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

Optimal. Leaf size=199 \[ -\frac {6 e^3 F^{c (a+b x)} \cos (d+e x)}{9 e^4+10 b^2 c^2 e^2 \log ^2(F)+b^4 c^4 \log ^4(F)}+\frac {6 b c e^2 F^{c (a+b x)} \log (F) \sin (d+e x)}{9 e^4+10 b^2 c^2 e^2 \log ^2(F)+b^4 c^4 \log ^4(F)}-\frac {3 e F^{c (a+b x)} \cos (d+e x) \sin ^2(d+e x)}{9 e^2+b^2 c^2 \log ^2(F)}+\frac {b c F^{c (a+b x)} \log (F) \sin ^3(d+e x)}{9 e^2+b^2 c^2 \log ^2(F)} \]

[Out]

-6*e^3*F^(c*(b*x+a))*cos(e*x+d)/(9*e^4+10*b^2*c^2*e^2*ln(F)^2+b^4*c^4*ln(F)^4)+6*b*c*e^2*F^(c*(b*x+a))*ln(F)*s
in(e*x+d)/(9*e^4+10*b^2*c^2*e^2*ln(F)^2+b^4*c^4*ln(F)^4)-3*e*F^(c*(b*x+a))*cos(e*x+d)*sin(e*x+d)^2/(9*e^2+b^2*
c^2*ln(F)^2)+b*c*F^(c*(b*x+a))*ln(F)*sin(e*x+d)^3/(9*e^2+b^2*c^2*ln(F)^2)

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Rubi [A]
time = 0.05, antiderivative size = 199, 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, 4517} \begin {gather*} \frac {b c \log (F) \sin ^3(d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+9 e^2}-\frac {3 e \sin ^2(d+e x) \cos (d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+9 e^2}+\frac {6 b c e^2 \log (F) \sin (d+e x) F^{c (a+b x)}}{b^4 c^4 \log ^4(F)+10 b^2 c^2 e^2 \log ^2(F)+9 e^4}-\frac {6 e^3 \cos (d+e x) F^{c (a+b x)}}{b^4 c^4 \log ^4(F)+10 b^2 c^2 e^2 \log ^2(F)+9 e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[F^(c*(a + b*x))*Sin[d + e*x]^3,x]

[Out]

(-6*e^3*F^(c*(a + b*x))*Cos[d + e*x])/(9*e^4 + 10*b^2*c^2*e^2*Log[F]^2 + b^4*c^4*Log[F]^4) + (6*b*c*e^2*F^(c*(
a + b*x))*Log[F]*Sin[d + e*x])/(9*e^4 + 10*b^2*c^2*e^2*Log[F]^2 + b^4*c^4*Log[F]^4) - (3*e*F^(c*(a + b*x))*Cos
[d + e*x]*Sin[d + e*x]^2)/(9*e^2 + b^2*c^2*Log[F]^2) + (b*c*F^(c*(a + b*x))*Log[F]*Sin[d + e*x]^3)/(9*e^2 + b^
2*c^2*Log[F]^2)

Rule 4517

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

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 ^3(d+e x) \, dx &=-\frac {3 e F^{c (a+b x)} \cos (d+e x) \sin ^2(d+e x)}{9 e^2+b^2 c^2 \log ^2(F)}+\frac {b c F^{c (a+b x)} \log (F) \sin ^3(d+e x)}{9 e^2+b^2 c^2 \log ^2(F)}+\frac {\left (6 e^2\right ) \int F^{c (a+b x)} \sin (d+e x) \, dx}{9 e^2+b^2 c^2 \log ^2(F)}\\ &=-\frac {6 e^3 F^{c (a+b x)} \cos (d+e x)}{9 e^4+10 b^2 c^2 e^2 \log ^2(F)+b^4 c^4 \log ^4(F)}+\frac {6 b c e^2 F^{c (a+b x)} \log (F) \sin (d+e x)}{9 e^4+10 b^2 c^2 e^2 \log ^2(F)+b^4 c^4 \log ^4(F)}-\frac {3 e F^{c (a+b x)} \cos (d+e x) \sin ^2(d+e x)}{9 e^2+b^2 c^2 \log ^2(F)}+\frac {b c F^{c (a+b x)} \log (F) \sin ^3(d+e x)}{9 e^2+b^2 c^2 \log ^2(F)}\\ \end {align*}

