3.3.73 \(\int x (a+b \sin (c+d (f+g x)^n))^2 \, dx\) [273]

Optimal. Leaf size=556 \[ -\frac {\left (2 a^2+b^2\right ) f x}{2 g}+\frac {\left (2 a^2+b^2\right ) (f+g x)^2}{4 g^2}-\frac {i a b e^{i c} f (f+g x) \left (-i d (f+g x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},-i d (f+g x)^n\right )}{g^2 n}+\frac {i a b e^{-i c} f (f+g x) \left (i d (f+g x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},i d (f+g x)^n\right )}{g^2 n}-\frac {2^{-2-\frac {1}{n}} b^2 e^{2 i c} f (f+g x) \left (-i d (f+g x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},-2 i d (f+g x)^n\right )}{g^2 n}-\frac {2^{-2-\frac {1}{n}} b^2 e^{-2 i c} f (f+g x) \left (i d (f+g x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},2 i d (f+g x)^n\right )}{g^2 n}+\frac {i a b e^{i c} (f+g x)^2 \left (-i d (f+g x)^n\right )^{-2/n} \Gamma \left (\frac {2}{n},-i d (f+g x)^n\right )}{g^2 n}-\frac {i a b e^{-i c} (f+g x)^2 \left (i d (f+g x)^n\right )^{-2/n} \Gamma \left (\frac {2}{n},i d (f+g x)^n\right )}{g^2 n}+\frac {4^{-1-\frac {1}{n}} b^2 e^{2 i c} (f+g x)^2 \left (-i d (f+g x)^n\right )^{-2/n} \Gamma \left (\frac {2}{n},-2 i d (f+g x)^n\right )}{g^2 n}+\frac {4^{-1-\frac {1}{n}} b^2 e^{-2 i c} (f+g x)^2 \left (i d (f+g x)^n\right )^{-2/n} \Gamma \left (\frac {2}{n},2 i d (f+g x)^n\right )}{g^2 n} \]

[Out]

-1/2*(2*a^2+b^2)*f*x/g+1/4*(2*a^2+b^2)*(g*x+f)^2/g^2-I*a*b*exp(I*c)*f*(g*x+f)*GAMMA(1/n,-I*d*(g*x+f)^n)/g^2/n/
((-I*d*(g*x+f)^n)^(1/n))+I*a*b*f*(g*x+f)*GAMMA(1/n,I*d*(g*x+f)^n)/exp(I*c)/g^2/n/((I*d*(g*x+f)^n)^(1/n))-2^(-2
-1/n)*b^2*exp(2*I*c)*f*(g*x+f)*GAMMA(1/n,-2*I*d*(g*x+f)^n)/g^2/n/((-I*d*(g*x+f)^n)^(1/n))-2^(-2-1/n)*b^2*f*(g*
x+f)*GAMMA(1/n,2*I*d*(g*x+f)^n)/exp(2*I*c)/g^2/n/((I*d*(g*x+f)^n)^(1/n))+I*a*b*exp(I*c)*(g*x+f)^2*GAMMA(2/n,-I
*d*(g*x+f)^n)/g^2/n/((-I*d*(g*x+f)^n)^(2/n))-I*a*b*(g*x+f)^2*GAMMA(2/n,I*d*(g*x+f)^n)/exp(I*c)/g^2/n/((I*d*(g*
x+f)^n)^(2/n))+4^(-1-1/n)*b^2*exp(2*I*c)*(g*x+f)^2*GAMMA(2/n,-2*I*d*(g*x+f)^n)/g^2/n/((-I*d*(g*x+f)^n)^(2/n))+
4^(-1-1/n)*b^2*(g*x+f)^2*GAMMA(2/n,2*I*d*(g*x+f)^n)/exp(2*I*c)/g^2/n/((I*d*(g*x+f)^n)^(2/n))

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Rubi [A]
time = 0.30, antiderivative size = 556, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 10, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3514, 3448, 3447, 2239, 3446, 3506, 6, 3505, 2250, 3504} \begin {gather*} \frac {i a b e^{i c} (f+g x)^2 \left (-i d (f+g x)^n\right )^{-2/n} \text {Gamma}\left (\frac {2}{n},-i d (f+g x)^n\right )}{g^2 n}-\frac {i a b e^{i c} f (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^2 n}+\frac {i a b e^{-i c} f (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^2 n}-\frac {i a b e^{-i c} (f+g x)^2 \left (i d (f+g x)^n\right )^{-2/n} \text {Gamma}\left (\frac {2}{n},i d (f+g x)^n\right )}{g^2 n}+\frac {b^2 e^{2 i c} 4^{-\frac {1}{n}-1} (f+g x)^2 \left (-i d (f+g x)^n\right )^{-2/n} \text {Gamma}\left (\frac {2}{n},-2 i d (f+g x)^n\right )}{g^2 n}-\frac {b^2 e^{2 i c} f 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^2 n}-\frac {b^2 e^{-2 i c} f 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^2 n}+\frac {b^2 e^{-2 i c} 4^{-\frac {1}{n}-1} (f+g x)^2 \left (i d (f+g x)^n\right )^{-2/n} \text {Gamma}\left (\frac {2}{n},2 i d (f+g x)^n\right )}{g^2 n}+\frac {\left (2 a^2+b^2\right ) (f+g x)^2}{4 g^2}-\frac {f x \left (2 a^2+b^2\right )}{2 g} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

