3.3.74 \(\int (a+b \sin (c+d (f+g x)^n))^2 \, dx\) [274]

Optimal. Leaf size=261 \[ \frac {1}{2} \left (2 a^2+b^2\right ) x+\frac {i a b e^{i c} (f+g x) \left (-i d (f+g x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},-i d (f+g x)^n\right )}{g n}-\frac {i a b e^{-i c} (f+g x) \left (i d (f+g x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},i d (f+g x)^n\right )}{g n}+\frac {2^{-2-\frac {1}{n}} b^2 e^{2 i c} (f+g x) \left (-i d (f+g x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},-2 i d (f+g x)^n\right )}{g n}+\frac {2^{-2-\frac {1}{n}} b^2 e^{-2 i c} (f+g x) \left (i d (f+g x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},2 i d (f+g x)^n\right )}{g n} \]

[Out]

1/2*(2*a^2+b^2)*x+I*a*b*exp(I*c)*(g*x+f)*GAMMA(1/n,-I*d*(g*x+f)^n)/g/n/((-I*d*(g*x+f)^n)^(1/n))-I*a*b*(g*x+f)*
GAMMA(1/n,I*d*(g*x+f)^n)/exp(I*c)/g/n/((I*d*(g*x+f)^n)^(1/n))+2^(-2-1/n)*b^2*exp(2*I*c)*(g*x+f)*GAMMA(1/n,-2*I
*d*(g*x+f)^n)/g/n/((-I*d*(g*x+f)^n)^(1/n))+2^(-2-1/n)*b^2*(g*x+f)*GAMMA(1/n,2*I*d*(g*x+f)^n)/exp(2*I*c)/g/n/((
I*d*(g*x+f)^n)^(1/n))

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Rubi [A]
time = 0.10, antiderivative size = 261, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3448, 3447, 2239, 3446} \begin {gather*} \frac {i a b e^{i c} (f+g x) \left (-i d (f+g x)^n\right )^{-1/n} \text {Gamma}\left (\frac {1}{n},-i d (f+g x)^n\right )}{g n}-\frac {i a b e^{-i c} (f+g x) \left (i d (f+g x)^n\right )^{-1/n} \text {Gamma}\left (\frac {1}{n},i d (f+g x)^n\right )}{g n}+\frac {b^2 e^{2 i c} 2^{-\frac {1}{n}-2} (f+g x) \left (-i d (f+g x)^n\right )^{-1/n} \text {Gamma}\left (\frac {1}{n},-2 i d (f+g x)^n\right )}{g n}+\frac {b^2 e^{-2 i c} 2^{-\frac {1}{n}-2} (f+g x) \left (i d (f+g x)^n\right )^{-1/n} \text {Gamma}\left (\frac {1}{n},2 i d (f+g x)^n\right )}{g n}+\frac {1}{2} x \left (2 a^2+b^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*(f + g*x)^n])^2,x]

[Out]

((2*a^2 + b^2)*x)/2 + (I*a*b*E^(I*c)*(f + g*x)*Gamma[n^(-1), (-I)*d*(f + g*x)^n])/(g*n*((-I)*d*(f + g*x)^n)^n^
(-1)) - (I*a*b*(f + g*x)*Gamma[n^(-1), I*d*(f + g*x)^n])/(E^(I*c)*g*n*(I*d*(f + g*x)^n)^n^(-1)) + (2^(-2 - n^(
-1))*b^2*E^((2*I)*c)*(f + g*x)*Gamma[n^(-1), (-2*I)*d*(f + g*x)^n])/(g*n*((-I)*d*(f + g*x)^n)^n^(-1)) + (2^(-2
- n^(-1))*b^2*(f + g*x)*Gamma[n^(-1), (2*I)*d*(f + g*x)^n])/(E^((2*I)*c)*g*n*(I*d*(f + g*x)^n)^n^(-1))

Rule 2239

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_)), x_Symbol] :> Simp[(-F^a)*(c + d*x)*(Gamma[1/n, (-b)*(c + d
*x)^n*Log[F]]/(d*n*((-b)*(c + d*x)^n*Log[F])^(1/n))), x] /; FreeQ[{F, a, b, c, d, n}, x] &&  !IntegerQ[2/n]

Rule 3446

Int[Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)], x_Symbol] :> Dist[I/2, Int[E^((-c)*I - d*I*(e + f*x)^n), x],
x] - Dist[I/2, Int[E^(c*I + d*I*(e + f*x)^n), x], x] /; FreeQ[{c, d, e, f, n}, x]

Rule 3447

Int[Cos[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)], x_Symbol] :> Dist[1/2, Int[E^((-c)*I - d*I*(e + f*x)^n), x],
x] + Dist[1/2, Int[E^(c*I + d*I*(e + f*x)^n), x], x] /; FreeQ[{c, d, e, f, n}, x]

Rule 3448

Int[((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)])^(p_), x_Symbol] :> Int[ExpandTrigReduce[(a +
b*Sin[c + d*(e + f*x)^n])^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && IGtQ[p, 1]

