3.6.16 \(\int \frac {\cos ^5(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx\) [516]

Optimal. Leaf size=150 \[ -\frac {2 \left (a^2-b^2\right )^2}{b^5 d \sqrt {a+b \sin (c+d x)}}-\frac {8 a \left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}{b^5 d}+\frac {4 \left (3 a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}{3 b^5 d}-\frac {8 a (a+b \sin (c+d x))^{5/2}}{5 b^5 d}+\frac {2 (a+b \sin (c+d x))^{7/2}}{7 b^5 d} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2747, 711} \begin {gather*} \frac {4 \left (3 a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}{3 b^5 d}-\frac {8 a \left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}{b^5 d}-\frac {2 \left (a^2-b^2\right )^2}{b^5 d \sqrt {a+b \sin (c+d x)}}+\frac {2 (a+b \sin (c+d x))^{7/2}}{7 b^5 d}-\frac {8 a (a+b \sin (c+d x))^{5/2}}{5 b^5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5/(a + b*Sin[c + d*x])^(3/2),x]

[Out]

(-2*(a^2 - b^2)^2)/(b^5*d*Sqrt[a + b*Sin[c + d*x]]) - (8*a*(a^2 - b^2)*Sqrt[a + b*Sin[c + d*x]])/(b^5*d) + (4*
(3*a^2 - b^2)*(a + b*Sin[c + d*x])^(3/2))/(3*b^5*d) - (8*a*(a + b*Sin[c + d*x])^(5/2))/(5*b^5*d) + (2*(a + b*S
in[c + d*x])^(7/2))/(7*b^5*d)

Rule 711

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 2747

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 ^5(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx &=\frac {\text {Subst}\left (\int \frac {\left (b^2-x^2\right )^2}{(a+x)^{3/2}} \, dx,x,b \sin (c+d x)\right )}{b^5 d}\\ &=\frac {\text {Subst}\left (\int \left (\frac {\left (a^2-b^2\right )^2}{(a+x)^{3/2}}-\frac {4 \left (a^3-a b^2\right )}{\sqrt {a+x}}+2 \left (3 a^2-b^2\right ) \sqrt {a+x}-4 a (a+x)^{3/2}+(a+x)^{5/2}\right ) \, dx,x,b \sin (c+d x)\right )}{b^5 d}\\ &=-\frac {2 \left (a^2-b^2\right )^2}{b^5 d \sqrt {a+b \sin (c+d x)}}-\frac {8 a \left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}{b^5 d}+\frac {4 \left (3 a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}{3 b^5 d}-\frac {8 a (a+b \sin (c+d x))^{5/2}}{5 b^5 d}+\frac {2 (a+b \sin (c+d x))^{7/2}}{7 b^5 d}\\ \end {align*}

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Mathematica [A]
time = 0.69, size = 165, normalized size = 1.10 \begin {gather*} \frac {-3072 a^4+4672 a^2 b^2-1075 b^4-4 \left (48 a^2 b^2-55 b^4\right ) \cos (2 (c+d x))+15 b^4 \cos (4 (c+d x))-1536 a^3 b \sin (c+d x)+2096 a b^3 \sin (c+d x)+2232 a^4 \sqrt {1+\frac {b \sin (c+d x)}{a}}-2644 a^2 b^2 \sqrt {1+\frac {b \sin (c+d x)}{a}}+48 a b^3 \sin (3 (c+d x))}{420 b^5 d \sqrt {a+b \sin (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5/(a + b*Sin[c + d*x])^(3/2),x]

[Out]

(-3072*a^4 + 4672*a^2*b^2 - 1075*b^4 - 4*(48*a^2*b^2 - 55*b^4)*Cos[2*(c + d*x)] + 15*b^4*Cos[4*(c + d*x)] - 15
36*a^3*b*Sin[c + d*x] + 2096*a*b^3*Sin[c + d*x] + 2232*a^4*Sqrt[1 + (b*Sin[c + d*x])/a] - 2644*a^2*b^2*Sqrt[1
+ (b*Sin[c + d*x])/a] + 48*a*b^3*Sin[3*(c + d*x)])/(420*b^5*d*Sqrt[a + b*Sin[c + d*x]])

