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

Optimal. Leaf size=190 \[ -\frac {6 a \left (a^2-b^2\right )^2}{b^7 d (a+b \sin (c+d x))}+\frac {\left (a^2-b^2\right )^3}{2 b^7 d (a+b \sin (c+d x))^2}+\frac {a \left (10 a^2-9 b^2\right ) \sin (c+d x)}{b^6 d}-\frac {3 \left (2 a^2-b^2\right ) \sin ^2(c+d x)}{2 b^5 d}-\frac {3 \left (5 a^4-6 a^2 b^2+b^4\right ) \log (a+b \sin (c+d x))}{b^7 d}+\frac {a \sin ^3(c+d x)}{b^4 d}-\frac {\sin ^4(c+d x)}{4 b^3 d} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 190, 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} \[ -\frac {3 \left (2 a^2-b^2\right ) \sin ^2(c+d x)}{2 b^5 d}+\frac {a \left (10 a^2-9 b^2\right ) \sin (c+d x)}{b^6 d}-\frac {6 a \left (a^2-b^2\right )^2}{b^7 d (a+b \sin (c+d x))}+\frac {\left (a^2-b^2\right )^3}{2 b^7 d (a+b \sin (c+d x))^2}-\frac {3 \left (-6 a^2 b^2+5 a^4+b^4\right ) \log (a+b \sin (c+d x))}{b^7 d}+\frac {a \sin ^3(c+d x)}{b^4 d}-\frac {\sin ^4(c+d x)}{4 b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(5*a^4 - 6*a^2*b^2 + b^4)*Log[a + b*Sin[c + d*x]])/(b^7*d) + (a*(10*a^2 - 9*b^2)*Sin[c + d*x])/(b^6*d) - (
3*(2*a^2 - b^2)*Sin[c + d*x]^2)/(2*b^5*d) + (a*Sin[c + d*x]^3)/(b^4*d) - Sin[c + d*x]^4/(4*b^3*d) + (a^2 - b^2
)^3/(2*b^7*d*(a + b*Sin[c + d*x])^2) - (6*a*(a^2 - b^2)^2)/(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))^3} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (b^2-x^2\right )^3}{(a+x)^3} \, dx,x,b \sin (c+d x)\right )}{b^7 d}\\ &=\frac {\operatorname {Subst}\left (\int \left (10 a^3 \left (1-\frac {9 b^2}{10 a^2}\right )-3 \left (2 a^2-b^2\right ) x+3 a x^2-x^3-\frac {\left (a^2-b^2\right )^3}{(a+x)^3}+\frac {6 a \left (a^2-b^2\right )^2}{(a+x)^2}-\frac {3 \left (5 a^4-6 a^2 b^2+b^4\right )}{a+x}\right ) \, dx,x,b \sin (c+d x)\right )}{b^7 d}\\ &=-\frac {3 \left (5 a^4-6 a^2 b^2+b^4\right ) \log (a+b \sin (c+d x))}{b^7 d}+\frac {a \left (10 a^2-9 b^2\right ) \sin (c+d x)}{b^6 d}-\frac {3 \left (2 a^2-b^2\right ) \sin ^2(c+d x)}{2 b^5 d}+\frac {a \sin ^3(c+d x)}{b^4 d}-\frac {\sin ^4(c+d x)}{4 b^3 d}+\frac {\left (a^2-b^2\right )^3}{2 b^7 d (a+b \sin (c+d x))^2}-\frac {6 a \left (a^2-b^2\right )^2}{b^7 d (a+b \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.64, size = 282, normalized size = 1.48 \[ \frac {b^4 \cos ^4(c+d x) \left (-a^2+2 a b \sin (c+d x)+3 b^2\right )-2 \left (2 a^2 b^4 \sin ^4(c+d x)-10 a b^3 \left (a^2-b^2\right ) \sin ^3(c+d x)+2 b^2 \sin ^2(c+d x) \left (-13 a^4+10 a^2 b^2+3 \left (5 a^4-6 a^2 b^2+b^4\right ) \log (a+b \sin (c+d x))\right )+2 a b \sin (c+d x) \left (4 a^4-17 a^2 b^2+6 \left (5 a^4-6 a^2 b^2+b^4\right ) \log (a+b \sin (c+d x))+11 b^4\right )+\left (a^2-b^2\right ) \left (19 a^4+6 a^2 \left (5 a^2-b^2\right ) \log (a+b \sin (c+d x))-16 a^2 b^2-3 b^4\right )\right )+b^6 \cos ^6(c+d x)}{4 b^7 d (a+b \sin (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^6*Cos[c + d*x]^6 + b^4*Cos[c + d*x]^4*(-a^2 + 3*b^2 + 2*a*b*Sin[c + d*x]) - 2*((a^2 - b^2)*(19*a^4 - 16*a^2
*b^2 - 3*b^4 + 6*a^2*(5*a^2 - b^2)*Log[a + b*Sin[c + d*x]]) + 2*a*b*(4*a^4 - 17*a^2*b^2 + 11*b^4 + 6*(5*a^4 -
6*a^2*b^2 + b^4)*Log[a + b*Sin[c + d*x]])*Sin[c + d*x] + 2*b^2*(-13*a^4 + 10*a^2*b^2 + 3*(5*a^4 - 6*a^2*b^2 +
b^4)*Log[a + b*Sin[c + d*x]])*Sin[c + d*x]^2 - 10*a*b^3*(a^2 - b^2)*Sin[c + d*x]^3 + 2*a^2*b^4*Sin[c + d*x]^4)
)/(4*b^7*d*(a + b*Sin[c + d*x])^2)

