}}{2} - \frac {\sin {\left (d + e x \right )} \cos {\left (d + e x \right )}}{2 e}\right ) & \text {for}\: b = 0 \\\frac {x \sin ^{2}{\left (d + e x \right )}}{2} + \frac {x \cos ^{2}{\left (d + e x \right )}}{2} - \frac {\sin {\left (d + e x \right )} \cos {\left (d + e x \right )}}{2 e} & \text {for}\: c = 0 \\\frac {F^{a c} F^{b c x} b^{2} c^{2} \log {\left (F \right )}^{2} \sin ^{2}{\left (d + e x \right )}}{b^{3} c^{3} \log {\left (F \right )}^{3} + 4 b c e^{2} \log {\left (F \right )}} - \frac {2 F^{a c} F^{b c x} b c e \log {\left (F \right )} \sin {\left (d + e x \right )} \cos {\left (d + e x \right )}}{b^{3} c^{3} \log {\left (F \right )}^{3} + 4 b c e^{2} \log {\left (F \right )}} + \frac {2 F^{a c} F^{b c x} e^{2} \sin ^{2}{\left (d + e x \right )}}{b^{3} c^{3} \log {\left (F \right )}^{3} + 4 b c e^{2} \log {\left (F \right )}} + \frac {2 F^{a c} F^{b c x} e^{2} \cos ^{2}{\left (d + e x \right )}}{b^{3} c^{3} \log {\left (F \right )}^{3} + 4 b c e^{2} \log {\left (F \right )}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(c*(b*x+a))*sin(e*x+d)**2,x)

[Out]

Piecewise((x*sin(d + e*x)**2/2 + x*cos(d + e*x)**2/2 - sin(d + e*x)*cos(d + e*x)/(2*e), Eq(F, 1)), (b**2*c**2*
exp(-2*I*e/(b*c))**(a*c)*exp(-2*I*e/(b*c))**(b*c*x)*log(exp(-2*I*e/(b*c)))**2*sin(d + e*x)**2/(b**3*c**3*log(e
xp(-2*I*e/(b*c)))**3 + 4*b*c*e**2*log(exp(-2*I*e/(b*c)))) - 2*b*c*e*exp(-2*I*e/(b*c))**(a*c)*exp(-2*I*e/(b*c))
**(b*c*x)*log(exp(-2*I*e/(b*c)))*sin(d + e*x)*cos(d + e*x)/(b**3*c**3*log(exp(-2*I*e/(b*c)))**3 + 4*b*c*e**2*l
og(exp(-2*I*e/(b*c)))) + 2*e**2*exp(-2*I*e/(b*c))**(a*c)*exp(-2*I*e/(b*c))**(b*c*x)*sin(d + e*x)**2/(b**3*c**3
