\(\int \sin ^2(c+d x) (a+a \sin (c+d x))^n \, dx\) [142]

Optimal result

Integrand size = 21, antiderivative size = 156 \[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^n \, dx=\frac {\cos (c+d x) (a+a \sin (c+d x))^n}{d \left (2+3 n+n^2\right )}-\frac {2^{\frac {1}{2}+n} \left (1+n+n^2\right ) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-n,\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{-\frac {1}{2}-n} (a+a \sin (c+d x))^n}{d (1+n) (2+n)}-\frac {\cos (c+d x) (a+a \sin (c+d x))^{1+n}}{a d (2+n)} \]

[Out]

cos(d*x+c)*(a+a*sin(d*x+c))^n/d/(n^2+3*n+2)-2^(1/2+n)*(n^2+n+1)*cos(d*x+c)*hypergeom([1/2, 1/2-n],[3/2],1/2-1/
2*sin(d*x+c))*(1+sin(d*x+c))^(-1/2-n)*(a+a*sin(d*x+c))^n/d/(n^2+3*n+2)-cos(d*x+c)*(a+a*sin(d*x+c))^(1+n)/a/d/(
2+n)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2838, 2830, 2731, 2730} \[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^n \, dx=-\frac {2^{n+\frac {1}{2}} \left (n^2+n+1\right ) \cos (c+d x) (\sin (c+d x)+1)^{-n-\frac {1}{2}} (a \sin (c+d x)+a)^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-n,\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right )}{d (n+1) (n+2)}+\frac {\cos (c+d x) (a \sin (c+d x)+a)^n}{d \left (n^2+3 n+2\right )}-\frac {\cos (c+d x) (a \sin (c+d x)+a)^{n+1}}{a d (n+2)} \]

[In]

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

[Out]

(Cos[c + d*x]*(a + a*Sin[c + d*x])^n)/(d*(2 + 3*n + n^2)) - (2^(1/2 + n)*(1 + n + n^2)*Cos[c + d*x]*Hypergeome
tric2F1[1/2, 1/2 - n, 3/2, (1 - Sin[c + d*x])/2]*(1 + Sin[c + d*x])^(-1/2 - n)*(a + a*Sin[c + d*x])^n)/(d*(1 +
n)*(2 + n)) - (Cos[c + d*x]*(a + a*Sin[c + d*x])^(1 + n))/(a*d*(2 + n))

Rule 2730

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2^(n + 1/2))*a^(n - 1/2)*b*(Cos[c + d*x]/
(d*Sqrt[a + b*Sin[c + d*x]]))*Hypergeometric2F1[1/2, 1/2 - n, 3/2, (1/2)*(1 - b*(Sin[c + d*x]/a))], x] /; Free
Q[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && GtQ[a, 0]

Rule 2731

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[a^IntPart[n]*((a + b*Sin[c + d*x])^FracPart
[n]/(1 + (b/a)*Sin[c + d*x])^FracPart[n]), Int[(1 + (b/a)*Sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, n}, x]
&& EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] &&  !GtQ[a, 0]

Rule 2830

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d
)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(f*(m + 1))), x] + Dist[(a*d*m + b*c*(m + 1))/(b*(m + 1)), Int[(a + b*S
in[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&  !LtQ[m
, -2^(-1)]

Rule 2838

Int[sin[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(-Cos[e + f*x])*(
(a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[1/(b*(m + 2)), Int[(a + b*Sin[e + f*x])^m*(b*(m + 1) -
a*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)]

