3.337 \(\int (e \cos (c+d x))^p (a+a \sin (c+d x))^{3/2} \, dx\)

Optimal. Leaf size=103 \[ -\frac {a^2 2^{\frac {p}{2}+2} (\sin (c+d x)+1)^{-p/2} (e \cos (c+d x))^{p+1} \, _2F_1\left (\frac {1}{2} (-p-2),\frac {p+1}{2};\frac {p+3}{2};\frac {1}{2} (1-\sin (c+d x))\right )}{d e (p+1) \sqrt {a \sin (c+d x)+a}} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2689, 70, 69} \[ -\frac {a^2 2^{\frac {p}{2}+2} (\sin (c+d x)+1)^{-p/2} (e \cos (c+d x))^{p+1} \, _2F_1\left (\frac {1}{2} (-p-2),\frac {p+1}{2};\frac {p+3}{2};\frac {1}{2} (1-\sin (c+d x))\right )}{d e (p+1) \sqrt {a \sin (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[(e*Cos[c + d*x])^p*(a + a*Sin[c + d*x])^(3/2),x]

[Out]

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

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]))

Rule 70

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 2689

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

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Mathematica [A]  time = 0.19, size = 101, normalized size = 0.98 \[ -\frac {2^{\frac {p}{2}+2} \cos (c+d x) (a (\sin (c+d x)+1))^{3/2} (\sin (c+d x)+1)^{-\frac {p}{2}-2} (e \cos (c+d x))^p \, _2F_1\left (-\frac {p}{2}-1,\frac {p+1}{2};\frac {p+3}{2};\frac {1}{2} (1-\sin (c+d x))\right )}{d (p+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Cos[c + d*x])^p*(a + a*Sin[c + d*x])^(3/2),x]

[Out]

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

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \left (e \cos \left (d x + c\right )\right )^{p}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^p*(a+a*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^p*(a+a*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(cos((d*x+c)/2-pi/4))]Simplification assuming c near 0Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to c
heck sign: (2*pi/x/2)>(-2*pi/x/2)Unable to divide, perhaps due to rounding error%%%{-64*i,[0,2,0,2,2,1,1,1,1]%
%%} / %%%{128*i,[0,2,0,2,2,0,0,0,0]%%%} Error: Bad Argument Value

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maple [F]  time = 0.21, size = 0, normalized size = 0.00 \[ \int \left (e \cos \left (d x +c \right )\right )^{p} \left (a +a \sin \left (d x +c \right )\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \left (e \cos \left (d x + c\right )\right )^{p}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^p*(a+a*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (e\,\cos \left (c+d\,x\right )\right )}^p\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}} \left (e \cos {\left (c + d x \right )}\right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))**p*(a+a*sin(d*x+c))**(3/2),x)

[Out]

Integral((a*(sin(c + d*x) + 1))**(3/2)*(e*cos(c + d*x))**p, x)

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