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fricas [A]  time = 0.55, size = 304, normalized size = 1.60 \[ -\frac {8 \, b^{6} \cos \left (d x + c\right )^{6} - 176 \, a^{6} + 928 \, a^{4} b^{2} - 685 \, a^{2} b^{4} + 3 \, b^{6} - 8 \, {\left (5 \, a^{2} b^{4} - 3 \, b^{6}\right )} \cos \left (d x + c\right )^{4} - {\left (544 \, a^{4} b^{2} - 560 \, a^{2} b^{4} + 51 \, b^{6}\right )} \cos \left (d x + c\right )^{2} - 96 \, {\left (5 \, a^{6} - a^{4} b^{2} - 5 \, a^{2} b^{4} + b^{6} - {\left (5 \, a^{4} b^{2} - 6 \, a^{2} b^{4} + b^{6}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (5 \, a^{5} b - 6 \, a^{3} b^{3} + a b^{5}\right )} \sin \left (d x + c\right )\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) + 2 \, {\left (8 \, a b^{5} \cos \left (d x + c\right )^{4} + 64 \, a^{5} b + 176 \, a^{3} b^{3} - 205 \, a b^{5} - 80 \, {\left (a^{3} b^{3} - a b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{32 \, {\left (b^{9} d \cos \left (d x + c\right )^{2} - 2 \, a b^{8} d \sin \left (d x + c\right ) - {\left (a^{2} b^{7} + b^{9}\right )} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*(8*b^6*cos(d*x + c)^6 - 176*a^6 + 928*a^4*b^2 - 685*a^2*b^4 + 3*b^6 - 8*(5*a^2*b^4 - 3*b^6)*cos(d*x + c)
^4 - (544*a^4*b^2 - 560*a^2*b^4 + 51*b^6)*cos(d*x + c)^2 - 96*(5*a^6 - a^4*b^2 - 5*a^2*b^4 + b^6 - (5*a^4*b^2
- 6*a^2*b^4 + b^6)*cos(d*x + c)^2 + 2*(5*a^5*b - 6*a^3*b^3 + a*b^5)*sin(d*x + c))*log(b*sin(d*x + c) + a) + 2*
(8*a*b^5*cos(d*x + c)^4 + 64*a^5*b + 176*a^3*b^3 - 205*a*b^5 - 80*(a^3*b^3 - a*b^5)*cos(d*x + c)^2)*sin(d*x +
c))/(b^9*d*cos(d*x + c)^2 - 2*a*b^8*d*sin(d*x + c) - (a^2*b^7 + b^9)*d)

