\(\int (d \sin (e+f x))^m (a+b \sin (e+f x)) \, dx\) [215]

Optimal result

Integrand size = 21, antiderivative size = 139 \[ \int (d \sin (e+f x))^m (a+b \sin (e+f x)) \, dx=\frac {a \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\sin ^2(e+f x)\right ) (d \sin (e+f x))^{1+m}}{d f (1+m) \sqrt {\cos ^2(e+f x)}}+\frac {b \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},\sin ^2(e+f x)\right ) (d \sin (e+f x))^{2+m}}{d^2 f (2+m) \sqrt {\cos ^2(e+f x)}} \]

[Out]

a*cos(f*x+e)*hypergeom([1/2, 1/2+1/2*m],[3/2+1/2*m],sin(f*x+e)^2)*(d*sin(f*x+e))^(1+m)/d/f/(1+m)/(cos(f*x+e)^2
)^(1/2)+b*cos(f*x+e)*hypergeom([1/2, 1+1/2*m],[2+1/2*m],sin(f*x+e)^2)*(d*sin(f*x+e))^(2+m)/d^2/f/(2+m)/(cos(f*
x+e)^2)^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2827, 2722} \[ \int (d \sin (e+f x))^m (a+b \sin (e+f x)) \, dx=\frac {a \cos (e+f x) (d \sin (e+f x))^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},\sin ^2(e+f x)\right )}{d f (m+1) \sqrt {\cos ^2(e+f x)}}+\frac {b \cos (e+f x) (d \sin (e+f x))^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},\sin ^2(e+f x)\right )}{d^2 f (m+2) \sqrt {\cos ^2(e+f x)}} \]

[In]

Int[(d*Sin[e + f*x])^m*(a + b*Sin[e + f*x]),x]

[Out]

(a*Cos[e + f*x]*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, Sin[e + f*x]^2]*(d*Sin[e + f*x])^(1 + m))/(d*f*(1
+ m)*Sqrt[Cos[e + f*x]^2]) + (b*Cos[e + f*x]*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, Sin[e + f*x]^2]*(d*
Sin[e + f*x])^(2 + m))/(d^2*f*(2 + m)*Sqrt[Cos[e + f*x]^2])

Rule 2722

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

Rule 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rubi steps \begin{align*} \text {integral}& = a \int (d \sin (e+f x))^m \, dx+\frac {b \int (d \sin (e+f x))^{1+m} \, dx}{d} \\ & = \frac {a \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\sin ^2(e+f x)\right ) (d \sin (e+f x))^{1+m}}{d f (1+m) \sqrt {\cos ^2(e+f x)}}+\frac {b \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},\sin ^2(e+f x)\right ) (d \sin (e+f x))^{2+m}}{d^2 f (2+m) \sqrt {\cos ^2(e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.80 \[ \int (d \sin (e+f x))^m (a+b \sin (e+f x)) \, dx=\frac {\sqrt {\cos ^2(e+f x)} (d \sin (e+f x))^m \left (a (2+m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\sin ^2(e+f x)\right )+b (1+m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},\sin ^2(e+f x)\right ) \sin (e+f x)\right ) \tan (e+f x)}{f (1+m) (2+m)} \]

[In]

Integrate[(d*Sin[e + f*x])^m*(a + b*Sin[e + f*x]),x]

[Out]

(Sqrt[Cos[e + f*x]^2]*(d*Sin[e + f*x])^m*(a*(2 + m)*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, Sin[e + f*x]^
2] + b*(1 + m)*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, Sin[e + f*x]^2]*Sin[e + f*x])*Tan[e + f*x])/(f*(1
+ m)*(2 + m))

Maple [F]

\[\int \left (d \sin \left (f x +e \right )\right )^{m} \left (a +b \sin \left (f x +e \right )\right )d x\]

[In]

int((d*sin(f*x+e))^m*(a+b*sin(f*x+e)),x)

[Out]

int((d*sin(f*x+e))^m*(a+b*sin(f*x+e)),x)

Fricas [F]

\[ \int (d \sin (e+f x))^m (a+b \sin (e+f x)) \, dx=\int { {\left (b \sin \left (f x + e\right ) + a\right )} \left (d \sin \left (f x + e\right )\right )^{m} \,d x } \]

[In]

integrate((d*sin(f*x+e))^m*(a+b*sin(f*x+e)),x, algorithm="fricas")

[Out]

integral((b*sin(f*x + e) + a)*(d*sin(f*x + e))^m, x)

Sympy [F]

\[ \int (d \sin (e+f x))^m (a+b \sin (e+f x)) \, dx=\int \left (d \sin {\left (e + f x \right )}\right )^{m} \left (a + b \sin {\left (e + f x \right )}\right )\, dx \]

[In]

integrate((d*sin(f*x+e))**m*(a+b*sin(f*x+e)),x)

[Out]

Integral((d*sin(e + f*x))**m*(a + b*sin(e + f*x)), x)

Maxima [F]

\[ \int (d \sin (e+f x))^m (a+b \sin (e+f x)) \, dx=\int { {\left (b \sin \left (f x + e\right ) + a\right )} \left (d \sin \left (f x + e\right )\right )^{m} \,d x } \]

[In]

integrate((d*sin(f*x+e))^m*(a+b*sin(f*x+e)),x, algorithm="maxima")

[Out]

integrate((b*sin(f*x + e) + a)*(d*sin(f*x + e))^m, x)

Giac [F]

\[ \int (d \sin (e+f x))^m (a+b \sin (e+f x)) \, dx=\int { {\left (b \sin \left (f x + e\right ) + a\right )} \left (d \sin \left (f x + e\right )\right )^{m} \,d x } \]

[In]

integrate((d*sin(f*x+e))^m*(a+b*sin(f*x+e)),x, algorithm="giac")

[Out]

integrate((b*sin(f*x + e) + a)*(d*sin(f*x + e))^m, x)

Mupad [F(-1)]

Timed out. \[ \int (d \sin (e+f x))^m (a+b \sin (e+f x)) \, dx=\int {\left (d\,\sin \left (e+f\,x\right )\right )}^m\,\left (a+b\,\sin \left (e+f\,x\right )\right ) \,d x \]

[In]

int((d*sin(e + f*x))^m*(a + b*sin(e + f*x)),x)

[Out]

int((d*sin(e + f*x))^m*(a + b*sin(e + f*x)), x)