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

Optimal. Leaf size=184 \[ \frac {\left (a^2-b^2\right )^3}{b^7 d (a+b \sin (c+d x))}+\frac {6 a \left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{b^7 d}+\frac {a \left (2 a^2-3 b^2\right ) \sin ^2(c+d x)}{b^5 d}-\frac {\left (a^2-b^2\right ) \sin ^3(c+d x)}{b^4 d}-\frac {\left (5 a^4-9 a^2 b^2+3 b^4\right ) \sin (c+d x)}{b^6 d}+\frac {a \sin ^4(c+d x)}{2 b^3 d}-\frac {\sin ^5(c+d x)}{5 b^2 d} \]

[Out]

6*a*(a^2-b^2)^2*ln(a+b*sin(d*x+c))/b^7/d-(5*a^4-9*a^2*b^2+3*b^4)*sin(d*x+c)/b^6/d+a*(2*a^2-3*b^2)*sin(d*x+c)^2
/b^5/d-(a^2-b^2)*sin(d*x+c)^3/b^4/d+1/2*a*sin(d*x+c)^4/b^3/d-1/5*sin(d*x+c)^5/b^2/d+(a^2-b^2)^3/b^7/d/(a+b*sin
(d*x+c))

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Rubi [A]  time = 0.17, antiderivative size = 184, 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 = {2668, 697} \[ -\frac {\left (a^2-b^2\right ) \sin ^3(c+d x)}{b^4 d}+\frac {a \left (2 a^2-3 b^2\right ) \sin ^2(c+d x)}{b^5 d}-\frac {\left (-9 a^2 b^2+5 a^4+3 b^4\right ) \sin (c+d x)}{b^6 d}+\frac {\left (a^2-b^2\right )^3}{b^7 d (a+b \sin (c+d x))}+\frac {6 a \left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{b^7 d}+\frac {a \sin ^4(c+d x)}{2 b^3 d}-\frac {\sin ^5(c+d x)}{5 b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*a*(a^2 - b^2)^2*Log[a + b*Sin[c + d*x]])/(b^7*d) - ((5*a^4 - 9*a^2*b^2 + 3*b^4)*Sin[c + d*x])/(b^6*d) + (a*
(2*a^2 - 3*b^2)*Sin[c + d*x]^2)/(b^5*d) - ((a^2 - b^2)*Sin[c + d*x]^3)/(b^4*d) + (a*Sin[c + d*x]^4)/(2*b^3*d)
- Sin[c + d*x]^5/(5*b^2*d) + (a^2 - b^2)^3/(b^7*d*(a + b*Sin[c + d*x]))

Rule 697

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + c*
x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0]