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giac [A]  time = 0.47, size = 245, normalized size = 1.29 \[ -\frac {\frac {12 \, {\left (5 \, a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{7}} - \frac {2 \, {\left (45 \, a^{4} b^{2} \sin \left (d x + c\right )^{2} - 54 \, a^{2} b^{4} \sin \left (d x + c\right )^{2} + 9 \, b^{6} \sin \left (d x + c\right )^{2} + 78 \, a^{5} b \sin \left (d x + c\right ) - 84 \, a^{3} b^{3} \sin \left (d x + c\right ) + 6 \, a b^{5} \sin \left (d x + c\right ) + 34 \, a^{6} - 33 \, a^{4} b^{2} - b^{6}\right )}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{2} b^{7}} + \frac {b^{9} \sin \left (d x + c\right )^{4} - 4 \, a b^{8} \sin \left (d x + c\right )^{3} + 12 \, a^{2} b^{7} \sin \left (d x + c\right )^{2} - 6 \, b^{9} \sin \left (d x + c\right )^{2} - 40 \, a^{3} b^{6} \sin \left (d x + c\right ) + 36 \, a b^{8} \sin \left (d x + c\right )}{b^{12}}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(12*(5*a^4 - 6*a^2*b^2 + b^4)*log(abs(b*sin(d*x + c) + a))/b^7 - 2*(45*a^4*b^2*sin(d*x + c)^2 - 54*a^2*b^
4*sin(d*x + c)^2 + 9*b^6*sin(d*x + c)^2 + 78*a^5*b*sin(d*x + c) - 84*a^3*b^3*sin(d*x + c) + 6*a*b^5*sin(d*x +
c) + 34*a^6 - 33*a^4*b^2 - b^6)/((b*sin(d*x + c) + a)^2*b^7) + (b^9*sin(d*x + c)^4 - 4*a*b^8*sin(d*x + c)^3 +
12*a^2*b^7*sin(d*x + c)^2 - 6*b^9*sin(d*x + c)^2 - 40*a^3*b^6*sin(d*x + c) + 36*a*b^8*sin(d*x + c))/b^12)/d

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maple [A]  time = 0.28, size = 320, normalized size = 1.68 \[ -\frac {\sin ^{4}\left (d x +c \right )}{4 b^{3} d}+\frac {a \left (\sin ^{3}\left (d x +c \right )\right )}{b^{4} d}-\frac {3 \left (\sin ^{2}\left (d x +c \right )\right ) a^{2}}{d \,b^{5}}+\frac {3 \left (\sin ^{2}\left (d x +c \right )\right )}{2 b^{3} d}+\frac {10 \sin \left (d x +c \right ) a^{3}}{d \,b^{6}}-\frac {9 a \sin \left (d x +c \right )}{b^{4} d}-\frac {15 \ln \left (a +b \sin \left (d x +c \right )\right ) a^{4}}{d \,b^{7}}+\frac {18 \ln \left (a +b \sin \left (d x +c \right )\right ) a^{2}}{d \,b^{5}}-\frac {3 \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{3} d}-\frac {6 a^{5}}{d \,b^{7} \left (a +b \sin \left (d x +c \right )\right )}+\frac {12 a^{3}}{d \,b^{5} \left (a +b \sin \left (d x +c \right )\right )}-\frac {6 a}{b^{3} d \left (a +b \sin \left (d x +c \right )\right )}+\frac {a^{6}}{2 d \,b^{7} \left (a +b \sin \left (d x +c \right )\right )^{2}}-\frac {3 a^{4}}{2 d \,b^{5} \left (a +b \sin \left (d x +c \right )\right )^{2}}+\frac {3 a^{2}}{2 d \,b^{3} \left (a +b \sin \left (d x +c \right )\right )^{2}}-\frac {1}{2 b d \left (a +b \sin \left (d x +c \right )\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*sin(d*x+c)^4/b^3/d+a*sin(d*x+c)^3/b^4/d-3/d/b^5*sin(d*x+c)^2*a^2+3/2*sin(d*x+c)^2/b^3/d+10/d/b^6*sin(d*x+
c)*a^3-9*a*sin(d*x+c)/b^4/d-15/d/b^7*ln(a+b*sin(d*x+c))*a^4+18/d/b^5*ln(a+b*sin(d*x+c))*a^2-3*ln(a+b*sin(d*x+c
))/b^3/d-6/d*a^5/b^7/(a+b*sin(d*x+c))+12/d*a^3/b^5/(a+b*sin(d*x+c))-6*a/b^3/d/(a+b*sin(d*x+c))+1/2/d/b^7/(a+b*
sin(d*x+c))^2*a^6-3/2/d/b^5/(a+b*sin(d*x+c))^2*a^4+3/2/d/b^3/(a+b*sin(d*x+c))^2*a^2-1/2/b/d/(a+b*sin(d*x+c))^2

