3.76 \(\int \frac {\cos ^7(c+d x)}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=23 \[ -\frac {(a-a \sin (c+d x))^4}{4 a^7 d} \]

[Out]

-1/4*(a-a*sin(d*x+c))^4/a^7/d

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Rubi [A]  time = 0.04, antiderivative size = 23, 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 = {2667, 32} \[ -\frac {(a-a \sin (c+d x))^4}{4 a^7 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^7/(a + a*Sin[c + d*x])^3,x]

[Out]

-(a - a*Sin[c + d*x])^4/(4*a^7*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 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 \frac {\cos ^7(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac {\operatorname {Subst}\left (\int (a-x)^3 \, dx,x,a \sin (c+d x)\right )}{a^7 d}\\ &=-\frac {(a-a \sin (c+d x))^4}{4 a^7 d}\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 44, normalized size = 1.91 \[ -\frac {\sin (c+d x) \left (\sin ^3(c+d x)-4 \sin ^2(c+d x)+6 \sin (c+d x)-4\right )}{4 a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^7/(a + a*Sin[c + d*x])^3,x]

[Out]

-1/4*(Sin[c + d*x]*(-4 + 6*Sin[c + d*x] - 4*Sin[c + d*x]^2 + Sin[c + d*x]^3))/(a^3*d)

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fricas [B]  time = 0.55, size = 45, normalized size = 1.96 \[ -\frac {\cos \left (d x + c\right )^{4} - 8 \, \cos \left (d x + c\right )^{2} + 4 \, {\left (\cos \left (d x + c\right )^{2} - 2\right )} \sin \left (d x + c\right )}{4 \, a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.74, size = 45, normalized size = 1.96 \[ -\frac {\sin \left (d x + c\right )^{4} - 4 \, \sin \left (d x + c\right )^{3} + 6 \, \sin \left (d x + c\right )^{2} - 4 \, \sin \left (d x + c\right )}{4 \, a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.18, size = 19, normalized size = 0.83 \[ -\frac {\left (\sin \left (d x +c \right )-1\right )^{4}}{4 d \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^7/(a+a*sin(d*x+c))^3,x)

[Out]

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

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maxima [B]  time = 0.45, size = 45, normalized size = 1.96 \[ -\frac {\sin \left (d x + c\right )^{4} - 4 \, \sin \left (d x + c\right )^{3} + 6 \, \sin \left (d x + c\right )^{2} - 4 \, \sin \left (d x + c\right )}{4 \, a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.55, size = 53, normalized size = 2.30 \[ \frac {\frac {\sin \left (c+d\,x\right )}{a^3}-\frac {3\,{\sin \left (c+d\,x\right )}^2}{2\,a^3}+\frac {{\sin \left (c+d\,x\right )}^3}{a^3}-\frac {{\sin \left (c+d\,x\right )}^4}{4\,a^3}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^7/(a + a*sin(c + d*x))^3,x)

[Out]

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

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sympy [A]  time = 161.61, size = 654, normalized size = 28.43 \[ \begin {cases} \frac {2 \tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a^{3} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a^{3} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d} - \frac {6 \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a^{3} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a^{3} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d} + \frac {14 \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a^{3} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a^{3} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d} - \frac {16 \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a^{3} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a^{3} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d} + \frac {14 \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a^{3} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a^{3} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d} - \frac {6 \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a^{3} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a^{3} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d} + \frac {2 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a^{3} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a^{3} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \cos ^{7}{\relax (c )}}{\left (a \sin {\relax (c )} + a\right )^{3}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**7/(a+a*sin(d*x+c))**3,x)

[Out]

Piecewise((2*tan(c/2 + d*x/2)**7/(a**3*d*tan(c/2 + d*x/2)**8 + 4*a**3*d*tan(c/2 + d*x/2)**6 + 6*a**3*d*tan(c/2
+ d*x/2)**4 + 4*a**3*d*tan(c/2 + d*x/2)**2 + a**3*d) - 6*tan(c/2 + d*x/2)**6/(a**3*d*tan(c/2 + d*x/2)**8 + 4*
a**3*d*tan(c/2 + d*x/2)**6 + 6*a**3*d*tan(c/2 + d*x/2)**4 + 4*a**3*d*tan(c/2 + d*x/2)**2 + a**3*d) + 14*tan(c/
2 + d*x/2)**5/(a**3*d*tan(c/2 + d*x/2)**8 + 4*a**3*d*tan(c/2 + d*x/2)**6 + 6*a**3*d*tan(c/2 + d*x/2)**4 + 4*a*
*3*d*tan(c/2 + d*x/2)**2 + a**3*d) - 16*tan(c/2 + d*x/2)**4/(a**3*d*tan(c/2 + d*x/2)**8 + 4*a**3*d*tan(c/2 + d
*x/2)**6 + 6*a**3*d*tan(c/2 + d*x/2)**4 + 4*a**3*d*tan(c/2 + d*x/2)**2 + a**3*d) + 14*tan(c/2 + d*x/2)**3/(a**
3*d*tan(c/2 + d*x/2)**8 + 4*a**3*d*tan(c/2 + d*x/2)**6 + 6*a**3*d*tan(c/2 + d*x/2)**4 + 4*a**3*d*tan(c/2 + d*x
/2)**2 + a**3*d) - 6*tan(c/2 + d*x/2)**2/(a**3*d*tan(c/2 + d*x/2)**8 + 4*a**3*d*tan(c/2 + d*x/2)**6 + 6*a**3*d
*tan(c/2 + d*x/2)**4 + 4*a**3*d*tan(c/2 + d*x/2)**2 + a**3*d) + 2*tan(c/2 + d*x/2)/(a**3*d*tan(c/2 + d*x/2)**8
+ 4*a**3*d*tan(c/2 + d*x/2)**6 + 6*a**3*d*tan(c/2 + d*x/2)**4 + 4*a**3*d*tan(c/2 + d*x/2)**2 + a**3*d), Ne(d,
0)), (x*cos(c)**7/(a*sin(c) + a)**3, True))

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