3.2.59 \(\int \frac {\cos ^4(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx\) [159]

Optimal. Leaf size=63 \[ -\frac {8 a^2 \cos ^5(c+d x)}{35 d (a+a \sin (c+d x))^{5/2}}-\frac {2 a \cos ^5(c+d x)}{7 d (a+a \sin (c+d x))^{3/2}} \]

[Out]

-8/35*a^2*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(5/2)-2/7*a*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(3/2)

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Rubi [A]
time = 0.08, antiderivative size = 63, 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 = {2753, 2752} \begin {gather*} -\frac {8 a^2 \cos ^5(c+d x)}{35 d (a \sin (c+d x)+a)^{5/2}}-\frac {2 a \cos ^5(c+d x)}{7 d (a \sin (c+d x)+a)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4/Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(-8*a^2*Cos[c + d*x]^5)/(35*d*(a + a*Sin[c + d*x])^(5/2)) - (2*a*Cos[c + d*x]^5)/(7*d*(a + a*Sin[c + d*x])^(3/
2))

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 \frac {\cos ^4(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx &=-\frac {2 a \cos ^5(c+d x)}{7 d (a+a \sin (c+d x))^{3/2}}+\frac {1}{7} (4 a) \int \frac {\cos ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\\ &=-\frac {8 a^2 \cos ^5(c+d x)}{35 d (a+a \sin (c+d x))^{5/2}}-\frac {2 a \cos ^5(c+d x)}{7 d (a+a \sin (c+d x))^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 49, normalized size = 0.78 \begin {gather*} -\frac {2 \cos ^5(c+d x) (9+5 \sin (c+d x))}{35 d (1+\sin (c+d x))^2 \sqrt {a (1+\sin (c+d x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4/Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(-2*Cos[c + d*x]^5*(9 + 5*Sin[c + d*x]))/(35*d*(1 + Sin[c + d*x])^2*Sqrt[a*(1 + Sin[c + d*x])])

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Maple [A]
time = 0.46, size = 54, normalized size = 0.86

method	result	size
default	\(\frac {2 \left (1+\sin \left (d x +c \right )\right ) \left (\sin \left (d x +c \right )-1\right )^{3} \left (5 \sin \left (d x +c \right )+9\right )}{35 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4/(a+a*sin(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/35*(1+sin(d*x+c))*(sin(d*x+c)-1)^3*(5*sin(d*x+c)+9)/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))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 115 vs. \(2 (55) = 110\).
time = 0.34, size = 115, normalized size = 1.83 \begin {gather*} \frac {2 \, {\left (5 \, \cos \left (d x + c\right )^{4} - \cos \left (d x + c\right )^{3} + 2 \, \cos \left (d x + c\right )^{2} + {\left (5 \, \cos \left (d x + c\right )^{3} + 6 \, \cos \left (d x + c\right )^{2} + 8 \, \cos \left (d x + c\right ) + 16\right )} \sin \left (d x + c\right ) - 8 \, \cos \left (d x + c\right ) - 16\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{35 \, {\left (a d \cos \left (d x + c\right ) + a d \sin \left (d x + c\right ) + a 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))^(1/2),x, algorithm="fricas")

[Out]

2/35*(5*cos(d*x + c)^4 - cos(d*x + c)^3 + 2*cos(d*x + c)^2 + (5*cos(d*x + c)^3 + 6*cos(d*x + c)^2 + 8*cos(d*x
+ c) + 16)*sin(d*x + c) - 8*cos(d*x + c) - 16)*sqrt(a*sin(d*x + c) + a)/(a*d*cos(d*x + c) + a*d*sin(d*x + c) +
a*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\cos ^{4}{\left (c + d x \right )}}{\sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )}}\, dx \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))**(1/2),x)

[Out]

Integral(cos(c + d*x)**4/sqrt(a*(sin(c + d*x) + 1)), x)

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Giac [A]
time = 5.93, size = 65, normalized size = 1.03 \begin {gather*} -\frac {16 \, \sqrt {2} {\left (5 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 7 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}\right )}}{35 \, a d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\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))^(1/2),x, algorithm="giac")

[Out]

-16/35*sqrt(2)*(5*sqrt(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^7 - 7*sqrt(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^5)/(a*d*
sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\cos \left (c+d\,x\right )}^4}{\sqrt {a+a\,\sin \left (c+d\,x\right )}} \,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))^(1/2),x)

[Out]

int(cos(c + d*x)^4/(a + a*sin(c + d*x))^(1/2), x)

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