Rubi steps \begin{align*} \text {integral}& = -\frac {\cos (c+d x) (a+a \sin (c+d x))^{1+n}}{a d (2+n)}+\frac {\int (a (1+n)-a \sin (c+d x)) (a+a \sin (c+d x))^n \, dx}{a (2+n)} \\ & = \frac {\cos (c+d x) (a+a \sin (c+d x))^n}{d \left (2+3 n+n^2\right )}-\frac {\cos (c+d x) (a+a \sin (c+d x))^{1+n}}{a d (2+n)}+\frac {\left (1+n+n^2\right ) \int (a+a \sin (c+d x))^n \, dx}{(1+n) (2+n)} \\ & = \frac {\cos (c+d x) (a+a \sin (c+d x))^n}{d \left (2+3 n+n^2\right )}-\frac {\cos (c+d x) (a+a \sin (c+d x))^{1+n}}{a d (2+n)}+\frac {\left (\left (1+n+n^2\right ) (1+\sin (c+d x))^{-n} (a+a \sin (c+d x))^n\right ) \int (1+\sin (c+d x))^n \, dx}{(1+n) (2+n)} \\ & = \frac {\cos (c+d x) (a+a \sin (c+d x))^n}{d \left (2+3 n+n^2\right )}-\frac {2^{\frac {1}{2}+n} \left (1+n+n^2\right ) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-n,\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{-\frac {1}{2}-n} (a+a \sin (c+d x))^n}{d (1+n) (2+n)}-\frac {\cos (c+d x) (a+a \sin (c+d x))^{1+n}}{a d (2+n)} \\ \end{align*}

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(1732\) vs. \(2(156)=312\).