*log(exp(-2*I*e/(b*c)))**3 + 4*b*c*e**2*log(exp(-2*I*e/(b*c)))) + 2*e**2*exp(-2*I*e/(b*c))**(a*c)*exp(-2*I*e/(
b*c))**(b*c*x)*cos(d + e*x)**2/(b**3*c**3*log(exp(-2*I*e/(b*c)))**3 + 4*b*c*e**2*log(exp(-2*I*e/(b*c)))), Eq(F
, exp(-2*I*e/(b*c)))), (b**2*c**2*exp(2*I*e/(b*c))**(a*c)*exp(2*I*e/(b*c))**(b*c*x)*log(exp(2*I*e/(b*c)))**2*s
in(d + e*x)**2/(b**3*c**3*log(exp(2*I*e/(b*c)))**3 + 4*b*c*e**2*log(exp(2*I*e/(b*c)))) - 2*b*c*e*exp(2*I*e/(b*
c))**(a*c)*exp(2*I*e/(b*c))**(b*c*x)*log(exp(2*I*e/(b*c)))*sin(d + e*x)*cos(d + e*x)/(b**3*c**3*log(exp(2*I*e/
(b*c)))**3 + 4*b*c*e**2*log(exp(2*I*e/(b*c)))) + 2*e**2*exp(2*I*e/(b*c))**(a*c)*exp(2*I*e/(b*c))**(b*c*x)*sin(
d + e*x)**2/(b**3*c**3*log(exp(2*I*e/(b*c)))**3 + 4*b*c*e**2*log(exp(2*I*e/(b*c)))) + 2*e**2*exp(2*I*e/(b*c))*
*(a*c)*exp(2*I*e/(b*c))**(b*c*x)*cos(d + e*x)**2/(b**3*c**3*log(exp(2*I*e/(b*c)))**3 + 4*b*c*e**2*log(exp(2*I*
e/(b*c)))), Eq(F, exp(2*I*e/(b*c)))), (F**(a*c)*(x*sin(d + e*x)**2/2 + x*cos(d + e*x)**2/2 - sin(d + e*x)*cos(
d + e*x)/(2*e)), Eq(b, 0)), (x*sin(d + e*x)**2/2 + x*cos(d + e*x)**2/2 - sin(d + e*x)*cos(d + e*x)/(2*e), Eq(c
, 0)), (F**(a*c)*F**(b*c*x)*b**2*c**2*log(F)**2*sin(d + e*x)**2/(b**3*c**3*log(F)**3 + 4*b*c*e**2*log(F)) - 2*
F**(a*c)*F**(b*c*x)*b*c*e*log(F)*sin(d + e*x)*cos(d + e*x)/(b**3*c**3*log(F)**3 + 4*b*c*e**2*log(F)) + 2*F**(a
*c)*F**(b*c*x)*e**2*sin(d + e*x)**2/(b**3*c**3*log(F)**3 + 4*b*c*e**2*log(F)) + 2*F**(a*c)*F**(b*c*x)*e**2*cos
(d + e*x)**2/(b**3*c**3*log(F)**3 + 4*b*c*e**2*log(F)), True))

