3.63 \(\int \frac {\cos ^7(c+d x)}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=47 \[ \frac {(a-a \sin (c+d x))^5}{5 a^7 d}-\frac {(a-a \sin (c+d x))^4}{2 a^6 d} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 47, 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} \[ \frac {(a-a \sin (c+d x))^5}{5 a^7 d}-\frac {(a-a \sin (c+d x))^4}{2 a^6 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^7/(a + a*Sin[c + d*x])^2,x]

[Out]

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

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Mathematica [A]  time = 0.16, size = 46, normalized size = 0.98 \[ -\frac {\sin (c+d x) \left (2 \sin ^4(c+d x)-5 \sin ^3(c+d x)+10 \sin (c+d x)-10\right )}{10 a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^7/(a + a*Sin[c + d*x])^2,x]

[Out]

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

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fricas [A]  time = 0.73, size = 47, normalized size = 1.00 \[ \frac {5 \, \cos \left (d x + c\right )^{4} - 2 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right )}{10 \, a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 1.01, size = 47, normalized size = 1.00 \[ -\frac {2 \, \sin \left (d x + c\right )^{5} - 5 \, \sin \left (d x + c\right )^{4} + 10 \, \sin \left (d x + c\right )^{2} - 10 \, \sin \left (d x + c\right )}{10 \, a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.19, size = 45, normalized size = 0.96 \[ \frac {-\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{2}-\left (\sin ^{2}\left (d x +c \right )\right )+\sin \left (d x +c \right )}{d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^7/(a+a*sin(d*x+c))^2,x)

[Out]

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

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maxima [A]  time = 0.35, size = 47, normalized size = 1.00 \[ -\frac {2 \, \sin \left (d x + c\right )^{5} - 5 \, \sin \left (d x + c\right )^{4} + 10 \, \sin \left (d x + c\right )^{2} - 10 \, \sin \left (d x + c\right )}{10 \, a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.66, size = 54, normalized size = 1.15 \[ \frac {\frac {\sin \left (c+d\,x\right )}{a^2}-\frac {{\sin \left (c+d\,x\right )}^2}{a^2}+\frac {{\sin \left (c+d\,x\right )}^4}{2\,a^2}-\frac {{\sin \left (c+d\,x\right )}^5}{5\,a^2}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^7/(a + a*sin(c + d*x))^2,x)

[Out]

(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*a^2))/d

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sympy [A]  time = 92.86, size = 1037, normalized size = 22.06 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**7/(a+a*sin(d*x+c))**2,x)

[Out]

Piecewise((10*tan(c/2 + d*x/2)**9/(5*a**2*d*tan(c/2 + d*x/2)**10 + 25*a**2*d*tan(c/2 + d*x/2)**8 + 50*a**2*d*t
an(c/2 + d*x/2)**6 + 50*a**2*d*tan(c/2 + d*x/2)**4 + 25*a**2*d*tan(c/2 + d*x/2)**2 + 5*a**2*d) - 20*tan(c/2 +
d*x/2)**8/(5*a**2*d*tan(c/2 + d*x/2)**10 + 25*a**2*d*tan(c/2 + d*x/2)**8 + 50*a**2*d*tan(c/2 + d*x/2)**6 + 50*
a**2*d*tan(c/2 + d*x/2)**4 + 25*a**2*d*tan(c/2 + d*x/2)**2 + 5*a**2*d) + 40*tan(c/2 + d*x/2)**7/(5*a**2*d*tan(
c/2 + d*x/2)**10 + 25*a**2*d*tan(c/2 + d*x/2)**8 + 50*a**2*d*tan(c/2 + d*x/2)**6 + 50*a**2*d*tan(c/2 + d*x/2)*
*4 + 25*a**2*d*tan(c/2 + d*x/2)**2 + 5*a**2*d) - 20*tan(c/2 + d*x/2)**6/(5*a**2*d*tan(c/2 + d*x/2)**10 + 25*a*
*2*d*tan(c/2 + d*x/2)**8 + 50*a**2*d*tan(c/2 + d*x/2)**6 + 50*a**2*d*tan(c/2 + d*x/2)**4 + 25*a**2*d*tan(c/2 +
d*x/2)**2 + 5*a**2*d) + 28*tan(c/2 + d*x/2)**5/(5*a**2*d*tan(c/2 + d*x/2)**10 + 25*a**2*d*tan(c/2 + d*x/2)**8
+ 50*a**2*d*tan(c/2 + d*x/2)**6 + 50*a**2*d*tan(c/2 + d*x/2)**4 + 25*a**2*d*tan(c/2 + d*x/2)**2 + 5*a**2*d) -
20*tan(c/2 + d*x/2)**4/(5*a**2*d*tan(c/2 + d*x/2)**10 + 25*a**2*d*tan(c/2 + d*x/2)**8 + 50*a**2*d*tan(c/2 + d
*x/2)**6 + 50*a**2*d*tan(c/2 + d*x/2)**4 + 25*a**2*d*tan(c/2 + d*x/2)**2 + 5*a**2*d) + 40*tan(c/2 + d*x/2)**3/
(5*a**2*d*tan(c/2 + d*x/2)**10 + 25*a**2*d*tan(c/2 + d*x/2)**8 + 50*a**2*d*tan(c/2 + d*x/2)**6 + 50*a**2*d*tan
(c/2 + d*x/2)**4 + 25*a**2*d*tan(c/2 + d*x/2)**2 + 5*a**2*d) - 20*tan(c/2 + d*x/2)**2/(5*a**2*d*tan(c/2 + d*x/
2)**10 + 25*a**2*d*tan(c/2 + d*x/2)**8 + 50*a**2*d*tan(c/2 + d*x/2)**6 + 50*a**2*d*tan(c/2 + d*x/2)**4 + 25*a*
*2*d*tan(c/2 + d*x/2)**2 + 5*a**2*d) + 10*tan(c/2 + d*x/2)/(5*a**2*d*tan(c/2 + d*x/2)**10 + 25*a**2*d*tan(c/2
+ d*x/2)**8 + 50*a**2*d*tan(c/2 + d*x/2)**6 + 50*a**2*d*tan(c/2 + d*x/2)**4 + 25*a**2*d*tan(c/2 + d*x/2)**2 +
5*a**2*d), Ne(d, 0)), (x*cos(c)**7/(a*sin(c) + a)**2, True))

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