Time = 21.41 (sec) , antiderivative size = 1732, normalized size of antiderivative = 11.10 \[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^n \, dx=-\frac {2 \sin ^2(c+d x) (a+a \sin (c+d x))^n \left (a+\frac {a \tan (c+d x)}{\sqrt {\sec ^2(c+d x)}}\right )^n \left (\sec ^2(c+d x)+\sqrt {\sec ^2(c+d x)} \tan (c+d x)\right ) \left (1+\left (\sqrt {\sec ^2(c+d x)}+\tan (c+d x)\right )^2\right )^n \left (-\left ((3+2 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2}+n,1+n,\frac {3}{2}+n,-\left (\sqrt {\sec ^2(c+d x)}+\tan (c+d x)\right )^2\right )\right )+4 (1+2 n) \operatorname {Hypergeometric2F1}\left (\frac {3}{2}+n,3+n,\frac {5}{2}+n,-\left (\sqrt {\sec ^2(c+d x)}+\tan (c+d x)\right )^2\right ) \left (1+2 \sqrt {\sec ^2(c+d x)} \tan (c+d x)+2 \tan ^2(c+d x)\right )\right )}{d \left (3+8 n+4 n^2\right ) \sqrt {\sec ^2(c+d x)} \left (-\frac {4 n \left (\sqrt {\sec ^2(c+d x)}+\tan (c+d x)\right ) \left (a+\frac {a \tan (c+d x)}{\sqrt {\sec ^2(c+d x)}}\right )^n \left (\sec ^2(c+d x)+\sqrt {\sec ^2(c+d x)} \tan (c+d x)\right )^2 \left (1+\left (\sqrt {\sec ^2(c+d x)}+\tan (c+d x)\right )^2\right )^{-1+n} \left (-\left ((3+2 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2}+n,1+n,\frac {3}{2}+n,-\left (\sqrt {\sec ^2(c+d x)}+\tan (c+d x)\right )^2\right )\right )+4 (1+2 n) \operatorname {Hypergeometric2F1}\left (\frac {3}{2}+n,3+n,\frac {5}{2}+n,-\left (\sqrt {\sec ^2(c+d x)}+\tan (c+d x)\right )^2\right ) \left (1+2 \sqrt {\sec ^2(c+d x)} \tan (c+d x)+2 \tan ^2(c+d x)\right )\right )}{\left (3+8 n+4 n^2\right ) \sqrt {\sec ^2(c+d x)}}+\frac {2 \tan (c+d x) \left (a+\frac {a \tan (c+d x)}{\sqrt {\sec ^2(c+d x)}}\right )^n \left (\sec ^2(c+d x)+\sqrt {\sec ^2(c+d x)} \tan (c+d x)\right ) \left (1+\left (\sqrt {\sec ^2(c+d x)}+\tan (c+d x)\right )^2\right )^n \left (-\left ((3+2 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2}+n,1+n,\frac {3}{2}+n,-\left (\sqrt {\sec ^2(c+d x)}+\tan (c+d x)\right )^2\right )\right )+4 (1+2 n) \operatorname {Hypergeometric2F1}\left (\frac {3}{2}+n,3+n,\frac {5}{2}+n,-\left (\sqrt {\sec ^2(c+d x)}+\tan (c+d x)\right )^2\right ) \left (1+2 \sqrt {\sec ^2(c+d x)} \tan (c+d x)+2 \tan ^2(c+d x)\right )\right )}{\left (3+8 n+4 n^2\right ) \sqrt {\sec ^2(c+d x)}}-\frac {2 n \left (a+\frac {a \tan (c+d x)}{\sqrt {\sec ^2(c+d x)}}\right )^{-1+n} \left (\sec ^2(c+d x)+\sqrt {\sec ^2(c+d x)} \tan (c+d x)\right ) \left (a \sqrt {\sec ^2(c+d x)}-\frac {a \tan ^2(c+d x)}{\sqrt {\sec ^2(c+d x)}}\right ) \left (1+\left (\sqrt {\sec ^2(c+d x)}+\tan (c+d