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maxima [A]  time = 0.33, size = 200, normalized size = 1.05 \[ -\frac {\frac {2 \, {\left (11 \, a^{6} - 21 \, a^{4} b^{2} + 9 \, a^{2} b^{4} + b^{6} + 12 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} \sin \left (d x + c\right )\right )}}{b^{9} \sin \left (d x + c\right )^{2} + 2 \, a b^{8} \sin \left (d x + c\right ) + a^{2} b^{7}} + \frac {b^{3} \sin \left (d x + c\right )^{4} - 4 \, a b^{2} \sin \left (d x + c\right )^{3} + 6 \, {\left (2 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )^{2} - 4 \, {\left (10 \, a^{3} - 9 \, a b^{2}\right )} \sin \left (d x + c\right )}{b^{6}} + \frac {12 \, {\left (5 \, a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{7}}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*(11*a^6 - 21*a^4*b^2 + 9*a^2*b^4 + b^6 + 12*(a^5*b - 2*a^3*b^3 + a*b^5)*sin(d*x + c))/(b^9*sin(d*x + c
)^2 + 2*a*b^8*sin(d*x + c) + a^2*b^7) + (b^3*sin(d*x + c)^4 - 4*a*b^2*sin(d*x + c)^3 + 6*(2*a^2*b - b^3)*sin(d
*x + c)^2 - 4*(10*a^3 - 9*a*b^2)*sin(d*x + c))/b^6 + 12*(5*a^4 - 6*a^2*b^2 + b^4)*log(b*sin(d*x + c) + a)/b^7)
/d

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mupad [B]  time = 0.12, size = 234, normalized size = 1.23 \[ \frac {{\sin \left (c+d\,x\right )}^2\,\left (\frac {3}{2\,b^3}-\frac {3\,a^2}{b^5}\right )}{d}-\frac {{\sin \left (c+d\,x\right )}^4}{4\,b^3\,d}-\frac {\sin \left (c+d\,x\right )\,\left (\frac {8\,a^3}{b^6}+\frac {3\,a\,\left (\frac {3}{b^3}-\frac {6\,a^2}{b^5}\right )}{b}\right )}{d}-\frac {\frac {11\,a^6-21\,a^4\,b^2+9\,a^2\,b^4+b^6}{2\,b}+\sin \left (c+d\,x\right )\,\left (6\,a^5-12\,a^3\,b^2+6\,a\,b^4\right )}{d\,\left (a^2\,b^6+2\,a\,b^7\,\sin \left (c+d\,x\right )+b^8\,{\sin \left (c+d\,x\right )}^2\right )}+\frac {a\,{\sin \left (c+d\,x\right )}^3}{b^4\,d}-\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (15\,a^4-18\,a^2\,b^2+3\,b^4\right )}{b^7\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(sin(c + d*x)^2*(3/(2*b^3) - (3*a^2)/b^5))/d - sin(c + d*x)^4/(4*b^3*d) - (sin(c + d*x)*((8*a^3)/b^6 + (3*a*(3
/b^3 - (6*a^2)/b^5))/b))/d - ((11*a^6 + b^6 + 9*a^2*b^4 - 21*a^4*b^2)/(2*b) + sin(c + d*x)*(6*a*b^4 + 6*a^5 -
12*a^3*b^2))/(d*(a^2*b^6 + b^8*sin(c + d*x)^2 + 2*a*b^7*sin(c + d*x))) + (a*sin(c + d*x)^3)/(b^4*d) - (log(a +
b*sin(c + d*x))*(15*a^4 + 3*b^4 - 18*a^2*b^2))/(b^7*d)

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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))**3,x)

[Out]

Timed out

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