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

Optimal. Leaf size=136 \[ \frac {\sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{3 b}+\frac {5 \sin (a+b x) \sqrt {\sin (2 a+2 b x)}}{8 b}-\frac {5 \sin ^{-1}(\cos (a+b x)-\sin (a+b x))}{16 b}-\frac {5 \sin ^{\frac {3}{2}}(2 a+2 b x) \cos (a+b x)}{12 b}-\frac {5 \log \left (\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}+\cos (a+b x)\right )}{16 b} \]

[Out]

-5/16*arcsin(cos(b*x+a)-sin(b*x+a))/b-5/16*ln(cos(b*x+a)+sin(b*x+a)+sin(2*b*x+2*a)^(1/2))/b-5/12*cos(b*x+a)*si
n(2*b*x+2*a)^(3/2)/b+1/3*sin(b*x+a)*sin(2*b*x+2*a)^(5/2)/b+5/8*sin(b*x+a)*sin(2*b*x+2*a)^(1/2)/b

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Rubi [A]  time = 0.12, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4308, 4301, 4302, 4306} \[ \frac {\sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{3 b}+\frac {5 \sin (a+b x) \sqrt {\sin (2 a+2 b x)}}{8 b}-\frac {5 \sin ^{\frac {3}{2}}(2 a+2 b x) \cos (a+b x)}{12 b}-\frac {5 \sin ^{-1}(\cos (a+b x)-\sin (a+b x))}{16 b}-\frac {5 \log \left (\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}+\cos (a+b x)\right )}{16 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*ArcSin[Cos[a + b*x] - Sin[a + b*x]])/(16*b) - (5*Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt[Sin[2*a + 2*b*x]]]
)/(16*b) + (5*Sin[a + b*x]*Sqrt[Sin[2*a + 2*b*x]])/(8*b) - (5*Cos[a + b*x]*Sin[2*a + 2*b*x]^(3/2))/(12*b) + (S
in[a + b*x]*Sin[2*a + 2*b*x]^(5/2))/(3*b)

Rule 4301

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 4302

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 4306

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 4308

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 (a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x) \, dx &=2 \int \cos (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx\\ &=\frac {\sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{3 b}+\frac {5}{3} \int \sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx\\ &=-\frac {5 \cos (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{12 b}+\frac {\sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{3 b}+\frac {5}{4} \int \cos (a+b x) \sqrt {\sin (2 a+2 b x)} \, dx\\ &=\frac {5 \sin (a+b x) \sqrt {\sin (2 a+2 b x)}}{8 b}-\frac {5 \cos (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{12 b}+\frac {\sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{3 b}+\frac {5}{8} \int \frac {\sin (a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx\\ &=-\frac {5 \sin ^{-1}(\cos (a+b x)-\sin (a+b x))}{16 b}-\frac {5 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}\right )}{16 b}+\frac {5 \sin (a+b x) \sqrt {\sin (2 a+2 b x)}}{8 b}-\frac {5 \cos (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{12 b}+\frac {\sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{3 b}\\ \end {align*}

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Mathematica [A]  time = 0.33, size = 98, normalized size = 0.72 \[ \frac {\frac {2}{3} \sqrt {\sin (2 (a+b x))} (14 \sin (a+b x)-3 \sin (3 (a+b x))-2 \sin (5 (a+b x)))-5 \left (\sin ^{-1}(\cos (a+b x)-\sin (a+b x))+\log \left (\sin (a+b x)+\sqrt {\sin (2 (a+b x))}+\cos (a+b x)\right )\right )}{16 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*(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*Sqr
t[Sin[2*(a + b*x)]]*(14*Sin[a + b*x] - 3*Sin[3*(a + b*x)] - 2*Sin[5*(a + b*x)]))/3)/(16*b)

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fricas [B]  time = 0.45, size = 290, normalized size = 2.13 \[ -\frac {8 \, \sqrt {2} {\left (32 \, \cos \left (b x + a\right )^{4} - 12 \, \cos \left (b x + a\right )^{2} - 15\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} \sin \left (b x + a\right ) - 30 \, \arctan \left (-\frac {\sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} + \cos \left (b x + a\right ) \sin \left (b x + a\right )}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 1}\right ) + 30 \, \arctan \left (-\frac {2 \, \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} - \cos \left (b x + a\right ) - \sin \left (b x + a\right )}{\cos \left (b x + a\right ) - \sin \left (b x + a\right )}\right ) - 15 \, \log \left (-32 \, \cos \left (b x + a\right )^{4} + 4 \, \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{3} - {\left (4 \, \cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right ) - 5 \, \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 32 \, \cos \left (b x + a\right )^{2} + 16 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 1\right )}{192 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/192*(8*sqrt(2)*(32*cos(b*x + a)^4 - 12*cos(b*x + a)^2 - 15)*sqrt(cos(b*x + a)*sin(b*x + a))*sin(b*x + a) -
30*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)) + 30*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))) - 15*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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giac [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(csc(b*x+a)*sin(2*b*x+2*a)^(7/2),x, algorithm="giac")

[Out]