x)\right )^2\right )^n \left (-\left ((3+2 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2}+n,1+n,\frac {3}{2}+n,-\left (\sqrt {\sec ^2(c+d x)}+\tan (c+d x)\right )^2\right )\right )+4 (1+2 n) \operatorname {Hypergeometric2F1}\left (\frac {3}{2}+n,3+n,\frac {5}{2}+n,-\left (\sqrt {\sec ^2(c+d x)}+\tan (c+d x)\right )^2\right ) \left (1+2 \sqrt {\sec ^2(c+d x)} \tan (c+d x)+2 \tan ^2(c+d x)\right )\right )}{\left (3+8 n+4 n^2\right ) \sqrt {\sec ^2(c+d x)}}-\frac {2 \left (a+\frac {a \tan (c+d x)}{\sqrt {\sec ^2(c+d x)}}\right )^n \left (\sec ^2(c+d x)^{3/2}+2 \sec ^2(c+d x) \tan (c+d x)+\sqrt {\sec ^2(c+d x)} \tan ^2(c+d x)\right ) \left (1+\left (\sqrt {\sec ^2(c+d x)}+\tan (c+d x)\right )^2\right )^n \left (-\left ((3+2 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2}+n,1+n,\frac {3}{2}+n,-\left (\sqrt {\sec ^2(c+d x)}+\tan (c+d x)\right )^2\right )\right )+4 (1+2 n) \operatorname {Hypergeometric2F1}\left (\frac {3}{2}+n,3+n,\frac {5}{2}+n,-\left (\sqrt {\sec ^2(c+d x)}+\tan (c+d x)\right )^2\right ) \left (1+2 \sqrt {\sec ^2(c+d x)} \tan (c+d x)+2 \tan ^2(c+d x)\right )\right )}{\left (3+8 n+4 n^2\right ) \sqrt {\sec ^2(c+d x)}}-\frac {2 \left (a+\frac {a \tan (c+d x)}{\sqrt {\sec ^2(c+d x)}}\right )^n \left (\sec ^2(c+d x)+\sqrt {\sec ^2(c+d x)} \tan (c+d x)\right ) \left (1+\left (\sqrt {\sec ^2(c+d x)}+\tan (c+d x)\right )^2\right )^n \left (4 (1+2 n) \operatorname {Hypergeometric2F1}\left (\frac {3}{2}+n,3+n,\frac {5}{2}+n,-\left (\sqrt {\sec ^2(c+d x)}+\tan (c+d x)\right )^2\right ) \left (2 \sec ^2(c+d x)^{3/2}+4 \sec ^2(c+d x) \tan (c+d x)+2 \sqrt {\sec ^2(c+d x)} \tan ^2(c+d x)\right )+\frac {8 \left (\frac {3}{2}+n\right ) (1+2 n) \left (\sec ^2(c+d x)+\sqrt {\sec ^2(c+d x)} \tan (c+d x)\right ) \left (1+2 \sqrt {\sec ^2(c+d x)} \tan (c+d x)+2 \tan ^2(c+d x)\right ) \left (-\operatorname {Hypergeometric2F1}\left (\frac {3}{2}+n,3+n,\frac {5}{2}+n,-\left (\sqrt {\sec ^2(c+d x)}+\tan (c+d x)\right )^2\right )+\left (1+\left (\sqrt {\sec ^2(c+d x)}+\tan (c+d x)\right )^2\right )^{-3-n}\right )}{\sqrt {\sec ^2(c+d x)}+\tan (c+d x)}-\frac {2 \left (\frac {1}{2}+n\right ) (3+2 n) \left (\sec ^2(c+d x)+\sqrt {\sec ^2(c+d x)} \tan (c+d x)\right ) \left (-\operatorname {Hypergeometric2F1}\left (\frac {1}{2}+n,1+n,\frac {3}{2}+n,-\left (\sqrt {\sec ^2(c+d x)}+\tan (c+d x)\right )^2\right )+\left (1+\left (\sqrt {\sec ^2(c+d x)}+\tan (c+d x)\right )^2\right )^{-1-n}\right )}{\sqrt {\sec ^2(c+d x)}+\tan (c+d x)}\right )}{\left (3+8 n+4 n^2\right ) \sqrt {\sec ^2(c+d x)}}\right )} \]

