3.2.4 \(\int \frac {\csc (a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx\) [104]

Optimal. Leaf size=105 \[ -\frac {2 \cos (a+b x)}{7 b \sin ^{\frac {7}{2}}(2 a+2 b x)}+\frac {12 \sin (a+b x)}{35 b \sin ^{\frac {5}{2}}(2 a+2 b x)}-\frac {16 \cos (a+b x)}{35 b \sin ^{\frac {3}{2}}(2 a+2 b x)}+\frac {32 \sin (a+b x)}{35 b \sqrt {\sin (2 a+2 b x)}} \]

[Out]

-2/7*cos(b*x+a)/b/sin(2*b*x+2*a)^(7/2)+12/35*sin(b*x+a)/b/sin(2*b*x+2*a)^(5/2)-16/35*cos(b*x+a)/b/sin(2*b*x+2*
a)^(3/2)+32/35*sin(b*x+a)/b/sin(2*b*x+2*a)^(1/2)

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Rubi [A]
time = 0.09, antiderivative size = 105, 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 = {4393, 4388, 4389, 4377} \begin {gather*} \frac {12 \sin (a+b x)}{35 b \sin ^{\frac {5}{2}}(2 a+2 b x)}+\frac {32 \sin (a+b x)}{35 b \sqrt {\sin (2 a+2 b x)}}-\frac {16 \cos (a+b x)}{35 b \sin ^{\frac {3}{2}}(2 a+2 b x)}-\frac {2 \cos (a+b x)}{7 b \sin ^{\frac {7}{2}}(2 a+2 b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Cos[a + b*x])/(7*b*Sin[2*a + 2*b*x]^(7/2)) + (12*Sin[a + b*x])/(35*b*Sin[2*a + 2*b*x]^(5/2)) - (16*Cos[a +
b*x])/(35*b*Sin[2*a + 2*b*x]^(3/2)) + (32*Sin[a + b*x])/(35*b*Sqrt[Sin[2*a + 2*b*x]])

Rule 4377

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)/(b*g*m)), x] /; FreeQ[{a, b, c, d, e, g, m, p}, x] && EqQ[b*c - a*d, 0] && Eq
Q[d/b, 2] &&  !IntegerQ[p] && EqQ[m + 2*p + 2, 0]

Rule 4388

Int[cos[(a_.) + (b_.)*(x_)]*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[Cos[a + b*x]*((g*Sin[c + d
*x])^(p + 1)/(2*b*g*(p + 1))), x] + Dist[(2*p + 3)/(2*g*(p + 1)), Int[Sin[a + b*x]*(g*Sin[c + d*x])^(p + 1), x
], x] /; FreeQ[{a, b, c, d, g}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] &&  !IntegerQ[p] && LtQ[p, -1] && Integ
erQ[2*p]

Rule 4389

Int[sin[(a_.) + (b_.)*(x_)]*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[(-Sin[a + b*x])*((g*Sin[c
+ d*x])^(p + 1)/(2*b*g*(p + 1))), x] + Dist[(2*p + 3)/(2*g*(p + 1)), Int[Cos[a + b*x]*(g*Sin[c + d*x])^(p + 1)
, x], x] /; FreeQ[{a, b, c, d, g}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] &&  !IntegerQ[p] && LtQ[p, -1] && In
tegerQ[2*p]

