3.1.16 \(\int \cos ^3(c+d x) (a+a \sin (c+d x))^2 \, dx\) [16]

Optimal. Leaf size=45 \[ \frac {(a+a \sin (c+d x))^4}{2 a^2 d}-\frac {(a+a \sin (c+d x))^5}{5 a^3 d} \]

[Out]

1/2*(a+a*sin(d*x+c))^4/a^2/d-1/5*(a+a*sin(d*x+c))^5/a^3/d

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Rubi [A]
time = 0.03, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2746, 45} \begin {gather*} \frac {(a \sin (c+d x)+a)^4}{2 a^2 d}-\frac {(a \sin (c+d x)+a)^5}{5 a^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a + a*Sin[c + d*x])^4/(2*a^2*d) - (a + a*Sin[c + d*x])^5/(5*a^3*d)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 ^3(c+d x) (a+a \sin (c+d x))^2 \, dx &=\frac {\text {Subst}\left (\int (a-x) (a+x)^3 \, dx,x,a \sin (c+d x)\right )}{a^3 d}\\ &=\frac {\text {Subst}\left (\int \left (2 a (a+x)^3-(a+x)^4\right ) \, dx,x,a \sin (c+d x)\right )}{a^3 d}\\ &=\frac {(a+a \sin (c+d x))^4}{2 a^2 d}-\frac {(a+a \sin (c+d x))^5}{5 a^3 d}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 46, normalized size = 1.02 \begin {gather*} -\frac {a^2 \sin (c+d x) \left (-10-10 \sin (c+d x)+5 \sin ^3(c+d x)+2 \sin ^4(c+d x)\right )}{10 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/10*(a^2*Sin[c + d*x]*(-10 - 10*Sin[c + d*x] + 5*Sin[c + d*x]^3 + 2*Sin[c + d*x]^4))/d

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Maple [A]
time = 0.18, size = 79, normalized size = 1.76

method	result	size
derivativedivides	\(\frac {a^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {\left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{15}\right )-\frac {\left (\cos ^{4}\left (d x +c \right )\right ) a^{2}}{2}+\frac {a^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}}{d}\)	\(79\)
default	\(\frac {a^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {\left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{15}\right )-\frac {\left (\cos ^{4}\left (d x +c \right )\right ) a^{2}}{2}+\frac {a^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}}{d}\)	\(79\)
risch	\(\frac {7 a^{2} \sin \left (d x +c \right )}{8 d}-\frac {a^{2} \sin \left (5 d x +5 c \right )}{80 d}-\frac {a^{2} \cos \left (4 d x +4 c \right )}{16 d}+\frac {a^{2} \sin \left (3 d x +3 c \right )}{16 d}-\frac {a^{2} \cos \left (2 d x +2 c \right )}{4 d}\)	\(84\)
norman	\(\frac {\frac {4 a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a^{2} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {8 a^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {28 a^{2} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}+\frac {8 a^{2} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 a^{2} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a^{2} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a^{2} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}\)	\(187\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/d*(a^2*(-1/5*sin(d*x+c)*cos(d*x+c)^4+1/15*(2+cos(d*x+c)^2)*sin(d*x+c))-1/2*cos(d*x+c)^4*a^2+1/3*a^2*(2+cos(d
*x+c)^2)*sin(d*x+c))

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Maxima [A]
time = 0.29, size = 56, normalized size = 1.24 \begin {gather*} -\frac {2 \, a^{2} \sin \left (d x + c\right )^{5} + 5 \, a^{2} \sin \left (d x + c\right )^{4} - 10 \, a^{2} \sin \left (d x + c\right )^{2} - 10 \, a^{2} \sin \left (d x + c\right )}{10 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/10*(2*a^2*sin(d*x + c)^5 + 5*a^2*sin(d*x + c)^4 - 10*a^2*sin(d*x + c)^2 - 10*a^2*sin(d*x + c))/d

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Fricas [A]
time = 0.35, size = 58, normalized size = 1.29 \begin {gather*} -\frac {5 \, a^{2} \cos \left (d x + c\right )^{4} + 2 \, {\left (a^{2} \cos \left (d x + c\right )^{4} - 2 \, a^{2} \cos \left (d x + c\right )^{2} - 4 \, a^{2}\right )} \sin \left (d x + c\right )}{10 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/10*(5*a^2*cos(d*x + c)^4 + 2*(a^2*cos(d*x + c)^4 - 2*a^2*cos(d*x + c)^2 - 4*a^2)*sin(d*x + c))/d

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 107 vs. \(2 (36) = 72\).
time = 0.26, size = 107, normalized size = 2.38 \begin {gather*} \begin {cases} \frac {2 a^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {2 a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {a^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} - \frac {a^{2} \cos ^{4}{\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right )^{2} \cos ^{3}{\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)**3*(a+a*sin(d*x+c))**2,x)

[Out]

Piecewise((2*a**2*sin(c + d*x)**5/(15*d) + a**2*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) + 2*a**2*sin(c + d*x)**3
/(3*d) + a**2*sin(c + d*x)*cos(c + d*x)**2/d - a**2*cos(c + d*x)**4/(2*d), Ne(d, 0)), (x*(a*sin(c) + a)**2*cos
(c)**3, True))

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Giac [A]
time = 7.27, size = 56, normalized size = 1.24 \begin {gather*} -\frac {2 \, a^{2} \sin \left (d x + c\right )^{5} + 5 \, a^{2} \sin \left (d x + c\right )^{4} - 10 \, a^{2} \sin \left (d x + c\right )^{2} - 10 \, a^{2} \sin \left (d x + c\right )}{10 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/10*(2*a^2*sin(d*x + c)^5 + 5*a^2*sin(d*x + c)^4 - 10*a^2*sin(d*x + c)^2 - 10*a^2*sin(d*x + c))/d

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Mupad [B]
time = 0.06, size = 53, normalized size = 1.18 \begin {gather*} \frac {-\frac {a^2\,{\sin \left (c+d\,x\right )}^5}{5}-\frac {a^2\,{\sin \left (c+d\,x\right )}^4}{2}+a^2\,{\sin \left (c+d\,x\right )}^2+a^2\,\sin \left (c+d\,x\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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