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

Optimal. Leaf size=24 \[ \frac{2 (a \sin (c+d x)+a)^{5/2}}{5 a d} \]

[Out]

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

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Rubi [A]  time = 0.0342689, antiderivative size = 24, normalized size of antiderivative = 1., 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} \[ \frac{2 (a \sin (c+d x)+a)^{5/2}}{5 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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]

Rubi steps

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

Mathematica [A]  time = 0.0412181, size = 24, normalized size = 1. \[ \frac{2 (a \sin (c+d x)+a)^{5/2}}{5 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.009, size = 21, normalized size = 0.9 \begin{align*}{\frac{2}{5\,da} \left ( a+a\sin \left ( dx+c \right ) \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.941551, size = 27, normalized size = 1.12 \begin{align*} \frac{2 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}{5 \, a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(a*sin(d*x + c) + a)^(5/2)/(a*d)

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Fricas [A]  time = 1.57513, size = 104, normalized size = 4.33 \begin{align*} -\frac{2 \,{\left (a \cos \left (d x + c\right )^{2} - 2 \, a \sin \left (d x + c\right ) - 2 \, a\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{5 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 26.2348, size = 90, normalized size = 3.75 \begin{align*} \begin{cases} \frac{2 a \sqrt{a \sin{\left (c + d x \right )} + a} \sin ^{2}{\left (c + d x \right )}}{5 d} + \frac{4 a \sqrt{a \sin{\left (c + d x \right )} + a} \sin{\left (c + d x \right )}}{5 d} + \frac{2 a \sqrt{a \sin{\left (c + d x \right )} + a}}{5 d} & \text{for}\: d \neq 0 \\x \left (a \sin{\left (c \right )} + a\right )^{\frac{3}{2}} \cos{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((2*a*sqrt(a*sin(c + d*x) + a)*sin(c + d*x)**2/(5*d) + 4*a*sqrt(a*sin(c + d*x) + a)*sin(c + d*x)/(5*d
) + 2*a*sqrt(a*sin(c + d*x) + a)/(5*d), Ne(d, 0)), (x*(a*sin(c) + a)**(3/2)*cos(c), True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \cos \left (d x + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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