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Mathematica [A]
time = 0.70, size = 154, normalized size = 0.77 \begin {gather*} \frac {F^{c (a+b x)} \left (-3 e \cos (d+e x) \left (9 e^2+b^2 c^2 \log ^2(F)\right )+3 \cos (3 (d+e x)) \left (e^3+b^2 c^2 e \log ^2(F)\right )-2 b c \log (F) \left (-13 e^2-b^2 c^2 \log ^2(F)+\cos (2 (d+e x)) \left (e^2+b^2 c^2 \log ^2(F)\right )\right ) \sin (d+e x)\right )}{4 \left (9 e^4+10 b^2 c^2 e^2 \log ^2(F)+b^4 c^4 \log ^4(F)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[F^(c*(a + b*x))*Sin[d + e*x]^3,x]

[Out]

(F^(c*(a + b*x))*(-3*e*Cos[d + e*x]*(9*e^2 + b^2*c^2*Log[F]^2) + 3*Cos[3*(d + e*x)]*(e^3 + b^2*c^2*e*Log[F]^2)
- 2*b*c*Log[F]*(-13*e^2 - b^2*c^2*Log[F]^2 + Cos[2*(d + e*x)]*(e^2 + b^2*c^2*Log[F]^2))*Sin[d + e*x]))/(4*(9*
e^4 + 10*b^2*c^2*e^2*Log[F]^2 + b^4*c^4*Log[F]^4))

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Maple [A]
time = 0.64, size = 377, normalized size = 1.89

method	result	size
risch	\(-\frac {3 e \,F^{c \left (b x +a \right )} \cos \left (e x +d \right )}{4 \left (e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}+\frac {3 b c \,F^{c \left (b x +a \right )} \ln \left (F \right ) \sin \left (e x +d \right )}{4 \left (e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}+\frac {3 e \,F^{c \left (b x +a \right )} \cos \left (3 e x +3 d \right )}{4 \left (9 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}-\frac {c b \ln \left (F \right ) F^{c \left (b x +a \right )} \sin \left (3 e x +3 d \right )}{4 \left (9 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}\)	\(158\)
default	\(-\frac {F^{a c} \left (\frac {\frac {4 e \,{\mathrm e}^{b c x \ln \left (F \right )}}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}-\frac {4 e \,{\mathrm e}^{b c x \ln \left (F \right )} \left (\tan ^{2}\left (\frac {d}{2}+\frac {e x}{2}\right )\right )}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}-\frac {8 b c \ln \left (F \right ) {\mathrm e}^{b c x \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}}{1+\tan ^{2}\left (\frac {d}{2}+\frac {e x}{2}\right )}+\frac {-\frac {3 e \,{\mathrm e}^{b c x \ln \left (F \right )}}{9 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {3 e \,{\mathrm e}^{b c x \ln \left (F \right )} \left (\tan ^{2}\left (\frac {3 e x}{2}+\frac {3 d}{2}\right )\right )}{9 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {2 b c \ln \left (F \right ) {\mathrm e}^{b c x \ln \left (F \right )} \tan \left (\frac {3 e x}{2}+\frac {3 d}{2}\right )}{9 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}}{1+\tan ^{2}\left (\frac {3 e x}{2}+\frac {3 d}{2}\right )}+\frac {\frac {e \,{\mathrm e}^{b c x \ln \left (F \right )} \left (\tan ^{2}\left (\frac {d}{2}+\frac {e x}{2}\right )\right )}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}-\frac {e \,{\mathrm e}^{b c x \ln \left (F \right )}}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {2 b c \ln \left (F \right ) {\mathrm e}^{b c x \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}}{1+\tan ^{2}\left (\frac {d}{2}+\frac {e x}{2}\right )}\right )}{4}\)	\(377\)
norman	\(\frac {-\frac {6 e^{3} {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{9 e^{4}+10 b^{2} c^{2} e^{2} \ln \left (F \right )^{2}+b^{4} c^{4} \ln \left (F \right )^{4}}+\frac {6 e^{3} {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \left (\tan ^{6}\left (\frac {d}{2}+\frac {e x}{2}\right )\right )}{9 e^{4}+10 b^{2} c^{2} e^{2} \ln \left (F \right )^{2}+b^{4} c^{4} \ln \left (F \right )^{4}}-\frac {6 e \left (2 b^{2} c^{2} \ln \left (F \right )^{2}+3 e^{2}\right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \left (\tan ^{2}\left (\frac {d}{2}+\frac {e x}{2}\right )\right )}{9 e^{4}+10 b^{2} c^{2} e^{2} \ln \left (F \right )^{2}+b^{4} c^{4} \ln \left (F \right )^{4}}+\frac {6 e \left (2 b^{2} c^{2} \ln \left (F \right )^{2}+3 e^{2}\right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \left (\tan ^{4}\left (\frac {d}{2}+\frac {e x}{2}\right )\right )}{9 e^{4}+10 b^{2} c^{2} e^{2} \ln \left (F \right )^{2}+b^{4} c^{4} \ln \left (F \right )^{4}}+\frac {12 e^{2} b c \ln \left (F \right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )}{9 e^{4}+10 b^{2} c^{2} e^{2} \ln \left (F \right )^{2}+b^{4} c^{4} \ln \left (F \right )^{4}}+\frac {12 e^{2} b c \ln \left (F \right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \left (\tan ^{5}\left (\frac {d}{2}+\frac {e x}{2}\right )\right )}{9 e^{4}+10 b^{2} c^{2} e^{2} \ln \left (F \right )^{2}+b^{4} c^{4} \ln \left (F \right )^{4}}+\frac {8 \ln \left (F \right ) b c \left (4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \left (\tan ^{3}\left (\frac {d}{2}+\frac {e x}{2}\right )\right )}{9 e^{4}+10 b^{2} c^{2} e^{2} \ln \left (F \right )^{2}+b^{4} c^{4} \ln \left (F \right )^{4}}}{\left (1+\tan ^{2}\left (\frac {d}{2}+\frac {e x}{2}\right )\right )^{3}}\)	\(483\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(c*(b*x+a))*sin(e*x+d)^3,x,method=_RETURNVERBOSE)

