3.2.31 \(\int \cos (c+d x) (a+a \sin (c+d x))^{5/2} \, dx\) [131]

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

[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 = {2746, 32} \begin {gather*} \frac {2 (a \sin (c+d x)+a)^{7/2}}{7 a d} \end {gather*}

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 2746

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 {\text {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 = 0.04, size = 24, normalized size = 1.00 \begin {gather*} \frac {2 (a+a \sin (c+d x))^{7/2}}{7 a d} \end {gather*}

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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Maple [A]
time = 0.04, size = 21, normalized size = 0.88

method	result	size
derivativedivides	\(\frac {2 \left (a +a \sin \left (d x +c \right )\right )^{\frac {7}{2}}}{7 d a}\)	\(21\)
default	\(\frac {2 \left (a +a \sin \left (d x +c \right )\right )^{\frac {7}{2}}}{7 d a}\)	\(21\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 20, normalized size = 0.83 \begin {gather*} \frac {2 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}}}{7 \, a d} \end {gather*}

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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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (20) = 40\).
time = 0.34, size = 61, normalized size = 2.54 \begin {gather*} -\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} \end {gather*}

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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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 126 vs. \(2 (19) = 38\).
time = 16.92, size = 126, normalized size = 5.25 \begin {gather*} \begin {cases} \frac {2 a^{2} \sqrt {a \sin {\left (c + d x \right )} + a} \sin ^{3}{\left (c + d x \right )}}{7 d} + \frac {6 a^{2} \sqrt {a \sin {\left (c + d x \right )} + a} \sin ^{2}{\left (c + d x \right )}}{7 d} + \frac {6 a^{2} \sqrt {a \sin {\left (c + d x \right )} + a} \sin {\left (c + d x \right )}}{7 d} + \frac {2 a^{2} \sqrt {a \sin {\left (c + d x \right )} + a}}{7 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right )^{\frac {5}{2}} \cos {\left (c \right )} & \text {otherwise} \end {cases} \end {gather*}

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]

Piecewise((2*a**2*sqrt(a*sin(c + d*x) + a)*sin(c + d*x)**3/(7*d) + 6*a**2*sqrt(a*sin(c + d*x) + a)*sin(c + d*x
)**2/(7*d) + 6*a**2*sqrt(a*sin(c + d*x) + a)*sin(c + d*x)/(7*d) + 2*a**2*sqrt(a*sin(c + d*x) + a)/(7*d), Ne(d,
0)), (x*(a*sin(c) + a)**(5/2)*cos(c), True))

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Giac [A]
time = 4.89, size = 38, normalized size = 1.58 \begin {gather*} \frac {16 \, \sqrt {2} a^{\frac {5}{2}} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{7 \, d} \end {gather*}

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]

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

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

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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