3.2.80 \(\int \frac {\cos ^3(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx\) [180]

Optimal. Leaf size=82 \[ \frac {\text {ArcSin}(\cos (a+b x)-\sin (a+b x))}{4 b}+\frac {\log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}\right )}{4 b}-\frac {\cos (a+b x)}{b \sqrt {\sin (2 a+2 b x)}} \]

[Out]

1/4*arcsin(cos(b*x+a)-sin(b*x+a))/b+1/4*ln(cos(b*x+a)+sin(b*x+a)+sin(2*b*x+2*a)^(1/2))/b-cos(b*x+a)/b/sin(2*b*
x+2*a)^(1/2)

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Rubi [A]
time = 0.06, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {4378, 4392, 4391} \begin {gather*} \frac {\text {ArcSin}(\cos (a+b x)-\sin (a+b x))}{4 b}-\frac {\cos (a+b x)}{b \sqrt {\sin (2 a+2 b x)}}+\frac {\log \left (\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}+\cos (a+b x)\right )}{4 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

ArcSin[Cos[a + b*x] - Sin[a + b*x]]/(4*b) + Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt[Sin[2*a + 2*b*x]]]/(4*b) -
Cos[a + b*x]/(b*Sqrt[Sin[2*a + 2*b*x]])

Rule 4378

Int[(cos[(a_.) + (b_.)*(x_)]*(e_.))^(m_)*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[e^2*(e*Cos[a
+ b*x])^(m - 2)*((g*Sin[c + d*x])^(p + 1)/(2*b*g*(p + 1))), x] + Dist[e^4*((m + p - 1)/(4*g^2*(p + 1))), Int[(
e*Cos[a + b*x])^(m - 4)*(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, 2] && LtQ[p, -1] && (GtQ[m, 3] || EqQ[p, -3/2]) && IntegersQ[2*m, 2*
p]

Rule 4391

Int[sin[(a_.) + (b_.)*(x_)]/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-ArcSin[Cos[a + b*x] - Sin[a + b*
x]]/d, x] - Simp[Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt[Sin[c + d*x]]]/d, x] /; FreeQ[{a, b, c, d}, x] && EqQ[
b*c - a*d, 0] && EqQ[d/b, 2]

Rule 4392

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

Rubi steps

\begin {align*} \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx &=-\frac {\cos (a+b x)}{b \sqrt {\sin (2 a+2 b x)}}-\frac {1}{4} \int \sec (a+b x) \sqrt {\sin (2 a+2 b x)} \, dx\\ &=-\frac {\cos (a+b x)}{b \sqrt {\sin (2 a+2 b x)}}-\frac {1}{2} \int \frac {\sin (a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx\\ &=\frac {\sin ^{-1}(\cos (a+b x)-\sin (a+b x))}{4 b}+\frac {\log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}\right )}{4 b}-\frac {\cos (a+b x)}{b \sqrt {\sin (2 a+2 b x)}}\\ \end {align*}

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Mathematica [A]
time = 0.11, size = 70, normalized size = 0.85 \begin {gather*} \frac {\text {ArcSin}(\cos (a+b x)-\sin (a+b x))+\log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 (a+b x))}\right )-2 \csc (a+b x) \sqrt {\sin (2 (a+b x))}}{4 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(ArcSin[Cos[a + b*x] - Sin[a + b*x]] + Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt[Sin[2*(a + b*x)]]] - 2*Csc[a + b
*x]*Sqrt[Sin[2*(a + b*x)]])/(4*b)

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Maple [B] result has leaf size over 500,000. Avoiding possible recursion issues.
time = 23.23, size = 179323150, normalized size = 2186867.68

method	result	size
default	\(\text {Expression too large to display}\)	\(179323150\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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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)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 295 vs. \(2 (74) = 148\).
time = 3.66, size = 295, normalized size = 3.60 \begin {gather*} -\frac {2 \, \arctan \left (-\frac {\sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} + \cos \left (b x + a\right ) \sin \left (b x + a\right )}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 1}\right ) \sin \left (b x + a\right ) - 2 \, \arctan \left (-\frac {2 \, \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} - \cos \left (b x + a\right ) - \sin \left (b x + a\right )}{\cos \left (b x + a\right ) - \sin \left (b x + a\right )}\right ) \sin \left (b x + a\right ) + \log \left (-32 \, \cos \left (b x + a\right )^{4} + 4 \, \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{3} - {\left (4 \, \cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right ) - 5 \, \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 32 \, \cos \left (b x + a\right )^{2} + 16 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) + 8 \, \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 8 \, \sin \left (b x + a\right )}{16 \, b \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)^(3/2),x, algorithm="fricas")

[Out]

-1/16*(2*arctan(-(sqrt(2)*sqrt(cos(b*x + a)*sin(b*x + a))*(cos(b*x + a) - sin(b*x + a)) + cos(b*x + a)*sin(b*x
+ a))/(cos(b*x + a)^2 + 2*cos(b*x + a)*sin(b*x + a) - 1))*sin(b*x + a) - 2*arctan(-(2*sqrt(2)*sqrt(cos(b*x +
a)*sin(b*x + a)) - cos(b*x + a) - sin(b*x + a))/(cos(b*x + a) - sin(b*x + a)))*sin(b*x + a) + log(-32*cos(b*x
+ a)^4 + 4*sqrt(2)*(4*cos(b*x + a)^3 - (4*cos(b*x + a)^2 + 1)*sin(b*x + a) - 5*cos(b*x + a))*sqrt(cos(b*x + a)
*sin(b*x + a)) + 32*cos(b*x + a)^2 + 16*cos(b*x + a)*sin(b*x + a) + 1)*sin(b*x + a) + 8*sqrt(2)*sqrt(cos(b*x +
a)*sin(b*x + a)) + 8*sin(b*x + a))/(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)**(3/2),x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not 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)^(3/2),x, algorithm="giac")

[Out]

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

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

[Out]

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

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