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Giac [C] Result contains complex when optimal does not.
time = 0.45, size = 915, normalized size = 7.15 \begin {gather*} -\frac {1}{2} \, {\left (\frac {2 \, b c \cos \left (\frac {1}{2} \, \pi b c x \mathrm {sgn}\left (F\right ) - \frac {1}{2} \, \pi b c x + \frac {1}{2} \, \pi a c \mathrm {sgn}\left (F\right ) - \frac {1}{2} \, \pi a c + 2 \, e x + 2 \, d\right ) \log \left ({\left | F \right |}\right )}{4 \, b^{2} c^{2} \log \left ({\left | F \right |}\right )^{2} + {\left (\pi b c \mathrm {sgn}\left (F\right ) - \pi b c + 4 \, e\right )}^{2}} + \frac {{\left (\pi b c \mathrm {sgn}\left (F\right ) - \pi b c + 4 \, e\right )} \sin \left (\frac {1}{2} \, \pi b c x \mathrm {sgn}\left (F\right ) - \frac {1}{2} \, \pi b c x + \frac {1}{2} \, \pi a c \mathrm {sgn}\left (F\right ) - \frac {1}{2} \, \pi a c + 2 \, e x + 2 \, d\right )}{4 \, b^{2} c^{2} \log \left ({\left | F \right |}\right )^{2} + {\left (\pi b c \mathrm {sgn}\left (F\right ) - \pi b c + 4 \, e\right )}^{2}}\right )} e^{\left (b c x \log \left ({\left | F \right |}\right ) + a c \log \left ({\left | F \right |}\right )\right )} - \frac {1}{2} \, {\left (\frac {2 \, b c \cos \left (\frac {1}{2} \, \pi b c x \mathrm {sgn}\left (F\right ) - \frac {1}{2} \, \pi b c x + \frac {1}{2} \, \pi a c \mathrm {sgn}\left (F\right ) - \frac {1}{2} \, \pi a c - 2 \, e x - 2 \, d\right ) \log \left ({\left | F \right |}\right )}{4 \, b^{2} c^{2} \log \left ({\left | F \right |}\right )^{2} + {\left (\pi b c \mathrm {sgn}\left (F\right ) - \pi b c - 4 \, e\right )}^{2}} + \frac {{\left (\pi b c \mathrm {sgn}\left (F\right ) - \pi b c - 4 \, e\right )} \sin \left (\frac {1}{2} \, \pi b c x \mathrm {sgn}\left (F\right ) - \frac {1}{2} \, \pi b c x + \frac {1}{2} \, \pi a c \mathrm {sgn}\left (F\right ) - \frac {1}{2} \, \pi a c - 2 \, e x - 2 \, d\right )}{4 \, b^{2} c^{2} \log \left ({\left | F \right |}\right )^{2} + {\left (\pi b c \mathrm {sgn}\left (F\right ) - \pi b c - 4 \, e\right )}^{2}}\right )} e^{\left (b c x \log \left ({\left | F \right |}\right ) + a c \log \left ({\left | F \right |}\right )\right )} + {\left (\frac {2 \, b c \cos \left (-\frac {1}{2} \, \pi b c x \mathrm {sgn}\left (F\right ) + \frac {1}{2} \, \pi b c x - \frac {1}{2} \, \pi a c \mathrm {sgn}\left (F\right ) + \frac {1}{2} \, \pi a c\right ) \log \left ({\left | F \right |}\right )}{4 \, b^{2} c^{2} \log \left ({\left | F \right |}\right )^{2} + {\left (\pi b c \mathrm {sgn}\left (F\right ) - \pi b c\right )}^{2}} - \frac {{\left (\pi b c \mathrm {sgn}\left (F\right ) - \pi b c\right )} \sin \left (-\frac {1}{2} \, \pi b c x \mathrm {sgn}\left (F\right ) + \frac {1}{2} \, \pi b c x - \frac {1}{2} \, \pi a c \mathrm {sgn}\left (F\right ) + \frac {1}{2} \, \pi a c\right )}{4 \, b^{2} c^{2} \log \left ({\left | F \right |}\right )^{2} + {\left (\pi b c \mathrm {sgn}\left (F\right ) - \pi b c\right )}^{2}}\right )} e^{\left (b c x \log \left ({\left | F \right |}\right ) + a c \log \left ({\left | F \right |}\right )\right )} + i \, {\left (-\frac {i \, e^{\left (\frac {1}{2} i \, \pi b c x \mathrm {sgn}\left (F\right ) - \frac {1}{2} i \, \pi b c x + \frac {1}{2} i \, \pi a c \mathrm {sgn}\left (F\right ) - \frac {1}{2} i \, \pi a c + 2 i \, e x + 2 i \, d\right )}}{4 i \, \pi b c \mathrm {sgn}\left (F\right ) - 4 i \, \pi b c + 8 \, b c \log \left ({\left | F \right |}\right ) + 16 i \, e} + \frac {i \, e^{\left (-\frac {1}{2} i \, \pi b c x \mathrm {sgn}\left (F\right ) + \frac {1}{2} i \, \pi b c x - \frac {1}{2} i \, \pi a c \mathrm {sgn}\left (F\right ) + \frac {1}{2} i \, \pi a c - 2 i \, e x - 2 i \, d\right )}}{-4 i \, \pi b c \mathrm {sgn}\left (F\right ) + 4 i \, \pi b c + 8 \, b c \log \left ({\left | F \right |}\right ) - 16 i \, e}\right )} e^{\left (b c x \log \left ({\left | F \right |}\right ) + a c \log \left ({\left | F \right |}\right )\right )} + i \, {\left (-\frac {i \, e^{\left (\frac {1}{2} i \, \pi b c x \mathrm {sgn}\left (F\right ) - \frac {1}{2} i \, \pi b c x + \frac {1}{2} i \, \pi a c \mathrm {sgn}\left (F\right ) - \frac {1}{2} i \, \pi a c - 2 i \, e x - 2 i \, d\right )}}{4 i \, \pi b c \mathrm {sgn}\left (F\right ) - 4 i \, \pi b c + 8 \, b c \log \left ({\left | F \right |}\right ) - 16 i \, e} + \frac {i \, e^{\left (-\frac {1}{2} i \, \pi b c x \mathrm {sgn}\left (F\right ) + \frac {1}{2} i \, \pi b c x - \frac {1}{2} i \, \pi a c \mathrm {sgn}\left (F\right ) + \frac {1}{2} i \, \pi a c + 2 i \, e x + 2 i \, d\right )}}{-4 i \, \pi b c \mathrm {sgn}\left (F\right ) + 4 i \, \pi b c + 8 \, b c \log \left ({\left | F \right |}\right ) + 16 i \, e}\right )} e^{\left (b c x \log \left ({\left | F \right |}\right ) + a c \log \left ({\left | F \right |}\right )\right )} + i \, {\left (\frac {i \, e^{\left (\frac {1}{2} i \, \pi b c x \mathrm {sgn}\left (F\right ) - \frac {1}{2} i \, \pi b c x + \frac {1}{2} i \, \pi a c \mathrm {sgn}\left (F\right ) - \frac {1}{2} i \, \pi a c\right )}}{2 i \, \pi b c \mathrm {sgn}\left (F\right ) - 2 i \, \pi b c + 4 \, b c \log \left ({\left | F \right |}\right )} - \frac {i \, e^{\left (-\frac {1}{2} i \, \pi b c x \mathrm {sgn}\left (F\right ) + \frac {1}{2} i \, \pi b c x - \frac {1}{2} i \, \pi a c \mathrm {sgn}\left (F\right ) + \frac {1}{2} i \, \pi a c\right )}}{-2 i \, \pi b c \mathrm {sgn}\left (F\right ) + 2 i \, \pi b c + 4 \, b c \log \left ({\left | F \right |}\right )}\right )} e^{\left (b c x \log \left ({\left | F \right |}\right ) + a c \log \left ({\left | F \right |}\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))*sin(e*x+d)^2,x, algorithm="giac")