Rule 2668

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*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \frac {\cos ^7(c+d x)}{(a+b \sin (c+d x))^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (b^2-x^2\right )^3}{(a+x)^2} \, dx,x,b \sin (c+d x)\right )}{b^7 d}\\ &=\frac {\operatorname {Subst}\left (\int \left (-5 a^4 \left (1+\frac {3 b^2 \left (-3 a^2+b^2\right )}{5 a^4}\right )+2 a \left (2 a^2-3 b^2\right ) x-3 \left (a^2-b^2\right ) x^2+2 a x^3-x^4-\frac {\left (a^2-b^2\right )^3}{(a+x)^2}+\frac {6 a \left (a^2-b^2\right )^2}{a+x}\right ) \, dx,x,b \sin (c+d x)\right )}{b^7 d}\\ &=\frac {6 a \left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{b^7 d}-\frac {\left (5 a^4-9 a^2 b^2+3 b^4\right ) \sin (c+d x)}{b^6 d}+\frac {a \left (2 a^2-3 b^2\right ) \sin ^2(c+d x)}{b^5 d}-\frac {\left (a^2-b^2\right ) \sin ^3(c+d x)}{b^4 d}+\frac {a \sin ^4(c+d x)}{2 b^3 d}-\frac {\sin ^5(c+d x)}{5 b^2 d}+\frac {\left (a^2-b^2\right )^3}{b^7 d (a+b \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.48, size = 235, normalized size = 1.28 \[ \frac {-4 a^2 b^4 \sin ^4(c+d x)+4 \left (a^2-b^2\right )^2 \left (15 a^2 \log (a+b \sin (c+d x))+4 a^2-4 b^2\right )+b^4 \cos ^4(c+d x) \left (-a^2+3 a b \sin (c+d x)+4 b^2\right )+2 a b^3 \left (5 a^2-7 b^2\right ) \sin ^3(c+d x)-2 b^2 \left (15 a^4-29 a^2 b^2+8 b^4\right ) \sin ^2(c+d x)+4 a b \sin (c+d x) \left (-11 a^4+15 \left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))+18 a^2 b^2-4 b^4\right )+2 b^6 \cos ^6(c+d x)}{10 b^7 d (a+b \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b^6*Cos[c + d*x]^6 + 4*(a^2 - b^2)^2*(4*a^2 - 4*b^2 + 15*a^2*Log[a + b*Sin[c + d*x]]) + 4*a*b*(-11*a^4 + 18
*a^2*b^2 - 4*b^4 + 15*(a^2 - b^2)^2*Log[a + b*Sin[c + d*x]])*Sin[c + d*x] - 2*b^2*(15*a^4 - 29*a^2*b^2 + 8*b^4
)*Sin[c + d*x]^2 + 2*a*b^3*(5*a^2 - 7*b^2)*Sin[c + d*x]^3 - 4*a^2*b^4*Sin[c + d*x]^4 + b^4*Cos[c + d*x]^4*(-a^
2 + 4*b^2 + 3*a*b*Sin[c + d*x]))/(10*b^7*d*(a + b*Sin[c + d*x]))

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fricas [A]  time = 0.54, size = 243, normalized size = 1.32 \[ \frac {16 \, b^{6} \cos \left (d x + c\right )^{6} + 80 \, a^{6} - 560 \, a^{4} b^{2} + 785 \, a^{2} b^{4} - 256 \, b^{6} - 8 \, {\left (5 \, a^{2} b^{4} - 4 \, b^{6}\right )} \cos \left (d x + c\right )^{4} + 16 \, {\left (15 \, a^{4} b^{2} - 25 \, a^{2} b^{4} + 8 \, b^{6}\right )} \cos \left (d x + c\right )^{2} + 480 \, {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4} + {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} \sin \left (d x + c\right )\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) + {\left (24 \, a b^{5} \cos \left (d x + c\right )^{4} - 400 \, a^{5} b + 720 \, a^{3} b^{3} - 271 \, a b^{5} - 16 \, {\left (5 \, a^{3} b^{3} - 7 \, a b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{80 \, {\left (b^{8} d \sin \left (d x + c\right ) + a b^{7} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/80*(16*b^6*cos(d*x + c)^6 + 80*a^6 - 560*a^4*b^2 + 785*a^2*b^4 - 256*b^6 - 8*(5*a^2*b^4 - 4*b^6)*cos(d*x + c
)^4 + 16*(15*a^4*b^2 - 25*a^2*b^4 + 8*b^6)*cos(d*x + c)^2 + 480*(a^6 - 2*a^4*b^2 + a^2*b^4 + (a^5*b - 2*a^3*b^
3 + a*b^5)*sin(d*x + c))*log(b*sin(d*x + c) + a) + (24*a*b^5*cos(d*x + c)^4 - 400*a^5*b + 720*a^3*b^3 - 271*a*
b^5 - 16*(5*a^3*b^3 - 7*a*b^5)*cos(d*x + c)^2)*sin(d*x + c))/(b^8*d*sin(d*x + c) + a*b^7*d)

