3.2.17 \(\int \csc ^3(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx\) [117]

Optimal. Leaf size=127 \[ -\frac {3 \text {ArcSin}(\cos (a+b x)-\sin (a+b x))}{b}+\frac {3 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}\right )}{b}-\frac {6 \cos (a+b x) \sqrt {\sin (2 a+2 b x)}}{b}+\frac {4 \sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{b}+\frac {\csc ^3(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x)}{b} \]

[Out]

-3*arcsin(cos(b*x+a)-sin(b*x+a))/b+3*ln(cos(b*x+a)+sin(b*x+a)+sin(2*b*x+2*a)^(1/2))/b+4*sin(b*x+a)*sin(2*b*x+2
*a)^(3/2)/b+csc(b*x+a)^3*sin(2*b*x+2*a)^(7/2)/b-6*cos(b*x+a)*sin(2*b*x+2*a)^(1/2)/b

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Rubi [A]
time = 0.09, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {4385, 4393, 4386, 4387, 4390} \begin {gather*} -\frac {3 \text {ArcSin}(\cos (a+b x)-\sin (a+b x))}{b}+\frac {4 \sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{b}-\frac {6 \sqrt {\sin (2 a+2 b x)} \cos (a+b x)}{b}+\frac {\sin ^{\frac {7}{2}}(2 a+2 b x) \csc ^3(a+b x)}{b}+\frac {3 \log \left (\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}+\cos (a+b x)\right )}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*ArcSin[Cos[a + b*x] - Sin[a + b*x]])/b + (3*Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt[Sin[2*a + 2*b*x]]])/b -
(6*Cos[a + b*x]*Sqrt[Sin[2*a + 2*b*x]])/b + (4*Sin[a + b*x]*Sin[2*a + 2*b*x]^(3/2))/b + (Csc[a + b*x]^3*Sin[2
*a + 2*b*x]^(7/2))/b

Rule 4385

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

Rule 4386

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

Rule 4387

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

Rule 4390

Int[cos[(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 4393

Int[((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_)/sin[(a_.) + (b_.)*(x_)], x_Symbol] :> Dist[2*g, Int[Cos[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 \csc ^3(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx &=\frac {\csc ^3(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x)}{b}+8 \int \csc (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx\\ &=\frac {\csc ^3(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x)}{b}+16 \int \cos (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx\\ &=\frac {4 \sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{b}+\frac {\csc ^3(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x)}{b}+12 \int \sin (a+b x) \sqrt {\sin (2 a+2 b x)} \, dx\\ &=-\frac {6 \cos (a+b x) \sqrt {\sin (2 a+2 b x)}}{b}+\frac {4 \sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{b}+\frac {\csc ^3(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x)}{b}+6 \int \frac {\cos (a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx\\ &=-\frac {3 \sin ^{-1}(\cos (a+b x)-\sin (a+b x))}{b}+\frac {3 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}\right )}{b}-\frac {6 \cos (a+b x) \sqrt {\sin (2 a+2 b x)}}{b}+\frac {4 \sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{b}+\frac {\csc ^3(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x)}{b}\\ \end {align*}

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Mathematica [A]
time = 0.15, size = 70, normalized size = 0.55 \begin {gather*} \frac {-3 \text {ArcSin}(\cos (a+b x)-\sin (a+b x))+3 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 (a+b x))}\right )+\csc (a+b x) \sin ^{\frac {3}{2}}(2 (a+b x))}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*ArcSin[Cos[a + b*x] - Sin[a + b*x]] + 3*Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt[Sin[2*(a + b*x)]]] + Csc[a
+ b*x]*Sin[2*(a + b*x)]^(3/2))/b

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 33.22, size = 243, normalized size = 1.91

method	result	size
default	\(\frac {16 \sqrt {-\frac {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}{\tan ^{2}\left (\frac {a}{2}+\frac {x b}{2}\right )-1}}\, \left (\sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )+1}\, \sqrt {-2 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )+2}\, \sqrt {-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}\, \EllipticF \left (\sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )+1}, \frac {\sqrt {2}}{2}\right ) \left (\tan ^{2}\left (\frac {a}{2}+\frac {x b}{2}\right )\right )-\sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )+1}\, \sqrt {-2 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )+2}\, \sqrt {-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}\, \EllipticF \left (\sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )+1}, \frac {\sqrt {2}}{2}\right )-\left (\tan ^{3}\left (\frac {a}{2}+\frac {x b}{2}\right )\right )-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )\right )}{3 \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \left (\tan ^{2}\left (\frac {a}{2}+\frac {x b}{2}\right )-1\right )}\, \sqrt {\tan ^{3}\left (\frac {a}{2}+\frac {x b}{2}\right )-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}\, b}\)	\(243\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

16/3*(-tan(1/2*a+1/2*x*b)/(tan(1/2*a+1/2*x*b)^2-1))^(1/2)*((tan(1/2*a+1/2*x*b)+1)^(1/2)*(-2*tan(1/2*a+1/2*x*b)
+2)^(1/2)*(-tan(1/2*a+1/2*x*b))^(1/2)*EllipticF((tan(1/2*a+1/2*x*b)+1)^(1/2),1/2*2^(1/2))*tan(1/2*a+1/2*x*b)^2
-(tan(1/2*a+1/2*x*b)+1)^(1/2)*(-2*tan(1/2*a+1/2*x*b)+2)^(1/2)*(-tan(1/2*a+1/2*x*b))^(1/2)*EllipticF((tan(1/2*a
+1/2*x*b)+1)^(1/2),1/2*2^(1/2))-tan(1/2*a+1/2*x*b)^3-tan(1/2*a+1/2*x*b))/(tan(1/2*a+1/2*x*b)*(tan(1/2*a+1/2*x*
b)^2-1))^(1/2)/(tan(1/2*a+1/2*x*b)^3-tan(1/2*a+1/2*x*b))^(1/2)/b

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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(csc(b*x+a)^3*sin(2*b*x+2*a)^(5/2),x, algorithm="maxima")

[Out]

integrate(csc(b*x + a)^3*sin(2*b*x + 2*a)^(5/2), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 268 vs. \(2 (119) = 238\).
time = 2.46, size = 268, normalized size = 2.11 \begin {gather*} \frac {8 \, \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} \cos \left (b x + a\right ) + 6 \, \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 ) - 6 \, \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 ) - 3 \, \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 )}{4 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(8*sqrt(2)*sqrt(cos(b*x + a)*sin(b*x + a))*cos(b*x + a) + 6*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)) - 6*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))) - 3*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))/b

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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(csc(b*x+a)**3*sin(2*b*x+2*a)**(5/2),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(csc(b*x+a)^3*sin(2*b*x+2*a)^(5/2),x, algorithm="giac")

[Out]

integrate(csc(b*x + a)^3*sin(2*b*x + 2*a)^(5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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