3.2.28 \(\int \cos ^4(c+d x) (a+a \sin (c+d x))^{5/2} \, dx\) [128]

Optimal. Leaf size=159 \[ -\frac {4096 a^5 \cos ^5(c+d x)}{15015 d (a+a \sin (c+d x))^{5/2}}-\frac {1024 a^4 \cos ^5(c+d x)}{3003 d (a+a \sin (c+d x))^{3/2}}-\frac {128 a^3 \cos ^5(c+d x)}{429 d \sqrt {a+a \sin (c+d x)}}-\frac {32 a^2 \cos ^5(c+d x) \sqrt {a+a \sin (c+d x)}}{143 d}-\frac {2 a \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{13 d} \]

[Out]

-4096/15015*a^5*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(5/2)-1024/3003*a^4*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(3/2)-2/13
*a*cos(d*x+c)^5*(a+a*sin(d*x+c))^(3/2)/d-128/429*a^3*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(1/2)-32/143*a^2*cos(d*x+
c)^5*(a+a*sin(d*x+c))^(1/2)/d

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Rubi [A]
time = 0.20, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2753, 2752} \begin {gather*} -\frac {4096 a^5 \cos ^5(c+d x)}{15015 d (a \sin (c+d x)+a)^{5/2}}-\frac {1024 a^4 \cos ^5(c+d x)}{3003 d (a \sin (c+d x)+a)^{3/2}}-\frac {128 a^3 \cos ^5(c+d x)}{429 d \sqrt {a \sin (c+d x)+a}}-\frac {32 a^2 \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{143 d}-\frac {2 a \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*(a + a*Sin[c + d*x])^(5/2),x]

[Out]

(-4096*a^5*Cos[c + d*x]^5)/(15015*d*(a + a*Sin[c + d*x])^(5/2)) - (1024*a^4*Cos[c + d*x]^5)/(3003*d*(a + a*Sin
[c + d*x])^(3/2)) - (128*a^3*Cos[c + d*x]^5)/(429*d*Sqrt[a + a*Sin[c + d*x]]) - (32*a^2*Cos[c + d*x]^5*Sqrt[a
+ a*Sin[c + d*x]])/(143*d) - (2*a*Cos[c + d*x]^5*(a + a*Sin[c + d*x])^(3/2))/(13*d)

Rule 2752

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

Rule 2753

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[a*((2*m + p - 1)/(m + p)), Int
[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2,
0] && IGtQ[Simplify[(2*m + p - 1)/2], 0] && NeQ[m + p, 0]

Rubi steps

\begin {align*} \int \cos ^4(c+d x) (a+a \sin (c+d x))^{5/2} \, dx &=-\frac {2 a \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{13 d}+\frac {1}{13} (16 a) \int \cos ^4(c+d x) (a+a \sin (c+d x))^{3/2} \, dx\\ &=-\frac {32 a^2 \cos ^5(c+d x) \sqrt {a+a \sin (c+d x)}}{143 d}-\frac {2 a \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{13 d}+\frac {1}{143} \left (192 a^2\right ) \int \cos ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx\\ &=-\frac {128 a^3 \cos ^5(c+d x)}{429 d \sqrt {a+a \sin (c+d x)}}-\frac {32 a^2 \cos ^5(c+d x) \sqrt {a+a \sin (c+d x)}}{143 d}-\frac {2 a \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{13 d}+\frac {1}{429} \left (512 a^3\right ) \int \frac {\cos ^4(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx\\ &=-\frac {1024 a^4 \cos ^5(c+d x)}{3003 d (a+a \sin (c+d x))^{3/2}}-\frac {128 a^3 \cos ^5(c+d x)}{429 d \sqrt {a+a \sin (c+d x)}}-\frac {32 a^2 \cos ^5(c+d x) \sqrt {a+a \sin (c+d x)}}{143 d}-\frac {2 a \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{13 d}+\frac {\left (2048 a^4\right ) \int \frac {\cos ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx}{3003}\\ &=-\frac {4096 a^5 \cos ^5(c+d x)}{15015 d (a+a \sin (c+d x))^{5/2}}-\frac {1024 a^4 \cos ^5(c+d x)}{3003 d (a+a \sin (c+d x))^{3/2}}-\frac {128 a^3 \cos ^5(c+d x)}{429 d \sqrt {a+a \sin (c+d x)}}-\frac {32 a^2 \cos ^5(c+d x) \sqrt {a+a \sin (c+d x)}}{143 d}-\frac {2 a \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{13 d}\\ \end {align*}