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 2250

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(-F^a)*((e +
f*x)^(m + 1)/(f*n*((-b)*(c + d*x)^n*Log[F])^((m + 1)/n)))*Gamma[(m + 1)/n, (-b)*(c + d*x)^n*Log[F]], x] /; Fre
eQ[{F, a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0]

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]

Rule 3504

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

Rule 3505

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

Rule 3506

Int[((e_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_), x_Symbol] :> Int[ExpandTrigReduce[(e
*x)^m, (a + b*Sin[c + d*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]

Rule 3514

Int[((g_.) + (h_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)])^(p_.), x_Symbol] :
> Module[{k = If[FractionQ[n], Denominator[n], 1]}, Dist[k/f^(m + 1), Subst[Int[ExpandIntegrand[(a + b*Sin[c +
 d*x^(k*n)])^p, x^(k - 1)*(f*g - e*h + h*x^k)^m, x], x], x, (e + f*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, f,
g, h}, x] && IGtQ[p, 0] && IGtQ[m, 0]

Rubi steps

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

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Mathematica [A]
time = 27.06, size = 552, normalized size = 0.99 \begin {gather*} \frac {2 a^2 g^2 n x^2+b^2 g^2 n x^2-4 i a b (f+g x)^2 \left (i d (f+g x)^n\right )^{-2/n} \Gamma \left (\frac {2}{n},i d (f+g x)^n\right ) (\cos (c)-i \sin (c))+4 i a b (f+g x)^2 \left (-i d (f+g x)^n\right )^{-2/n} \Gamma \left (\frac {2}{n},-i d (f+g x)^n\right ) (\cos (c)+i \sin (c))+4 a b f (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 ) (-i \cos (c)+\sin (c))+4 a b f (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 ) (i \cos (c)+\sin (c))-b^2 f (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 ) (\cos (c)-i \sin (c))^2 \left (\cosh \left (\frac {\log (2)}{n}\right )-\sinh \left (\frac {\log (2)}{n}\right )\right )-b^2 f (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 ) (\cos (c)+i \sin (c))^2 \left (\cosh \left (\frac {\log (2)}{n}\right )-\sinh \left (\frac {\log (2)}{n}\right )\right )+b^2 (f+g x)^2 \left (i d (f+g x)^n\right )^{-2/n} \Gamma \left (\frac {2}{n},2 i d (f+g x)^n\right ) (\cos (c)-i \sin (c))^2 \left (\cosh \left (\frac {\log (4)}{n}\right )-\sinh \left (\frac {\log (4)}{n}\right )\right )+b^2 (f+g x)^2 \left (-i d (f+g x)^n\right )^{-2/n} \Gamma \left (\frac {2}{n},-2 i d (f+g x)^n\right ) (\cos (c)+i \sin (c))^2 \left (\cosh \left (\frac {\log (4)}{n}\right )-\sinh \left (\frac {\log (4)}{n}\right )\right )}{4 g^2 n} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*g^2*n*x^2 + b^2*g^2*n*x^2 - ((4*I)*a*b*(f + g*x)^2*Gamma[2/n, I*d*(f + g*x)^n]*(Cos[c] - I*Sin[c]))/(I*
d*(f + g*x)^n)^(2/n) + ((4*I)*a*b*(f + g*x)^2*Gamma[2/n, (-I)*d*(f + g*x)^n]*(Cos[c] + I*Sin[c]))/((-I)*d*(f +
 g*x)^n)^(2/n) + (4*a*b*f*(f + g*x)*Gamma[n^(-1), (-I)*d*(f + g*x)^n]*((-I)*Cos[c] + Sin[c]))/((-I)*d*(f + g*x
)^n)^n^(-1) + (4*a*b*f*(f + g*x)*Gamma[n^(-1), I*d*(f + g*x)^n]*(I*Cos[c] + Sin[c]))/(I*d*(f + g*x)^n)^n^(-1)
- (b^2*f*(f + g*x)*Gamma[n^(-1), (2*I)*d*(f + g*x)^n]*(Cos[c] - I*Sin[c])^2*(Cosh[Log[2]/n] - Sinh[Log[2]/n]))
/(I*d*(f + g*x)^n)^n^(-1) - (b^2*f*(f + g*x)*Gamma[n^(-1), (-2*I)*d*(f + g*x)^n]*(Cos[c] + I*Sin[c])^2*(Cosh[L
og[2]/n] - Sinh[Log[2]/n]))/((-I)*d*(f + g*x)^n)^n^(-1) + (b^2*(f + g*x)^2*Gamma[2/n, (2*I)*d*(f + g*x)^n]*(Co
s[c] - I*Sin[c])^2*(Cosh[Log[4]/n] - Sinh[Log[4]/n]))/(I*d*(f + g*x)^n)^(2/n) + (b^2*(f + g*x)^2*Gamma[2/n, (-
2*I)*d*(f + g*x)^n]*(Cos[c] + I*Sin[c])^2*(Cosh[Log[4]/n] - Sinh[Log[4]/n]))/((-I)*d*(f + g*x)^n)^(2/n))/(4*g^
2*n)

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Maple [F]
time = 0.20, size = 0, normalized size = 0.00 \[\int x \left (a +b \sin \left (c +d \left (g x +f \right )^{n}\right )\right )^{2}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

1/2*a^2*x^2 + 1/4*b^2*x^2 - 1/2*b^2*integrate(x*cos(2*(g*x + f)^n*d + 2*c), x) + 2*a*b*integrate(x*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(x*(a+b*sin(c+d*(g*x+f)^n))^2,x, algorithm="fricas")

[Out]

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

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

[Out]

Integral(x*(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(x*(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, x)

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

[Out]

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

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