3.328 \(\int (e \cos (c+d x))^p (a+a \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=95 \[ -\frac{a^3 2^{\frac{p}{2}+\frac{7}{2}} (\sin (c+d x)+1)^{\frac{1}{2} (-p-1)} (e \cos (c+d x))^{p+1} \, _2F_1\left (\frac{1}{2} (-p-5),\frac{p+1}{2};\frac{p+3}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{d e (p+1)} \]

[Out]

-((2^(7/2 + p/2)*a^3*(e*Cos[c + d*x])^(1 + p)*Hypergeometric2F1[(-5 - p)/2, (1 + p)/2, (3 + p)/2, (1 - Sin[c +
d*x])/2]*(1 + Sin[c + d*x])^((-1 - p)/2))/(d*e*(1 + p)))

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Rubi [A]  time = 0.0801343, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {2688, 69} \[ -\frac{a^3 2^{\frac{p}{2}+\frac{7}{2}} (\sin (c+d x)+1)^{\frac{1}{2} (-p-1)} (e \cos (c+d x))^{p+1} \, _2F_1\left (\frac{1}{2} (-p-5),\frac{p+1}{2};\frac{p+3}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{d e (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((2^(7/2 + p/2)*a^3*(e*Cos[c + d*x])^(1 + p)*Hypergeometric2F1[(-5 - p)/2, (1 + p)/2, (3 + p)/2, (1 - Sin[c +
d*x])/2]*(1 + Sin[c + d*x])^((-1 - p)/2))/(d*e*(1 + p)))

Rule 2688

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

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rubi steps

\begin{align*} \int (e \cos (c+d x))^p (a+a \sin (c+d x))^3 \, dx &=\frac{\left (a^3 (e \cos (c+d x))^{1+p} (1-\sin (c+d x))^{\frac{1}{2} (-1-p)} (1+\sin (c+d x))^{\frac{1}{2} (-1-p)}\right ) \operatorname{Subst}\left (\int (1-x)^{\frac{1}{2} (-1+p)} (1+x)^{3+\frac{1}{2} (-1+p)} \, dx,x,\sin (c+d x)\right )}{d e}\\ &=-\frac{2^{\frac{7}{2}+\frac{p}{2}} a^3 (e \cos (c+d x))^{1+p} \, _2F_1\left (\frac{1}{2} (-5-p),\frac{1+p}{2};\frac{3+p}{2};\frac{1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{\frac{1}{2} (-1-p)}}{d e (1+p)}\\ \end{align*}

Mathematica [A]  time = 0.105313, size = 94, normalized size = 0.99 \[ -\frac{a^3 2^{\frac{p+7}{2}} \cos (c+d x) (\sin (c+d x)+1)^{\frac{1}{2} (-p-1)} (e \cos (c+d x))^p \, _2F_1\left (\frac{1}{2} (-p-5),\frac{p+1}{2};\frac{p+3}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{d (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((2^((7 + p)/2)*a^3*Cos[c + d*x]*(e*Cos[c + d*x])^p*Hypergeometric2F1[(-5 - p)/2, (1 + p)/2, (3 + p)/2, (1 -
Sin[c + d*x])/2]*(1 + Sin[c + d*x])^((-1 - p)/2))/(d*(1 + p)))

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Maple [F]  time = 3.378, size = 0, normalized size = 0. \begin{align*} \int \left ( e\cos \left ( dx+c \right ) \right ) ^{p} \left ( a+a\sin \left ( dx+c \right ) \right ) ^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (d x + c\right ) + a\right )}^{3} \left (e \cos \left (d x + c\right )\right )^{p}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (3 \, a^{3} \cos \left (d x + c\right )^{2} - 4 \, a^{3} +{\left (a^{3} \cos \left (d x + c\right )^{2} - 4 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \left (e \cos \left (d x + c\right )\right )^{p}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(3*a^3*cos(d*x + c)^2 - 4*a^3 + (a^3*cos(d*x + c)^2 - 4*a^3)*sin(d*x + c))*(e*cos(d*x + c))^p, x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (d x + c\right ) + a\right )}^{3} \left (e \cos \left (d x + c\right )\right )^{p}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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