3.172 \(\int \frac {\cos ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=30 \[ -\frac {2 a \cos ^5(c+d x)}{5 d (a \sin (c+d x)+a)^{5/2}} \]

[Out]

-2/5*a*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(5/2)

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Rubi [A]  time = 0.06, antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {2673} \[ -\frac {2 a \cos ^5(c+d x)}{5 d (a \sin (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4/(a + a*Sin[c + d*x])^(3/2),x]

[Out]

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

Rule 2673

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 - 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]

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 42, normalized size = 1.40 \[ -\frac {2 \cos ^5(c+d x) \sqrt {a (\sin (c+d x)+1)}}{5 a^2 d (\sin (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4/(a + a*Sin[c + d*x])^(3/2),x]

[Out]

(-2*Cos[c + d*x]^5*Sqrt[a*(1 + Sin[c + d*x])])/(5*a^2*d*(1 + Sin[c + d*x])^3)

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fricas [B]  time = 0.71, size = 98, normalized size = 3.27 \[ \frac {2 \, {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) - 4\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 4\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{5 \, {\left (a^{2} d \cos \left (d x + c\right ) + a^{2} d \sin \left (d x + c\right ) + a^{2} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a+a*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 1.73, size = 199, normalized size = 6.63 \[ \frac {2 \, {\left (\frac {4 \, \sqrt {2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{\frac {3}{2}}} + \frac {{\left ({\left ({\left ({\left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} - \frac {5 \, a}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {10 \, a}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {10 \, a}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {5 \, a}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {a}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {5}{2}}}\right )}}{5 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a+a*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

2/5*(4*sqrt(2)*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^(3/2) + (((((a*tan(1/2*d*x + 1/2*c)/sgn(tan(1/2*d*x + 1/2*c) +
1) - 5*a/sgn(tan(1/2*d*x + 1/2*c) + 1))*tan(1/2*d*x + 1/2*c) + 10*a/sgn(tan(1/2*d*x + 1/2*c) + 1))*tan(1/2*d*x
+ 1/2*c) - 10*a/sgn(tan(1/2*d*x + 1/2*c) + 1))*tan(1/2*d*x + 1/2*c) + 5*a/sgn(tan(1/2*d*x + 1/2*c) + 1))*tan(
1/2*d*x + 1/2*c) - a/sgn(tan(1/2*d*x + 1/2*c) + 1))/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(5/2))/d

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maple [A]  time = 0.17, size = 47, normalized size = 1.57 \[ \frac {2 \left (1+\sin \left (d x +c \right )\right ) \left (\sin \left (d x +c \right )-1\right )^{3}}{5 a \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4/(a+a*sin(d*x+c))^(3/2),x)

[Out]

2/5/a*(1+sin(d*x+c))*(sin(d*x+c)-1)^3/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 \[ \int \frac {\cos \left (d x + c\right )^{4}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a+a*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate(cos(d*x + c)^4/(a*sin(d*x + c) + a)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {{\cos \left (c+d\,x\right )}^4}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^4/(a + a*sin(c + d*x))^(3/2),x)

[Out]

int(cos(c + d*x)^4/(a + a*sin(c + d*x))^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos ^{4}{\left (c + d x \right )}}{\left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4/(a+a*sin(d*x+c))**(3/2),x)

[Out]

Integral(cos(c + d*x)**4/(a*(sin(c + d*x) + 1))**(3/2), x)

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