3.4.42 \(\int (e \cos (c+d x))^p (a+a \sin (c+d x))^m \, dx\) [342]

Optimal. Leaf size=114 \[ -\frac {2^{\frac {1}{2}+m+\frac {p}{2}} a (e \cos (c+d x))^{1+p} \, _2F_1\left (\frac {1}{2} (1-2 m-p),\frac {1+p}{2};\frac {3+p}{2};\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{\frac {1}{2} (1-2 m-p)} (a+a \sin (c+d x))^{-1+m}}{d e (1+p)} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {2768, 72, 71} \begin {gather*} -\frac {a 2^{m+\frac {p}{2}+\frac {1}{2}} (a \sin (c+d x)+a)^{m-1} (e \cos (c+d x))^{p+1} (\sin (c+d x)+1)^{\frac {1}{2} (-2 m-p+1)} \, _2F_1\left (\frac {1}{2} (-2 m-p+1),\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])^m,x]

[Out]

-((2^(1/2 + m + p/2)*a*(e*Cos[c + d*x])^(1 + p)*Hypergeometric2F1[(1 - 2*m - p)/2, (1 + p)/2, (3 + p)/2, (1 -
Sin[c + d*x])/2]*(1 + Sin[c + d*x])^((1 - 2*m - p)/2)*(a + a*Sin[c + d*x])^(-1 + m))/(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 72

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

Rule 2768

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[a^2*(
(g*Cos[e + f*x])^(p + 1)/(f*g*(a + b*Sin[e + f*x])^((p + 1)/2)*(a - b*Sin[e + f*x])^((p + 1)/2))), Subst[Int[(
a + b*x)^(m + (p - 1)/2)*(a - b*x)^((p - 1)/2), x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, g, m, 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))^m \, dx &=\frac {\left (a^2 (e \cos (c+d x))^{1+p} (a-a \sin (c+d x))^{\frac {1}{2} (-1-p)} (a+a \sin (c+d x))^{\frac {1}{2} (-1-p)}\right ) \text {Subst}\left (\int (a-a x)^{\frac {1}{2} (-1+p)} (a+a x)^{m+\frac {1}{2} (-1+p)} \, dx,x,\sin (c+d x)\right )}{d e}\\ &=\frac {\left (2^{-\frac {1}{2}+m+\frac {p}{2}} a^2 (e \cos (c+d x))^{1+p} (a-a \sin (c+d x))^{\frac {1}{2} (-1-p)} (a+a \sin (c+d x))^{-\frac {1}{2}+m+\frac {1}{2} (-1-p)+\frac {p}{2}} \left (\frac {a+a \sin (c+d x)}{a}\right )^{\frac {1}{2}-m-\frac {p}{2}}\right ) \text {Subst}\left (\int \left (\frac {1}{2}+\frac {x}{2}\right )^{m+\frac {1}{2} (-1+p)} (a-a x)^{\frac {1}{2} (-1+p)} \, dx,x,\sin (c+d x)\right )}{d e}\\ &=-\frac {2^{\frac {1}{2}+m+\frac {p}{2}} a (e \cos (c+d x))^{1+p} \, _2F_1\left (\frac {1}{2} (1-2 m-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} (1-2 m-p)} (a+a \sin (c+d x))^{-1+m}}{d e (1+p)}\\ \end {align*}

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Mathematica [A]
time = 0.19, size = 112, normalized size = 0.98 \begin {gather*} -\frac {2^{\frac {1}{2} (1+2 m+p)} \cos (c+d x) (e \cos (c+d x))^p \, _2F_1\left (\frac {1}{2} (1-2 m-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} (-1-2 m-p)} (a (1+\sin (c+d x)))^m}{d (1+p)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.13, size = 0, normalized size = 0.00 \[\int \left (e \cos \left (d x +c \right )\right )^{p} \left (a +a \sin \left (d x +c \right )\right )^{m}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

integral((cos(d*x + c)*e)^p*(a*sin(d*x + c) + a)^m, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{m} \left (e \cos {\left (c + d x \right )}\right )^{p}\, dx \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))**m,x)

[Out]

Integral((a*(sin(c + d*x) + 1))**m*(e*cos(c + d*x))**p, 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))^m,x, algorithm="giac")

[Out]

integrate((cos(d*x + c)*e)^p*(a*sin(d*x + c) + a)^m, 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 )}^m \,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))^m,x)

[Out]

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

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