[Out]

-1/4*F^(a*c)*((4/(e^2+b^2*c^2*ln(F)^2)*e*exp(b*c*x*ln(F))-4/(e^2+b^2*c^2*ln(F)^2)*e*exp(b*c*x*ln(F))*tan(1/2*d
+1/2*e*x)^2-8*b*c*ln(F)/(e^2+b^2*c^2*ln(F)^2)*exp(b*c*x*ln(F))*tan(1/2*d+1/2*e*x))/(1+tan(1/2*d+1/2*e*x)^2)+(-
3/(9*e^2+b^2*c^2*ln(F)^2)*e*exp(b*c*x*ln(F))+3/(9*e^2+b^2*c^2*ln(F)^2)*e*exp(b*c*x*ln(F))*tan(3/2*e*x+3/2*d)^2
+2*b*c*ln(F)/(9*e^2+b^2*c^2*ln(F)^2)*exp(b*c*x*ln(F))*tan(3/2*e*x+3/2*d))/(1+tan(3/2*e*x+3/2*d)^2)+(1/(e^2+b^2
*c^2*ln(F)^2)*e*exp(b*c*x*ln(F))*tan(1/2*d+1/2*e*x)^2-1/(e^2+b^2*c^2*ln(F)^2)*e*exp(b*c*x*ln(F))+2*b*c*ln(F)/(
e^2+b^2*c^2*ln(F)^2)*exp(b*c*x*ln(F))*tan(1/2*d+1/2*e*x))/(1+tan(1/2*d+1/2*e*x)^2))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 761 vs. \(2 (197) = 394\).
time = 0.34, size = 761, normalized size = 3.82 \begin {gather*} \frac {{\left (3 \, {\left (F^{a c} b^{2} c^{2} e \log \left (F\right )^{2} + F^{a c} e^{3}\right )} \cos \left (3 \, d\right ) - {\left (F^{a c} b^{3} c^{3} \log \left (F\right )^{3} + F^{a c} b c e^{2} \log \left (F\right )\right )} \sin \left (3 \, d\right )\right )} F^{b c x} \cos \left (3 \, x e\right ) + {\left (3 \, {\left (F^{a c} b^{2} c^{2} e \log \left (F\right )^{2} + F^{a c} e^{3}\right )} \cos \left (3 \, d\right ) + {\left (F^{a c} b^{3} c^{3} \log \left (F\right )^{3} + F^{a c} b c e^{2} \log \left (F\right )\right )} \sin \left (3 \, d\right )\right )} F^{b c x} \cos \left (3 \, x e + 6 \, d\right ) - 3 \, {\left ({\left (F^{a c} b^{2} c^{2} e \log \left (F\right )^{2} + 9 \, F^{a c} e^{3}\right )} \cos \left (3 \, d\right ) + {\left (F^{a c} b^{3} c^{3} \log \left (F\right )^{3} + 9 \, F^{a c} b c e^{2} \log \left (F\right )\right )} \sin \left (3 \, d\right )\right )} F^{b c x} \cos \left (x e + 4 \, d\right ) - 3 \, {\left ({\left (F^{a c} b^{2} c^{2} e \log \left (F\right )^{2} + 9 \, F^{a c} e^{3}\right )} \cos \left (3 \, d\right ) - {\left (F^{a c} b^{3} c^{3} \log \left (F\right )^{3} + 9 \, F^{a c} b c e^{2} \log \left (F\right )\right )} \sin \left (3 \, d\right )\right )} F^{b c x} \cos \left (x e - 2 \, d\right ) - {\left ({\left (F^{a c} b^{3} c^{3} \log \left (F\right )^{3} + F^{a c} b c e^{2} \log \left (F\right )\right )} \cos \left (3 \, d\right ) + 3 \, {\left (F^{a c} b^{2} c^{2} e \log \left (F\right )^{2} + F^{a c} e^{3}\right )} \sin \left (3 \, d\right )\right )} F^{b c x} \sin \left (3 \, x e\right ) - {\left ({\left (F^{a c} b^{3} c^{3} \log \left (F\right )^{3} + F^{a c} b c e^{2} \log \left (F\right )\right )} \cos \left (3 \, d\right ) - 3 \, {\left (F^{a c} b^{2} c^{2} e \log \left (F\right )^{2} + F^{a c} e^{3}\right )} \sin \left (3 \, d\right )\right )} F^{b c x} \sin \left (3 \, x e + 6 \, d\right ) + 3 \, {\left ({\left (F^{a c} b^{3} c^{3} \log \left (F\right )^{3} + 9 \, F^{a c} b c e^{2} \log \left (F\right )\right )} \cos \left (3 \, d\right ) - {\left (F^{a c} b^{2} c^{2} e \log \left (F\right )^{2} + 9 \, F^{a c} e^{3}\right )} \sin \left (3 \, d\right )\right )} F^{b c x} \sin \left (x e + 4 \, d\right ) + 3 \, {\left ({\left (F^{a c} b^{3} c^{3} \log \left (F\right )^{3} + 9 \, F^{a c} b c e^{2} \log \left (F\right )\right )} \cos \left (3 \, d\right ) + {\left (F^{a c} b^{2} c^{2} e \log \left (F\right )^{2} + 9 \, F^{a c} e^{3}\right )} \sin \left (3 \, d\right )\right )} F^{b c x} \sin \left (x e - 2 \, d\right )}{8 \, {\left ({\left (b^{4} c^{4} \log \left (F\right )^{4} + 10 \, b^{2} c^{2} e^{2} \log \left (F\right )^{2} + 9 \, e^{4}\right )} \cos \left (3 \, d\right )^{2} + {\left (b^{4} c^{4} \log \left (F\right )^{4} + 10 \, b^{2} c^{2} e^{2} \log \left (F\right )^{2} + 9 \, e^{4}\right )} \sin \left (3 \, 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)^3,x, algorithm="maxima")

[Out]

