3.14 \(\int \cos ^5(c+d x) (a+a \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=67 \[ \frac {(a \sin (c+d x)+a)^7}{7 a^5 d}-\frac {2 (a \sin (c+d x)+a)^6}{3 a^4 d}+\frac {4 (a \sin (c+d x)+a)^5}{5 a^3 d} \]

[Out]

4/5*(a+a*sin(d*x+c))^5/a^3/d-2/3*(a+a*sin(d*x+c))^6/a^4/d+1/7*(a+a*sin(d*x+c))^7/a^5/d

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Rubi [A]  time = 0.06, antiderivative size = 67, 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 = {2667, 43} \[ \frac {(a \sin (c+d x)+a)^7}{7 a^5 d}-\frac {2 (a \sin (c+d x)+a)^6}{3 a^4 d}+\frac {4 (a \sin (c+d x)+a)^5}{5 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(a + a*Sin[c + d*x])^5)/(5*a^3*d) - (2*(a + a*Sin[c + d*x])^6)/(3*a^4*d) + (a + a*Sin[c + d*x])^7/(7*a^5*d)

Rule 43

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 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 ^5(c+d x) (a+a \sin (c+d x))^2 \, dx &=\frac {\operatorname {Subst}\left (\int (a-x)^2 (a+x)^4 \, dx,x,a \sin (c+d x)\right )}{a^5 d}\\ &=\frac {\operatorname {Subst}\left (\int \left (4 a^2 (a+x)^4-4 a (a+x)^5+(a+x)^6\right ) \, dx,x,a \sin (c+d x)\right )}{a^5 d}\\ &=\frac {4 (a+a \sin (c+d x))^5}{5 a^3 d}-\frac {2 (a+a \sin (c+d x))^6}{3 a^4 d}+\frac {(a+a \sin (c+d x))^7}{7 a^5 d}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 58, normalized size = 0.87 \[ -\frac {a^2 (\sin (c+d x)+1)^2 \left (15 \sin ^2(c+d x)-40 \sin (c+d x)+29\right ) \cos ^6(c+d x)}{105 d (\sin (c+d x)-1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/105*(a^2*Cos[c + d*x]^6*(1 + Sin[c + d*x])^2*(29 - 40*Sin[c + d*x] + 15*Sin[c + d*x]^2))/(d*(-1 + Sin[c + d
*x])^3)

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fricas [A]  time = 0.71, size = 71, normalized size = 1.06 \[ -\frac {35 \, a^{2} \cos \left (d x + c\right )^{6} + {\left (15 \, a^{2} \cos \left (d x + c\right )^{6} - 24 \, a^{2} \cos \left (d x + c\right )^{4} - 32 \, a^{2} \cos \left (d x + c\right )^{2} - 64 \, a^{2}\right )} \sin \left (d x + c\right )}{105 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/105*(35*a^2*cos(d*x + c)^6 + (15*a^2*cos(d*x + c)^6 - 24*a^2*cos(d*x + c)^4 - 32*a^2*cos(d*x + c)^2 - 64*a^
2)*sin(d*x + c))/d

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giac [A]  time = 2.27, size = 117, normalized size = 1.75 \[ -\frac {a^{2} \cos \left (6 \, d x + 6 \, c\right )}{96 \, d} - \frac {a^{2} \cos \left (4 \, d x + 4 \, c\right )}{16 \, d} - \frac {5 \, a^{2} \cos \left (2 \, d x + 2 \, c\right )}{32 \, d} - \frac {a^{2} \sin \left (7 \, d x + 7 \, c\right )}{448 \, d} + \frac {a^{2} \sin \left (5 \, d x + 5 \, c\right )}{320 \, d} + \frac {19 \, a^{2} \sin \left (3 \, d x + 3 \, c\right )}{192 \, d} + \frac {45 \, a^{2} \sin \left (d x + c\right )}{64 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96*a^2*cos(6*d*x + 6*c)/d - 1/16*a^2*cos(4*d*x + 4*c)/d - 5/32*a^2*cos(2*d*x + 2*c)/d - 1/448*a^2*sin(7*d*x
+ 7*c)/d + 1/320*a^2*sin(5*d*x + 5*c)/d + 19/192*a^2*sin(3*d*x + 3*c)/d + 45/64*a^2*sin(d*x + c)/d

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maple [A]  time = 0.17, size = 99, normalized size = 1.48 \[ \frac {a^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{6}\left (d x +c \right )\right )}{7}+\frac {\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{35}\right )-\frac {\left (\cos ^{6}\left (d x +c \right )\right ) a^{2}}{3}+\frac {a^{2} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^2*(-1/7*sin(d*x+c)*cos(d*x+c)^6+1/35*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c))-1/3*cos(d*x+c)^6*a
^2+1/5*a^2*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c))

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maxima [A]  time = 0.49, size = 95, normalized size = 1.42 \[ \frac {15 \, a^{2} \sin \left (d x + c\right )^{7} + 35 \, a^{2} \sin \left (d x + c\right )^{6} - 21 \, a^{2} \sin \left (d x + c\right )^{5} - 105 \, a^{2} \sin \left (d x + c\right )^{4} - 35 \, a^{2} \sin \left (d x + c\right )^{3} + 105 \, a^{2} \sin \left (d x + c\right )^{2} + 105 \, a^{2} \sin \left (d x + c\right )}{105 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*(15*a^2*sin(d*x + c)^7 + 35*a^2*sin(d*x + c)^6 - 21*a^2*sin(d*x + c)^5 - 105*a^2*sin(d*x + c)^4 - 35*a^2
*sin(d*x + c)^3 + 105*a^2*sin(d*x + c)^2 + 105*a^2*sin(d*x + c))/d

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mupad [B]  time = 4.54, size = 92, normalized size = 1.37 \[ \frac {\frac {a^2\,{\sin \left (c+d\,x\right )}^7}{7}+\frac {a^2\,{\sin \left (c+d\,x\right )}^6}{3}-\frac {a^2\,{\sin \left (c+d\,x\right )}^5}{5}-a^2\,{\sin \left (c+d\,x\right )}^4-\frac {a^2\,{\sin \left (c+d\,x\right )}^3}{3}+a^2\,{\sin \left (c+d\,x\right )}^2+a^2\,\sin \left (c+d\,x\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^5*(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)^3)/3 - a^2*sin(c + d*x)^4 - (a^2*sin(c + d*x)^5)/5
+ (a^2*sin(c + d*x)^6)/3 + (a^2*sin(c + d*x)^7)/7)/d

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sympy [A]  time = 8.34, size = 158, normalized size = 2.36 \[ \begin {cases} \frac {8 a^{2} \sin ^{7}{\left (c + d x \right )}}{105 d} + \frac {4 a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{15 d} + \frac {8 a^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{3 d} + \frac {4 a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {a^{2} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac {a^{2} \cos ^{6}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\relax (c )} + a\right )^{2} \cos ^{5}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((8*a**2*sin(c + d*x)**7/(105*d) + 4*a**2*sin(c + d*x)**5*cos(c + d*x)**2/(15*d) + 8*a**2*sin(c + d*x
)**5/(15*d) + a**2*sin(c + d*x)**3*cos(c + d*x)**4/(3*d) + 4*a**2*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) + a**2
*sin(c + d*x)*cos(c + d*x)**4/d - a**2*cos(c + d*x)**6/(3*d), Ne(d, 0)), (x*(a*sin(c) + a)**2*cos(c)**5, True)
)

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