3.131 \(\int \cos (c+d x) (a+a \sin (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=24 \[ \frac {2 (a \sin (c+d x)+a)^{7/2}}{7 a d} \]

[Out]

2/7*(a+a*sin(d*x+c))^(7/2)/a/d

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Rubi [A]  time = 0.03, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2667, 32} \[ \frac {2 (a \sin (c+d x)+a)^{7/2}}{7 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + a*Sin[c + d*x])^(7/2))/(7*a*d)

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 2667

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rubi steps

\begin {align*} \int \cos (c+d x) (a+a \sin (c+d x))^{5/2} \, dx &=\frac {\operatorname {Subst}\left (\int (a+x)^{5/2} \, dx,x,a \sin (c+d x)\right )}{a d}\\ &=\frac {2 (a+a \sin (c+d x))^{7/2}}{7 a d}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 24, normalized size = 1.00 \[ \frac {2 (a \sin (c+d x)+a)^{7/2}}{7 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + a*Sin[c + d*x])^(7/2))/(7*a*d)

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fricas [B]  time = 0.73, size = 61, normalized size = 2.54 \[ -\frac {2 \, {\left (3 \, a^{2} \cos \left (d x + c\right )^{2} - 4 \, a^{2} + {\left (a^{2} \cos \left (d x + c\right )^{2} - 4 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{7 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/7*(3*a^2*cos(d*x + c)^2 - 4*a^2 + (a^2*cos(d*x + c)^2 - 4*a^2)*sin(d*x + c))*sqrt(a*sin(d*x + c) + a)/d

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giac [B]  time = 0.56, size = 207, normalized size = 8.62 \[ -\frac {1}{420} \, \sqrt {2} {\left (\frac {140 \, a^{2} \cos \left (\frac {1}{4} \, \pi + \frac {3}{2} \, d x + \frac {3}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} + \frac {84 \, a^{2} \cos \left (-\frac {1}{4} \, \pi + \frac {5}{2} \, d x + \frac {5}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} + \frac {21 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{4} \, \pi + \frac {5}{2} \, d x + \frac {5}{2} \, c\right )}{d} - \frac {525 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{d} + \frac {15 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {7}{2} \, d x + \frac {7}{2} \, c\right )}{d} - \frac {175 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {3}{2} \, d x + \frac {3}{2} \, c\right )}{d}\right )} \sqrt {a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/420*sqrt(2)*(140*a^2*cos(1/4*pi + 3/2*d*x + 3/2*c)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d + 84*a^2*cos(-1/4*
pi + 5/2*d*x + 5/2*c)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d + 21*a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(1
/4*pi + 5/2*d*x + 5/2*c)/d - 525*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)/d + 15*
a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 7/2*d*x + 7/2*c)/d - 175*a^2*sgn(cos(-1/4*pi + 1/2*d*x +
1/2*c))*sin(-1/4*pi + 3/2*d*x + 3/2*c)/d)*sqrt(a)

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maple [A]  time = 0.04, size = 21, normalized size = 0.88 \[ \frac {2 \left (a +a \sin \left (d x +c \right )\right )^{\frac {7}{2}}}{7 d a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*(a+a*sin(d*x+c))^(7/2)/d/a

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maxima [A]  time = 0.56, size = 20, normalized size = 0.83 \[ \frac {2 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}}}{7 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*(a*sin(d*x + c) + a)^(7/2)/(a*d)

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mupad [B]  time = 4.78, size = 20, normalized size = 0.83 \[ \frac {2\,{\left (a\,\left (\sin \left (c+d\,x\right )+1\right )\right )}^{7/2}}{7\,a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(a*(sin(c + d*x) + 1))^(7/2))/(7*a*d)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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