Rubi steps

\begin {align*} \int \left (a+b \sin \left (c+d (f+g x)^n\right )\right )^2 \, dx &=\int \left (a^2+\frac {b^2}{2}-\frac {1}{2} b^2 \cos \left (2 c+2 d (f+g x)^n\right )+2 a b \sin \left (c+d (f+g x)^n\right )\right ) \, dx\\ &=\frac {1}{2} \left (2 a^2+b^2\right ) x+(2 a b) \int \sin \left (c+d (f+g x)^n\right ) \, dx-\frac {1}{2} b^2 \int \cos \left (2 c+2 d (f+g x)^n\right ) \, dx\\ &=\frac {1}{2} \left (2 a^2+b^2\right ) x+(i a b) \int e^{-i c-i d (f+g x)^n} \, dx-(i a b) \int e^{i c+i d (f+g x)^n} \, dx-\frac {1}{4} b^2 \int e^{-2 i c-2 i d (f+g x)^n} \, dx-\frac {1}{4} b^2 \int e^{2 i c+2 i d (f+g x)^n} \, dx\\ &=\frac {1}{2} \left (2 a^2+b^2\right ) x+\frac {i a b e^{i c} (f+g x) \left (-i d (f+g x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},-i d (f+g x)^n\right )}{g n}-\frac {i a b e^{-i c} (f+g x) \left (i d (f+g x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},i d (f+g x)^n\right )}{g n}+\frac {2^{-2-\frac {1}{n}} b^2 e^{2 i c} (f+g x) \left (-i d (f+g x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},-2 i d (f+g x)^n\right )}{g n}+\frac {2^{-2-\frac {1}{n}} b^2 e^{-2 i c} (f+g x) \left (i d (f+g x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},2 i d (f+g x)^n\right )}{g n}\\ \end {align*}

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Mathematica [A]
time = 0.53, size = 247, normalized size = 0.95 \begin {gather*} \frac {2 \left (2 a^2+b^2\right ) g n x+2^{-1/n} b^2 e^{2 i c} (f+g x) \left (-i d (f+g x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},-2 i d (f+g x)^n\right )+2^{-1/n} b^2 e^{-2 i c} (f+g x) \left (i d (f+g x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},2 i d (f+g x)^n\right )-4 i a b (f+g x) \left (i d (f+g x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},i d (f+g x)^n\right ) (\cos (c)-i \sin (c))+4 i a b (f+g x) \left (-i d (f+g x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},-i d (f+g x)^n\right ) (\cos (c)+i \sin (c))}{4 g n} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[c + d*(f + g*x)^n])^2,x]

[Out]

(2*(2*a^2 + b^2)*g*n*x + (b^2*E^((2*I)*c)*(f + g*x)*Gamma[n^(-1), (-2*I)*d*(f + g*x)^n])/(2^n^(-1)*((-I)*d*(f
+ g*x)^n)^n^(-1)) + (b^2*(f + g*x)*Gamma[n^(-1), (2*I)*d*(f + g*x)^n])/(2^n^(-1)*E^((2*I)*c)*(I*d*(f + g*x)^n)
^n^(-1)) - ((4*I)*a*b*(f + g*x)*Gamma[n^(-1), I*d*(f + g*x)^n]*(Cos[c] - I*Sin[c]))/(I*d*(f + g*x)^n)^n^(-1) +
((4*I)*a*b*(f + g*x)*Gamma[n^(-1), (-I)*d*(f + g*x)^n]*(Cos[c] + I*Sin[c]))/((-I)*d*(f + g*x)^n)^n^(-1))/(4*g
*n)

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Maple [F]
time = 0.08, size = 0, normalized size = 0.00 \[\int \left (a +b \sin \left (c +d \left (g x +f \right )^{n}\right )\right )^{2}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sin(c+d*(g*x+f)^n))^2,x)

[Out]

int((a+b*sin(c+d*(g*x+f)^n))^2,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(c+d*(g*x+f)^n))^2,x, algorithm="maxima")

[Out]

a^2*x + 1/2*b^2*x - 1/2*b^2*integrate(cos(2*(g*x + f)^n*d + 2*c), x) + 2*a*b*integrate(sin((g*x + f)^n*d + c),
x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(c+d*(g*x+f)^n))^2,x, algorithm="fricas")

[Out]

integral(-b^2*cos((g*x + f)^n*d + c)^2 + 2*a*b*sin((g*x + f)^n*d + c) + a^2 + b^2, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \sin {\left (c + d \left (f + g x\right )^{n} \right )}\right )^{2}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(c+d*(g*x+f)**n))**2,x)

[Out]

Integral((a + b*sin(c + d*(f + g*x)**n))**2, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(c+d*(g*x+f)^n))^2,x, algorithm="giac")

[Out]

integrate((b*sin((g*x + f)^n*d + c) + a)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\left (a+b\,\sin \left (c+d\,{\left (f+g\,x\right )}^n\right )\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(c + d*(f + g*x)^n))^2,x)

[Out]

int((a + b*sin(c + d*(f + g*x)^n))^2, x)

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