3.2.15 \(\int \csc ^3(a+b x) \sin ^{\frac {9}{2}}(2 a+2 b x) \, dx\) [115]

Optimal. Leaf size=190 \[ -\frac {7 \text {ArcSin}(\cos (a+b x)-\sin (a+b x))}{8 b}+\frac {7 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}\right )}{8 b}-\frac {7 \cos (a+b x) \sqrt {\sin (2 a+2 b x)}}{4 b}+\frac {7 \sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{6 b}-\frac {14 \cos (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{15 b}+\frac {4 \sin (a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x)}{5 b}+\frac {\csc ^3(a+b x) \sin ^{\frac {11}{2}}(2 a+2 b x)}{5 b} \]

[Out]

-7/8*arcsin(cos(b*x+a)-sin(b*x+a))/b+7/8*ln(cos(b*x+a)+sin(b*x+a)+sin(2*b*x+2*a)^(1/2))/b+7/6*sin(b*x+a)*sin(2
*b*x+2*a)^(3/2)/b-14/15*cos(b*x+a)*sin(2*b*x+2*a)^(5/2)/b+4/5*sin(b*x+a)*sin(2*b*x+2*a)^(7/2)/b+1/5*csc(b*x+a)
^3*sin(2*b*x+2*a)^(11/2)/b-7/4*cos(b*x+a)*sin(2*b*x+2*a)^(1/2)/b

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Rubi [A]
time = 0.14, antiderivative size = 190, normalized size of antiderivative = 1.00, number of steps used = 7, 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 {7 \text {ArcSin}(\cos (a+b x)-\sin (a+b x))}{8 b}+\frac {4 \sin (a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x)}{5 b}+\frac {7 \sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{6 b}-\frac {14 \sin ^{\frac {5}{2}}(2 a+2 b x) \cos (a+b x)}{15 b}-\frac {7 \sqrt {\sin (2 a+2 b x)} \cos (a+b x)}{4 b}+\frac {\sin ^{\frac {11}{2}}(2 a+2 b x) \csc ^3(a+b x)}{5 b}+\frac {7 \log \left (\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}+\cos (a+b x)\right )}{8 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^3*Sin[2*a + 2*b*x]^(9/2),x]

[Out]

(-7*ArcSin[Cos[a + b*x] - Sin[a + b*x]])/(8*b) + (7*Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt[Sin[2*a + 2*b*x]]])
/(8*b) - (7*Cos[a + b*x]*Sqrt[Sin[2*a + 2*b*x]])/(4*b) + (7*Sin[a + b*x]*Sin[2*a + 2*b*x]^(3/2))/(6*b) - (14*C
os[a + b*x]*Sin[2*a + 2*b*x]^(5/2))/(15*b) + (4*Sin[a + b*x]*Sin[2*a + 2*b*x]^(7/2))/(5*b) + (Csc[a + b*x]^3*S
in[2*a + 2*b*x]^(11/2))/(5*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 {9}{2}}(2 a+2 b x) \, dx &=\frac {\csc ^3(a+b x) \sin ^{\frac {11}{2}}(2 a+2 b x)}{5 b}+\frac {16}{5} \int \csc (a+b x) \sin ^{\frac {9}{2}}(2 a+2 b x) \, dx\\ &=\frac {\csc ^3(a+b x) \sin ^{\frac {11}{2}}(2 a+2 b x)}{5 b}+\frac {32}{5} \int \cos (a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x) \, dx\\ &=\frac {4 \sin (a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x)}{5 b}+\frac {\csc ^3(a+b x) \sin ^{\frac {11}{2}}(2 a+2 b x)}{5 b}+\frac {28}{5} \int \sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx\\ &=-\frac {14 \cos (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{15 b}+\frac {4 \sin (a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x)}{5 b}+\frac {\csc ^3(a+b x) \sin ^{\frac {11}{2}}(2 a+2 b x)}{5 b}+\frac {14}{3} \int \cos (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx\\ &=\frac {7 \sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{6 b}-\frac {14 \cos (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{15 b}+\frac {4 \sin (a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x)}{5 b}+\frac {\csc ^3(a+b x) \sin ^{\frac {11}{2}}(2 a+2 b x)}{5 b}+\frac {7}{2} \int \sin (a+b x) \sqrt {\sin (2 a+2 b x)} \, dx\\ &=-\frac {7 \cos (a+b x) \sqrt {\sin (2 a+2 b x)}}{4 b}+\frac {7 \sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{6 b}-\frac {14 \cos (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{15 b}+\frac {4 \sin (a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x)}{5 b}+\frac {\csc ^3(a+b x) \sin ^{\frac {11}{2}}(2 a+2 b x)}{5 b}+\frac {7}{4} \int \frac {\cos (a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx\\ &=-\frac {7 \sin ^{-1}(\cos (a+b x)-\sin (a+b x))}{8 b}+\frac {7 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}\right )}{8 b}-\frac {7 \cos (a+b x) \sqrt {\sin (2 a+2 b x)}}{4 b}+\frac {7 \sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{6 b}-\frac {14 \cos (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{15 b}+\frac {4 \sin (a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x)}{5 b}+\frac {\csc ^3(a+b x) \sin ^{\frac {11}{2}}(2 a+2 b x)}{5 b}\\ \end {align*}

