3.2.82 \(\int \frac {\cos ^3(a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx\) [182]

Optimal. Leaf size=55 \[ -\frac {\cos ^3(a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}-\frac {\cos (a+b x)}{5 b \sqrt {\sin (2 a+2 b x)}} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {4380, 4376} \begin {gather*} -\frac {\cos ^3(a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}-\frac {\cos (a+b x)}{5 b \sqrt {\sin (2 a+2 b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^3/Sin[2*a + 2*b*x]^(7/2),x]

[Out]

-1/5*Cos[a + b*x]^3/(b*Sin[2*a + 2*b*x]^(5/2)) - Cos[a + b*x]/(5*b*Sqrt[Sin[2*a + 2*b*x]])

Rule 4376

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] &&
EqQ[d/b, 2] &&  !IntegerQ[p] && EqQ[m + 2*p + 2, 0]

Rule 4380

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)/(2*b*g*(p + 1))), x] + Dist[e^2*((m + 2*p + 2)/(4*g^2*(p + 1))), Int[(e*Cos[a
+ b*x])^(m - 2)*(g*Sin[c + d*x])^(p + 2), x], x] /; FreeQ[{a, b, c, d, e, g}, x] && EqQ[b*c - a*d, 0] && EqQ[d
/b, 2] &&  !IntegerQ[p] && GtQ[m, 1] && LtQ[p, -1] && NeQ[m + 2*p + 2, 0] && (LtQ[p, -2] || EqQ[m, 2]) && Inte
gersQ[2*m, 2*p]

Rubi steps

\begin {align*} \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx &=-\frac {\cos ^3(a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}+\frac {1}{5} \int \frac {\cos (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx\\ &=-\frac {\cos ^3(a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}-\frac {\cos (a+b x)}{5 b \sqrt {\sin (2 a+2 b x)}}\\ \end {align*}

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Mathematica [A]
time = 0.11, size = 35, normalized size = 0.64 \begin {gather*} -\frac {\csc (a+b x) \left (4+\csc ^2(a+b x)\right ) \sqrt {\sin (2 (a+b x))}}{40 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^3/Sin[2*a + 2*b*x]^(7/2),x]

[Out]

-1/40*(Csc[a + b*x]*(4 + Csc[a + b*x]^2)*Sqrt[Sin[2*(a + b*x)]])/b

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {\cos ^{3}\left (b x +a \right )}{\sin \left (2 b x +2 a \right )^{\frac {7}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^3/sin(2*b*x+2*a)^(7/2),x)

[Out]

int(cos(b*x+a)^3/sin(2*b*x+2*a)^(7/2),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3/sin(2*b*x+2*a)^(7/2),x, algorithm="maxima")

[Out]

integrate(cos(b*x + a)^3/sin(2*b*x + 2*a)^(7/2), x)

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Fricas [A]
time = 3.16, size = 76, normalized size = 1.38 \begin {gather*} -\frac {\sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{2} - 5\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 4 \, {\left (\cos \left (b x + a\right )^{2} - 1\right )} \sin \left (b x + a\right )}{40 \, {\left (b \cos \left (b x + a\right )^{2} - b\right )} \sin \left (b x + a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3/sin(2*b*x+2*a)^(7/2),x, algorithm="fricas")

[Out]

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

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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(b*x+a)**3/sin(2*b*x+2*a)**(7/2),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(b*x+a)^3/sin(2*b*x+2*a)^(7/2),x, algorithm="giac")

[Out]

Timed out

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Mupad [B]
time = 3.07, size = 93, normalized size = 1.69 \begin {gather*} -\frac {{\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}}\,3{}\mathrm {i}+{\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}\,1{}\mathrm {i}+1{}\mathrm {i}\right )}{5\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}-1\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)^3/sin(2*a + 2*b*x)^(7/2),x)

[Out]

-(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*4i + b*x*4i)*1i -
exp(a*2i + b*x*2i)*3i + 1i))/(5*b*(exp(a*2i + b*x*2i) - 1)^3)

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