[In]

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

[Out]

(-2*Sin[c + d*x]^2*(a + a*Sin[c + d*x])^n*(a + (a*Tan[c + d*x])/Sqrt[Sec[c + d*x]^2])^n*(Sec[c + d*x]^2 + Sqrt
[Sec[c + d*x]^2]*Tan[c + d*x])*(1 + (Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2)^n*(-((3 + 2*n)*Hypergeometric2F1[
1/2 + n, 1 + n, 3/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2]) + 4*(1 + 2*n)*Hypergeometric2F1[3/2 + n, 3
+ n, 5/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2]*(1 + 2*Sqrt[Sec[c + d*x]^2]*Tan[c + d*x] + 2*Tan[c +
d*x]^2)))/(d*(3 + 8*n + 4*n^2)*Sqrt[Sec[c + d*x]^2]*((-4*n*(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])*(a + (a*Tan[c
+ d*x])/Sqrt[Sec[c + d*x]^2])^n*(Sec[c + d*x]^2 + Sqrt[Sec[c + d*x]^2]*Tan[c + d*x])^2*(1 + (Sqrt[Sec[c + d*x
]^2] + Tan[c + d*x])^2)^(-1 + n)*(-((3 + 2*n)*Hypergeometric2F1[1/2 + n, 1 + n, 3/2 + n, -(Sqrt[Sec[c + d*x]^2
] + Tan[c + d*x])^2]) + 4*(1 + 2*n)*Hypergeometric2F1[3/2 + n, 3 + n, 5/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan[c
+ d*x])^2]*(1 + 2*Sqrt[Sec[c + d*x]^2]*Tan[c + d*x] + 2*Tan[c + d*x]^2)))/((3 + 8*n + 4*n^2)*Sqrt[Sec[c + d*x]
^2]) + (2*Tan[c + d*x]*(a + (a*Tan[c + d*x])/Sqrt[Sec[c + d*x]^2])^n*(Sec[c + d*x]^2 + Sqrt[Sec[c + d*x]^2]*Ta
n[c + d*x])*(1 + (Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2)^n*(-((3 + 2*n)*Hypergeometric2F1[1/2 + n, 1 + n, 3/2
+ n, -(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2]) + 4*(1 + 2*n)*Hypergeometric2F1[3/2 + n, 3 + n, 5/2 + n, -(Sq
rt[Sec[c + d*x]^2] + Tan[c + d*x])^2]*(1 + 2*Sqrt[Sec[c + d*x]^2]*Tan[c + d*x] + 2*Tan[c + d*x]^2)))/((3 + 8*n
+ 4*n^2)*Sqrt[Sec[c + d*x]^2]) - (2*n*(a + (a*Tan[c + d*x])/Sqrt[Sec[c + d*x]^2])^(-1 + n)*(Sec[c + d*x]^2 +
Sqrt[Sec[c + d*x]^2]*Tan[c + d*x])*(a*Sqrt[Sec[c + d*x]^2] - (a*Tan[c + d*x]^2)/Sqrt[Sec[c + d*x]^2])*(1 + (Sq
rt[Sec[c + d*x]^2] + Tan[c + d*x])^2)^n*(-((3 + 2*n)*Hypergeometric2F1[1/2 + n, 1 + n, 3/2 + n, -(Sqrt[Sec[c +
d*x]^2] + Tan[c + d*x])^2]) + 4*(1 + 2*n)*Hypergeometric2F1[3/2 + n, 3 + n, 5/2 + n, -(Sqrt[Sec[c + d*x]^2] +
Tan[c + d*x])^2]*(1 + 2*Sqrt[Sec[c + d*x]^2]*Tan[c + d*x] + 2*Tan[c + d*x]^2)))/((3 + 8*n + 4*n^2)*Sqrt[Sec[c
+ d*x]^2]) - (2*(a + (a*Tan[c + d*x])/Sqrt[Sec[c + d*x]^2])^n*((Sec[c + d*x]^2)^(3/2) + 2*Sec[c + d*x]^2*Tan[
c + d*x] + Sqrt[Sec[c + d*x]^2]*Tan[c + d*x]^2)*(1 + (Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2)^n*(-((3 + 2*n)*H
ypergeometric2F1[1/2 + n, 1 + n, 3/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2]) + 4*(1 + 2*n)*Hypergeomet
ric2F1[3/2 + n, 3 + n, 5/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2]*(1 + 2*Sqrt[Sec[c + d*x]^2]*Tan[c +
d*x] + 2*Tan[c + d*x]^2)))/((3 + 8*n + 4*n^2)*Sqrt[Sec[c + d*x]^2]) - (2*(a + (a*Tan[c + d*x])/Sqrt[Sec[c + d*
x]^2])^n*(Sec[c + d*x]^2 + Sqrt[Sec[c + d*x]^2]*Tan[c + d*x])*(1 + (Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2)^n*
(4*(1 + 2*n)*Hypergeometric2F1[3/2 + n, 3 + n, 5/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2]*(2*(Sec[c +
d*x]^2)^(3/2) + 4*Sec[c + d*x]^2*Tan[c + d*x] + 2*Sqrt[Sec[c + d*x]^2]*Tan[c + d*x]^2) + (8*(3/2 + n)*(1 + 2*n
)*(Sec[c + d*x]^2 + Sqrt[Sec[c + d*x]^2]*Tan[c + d*x])*(1 + 2*Sqrt[Sec[c + d*x]^2]*Tan[c + d*x] + 2*Tan[c + d*
x]^2)*(-Hypergeometric2F1[3/2 + n, 3 + n, 5/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2] + (1 + (Sqrt[Sec[
c + d*x]^2] + Tan[c + d*x])^2)^(-3 - n)))/(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x]) - (2*(1/2 + n)*(3 + 2*n)*(Sec[
c + d*x]^2 + Sqrt[Sec[c + d*x]^2]*Tan[c + d*x])*(-Hypergeometric2F1[1/2 + n, 1 + n, 3/2 + n, -(Sqrt[Sec[c + d*
x]^2] + Tan[c + d*x])^2] + (1 + (Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2)^(-1 - n)))/(Sqrt[Sec[c + d*x]^2] + Ta
n[c + d*x])))/((3 + 8*n + 4*n^2)*Sqrt[Sec[c + d*x]^2])))

Maple [F]

\[\int \left (\sin ^{2}\left (d x +c \right )\right ) \left (a +a \sin \left (d x +c \right )\right )^{n}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^n \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right )^{2} \,d x } \]

[In]

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

[Out]

integral(-(cos(d*x + c)^2 - 1)*(a*sin(d*x + c) + a)^n, x)

Sympy [F]

\[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^n \, dx=\int \left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{n} \sin ^{2}{\left (c + d x \right )}\, dx \]

[In]

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

[Out]

Integral((a*(sin(c + d*x) + 1))**n*sin(c + d*x)**2, x)

Maxima [F]

\[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^n \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right )^{2} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^n \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right )^{2} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^n \, dx=\int {\sin \left (c+d\,x\right )}^2\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^n \,d x \]

[In]

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

[Out]

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