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Mathematica [A]
time = 0.37, size = 100, normalized size = 0.53 \begin {gather*} \frac {7 \left (-\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 )\right )-\frac {2}{3} (10 \cos (a+b x)+9 \cos (3 (a+b x))+2 \cos (5 (a+b x))) \sqrt {\sin (2 (a+b x))}}{8 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]^3*Sin[2*a + 2*b*x]^(9/2),x]

[Out]

(7*(-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*(10
*Cos[a + b*x] + 9*Cos[3*(a + b*x)] + 2*Cos[5*(a + b*x)])*Sqrt[Sin[2*(a + b*x)]])/3)/(8*b)

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 286.11, size = 441, normalized size = 2.32

method	result	size
default	\(-\frac {64 \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 ^{6}\left (\frac {a}{2}+\frac {x b}{2}\right )\right )-3 \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 ^{4}\left (\frac {a}{2}+\frac {x b}{2}\right )\right )+2 \left (\tan ^{7}\left (\frac {a}{2}+\frac {x b}{2}\right )\right )+3 \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 )+10 \left (\tan ^{5}\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 )+10 \left (\tan ^{3}\left (\frac {a}{2}+\frac {x b}{2}\right )\right )+2 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )\right )}{21 \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 )}\, \left (\tan \left (\frac {a}{2}+\frac {x b}{2}\right )+1\right )^{2} \sqrt {\tan ^{3}\left (\frac {a}{2}+\frac {x b}{2}\right )-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}\, \left (\tan \left (\frac {a}{2}+\frac {x b}{2}\right )-1\right )^{2} b}\)	\(441\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(b*x+a)^3*sin(2*b*x+2*a)^(9/2),x,method=_RETURNVERBOSE)

[Out]

-64/21*(-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)
^6-3*(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)^4+2*tan(1/2*a+1/2*x*b)^7+3*(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+10*tan(1/2*a+1/2*x*b)^5-(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))+10*tan(1/2*a+1/2*x*b)^3+2*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)+1)^2/(tan(1/2*a+1/2*x*b
)^3-tan(1/2*a+1/2*x*b))^(1/2)/(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)^(9/2),x, algorithm="maxima")

[Out]

integrate(csc(b*x + a)^3*sin(2*b*x + 2*a)^(9/2), x)

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Fricas [A]
time = 2.63, size = 291, normalized size = 1.53 \begin {gather*} -\frac {8 \, \sqrt {2} {\left (32 \, \cos \left (b x + a\right )^{5} - 4 \, \cos \left (b x + a\right )^{3} - 7 \, \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} - 42 \, \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 ) + 42 \, \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 ) + 21 \, \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 )}{96 \, 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)^(9/2),x, algorithm="fricas")

[Out]

-1/96*(8*sqrt(2)*(32*cos(b*x + a)^5 - 4*cos(b*x + a)^3 - 7*cos(b*x + a))*sqrt(cos(b*x + a)*sin(b*x + a)) - 42*
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))/(c
os(b*x + a)^2 + 2*cos(b*x + a)*sin(b*x + a) - 1)) + 42*arctan(-(2*sqrt(2)*sqrt(cos(b*x + a)*sin(b*x + a)) - co
s(b*x + a) - sin(b*x + a))/(cos(b*x + a) - sin(b*x + a))) + 21*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)**(9/2),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)^(9/2),x, algorithm="giac")

[Out]

integrate(csc(b*x + a)^3*sin(2*b*x + 2*a)^(9/2), 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 )}^{9/2}}{{\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)^(9/2)/sin(a + b*x)^3,x)

[Out]

int(sin(2*a + 2*b*x)^(9/2)/sin(a + b*x)^3, x)

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