3.4.28 \(\int (e \cos (c+d x))^p (a+a \sin (c+d x))^3 \, dx\) [328]

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

[Out]

-2^(7/2+1/2*p)*a^3*(e*cos(d*x+c))^(1+p)*hypergeom([-5/2-1/2*p, 1/2+1/2*p],[3/2+1/2*p],1/2-1/2*sin(d*x+c))*(1+s
in(d*x+c))^(-1/2-1/2*p)/d/e/(1+p)

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Rubi [A]
time = 0.06, antiderivative size = 95, normalized size of antiderivative = 1.00, 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 = {2767, 71} \begin {gather*} -\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)} \end {gather*}

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 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], 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]))

Rule 2767

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/a)*x)^(m + (p - 1)/2)*(1 - (b/a)*x)^((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]

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 ) \text {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*}

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Mathematica [A]
time = 0.11, size = 94, normalized size = 0.99 \begin {gather*} -\frac {2^{\frac {7+p}{2}} a^3 \cos (c+d x) (e \cos (c+d x))^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 (1+p)} \end {gather*}

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 = 0.56, size = 0, normalized size = 0.00 \[\int \left (e \cos \left (d x +c \right )\right )^{p} \left (a +a \sin \left (d x +c \right )\right )^{3}\, dx\]

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 = 0.00, 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((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*(cos(d*x + c)*e)^p, 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((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))*(cos(d*x + c)*e)^p, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a^{3} \left (\int \left (e \cos {\left (c + d x \right )}\right )^{p}\, dx + \int 3 \left (e \cos {\left (c + d x \right )}\right )^{p} \sin {\left (c + d x \right )}\, dx + \int 3 \left (e \cos {\left (c + d x \right )}\right )^{p} \sin ^{2}{\left (c + d x \right )}\, dx + \int \left (e \cos {\left (c + d x \right )}\right )^{p} \sin ^{3}{\left (c + d x \right )}\, dx\right ) \end {gather*}

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]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (e\,\cos \left (c+d\,x\right )\right )}^p\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^3 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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