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

Optimal. Leaf size=79 \[ \frac {4 \sin (a+b x)}{15 b \sin ^{\frac {3}{2}}(2 a+2 b x)}-\frac {\cos (a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}-\frac {8 \cos (a+b x)}{15 b \sqrt {\sin (2 a+2 b x)}} \]

[Out]

-1/5*cos(b*x+a)/b/sin(2*b*x+2*a)^(5/2)+4/15*sin(b*x+a)/b/sin(2*b*x+2*a)^(3/2)-8/15*cos(b*x+a)/b/sin(2*b*x+2*a)
^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4303, 4304, 4291} \[ \frac {4 \sin (a+b x)}{15 b \sin ^{\frac {3}{2}}(2 a+2 b x)}-\frac {\cos (a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}-\frac {8 \cos (a+b x)}{15 b \sqrt {\sin (2 a+2 b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Cos[a + b*x]/(5*b*Sin[2*a + 2*b*x]^(5/2)) + (4*Sin[a + b*x])/(15*b*Sin[2*a + 2*b*x]^(3/2)) - (8*Cos[a + b*x])
/(15*b*Sqrt[Sin[2*a + 2*b*x]])

Rule 4291

Int[(cos[(a_.) + (b_.)*(x_)]*(e_.))^(m_.)*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> -Simp[((e*Cos[a +
b*x])^m*(g*Sin[c + d*x])^(p + 1))/(b*g*m), x] /; FreeQ[{a, b, c, d, e, g, m, p}, x] && EqQ[b*c - a*d, 0] && E
qQ[d/b, 2] &&  !IntegerQ[p] && EqQ[m + 2*p + 2, 0]

Rule 4303

Int[cos[(a_.) + (b_.)*(x_)]*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[(Cos[a + b*x]*(g*Sin[c + d
*x])^(p + 1))/(2*b*g*(p + 1)), x] + Dist[(2*p + 3)/(2*g*(p + 1)), Int[Sin[a + b*x]*(g*Sin[c + d*x])^(p + 1), x
], x] /; FreeQ[{a, b, c, d, g}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] &&  !IntegerQ[p] && LtQ[p, -1] && Integ
erQ[2*p]

Rule 4304

Int[sin[(a_.) + (b_.)*(x_)]*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> -Simp[(Sin[a + b*x]*(g*Sin[c +
d*x])^(p + 1))/(2*b*g*(p + 1)), x] + Dist[(2*p + 3)/(2*g*(p + 1)), Int[Cos[a + b*x]*(g*Sin[c + d*x])^(p + 1),
x], x] /; FreeQ[{a, b, c, d, g}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] &&  !IntegerQ[p] && LtQ[p, -1] && Inte
gerQ[2*p]

Rubi steps

\begin {align*} \int \frac {\cos (a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx &=-\frac {\cos (a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}+\frac {4}{5} \int \frac {\sin (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx\\ &=-\frac {\cos (a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}+\frac {4 \sin (a+b x)}{15 b \sin ^{\frac {3}{2}}(2 a+2 b x)}+\frac {8}{15} \int \frac {\cos (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx\\ &=-\frac {\cos (a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}+\frac {4 \sin (a+b x)}{15 b \sin ^{\frac {3}{2}}(2 a+2 b x)}-\frac {8 \cos (a+b x)}{15 b \sqrt {\sin (2 a+2 b x)}}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 52, normalized size = 0.66 \[ -\frac {\sqrt {\sin (2 (a+b x))} \left (3 \csc ^3(a+b x)+27 \csc (a+b x)-5 \tan (a+b x) \sec (a+b x)\right )}{120 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/120*(Sqrt[Sin[2*(a + b*x)]]*(27*Csc[a + b*x] + 3*Csc[a + b*x]^3 - 5*Sec[a + b*x]*Tan[a + b*x]))/b

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fricas [A]  time = 0.46, size = 103, normalized size = 1.30 \[ -\frac {\sqrt {2} {\left (32 \, \cos \left (b x + a\right )^{4} - 40 \, \cos \left (b x + a\right )^{2} + 5\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 32 \, {\left (\cos \left (b x + a\right )^{4} - \cos \left (b x + a\right )^{2}\right )} \sin \left (b x + a\right )}{120 \, {\left (b \cos \left (b x + a\right )^{4} - b \cos \left (b x + a\right )^{2}\right )} \sin \left (b x + a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/120*(sqrt(2)*(32*cos(b*x + a)^4 - 40*cos(b*x + a)^2 + 5)*sqrt(cos(b*x + a)*sin(b*x + a)) + 32*(cos(b*x + a)
^4 - cos(b*x + a)^2)*sin(b*x + a))/((b*cos(b*x + a)^4 - b*cos(b*x + a)^2)*sin(b*x + a))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (b x + a\right )}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F(-1)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (b x +a \right )}{\sin \left (2 b x +2 a \right )^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (b x + a\right )}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.23, size = 136, normalized size = 1.72 \[ \frac {4\,{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,2{}\mathrm {i}+{\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}\,3{}\mathrm {i}+{\mathrm {e}}^{a\,6{}\mathrm {i}+b\,x\,6{}\mathrm {i}}\,2{}\mathrm {i}-{\mathrm {e}}^{a\,8{}\mathrm {i}+b\,x\,8{}\mathrm {i}}\,2{}\mathrm {i}-2{}\mathrm {i}\right )}{15\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}-1\right )}^3\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*exp(a*1i + b*x*1i)*((exp(- a*2i - b*x*2i)*1i)/2 - (exp(a*2i + b*x*2i)*1i)/2)^(1/2)*(exp(a*2i + b*x*2i)*2i +
exp(a*4i + b*x*4i)*3i + exp(a*6i + b*x*6i)*2i - exp(a*8i + b*x*8i)*2i - 2i))/(15*b*(exp(a*2i + b*x*2i) - 1)^3
*(exp(a*2i + b*x*2i) + 1)^2)

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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(b*x+a)/sin(2*b*x+2*a)**(7/2),x)

[Out]

Timed out

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