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giac [A]  time = 0.98, size = 251, normalized size = 1.36 \[ \frac {\frac {60 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{7}} - \frac {10 \, {\left (6 \, a^{5} b \sin \left (d x + c\right ) - 12 \, a^{3} b^{3} \sin \left (d x + c\right ) + 6 \, a b^{5} \sin \left (d x + c\right ) + 5 \, a^{6} - 9 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )}}{{\left (b \sin \left (d x + c\right ) + a\right )} b^{7}} - \frac {2 \, b^{8} \sin \left (d x + c\right )^{5} - 5 \, a b^{7} \sin \left (d x + c\right )^{4} + 10 \, a^{2} b^{6} \sin \left (d x + c\right )^{3} - 10 \, b^{8} \sin \left (d x + c\right )^{3} - 20 \, a^{3} b^{5} \sin \left (d x + c\right )^{2} + 30 \, a b^{7} \sin \left (d x + c\right )^{2} + 50 \, a^{4} b^{4} \sin \left (d x + c\right ) - 90 \, a^{2} b^{6} \sin \left (d x + c\right ) + 30 \, b^{8} \sin \left (d x + c\right )}{b^{10}}}{10 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*(60*(a^5 - 2*a^3*b^2 + a*b^4)*log(abs(b*sin(d*x + c) + a))/b^7 - 10*(6*a^5*b*sin(d*x + c) - 12*a^3*b^3*si
n(d*x + c) + 6*a*b^5*sin(d*x + c) + 5*a^6 - 9*a^4*b^2 + 3*a^2*b^4 + b^6)/((b*sin(d*x + c) + a)*b^7) - (2*b^8*s
in(d*x + c)^5 - 5*a*b^7*sin(d*x + c)^4 + 10*a^2*b^6*sin(d*x + c)^3 - 10*b^8*sin(d*x + c)^3 - 20*a^3*b^5*sin(d*
x + c)^2 + 30*a*b^7*sin(d*x + c)^2 + 50*a^4*b^4*sin(d*x + c) - 90*a^2*b^6*sin(d*x + c) + 30*b^8*sin(d*x + c))/
b^10)/d

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maple [A]  time = 0.25, size = 305, normalized size = 1.66 \[ -\frac {\sin ^{5}\left (d x +c \right )}{5 b^{2} d}+\frac {a \left (\sin ^{4}\left (d x +c \right )\right )}{2 b^{3} d}-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) a^{2}}{d \,b^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{b^{2} d}+\frac {2 \left (\sin ^{2}\left (d x +c \right )\right ) a^{3}}{d \,b^{5}}-\frac {3 a \left (\sin ^{2}\left (d x +c \right )\right )}{b^{3} d}-\frac {5 a^{4} \sin \left (d x +c \right )}{d \,b^{6}}+\frac {9 a^{2} \sin \left (d x +c \right )}{d \,b^{4}}-\frac {3 \sin \left (d x +c \right )}{b^{2} d}+\frac {6 a^{5} \ln \left (a +b \sin \left (d x +c \right )\right )}{d \,b^{7}}-\frac {12 a^{3} \ln \left (a +b \sin \left (d x +c \right )\right )}{d \,b^{5}}+\frac {6 a \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{3} d}+\frac {a^{6}}{d \,b^{7} \left (a +b \sin \left (d x +c \right )\right )}-\frac {3 a^{4}}{d \,b^{5} \left (a +b \sin \left (d x +c \right )\right )}+\frac {3 a^{2}}{d \,b^{3} \left (a +b \sin \left (d x +c \right )\right )}-\frac {1}{b d \left (a +b \sin \left (d x +c \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*sin(d*x+c)^5/b^2/d+1/2*a*sin(d*x+c)^4/b^3/d-1/d/b^4*sin(d*x+c)^3*a^2+sin(d*x+c)^3/b^2/d+2/d/b^5*sin(d*x+c
)^2*a^3-3*a*sin(d*x+c)^2/b^3/d-5/d/b^6*a^4*sin(d*x+c)+9/d/b^4*a^2*sin(d*x+c)-3*sin(d*x+c)/b^2/d+6/d*a^5/b^7*ln
(a+b*sin(d*x+c))-12/d*a^3/b^5*ln(a+b*sin(d*x+c))+6*a*ln(a+b*sin(d*x+c))/b^3/d+1/d/b^7/(a+b*sin(d*x+c))*a^6-3/d
/b^5/(a+b*sin(d*x+c))*a^4+3/d/b^3/(a+b*sin(d*x+c))*a^2-1/b/d/(a+b*sin(d*x+c))

