\(\int (e \cos (c+d x))^p (a+b \sin (c+d x))^3 \, dx\) [616]

Optimal result

Integrand size = 23, antiderivative size = 229 \[ \int (e \cos (c+d x))^p (a+b \sin (c+d x))^3 \, dx=-\frac {b \left (2 b^2 (2+p)+a^2 \left (11+6 p+p^2\right )\right ) (e \cos (c+d x))^{1+p}}{d e (1+p) (2+p) (3+p)}-\frac {a \left (3 b^2+a^2 (2+p)\right ) (e \cos (c+d x))^{1+p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+p}{2},\frac {3+p}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{d e (1+p) (2+p) \sqrt {\sin ^2(c+d x)}}-\frac {a b (5+p) (e \cos (c+d x))^{1+p} (a+b \sin (c+d x))}{d e (2+p) (3+p)}-\frac {b (e \cos (c+d x))^{1+p} (a+b \sin (c+d x))^2}{d e (3+p)} \]

[Out]

-b*(2*b^2*(2+p)+a^2*(p^2+6*p+11))*(e*cos(d*x+c))^(p+1)/d/e/(3+p)/(p^2+3*p+2)-a*b*(5+p)*(e*cos(d*x+c))^(p+1)*(a
+b*sin(d*x+c))/d/e/(2+p)/(3+p)-b*(e*cos(d*x+c))^(p+1)*(a+b*sin(d*x+c))^2/d/e/(3+p)-a*(3*b^2+a^2*(2+p))*(e*cos(
d*x+c))^(p+1)*hypergeom([1/2, 1/2+1/2*p],[3/2+1/2*p],cos(d*x+c)^2)*sin(d*x+c)/d/e/(p+1)/(2+p)/(sin(d*x+c)^2)^(
1/2)

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2771, 2941, 2748, 2722} \[ \int (e \cos (c+d x))^p (a+b \sin (c+d x))^3 \, dx=-\frac {a \left (a^2 (p+2)+3 b^2\right ) \sin (c+d x) (e \cos (c+d x))^{p+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {p+1}{2},\frac {p+3}{2},\cos ^2(c+d x)\right )}{d e (p+1) (p+2) \sqrt {\sin ^2(c+d x)}}-\frac {b \left (a^2 \left (p^2+6 p+11\right )+2 b^2 (p+2)\right ) (e \cos (c+d x))^{p+1}}{d e (p+1) (p+2) (p+3)}-\frac {b (a+b \sin (c+d x))^2 (e \cos (c+d x))^{p+1}}{d e (p+3)}-\frac {a b (p+5) (a+b \sin (c+d x)) (e \cos (c+d x))^{p+1}}{d e (p+2) (p+3)} \]

[In]

Int[(e*Cos[c + d*x])^p*(a + b*Sin[c + d*x])^3,x]

[Out]

-((b*(2*b^2*(2 + p) + a^2*(11 + 6*p + p^2))*(e*Cos[c + d*x])^(1 + p))/(d*e*(1 + p)*(2 + p)*(3 + p))) - (a*(3*b
^2 + a^2*(2 + p))*(e*Cos[c + d*x])^(1 + p)*Hypergeometric2F1[1/2, (1 + p)/2, (3 + p)/2, Cos[c + d*x]^2]*Sin[c
+ d*x])/(d*e*(1 + p)*(2 + p)*Sqrt[Sin[c + d*x]^2]) - (a*b*(5 + p)*(e*Cos[c + d*x])^(1 + p)*(a + b*Sin[c + d*x]
))/(d*e*(2 + p)*(3 + p)) - (b*(e*Cos[c + d*x])^(1 + p)*(a + b*Sin[c + d*x])^2)/(d*e*(3 + p))

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 2748

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-b)*((g*Co
s[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + Dist[a, Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x
] && (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])

Rule 2771

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(-b)*(
g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^(m - 1)/(f*g*(m + p))), x] + Dist[1/(m + p), Int[(g*Cos[e + f*x]
)^p*(a + b*Sin[e + f*x])^(m - 2)*(b^2*(m - 1) + a^2*(m + p) + a*b*(2*m + p - 1)*Sin[e + f*x]), x], x] /; FreeQ
[{a, b, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && NeQ[m + p, 0] && (IntegersQ[2*m, 2*p] || IntegerQ
[m])

Rule 2941

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

Rubi steps \begin{align*} \text {integral}& = -\frac {b (e \cos (c+d x))^{1+p} (a+b \sin (c+d x))^2}{d e (3+p)}+\frac {\int (e \cos (c+d x))^p (a+b \sin (c+d x)) \left (2 b^2+a^2 (3+p)+a b (5+p) \sin (c+d x)\right ) \, dx}{3+p} \\ & = -\frac {a b (5+p) (e \cos (c+d x))^{1+p} (a+b \sin (c+d x))}{d e (2+p) (3+p)}-\frac {b (e \cos (c+d x))^{1+p} (a+b \sin (c+d x))^2}{d e (3+p)}+\frac {\int (e \cos (c+d x))^p \left (a (3+p) \left (3 b^2+a^2 (2+p)\right )+b \left (2 b^2 (2+p)+a^2 \left (11+6 p+p^2\right )\right ) \sin (c+d x)\right ) \, dx}{6+5 p+p^2} \\ & = -\frac {b \left (2 b^2 (2+p)+a^2 \left (11+6 p+p^2\right )\right ) (e \cos (c+d x))^{1+p}}{d e (1+p) \left (6+5 p+p^2\right )}-\frac {a b (5+p) (e \cos (c+d x))^{1+p} (a+b \sin (c+d x))}{d e (2+p) (3+p)}-\frac {b (e \cos (c+d x))^{1+p} (a+b \sin (c+d x))^2}{d e (3+p)}+\left (a \left (a^2+\frac {3 b^2}{2+p}\right )\right ) \int (e \cos (c+d x))^p \, dx \\ & = -\frac {b \left (2 b^2 (2+p)+a^2 \left (11+6 p+p^2\right )\right ) (e \cos (c+d x))^{1+p}}{d e (1+p) \left (6+5 p+p^2\right )}-\frac {a \left (a^2+\frac {3 b^2}{2+p}\right ) (e \cos (c+d x))^{1+p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+p}{2},\frac {3+p}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{d e (1+p) \sqrt {\sin ^2(c+d x)}}-\frac {a b (5+p) (e \cos (c+d x))^{1+p} (a+b \sin (c+d x))}{d e (2+p) (3+p)}-\frac {b (e \cos (c+d x))^{1+p} (a+b \sin (c+d x))^2}{d e (3+p)} \\ \end{align*}

