3.2.3 \(\int \cos ^2(a+b x) \sin ^5(a+b x) \, dx\) [103]

Optimal. Leaf size=46 \[ -\frac {\cos ^3(a+b x)}{3 b}+\frac {2 \cos ^5(a+b x)}{5 b}-\frac {\cos ^7(a+b x)}{7 b} \]

[Out]

-1/3*cos(b*x+a)^3/b+2/5*cos(b*x+a)^5/b-1/7*cos(b*x+a)^7/b

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Rubi [A]
time = 0.03, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2645, 276} \begin {gather*} -\frac {\cos ^7(a+b x)}{7 b}+\frac {2 \cos ^5(a+b x)}{5 b}-\frac {\cos ^3(a+b x)}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^2*Sin[a + b*x]^5,x]

[Out]

-1/3*Cos[a + b*x]^3/b + (2*Cos[a + b*x]^5)/(5*b) - Cos[a + b*x]^7/(7*b)

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 2645

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[-(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rubi steps

\begin {align*} \int \cos ^2(a+b x) \sin ^5(a+b x) \, dx &=-\frac {\text {Subst}\left (\int x^2 \left (1-x^2\right )^2 \, dx,x,\cos (a+b x)\right )}{b}\\ &=-\frac {\text {Subst}\left (\int \left (x^2-2 x^4+x^6\right ) \, dx,x,\cos (a+b x)\right )}{b}\\ &=-\frac {\cos ^3(a+b x)}{3 b}+\frac {2 \cos ^5(a+b x)}{5 b}-\frac {\cos ^7(a+b x)}{7 b}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 37, normalized size = 0.80 \begin {gather*} \frac {\cos ^3(a+b x) (-157+108 \cos (2 (a+b x))-15 \cos (4 (a+b x)))}{840 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^2*Sin[a + b*x]^5,x]

[Out]

(Cos[a + b*x]^3*(-157 + 108*Cos[2*(a + b*x)] - 15*Cos[4*(a + b*x)]))/(840*b)

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Maple [A]
time = 0.07, size = 52, normalized size = 1.13

method	result	size
derivativedivides	\(\frac {-\frac {\left (\cos ^{3}\left (b x +a \right )\right ) \left (\sin ^{4}\left (b x +a \right )\right )}{7}-\frac {4 \left (\cos ^{3}\left (b x +a \right )\right ) \left (\sin ^{2}\left (b x +a \right )\right )}{35}-\frac {8 \left (\cos ^{3}\left (b x +a \right )\right )}{105}}{b}\)	\(52\)
default	\(\frac {-\frac {\left (\cos ^{3}\left (b x +a \right )\right ) \left (\sin ^{4}\left (b x +a \right )\right )}{7}-\frac {4 \left (\cos ^{3}\left (b x +a \right )\right ) \left (\sin ^{2}\left (b x +a \right )\right )}{35}-\frac {8 \left (\cos ^{3}\left (b x +a \right )\right )}{105}}{b}\)	\(52\)
risch	\(-\frac {5 \cos \left (b x +a \right )}{64 b}-\frac {\cos \left (7 b x +7 a \right )}{448 b}+\frac {3 \cos \left (5 b x +5 a \right )}{320 b}-\frac {\cos \left (3 b x +3 a \right )}{192 b}\)	\(55\)
norman	\(\frac {-\frac {16}{105 b}-\frac {32 \left (\tan ^{8}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{3 b}-\frac {16 \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{15 b}-\frac {16 \left (\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{5 b}+\frac {16 \left (\tan ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{3 b}}{\left (1+\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )^{7}}\)	\(87\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^2*sin(b*x+a)^5,x,method=_RETURNVERBOSE)

[Out]

1/b*(-1/7*cos(b*x+a)^3*sin(b*x+a)^4-4/35*cos(b*x+a)^3*sin(b*x+a)^2-8/105*cos(b*x+a)^3)

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Maxima [A]
time = 0.28, size = 36, normalized size = 0.78 \begin {gather*} -\frac {15 \, \cos \left (b x + a\right )^{7} - 42 \, \cos \left (b x + a\right )^{5} + 35 \, \cos \left (b x + a\right )^{3}}{105 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/105*(15*cos(b*x + a)^7 - 42*cos(b*x + a)^5 + 35*cos(b*x + a)^3)/b

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Fricas [A]
time = 0.37, size = 36, normalized size = 0.78 \begin {gather*} -\frac {15 \, \cos \left (b x + a\right )^{7} - 42 \, \cos \left (b x + a\right )^{5} + 35 \, \cos \left (b x + a\right )^{3}}{105 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/105*(15*cos(b*x + a)^7 - 42*cos(b*x + a)^5 + 35*cos(b*x + a)^3)/b

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Sympy [A]
time = 0.58, size = 68, normalized size = 1.48 \begin {gather*} \begin {cases} - \frac {\sin ^{4}{\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{3 b} - \frac {4 \sin ^{2}{\left (a + b x \right )} \cos ^{5}{\left (a + b x \right )}}{15 b} - \frac {8 \cos ^{7}{\left (a + b x \right )}}{105 b} & \text {for}\: b \neq 0 \\x \sin ^{5}{\left (a \right )} \cos ^{2}{\left (a \right )} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 3.89, size = 54, normalized size = 1.17 \begin {gather*} -\frac {\cos \left (7 \, b x + 7 \, a\right )}{448 \, b} + \frac {3 \, \cos \left (5 \, b x + 5 \, a\right )}{320 \, b} - \frac {\cos \left (3 \, b x + 3 \, a\right )}{192 \, b} - \frac {5 \, \cos \left (b x + a\right )}{64 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/448*cos(7*b*x + 7*a)/b + 3/320*cos(5*b*x + 5*a)/b - 1/192*cos(3*b*x + 3*a)/b - 5/64*cos(b*x + a)/b

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Mupad [B]
time = 0.37, size = 36, normalized size = 0.78 \begin {gather*} -\frac {15\,{\cos \left (a+b\,x\right )}^7-42\,{\cos \left (a+b\,x\right )}^5+35\,{\cos \left (a+b\,x\right )}^3}{105\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)^2*sin(a + b*x)^5,x)

[Out]

-(35*cos(a + b*x)^3 - 42*cos(a + b*x)^5 + 15*cos(a + b*x)^7)/(105*b)

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