Timed out

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maple [C]  time = 19.86, size = 973, normalized size = 7.15 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-16/5/b*(-tan(1/2*b*x+1/2*a)/(tan(1/2*b*x+1/2*a)^2-1))^(1/2)*(6*(tan(1/2*b*x+1/2*a)*(tan(1/2*b*x+1/2*a)-1)*(ta
n(1/2*b*x+1/2*a)+1))^(1/2)*(tan(1/2*b*x+1/2*a)+1)^(1/2)*(-2*tan(1/2*b*x+1/2*a)+2)^(1/2)*(-tan(1/2*b*x+1/2*a))^
(1/2)*EllipticE((tan(1/2*b*x+1/2*a)+1)^(1/2),1/2*2^(1/2))*tan(1/2*b*x+1/2*a)^4-3*(tan(1/2*b*x+1/2*a)*(tan(1/2*
b*x+1/2*a)-1)*(tan(1/2*b*x+1/2*a)+1))^(1/2)*(tan(1/2*b*x+1/2*a)+1)^(1/2)*(-2*tan(1/2*b*x+1/2*a)+2)^(1/2)*(-tan
(1/2*b*x+1/2*a))^(1/2)*EllipticF((tan(1/2*b*x+1/2*a)+1)^(1/2),1/2*2^(1/2))*tan(1/2*b*x+1/2*a)^4-12*(tan(1/2*b*
x+1/2*a)*(tan(1/2*b*x+1/2*a)-1)*(tan(1/2*b*x+1/2*a)+1))^(1/2)*(tan(1/2*b*x+1/2*a)+1)^(1/2)*(-2*tan(1/2*b*x+1/2
*a)+2)^(1/2)*(-tan(1/2*b*x+1/2*a))^(1/2)*EllipticE((tan(1/2*b*x+1/2*a)+1)^(1/2),1/2*2^(1/2))*tan(1/2*b*x+1/2*a
)^2+6*(tan(1/2*b*x+1/2*a)*(tan(1/2*b*x+1/2*a)-1)*(tan(1/2*b*x+1/2*a)+1))^(1/2)*(tan(1/2*b*x+1/2*a)+1)^(1/2)*(-
2*tan(1/2*b*x+1/2*a)+2)^(1/2)*(-tan(1/2*b*x+1/2*a))^(1/2)*EllipticF((tan(1/2*b*x+1/2*a)+1)^(1/2),1/2*2^(1/2))*
tan(1/2*b*x+1/2*a)^2+6*tan(1/2*b*x+1/2*a)^6*(tan(1/2*b*x+1/2*a)^3-tan(1/2*b*x+1/2*a))^(1/2)+6*(tan(1/2*b*x+1/2
*a)*(tan(1/2*b*x+1/2*a)-1)*(tan(1/2*b*x+1/2*a)+1))^(1/2)*(tan(1/2*b*x+1/2*a)+1)^(1/2)*(-2*tan(1/2*b*x+1/2*a)+2
)^(1/2)*(-tan(1/2*b*x+1/2*a))^(1/2)*EllipticE((tan(1/2*b*x+1/2*a)+1)^(1/2),1/2*2^(1/2))-3*(tan(1/2*b*x+1/2*a)*
(tan(1/2*b*x+1/2*a)-1)*(tan(1/2*b*x+1/2*a)+1))^(1/2)*(tan(1/2*b*x+1/2*a)+1)^(1/2)*(-2*tan(1/2*b*x+1/2*a)+2)^(1
/2)*(-tan(1/2*b*x+1/2*a))^(1/2)*EllipticF((tan(1/2*b*x+1/2*a)+1)^(1/2),1/2*2^(1/2))-4*(tan(1/2*b*x+1/2*a)*(tan
(1/2*b*x+1/2*a)-1)*(tan(1/2*b*x+1/2*a)+1))^(1/2)*tan(1/2*b*x+1/2*a)^4-12*tan(1/2*b*x+1/2*a)^4*(tan(1/2*b*x+1/2
*a)^3-tan(1/2*b*x+1/2*a))^(1/2)-4*(tan(1/2*b*x+1/2*a)*(tan(1/2*b*x+1/2*a)-1)*(tan(1/2*b*x+1/2*a)+1))^(1/2)*tan
(1/2*b*x+1/2*a)^2+6*tan(1/2*b*x+1/2*a)^2*(tan(1/2*b*x+1/2*a)^3-tan(1/2*b*x+1/2*a))^(1/2))/(tan(1/2*b*x+1/2*a)*
(tan(1/2*b*x+1/2*a)^2-1))^(1/2)/(tan(1/2*b*x+1/2*a)*(tan(1/2*b*x+1/2*a)-1)*(tan(1/2*b*x+1/2*a)+1))^(1/2)/(tan(
1/2*b*x+1/2*a)^3-tan(1/2*b*x+1/2*a))^(1/2)/(tan(1/2*b*x+1/2*a)-1)/(tan(1/2*b*x+1/2*a)+1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\sin \left (2\,a+2\,b\,x\right )}^{7/2}}{\sin \left (a+b\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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