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Maple [A]
time = 1.00, size = 116, normalized size = 0.77

method result size
default \(\frac {\frac {16 a \,b^{3} \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{35}+\frac {2 \left (-192 b \,a^{3}+256 a \,b^{3}\right ) \sin \left (d x +c \right )}{105}+\frac {2 b^{4} \left (\cos ^{4}\left (d x +c \right )\right )}{7}+\frac {2 \left (-48 a^{2} b^{2}+40 b^{4}\right ) \left (\cos ^{2}\left (d x +c \right )\right )}{105}-\frac {256 a^{4}}{35}+\frac {1216 a^{2} b^{2}}{105}-\frac {64 b^{4}}{21}}{b^{5} \sqrt {a +b \sin \left (d x +c \right )}\, d}\) \(116\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5/(a+b*sin(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

2/105/b^5*(24*a*b^3*cos(d*x+c)^2*sin(d*x+c)+(-192*a^3*b+256*a*b^3)*sin(d*x+c)+15*b^4*cos(d*x+c)^4+(-48*a^2*b^2
+40*b^4)*cos(d*x+c)^2-384*a^4+608*a^2*b^2-160*b^4)/(a+b*sin(d*x+c))^(1/2)/d

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Maxima [A]
time = 0.27, size = 124, normalized size = 0.83 \begin {gather*} \frac {2 \, {\left (\frac {15 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}} - 84 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a + 70 \, {\left (3 \, a^{2} - b^{2}\right )} {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} - 420 \, {\left (a^{3} - a b^{2}\right )} \sqrt {b \sin \left (d x + c\right ) + a}}{b^{4}} - \frac {105 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )}}{\sqrt {b \sin \left (d x + c\right ) + a} b^{4}}\right )}}{105 \, b d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5/(a+b*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

2/105*((15*(b*sin(d*x + c) + a)^(7/2) - 84*(b*sin(d*x + c) + a)^(5/2)*a + 70*(3*a^2 - b^2)*(b*sin(d*x + c) + a
)^(3/2) - 420*(a^3 - a*b^2)*sqrt(b*sin(d*x + c) + a))/b^4 - 105*(a^4 - 2*a^2*b^2 + b^4)/(sqrt(b*sin(d*x + c) +
 a)*b^4))/(b*d)

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Fricas [A]
time = 0.38, size = 125, normalized size = 0.83 \begin {gather*} \frac {2 \, {\left (15 \, b^{4} \cos \left (d x + c\right )^{4} - 384 \, a^{4} + 608 \, a^{2} b^{2} - 160 \, b^{4} - 8 \, {\left (6 \, a^{2} b^{2} - 5 \, b^{4}\right )} \cos \left (d x + c\right )^{2} + 8 \, {\left (3 \, a b^{3} \cos \left (d x + c\right )^{2} - 24 \, a^{3} b + 32 \, a b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {b \sin \left (d x + c\right ) + a}}{105 \, {\left (b^{6} d \sin \left (d x + c\right ) + a b^{5} d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5/(a+b*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

2/105*(15*b^4*cos(d*x + c)^4 - 384*a^4 + 608*a^2*b^2 - 160*b^4 - 8*(6*a^2*b^2 - 5*b^4)*cos(d*x + c)^2 + 8*(3*a
*b^3*cos(d*x + c)^2 - 24*a^3*b + 32*a*b^3)*sin(d*x + c))*sqrt(b*sin(d*x + c) + a)/(b^6*d*sin(d*x + c) + a*b^5*
d)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**5/(a+b*sin(d*x+c))**(3/2),x)

[Out]

Timed out

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5/(a+b*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\cos \left (c+d\,x\right )}^5}{{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^5/(a + b*sin(c + d*x))^(3/2),x)

[Out]

int(cos(c + d*x)^5/(a + b*sin(c + d*x))^(3/2), x)

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