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maxima [A]  time = 0.31, size = 190, normalized size = 1.03 \[ \frac {\frac {10 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )}}{b^{8} \sin \left (d x + c\right ) + a b^{7}} - \frac {2 \, b^{4} \sin \left (d x + c\right )^{5} - 5 \, a b^{3} \sin \left (d x + c\right )^{4} + 10 \, {\left (a^{2} b^{2} - b^{4}\right )} \sin \left (d x + c\right )^{3} - 10 \, {\left (2 \, a^{3} b - 3 \, a b^{3}\right )} \sin \left (d x + c\right )^{2} + 10 \, {\left (5 \, a^{4} - 9 \, a^{2} b^{2} + 3 \, b^{4}\right )} \sin \left (d x + c\right )}{b^{6}} + \frac {60 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{7}}}{10 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*(10*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)/(b^8*sin(d*x + c) + a*b^7) - (2*b^4*sin(d*x + c)^5 - 5*a*b^3*sin(
d*x + c)^4 + 10*(a^2*b^2 - b^4)*sin(d*x + c)^3 - 10*(2*a^3*b - 3*a*b^3)*sin(d*x + c)^2 + 10*(5*a^4 - 9*a^2*b^2
+ 3*b^4)*sin(d*x + c))/b^6 + 60*(a^5 - 2*a^3*b^2 + a*b^4)*log(b*sin(d*x + c) + a)/b^7)/d

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mupad [B]  time = 0.12, size = 259, normalized size = 1.41 \[ \frac {{\sin \left (c+d\,x\right )}^3\,\left (\frac {1}{b^2}-\frac {a^2}{b^4}\right )}{d}-\frac {{\sin \left (c+d\,x\right )}^5}{5\,b^2\,d}-\frac {{\sin \left (c+d\,x\right )}^2\,\left (\frac {a^3}{b^5}+\frac {a\,\left (\frac {3}{b^2}-\frac {3\,a^2}{b^4}\right )}{b}\right )}{d}-\frac {\sin \left (c+d\,x\right )\,\left (\frac {3}{b^2}+\frac {a^2\,\left (\frac {3}{b^2}-\frac {3\,a^2}{b^4}\right )}{b^2}-\frac {2\,a\,\left (\frac {2\,a^3}{b^5}+\frac {2\,a\,\left (\frac {3}{b^2}-\frac {3\,a^2}{b^4}\right )}{b}\right )}{b}\right )}{d}+\frac {a\,{\sin \left (c+d\,x\right )}^4}{2\,b^3\,d}+\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (6\,a^5-12\,a^3\,b^2+6\,a\,b^4\right )}{b^7\,d}+\frac {a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}{b\,d\,\left (\sin \left (c+d\,x\right )\,b^7+a\,b^6\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(sin(c + d*x)^3*(1/b^2 - a^2/b^4))/d - sin(c + d*x)^5/(5*b^2*d) - (sin(c + d*x)^2*(a^3/b^5 + (a*(3/b^2 - (3*a^
2)/b^4))/b))/d - (sin(c + d*x)*(3/b^2 + (a^2*(3/b^2 - (3*a^2)/b^4))/b^2 - (2*a*((2*a^3)/b^5 + (2*a*(3/b^2 - (3
*a^2)/b^4))/b))/b))/d + (a*sin(c + d*x)^4)/(2*b^3*d) + (log(a + b*sin(c + d*x))*(6*a*b^4 + 6*a^5 - 12*a^3*b^2)
)/(b^7*d) + (a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)/(b*d*(a*b^6 + b^7*sin(c + d*x)))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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