1/8*((3*(F^(a*c)*b^2*c^2*e*log(F)^2 + F^(a*c)*e^3)*cos(3*d) - (F^(a*c)*b^3*c^3*log(F)^3 + F^(a*c)*b*c*e^2*log(
F))*sin(3*d))*F^(b*c*x)*cos(3*x*e) + (3*(F^(a*c)*b^2*c^2*e*log(F)^2 + F^(a*c)*e^3)*cos(3*d) + (F^(a*c)*b^3*c^3
*log(F)^3 + F^(a*c)*b*c*e^2*log(F))*sin(3*d))*F^(b*c*x)*cos(3*x*e + 6*d) - 3*((F^(a*c)*b^2*c^2*e*log(F)^2 + 9*
F^(a*c)*e^3)*cos(3*d) + (F^(a*c)*b^3*c^3*log(F)^3 + 9*F^(a*c)*b*c*e^2*log(F))*sin(3*d))*F^(b*c*x)*cos(x*e + 4*
d) - 3*((F^(a*c)*b^2*c^2*e*log(F)^2 + 9*F^(a*c)*e^3)*cos(3*d) - (F^(a*c)*b^3*c^3*log(F)^3 + 9*F^(a*c)*b*c*e^2*
log(F))*sin(3*d))*F^(b*c*x)*cos(x*e - 2*d) - ((F^(a*c)*b^3*c^3*log(F)^3 + F^(a*c)*b*c*e^2*log(F))*cos(3*d) + 3
*(F^(a*c)*b^2*c^2*e*log(F)^2 + F^(a*c)*e^3)*sin(3*d))*F^(b*c*x)*sin(3*x*e) - ((F^(a*c)*b^3*c^3*log(F)^3 + F^(a
*c)*b*c*e^2*log(F))*cos(3*d) - 3*(F^(a*c)*b^2*c^2*e*log(F)^2 + F^(a*c)*e^3)*sin(3*d))*F^(b*c*x)*sin(3*x*e + 6*
d) + 3*((F^(a*c)*b^3*c^3*log(F)^3 + 9*F^(a*c)*b*c*e^2*log(F))*cos(3*d) - (F^(a*c)*b^2*c^2*e*log(F)^2 + 9*F^(a*
c)*e^3)*sin(3*d))*F^(b*c*x)*sin(x*e + 4*d) + 3*((F^(a*c)*b^3*c^3*log(F)^3 + 9*F^(a*c)*b*c*e^2*log(F))*cos(3*d)
+ (F^(a*c)*b^2*c^2*e*log(F)^2 + 9*F^(a*c)*e^3)*sin(3*d))*F^(b*c*x)*sin(x*e - 2*d))/((b^4*c^4*log(F)^4 + 10*b^
2*c^2*e^2*log(F)^2 + 9*e^4)*cos(3*d)^2 + (b^4*c^4*log(F)^4 + 10*b^2*c^2*e^2*log(F)^2 + 9*e^4)*sin(3*d)^2)

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Fricas [A]
time = 1.98, size = 174, normalized size = 0.87 \begin {gather*} \frac {{\left (3 \, \cos \left (x e + d\right )^{3} e^{3} + 3 \, {\left (b^{2} c^{2} \cos \left (x e + d\right )^{3} e - b^{2} c^{2} \cos \left (x e + d\right ) e\right )} \log \left (F\right )^{2} - 9 \, \cos \left (x e + d\right ) e^{3} - {\left ({\left (b^{3} c^{3} \cos \left (x e + d\right )^{2} - b^{3} c^{3}\right )} \log \left (F\right )^{3} + {\left (b c \cos \left (x e + d\right )^{2} e^{2} - 7 \, b c e^{2}\right )} \log \left (F\right )\right )} \sin \left (x e + d\right )\right )} F^{b c x + a c}}{b^{4} c^{4} \log \left (F\right )^{4} + 10 \, b^{2} c^{2} e^{2} \log \left (F\right )^{2} + 9 \, e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))*sin(e*x+d)^3,x, algorithm="fricas")

[Out]

(3*cos(x*e + d)^3*e^3 + 3*(b^2*c^2*cos(x*e + d)^3*e - b^2*c^2*cos(x*e + d)*e)*log(F)^2 - 9*cos(x*e + d)*e^3 -
((b^3*c^3*cos(x*e + d)^2 - b^3*c^3)*log(F)^3 + (b*c*cos(x*e + d)^2*e^2 - 7*b*c*e^2)*log(F))*sin(x*e + d))*F^(b
*c*x + a*c)/(b^4*c^4*log(F)^4 + 10*b^2*c^2*e^2*log(F)^2 + 9*e^4)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(c*(b*x+a))*sin(e*x+d)**3,x)

[Out]

Timed out

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Giac [C] Result contains complex when optimal does not.
time = 0.46, size = 1275, normalized size = 6.41 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))*sin(e*x+d)^3,x, algorithm="giac")

[Out]