[Out]

-1/2*(2*b*c*cos(1/2*pi*b*c*x*sgn(F) - 1/2*pi*b*c*x + 1/2*pi*a*c*sgn(F) - 1/2*pi*a*c + 2*e*x + 2*d)*log(abs(F))
/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F) - pi*b*c + 4*e)^2) + (pi*b*c*sgn(F) - pi*b*c + 4*e)*sin(1/2*pi*b*c*
x*sgn(F) - 1/2*pi*b*c*x + 1/2*pi*a*c*sgn(F) - 1/2*pi*a*c + 2*e*x + 2*d)/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn
(F) - pi*b*c + 4*e)^2))*e^(b*c*x*log(abs(F)) + a*c*log(abs(F))) - 1/2*(2*b*c*cos(1/2*pi*b*c*x*sgn(F) - 1/2*pi*
b*c*x + 1/2*pi*a*c*sgn(F) - 1/2*pi*a*c - 2*e*x - 2*d)*log(abs(F))/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F) -
pi*b*c - 4*e)^2) + (pi*b*c*sgn(F) - pi*b*c - 4*e)*sin(1/2*pi*b*c*x*sgn(F) - 1/2*pi*b*c*x + 1/2*pi*a*c*sgn(F) -
1/2*pi*a*c - 2*e*x - 2*d)/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F) - pi*b*c - 4*e)^2))*e^(b*c*x*log(abs(F))
+ a*c*log(abs(F))) + (2*b*c*cos(-1/2*pi*b*c*x*sgn(F) + 1/2*pi*b*c*x - 1/2*pi*a*c*sgn(F) + 1/2*pi*a*c)*log(abs(
F))/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F) - pi*b*c)^2) - (pi*b*c*sgn(F) - pi*b*c)*sin(-1/2*pi*b*c*x*sgn(F)
+ 1/2*pi*b*c*x - 1/2*pi*a*c*sgn(F) + 1/2*pi*a*c)/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F) - pi*b*c)^2))*e^(b
*c*x*log(abs(F)) + a*c*log(abs(F))) + I*(-I*e^(1/2*I*pi*b*c*x*sgn(F) - 1/2*I*pi*b*c*x + 1/2*I*pi*a*c*sgn(F) -
1/2*I*pi*a*c + 2*I*e*x + 2*I*d)/(4*I*pi*b*c*sgn(F) - 4*I*pi*b*c + 8*b*c*log(abs(F)) + 16*I*e) + I*e^(-1/2*I*pi
*b*c*x*sgn(F) + 1/2*I*pi*b*c*x - 1/2*I*pi*a*c*sgn(F) + 1/2*I*pi*a*c - 2*I*e*x - 2*I*d)/(-4*I*pi*b*c*sgn(F) + 4
*I*pi*b*c + 8*b*c*log(abs(F)) - 16*I*e))*e^(b*c*x*log(abs(F)) + a*c*log(abs(F))) + I*(-I*e^(1/2*I*pi*b*c*x*sgn
(F) - 1/2*I*pi*b*c*x + 1/2*I*pi*a*c*sgn(F) - 1/2*I*pi*a*c - 2*I*e*x - 2*I*d)/(4*I*pi*b*c*sgn(F) - 4*I*pi*b*c +
8*b*c*log(abs(F)) - 16*I*e) + I*e^(-1/2*I*pi*b*c*x*sgn(F) + 1/2*I*pi*b*c*x - 1/2*I*pi*a*c*sgn(F) + 1/2*I*pi*a
*c + 2*I*e*x + 2*I*d)/(-4*I*pi*b*c*sgn(F) + 4*I*pi*b*c + 8*b*c*log(abs(F)) + 16*I*e))*e^(b*c*x*log(abs(F)) + a
*c*log(abs(F))) + I*(I*e^(1/2*I*pi*b*c*x*sgn(F) - 1/2*I*pi*b*c*x + 1/2*I*pi*a*c*sgn(F) - 1/2*I*pi*a*c)/(2*I*pi
*b*c*sgn(F) - 2*I*pi*b*c + 4*b*c*log(abs(F))) - I*e^(-1/2*I*pi*b*c*x*sgn(F) + 1/2*I*pi*b*c*x - 1/2*I*pi*a*c*sg
n(F) + 1/2*I*pi*a*c)/(-2*I*pi*b*c*sgn(F) + 2*I*pi*b*c + 4*b*c*log(abs(F))))*e^(b*c*x*log(abs(F)) + a*c*log(abs
(F)))

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Mupad [B]
time = 3.02, size = 95, normalized size = 0.74 \begin {gather*} \frac {F^{a\,c+b\,c\,x}\,\left (2\,e^2+\frac {b^2\,c^2\,{\ln \left (F\right )}^2}{2}-\frac {b^2\,c^2\,{\ln \left (F\right )}^2\,\cos \left (2\,d+2\,e\,x\right )}{2}-b\,c\,e\,\ln \left (F\right )\,\sin \left (2\,d+2\,e\,x\right )\right )}{b\,c\,\ln \left (F\right )\,\left (b^2\,c^2\,{\ln \left (F\right )}^2+4\,e^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(c*(a + b*x))*sin(d + e*x)^2,x)

[Out]

(F^(a*c + b*c*x)*(2*e^2 + (b^2*c^2*log(F)^2)/2 - (b^2*c^2*log(F)^2*cos(2*d + 2*e*x))/2 - b*c*e*log(F)*sin(2*d
+ 2*e*x)))/(b*c*log(F)*(4*e^2 + b^2*c^2*log(F)^2))

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