Mathematica [A] (warning: unable to verify)

Time = 13.85 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.19 \[ \int (e \cos (c+d x))^p (a+b \sin (c+d x))^3 \, dx=\frac {8 (e \cos (c+d x))^p \sec ^2(c+d x)^{p/2} (a+b \sin (c+d x))^3 \left (a^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4+p}{2},\frac {3}{2},-\tan ^2(c+d x)\right ) \tan (c+d x)+\frac {1}{3} a \left (a^2+3 b^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {4+p}{2},\frac {5}{2},-\tan ^2(c+d x)\right ) \tan ^3(c+d x)-\frac {b \sec ^2(c+d x)^{-\frac {3}{2}-\frac {p}{2}} \left (3 a^2 (3+p) \sec ^2(c+d x)+b^2 \left (2+(3+p) \tan ^2(c+d x)\right )\right )}{(1+p) (3+p)}\right )}{d \left (8 a^3+12 a b^2-12 a b^2 \cos (2 (c+d x))+2 b \left (6 a^2+b^2\right ) \sqrt {\sec ^2(c+d x)} \sin (2 (c+d x))-b^3 \sqrt {\sec ^2(c+d x)} \sin (4 (c+d x))\right )} \]

[In]

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

[Out]

(8*(e*Cos[c + d*x])^p*(Sec[c + d*x]^2)^(p/2)*(a + b*Sin[c + d*x])^3*(a^3*Hypergeometric2F1[1/2, (4 + p)/2, 3/2
, -Tan[c + d*x]^2]*Tan[c + d*x] + (a*(a^2 + 3*b^2)*Hypergeometric2F1[3/2, (4 + p)/2, 5/2, -Tan[c + d*x]^2]*Tan
[c + d*x]^3)/3 - (b*(Sec[c + d*x]^2)^(-3/2 - p/2)*(3*a^2*(3 + p)*Sec[c + d*x]^2 + b^2*(2 + (3 + p)*Tan[c + d*x
]^2)))/((1 + p)*(3 + p))))/(d*(8*a^3 + 12*a*b^2 - 12*a*b^2*Cos[2*(c + d*x)] + 2*b*(6*a^2 + b^2)*Sqrt[Sec[c + d
*x]^2]*Sin[2*(c + d*x)] - b^3*Sqrt[Sec[c + d*x]^2]*Sin[4*(c + d*x)]))

Maple [F]

\[\int \left (e \cos \left (d x +c \right )\right )^{p} \left (a +b \sin \left (d x +c \right )\right )^{3}d x\]

[In]

int((e*cos(d*x+c))^p*(a+b*sin(d*x+c))^3,x)

[Out]

int((e*cos(d*x+c))^p*(a+b*sin(d*x+c))^3,x)

Fricas [F]

\[ \int (e \cos (c+d x))^p (a+b \sin (c+d x))^3 \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{3} \left (e \cos \left (d x + c\right )\right )^{p} \,d x } \]

[In]

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

[Out]

integral(-(3*a*b^2*cos(d*x + c)^2 - a^3 - 3*a*b^2 + (b^3*cos(d*x + c)^2 - 3*a^2*b - b^3)*sin(d*x + c))*(e*cos(
d*x + c))^p, x)

Sympy [F]

\[ \int (e \cos (c+d x))^p (a+b \sin (c+d x))^3 \, dx=\int \left (e \cos {\left (c + d x \right )}\right )^{p} \left (a + b \sin {\left (c + d x \right )}\right )^{3}\, dx \]

[In]

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

[Out]

Integral((e*cos(c + d*x))**p*(a + b*sin(c + d*x))**3, x)

Maxima [F]

\[ \int (e \cos (c+d x))^p (a+b \sin (c+d x))^3 \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{3} \left (e \cos \left (d x + c\right )\right )^{p} \,d x } \]

[In]

integrate((e*cos(d*x+c))^p*(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

integrate((b*sin(d*x + c) + a)^3*(e*cos(d*x + c))^p, x)

Giac [F]

\[ \int (e \cos (c+d x))^p (a+b \sin (c+d x))^3 \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{3} \left (e \cos \left (d x + c\right )\right )^{p} \,d x } \]

[In]

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

[Out]

integrate((b*sin(d*x + c) + a)^3*(e*cos(d*x + c))^p, x)

Mupad [F(-1)]

Timed out. \[ \int (e \cos (c+d x))^p (a+b \sin (c+d x))^3 \, dx=\int {\left (e\,\cos \left (c+d\,x\right )\right )}^p\,{\left (a+b\,\sin \left (c+d\,x\right )\right )}^3 \,d x \]

[In]

int((e*cos(c + d*x))^p*(a + b*sin(c + d*x))^3,x)

[Out]

int((e*cos(c + d*x))^p*(a + b*sin(c + d*x))^3, x)