-1/4*(2*b*c*log(abs(F))*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 + 3*e*x + 3*d)
/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F) - pi*b*c + 6*e)^2) - (pi*b*c*sgn(F) - pi*b*c + 6*e)*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 + 3*e*x + 3*d)/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn
(F) - pi*b*c + 6*e)^2))*e^(b*c*x*log(abs(F)) + a*c*log(abs(F))) + 3/4*(2*b*c*log(abs(F))*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 + e*x + d)/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F) - pi*b
*c + 2*e)^2) - (pi*b*c*sgn(F) - pi*b*c + 2*e)*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 + e*x + d)/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F) - pi*b*c + 2*e)^2))*e^(b*c*x*log(abs(F)) + a*c*lo
g(abs(F))) - 3/4*(2*b*c*log(abs(F))*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 -
e*x - d)/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F) - pi*b*c - 2*e)^2) - (pi*b*c*sgn(F) - pi*b*c - 2*e)*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 - e*x - d)/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c
*sgn(F) - pi*b*c - 2*e)^2))*e^(b*c*x*log(abs(F)) + a*c*log(abs(F))) + 1/4*(2*b*c*log(abs(F))*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 - 3*e*x - 3*d)/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F
) - pi*b*c - 6*e)^2) - (pi*b*c*sgn(F) - pi*b*c - 6*e)*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 - 3*e*x - 3*d)/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F) - pi*b*c - 6*e)^2))*e^(b*c*x*log(abs(
F)) + a*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*sgn(F) - 1/2*I*pi*a*c + 3
*I*e*x + 3*I*d)/(8*I*pi*b*c*sgn(F) - 8*I*pi*b*c + 16*b*c*log(abs(F)) + 48*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 - 3*I*e*x - 3*I*d)/(-8*I*pi*b*c*sgn(F) + 8*I*pi*b*c + 16*
b*c*log(abs(F)) - 48*I*e))*e^(b*c*x*log(abs(F)) + a*c*log(abs(F))) - 3*(-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 + I*e*x + I*d)/(8*I*pi*b*c*sgn(F) - 8*I*pi*b*c + 16*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 - I*e*x - I*d)
/(-8*I*pi*b*c*sgn(F) + 8*I*pi*b*c + 16*b*c*log(abs(F)) - 16*I*e))*e^(b*c*x*log(abs(F)) + a*c*log(abs(F))) - 3*
(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 - I*e*x - I*d)/(8*I*pi*b*c*s
gn(F) - 8*I*pi*b*c + 16*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 + I*e*x + I*d)/(-8*I*pi*b*c*sgn(F) + 8*I*pi*b*c + 16*b*c*log(abs(F)) + 16*I*e))*e^(b*c
*x*log(abs(F)) + a*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*sgn(F) - 1/2*
I*pi*a*c - 3*I*e*x - 3*I*d)/(8*I*pi*b*c*sgn(F) - 8*I*pi*b*c + 16*b*c*log(abs(F)) - 48*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 + 3*I*e*x + 3*I*d)/(-8*I*pi*b*c*sgn(F) + 8*I*
pi*b*c + 16*b*c*log(abs(F)) + 48*I*e))*e^(b*c*x*log(abs(F)) + a*c*log(abs(F)))

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Mupad [B]
time = 3.22, size = 190, normalized size = 0.95 \begin {gather*} -\frac {3\,F^{c\,\left (a+b\,x\right )}\,\left (\cos \left (e\,x\right )-\sin \left (e\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (d\right )-\sin \left (d\right )\,1{}\mathrm {i}\right )}{8\,\left (e+b\,c\,\ln \left (F\right )\,1{}\mathrm {i}\right )}+\frac {F^{c\,\left (a+b\,x\right )}\,\left (\cos \left (3\,e\,x\right )+\sin \left (3\,e\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (3\,d\right )+\sin \left (3\,d\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{8\,\left (b\,c\,\ln \left (F\right )+e\,3{}\mathrm {i}\right )}+\frac {F^{c\,\left (a+b\,x\right )}\,\left (\cos \left (3\,e\,x\right )-\sin \left (3\,e\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (3\,d\right )-\sin \left (3\,d\right )\,1{}\mathrm {i}\right )}{8\,\left (3\,e+b\,c\,\ln \left (F\right )\,1{}\mathrm {i}\right )}-\frac {F^{c\,\left (a+b\,x\right )}\,\left (\cos \left (e\,x\right )+\sin \left (e\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (d\right )+\sin \left (d\right )\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{8\,\left (b\,c\,\ln \left (F\right )+e\,1{}\mathrm {i}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(c*(a + b*x))*sin(d + e*x)^3,x)

[Out]

(F^(c*(a + b*x))*(cos(3*e*x) + sin(3*e*x)*1i)*(cos(3*d) + sin(3*d)*1i)*1i)/(8*(e*3i + b*c*log(F))) - (3*F^(c*(
a + b*x))*(cos(e*x) - sin(e*x)*1i)*(cos(d) - sin(d)*1i))/(8*(e + b*c*log(F)*1i)) + (F^(c*(a + b*x))*(cos(3*e*x
) - sin(3*e*x)*1i)*(cos(3*d) - sin(3*d)*1i))/(8*(3*e + b*c*log(F)*1i)) - (F^(c*(a + b*x))*(cos(e*x) + sin(e*x)
*1i)*(cos(d) + sin(d)*1i)*3i)/(8*(e*1i + b*c*log(F)))

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