3.1.76 \(\int \frac {\cos ^7(c+d x)}{(a+a \sin (c+d x))^3} \, dx\) [76]

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 = 0.03, 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 = {2746, 32} \begin {gather*} -\frac {(a-a \sin (c+d x))^4}{4 a^7 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(a - a*Sin[c + d*x])^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 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 \frac {\cos ^7(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac {\text {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 [B] Leaf count is larger than twice the leaf count of optimal. \(48\) vs. \(2(23)=46\).
time = 0.22, size = 48, normalized size = 2.09 \begin {gather*} -\frac {-28 \cos (2 (c+d x))+\cos (4 (c+d x))+8 (-7 \sin (c+d x)+\sin (3 (c+d x)))}{32 a^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/32*(-28*Cos[2*(c + d*x)] + Cos[4*(c + d*x)] + 8*(-7*Sin[c + d*x] + Sin[3*(c + d*x)]))/(a^3*d)

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Maple [A]
time = 0.13, size = 19, normalized size = 0.83

method	result	size
derivativedivides	\(-\frac {\left (\sin \left (d x +c \right )-1\right )^{4}}{4 d \,a^{3}}\)	\(19\)
default	\(-\frac {\left (\sin \left (d x +c \right )-1\right )^{4}}{4 d \,a^{3}}\)	\(19\)
risch	\(\frac {7 \sin \left (d x +c \right )}{4 a^{3} d}-\frac {\cos \left (4 d x +4 c \right )}{32 a^{3} d}-\frac {\sin \left (3 d x +3 c \right )}{4 a^{3} d}+\frac {7 \cos \left (2 d x +2 c \right )}{8 a^{3} d}\)	\(67\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (22) = 44\).
time = 0.29, size = 45, normalized size = 1.96 \begin {gather*} -\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} \end {gather*}

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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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (22) = 44\).
time = 0.38, size = 45, normalized size = 1.96 \begin {gather*} -\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} \end {gather*}

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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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 654 vs. \(2 (19) = 38\).
time = 75.63, size = 654, normalized size = 28.43 \begin {gather*} \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}{\left (c \right )}}{\left (a \sin {\left (c \right )} + a\right )^{3}} & \text {otherwise} \end {cases} \end {gather*}

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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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (22) = 44\).
time = 5.78, size = 45, normalized size = 1.96 \begin {gather*} -\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} \end {gather*}

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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Mupad [B]
time = 4.55, size = 53, normalized size = 2.30 \begin {gather*} \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} \end {gather*}

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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