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Mathematica [A]
time = 0.19, size = 79, normalized size = 0.50 \begin {gather*} -\frac {2 \cos ^5(c+d x) (a (1+\sin (c+d x)))^{5/2} \left (9683+16700 \sin (c+d x)+14210 \sin ^2(c+d x)+6300 \sin ^3(c+d x)+1155 \sin ^4(c+d x)\right )}{15015 d (1+\sin (c+d x))^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*(a + a*Sin[c + d*x])^(5/2),x]

[Out]

(-2*Cos[c + d*x]^5*(a*(1 + Sin[c + d*x]))^(5/2)*(9683 + 16700*Sin[c + d*x] + 14210*Sin[c + d*x]^2 + 6300*Sin[c
+ d*x]^3 + 1155*Sin[c + d*x]^4))/(15015*d*(1 + Sin[c + d*x])^5)

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Maple [A]
time = 0.38, size = 87, normalized size = 0.55

method	result	size
default	\(\frac {2 \left (1+\sin \left (d x +c \right )\right ) a^{3} \left (\sin \left (d x +c \right )-1\right )^{3} \left (1155 \left (\sin ^{4}\left (d x +c \right )\right )+6300 \left (\sin ^{3}\left (d x +c \right )\right )+14210 \left (\sin ^{2}\left (d x +c \right )\right )+16700 \sin \left (d x +c \right )+9683\right )}{15015 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(87\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*(a+a*sin(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/15015*(1+sin(d*x+c))*a^3*(sin(d*x+c)-1)^3*(1155*sin(d*x+c)^4+6300*sin(d*x+c)^3+14210*sin(d*x+c)^2+16700*sin(
d*x+c)+9683)/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

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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(cos(d*x+c)^4*(a+a*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((a*sin(d*x + c) + a)^(5/2)*cos(d*x + c)^4, x)

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Fricas [A]
time = 0.37, size = 219, normalized size = 1.38 \begin {gather*} \frac {2 \, {\left (1155 \, a^{2} \cos \left (d x + c\right )^{7} - 2835 \, a^{2} \cos \left (d x + c\right )^{6} - 6230 \, a^{2} \cos \left (d x + c\right )^{5} + 320 \, a^{2} \cos \left (d x + c\right )^{4} - 512 \, a^{2} \cos \left (d x + c\right )^{3} + 1024 \, a^{2} \cos \left (d x + c\right )^{2} - 4096 \, a^{2} \cos \left (d x + c\right ) - 8192 \, a^{2} - {\left (1155 \, a^{2} \cos \left (d x + c\right )^{6} + 3990 \, a^{2} \cos \left (d x + c\right )^{5} - 2240 \, a^{2} \cos \left (d x + c\right )^{4} - 2560 \, a^{2} \cos \left (d x + c\right )^{3} - 3072 \, a^{2} \cos \left (d x + c\right )^{2} - 4096 \, a^{2} \cos \left (d x + c\right ) - 8192 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{15015 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

2/15015*(1155*a^2*cos(d*x + c)^7 - 2835*a^2*cos(d*x + c)^6 - 6230*a^2*cos(d*x + c)^5 + 320*a^2*cos(d*x + c)^4
- 512*a^2*cos(d*x + c)^3 + 1024*a^2*cos(d*x + c)^2 - 4096*a^2*cos(d*x + c) - 8192*a^2 - (1155*a^2*cos(d*x + c)
^6 + 3990*a^2*cos(d*x + c)^5 - 2240*a^2*cos(d*x + c)^4 - 2560*a^2*cos(d*x + c)^3 - 3072*a^2*cos(d*x + c)^2 - 4
096*a^2*cos(d*x + c) - 8192*a^2)*sin(d*x + c))*sqrt(a*sin(d*x + c) + a)/(d*cos(d*x + c) + d*sin(d*x + c) + d)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*(a+a*sin(d*x+c))**(5/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3060 deep

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Giac [A]
time = 5.40, size = 172, normalized size = 1.08 \begin {gather*} \frac {128 \, \sqrt {2} {\left (1155 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} - 5460 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 10010 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 8580 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 3003 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}\right )} \sqrt {a}}{15015 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+a*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

128/15015*sqrt(2)*(1155*a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)^13 - 5460*a^2*s
gn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)^11 + 10010*a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1
/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)^9 - 8580*a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x +
1/2*c)^7 + 3003*a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)^5)*sqrt(a)/d

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\cos \left (c+d\,x\right )}^4\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^4*(a + a*sin(c + d*x))^(5/2),x)

[Out]

int(cos(c + d*x)^4*(a + a*sin(c + d*x))^(5/2), x)

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