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 \frac {\csc (a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx &=2 \int \frac {\cos (a+b x)}{\sin ^{\frac {9}{2}}(2 a+2 b x)} \, dx\\ &=-\frac {2 \cos (a+b x)}{7 b \sin ^{\frac {7}{2}}(2 a+2 b x)}+\frac {12}{7} \int \frac {\sin (a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx\\ &=-\frac {2 \cos (a+b x)}{7 b \sin ^{\frac {7}{2}}(2 a+2 b x)}+\frac {12 \sin (a+b x)}{35 b \sin ^{\frac {5}{2}}(2 a+2 b x)}+\frac {48}{35} \int \frac {\cos (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx\\ &=-\frac {2 \cos (a+b x)}{7 b \sin ^{\frac {7}{2}}(2 a+2 b x)}+\frac {12 \sin (a+b x)}{35 b \sin ^{\frac {5}{2}}(2 a+2 b x)}-\frac {16 \cos (a+b x)}{35 b \sin ^{\frac {3}{2}}(2 a+2 b x)}+\frac {32}{35} \int \frac {\sin (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx\\ &=-\frac {2 \cos (a+b x)}{7 b \sin ^{\frac {7}{2}}(2 a+2 b x)}+\frac {12 \sin (a+b x)}{35 b \sin ^{\frac {5}{2}}(2 a+2 b x)}-\frac {16 \cos (a+b x)}{35 b \sin ^{\frac {3}{2}}(2 a+2 b x)}+\frac {32 \sin (a+b x)}{35 b \sqrt {\sin (2 a+2 b x)}}\\ \end {align*}

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Mathematica [A]
time = 0.15, size = 67, normalized size = 0.64 \begin {gather*} \frac {(5-10 \cos (2 (a+b x))-4 \cos (4 (a+b x))+4 \cos (6 (a+b x))) \csc ^4(a+b x) \sec ^3(a+b x) \sqrt {\sin (2 (a+b x))}}{280 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((5 - 10*Cos[2*(a + b*x)] - 4*Cos[4*(a + b*x)] + 4*Cos[6*(a + b*x)])*Csc[a + b*x]^4*Sec[a + b*x]^3*Sqrt[Sin[2*
(a + b*x)]])/(280*b)

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 143.65, size = 222, normalized size = 2.11 \[\frac {\sqrt {-\frac {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}{\tan ^{2}\left (\frac {a}{2}+\frac {x b}{2}\right )-1}}\, \left (\tan ^{2}\left (\frac {a}{2}+\frac {x b}{2}\right )-1\right ) \left (3 \left (\tan ^{8}\left (\frac {a}{2}+\frac {x b}{2}\right )\right )+40 \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )+1}\, \sqrt {-2 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )+2}\, \sqrt {-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}\, \EllipticF \left (\sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )+1}, \frac {\sqrt {2}}{2}\right ) \left (\tan ^{3}\left (\frac {a}{2}+\frac {x b}{2}\right )\right )-26 \left (\tan ^{6}\left (\frac {a}{2}+\frac {x b}{2}\right )\right )+26 \left (\tan ^{2}\left (\frac {a}{2}+\frac {x b}{2}\right )\right )-3\right )}{1344 b \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3} \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \left (\tan ^{2}\left (\frac {a}{2}+\frac {x b}{2}\right )-1\right )}\, \sqrt {\tan ^{3}\left (\frac {a}{2}+\frac {x b}{2}\right )-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}}\]

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]

1/1344/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)^2-1)/tan(1/2*a+1/2*x*b)^3*(3
*tan(1/2*a+1/2*x*b)^8+40*(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)^3-26*tan(1/2*a+1/2*x*b)^6+26*tan(1/
2*a+1/2*x*b)^2-3)/(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)^3-tan(1/2*a+1/2*x*b)
)^(1/2)

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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)/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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Fricas [A]
time = 2.73, size = 118, normalized size = 1.12 \begin {gather*} \frac {128 \, \cos \left (b x + a\right )^{7} - 256 \, \cos \left (b x + a\right )^{5} + 128 \, \cos \left (b x + a\right )^{3} + \sqrt {2} {\left (128 \, \cos \left (b x + a\right )^{6} - 224 \, \cos \left (b x + a\right )^{4} + 84 \, \cos \left (b x + a\right )^{2} + 7\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )}}{280 \, {\left (b \cos \left (b x + a\right )^{7} - 2 \, b \cos \left (b x + a\right )^{5} + b \cos \left (b x + a\right )^{3}\right )}} \end {gather*}

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]

1/280*(128*cos(b*x + a)^7 - 256*cos(b*x + a)^5 + 128*cos(b*x + a)^3 + sqrt(2)*(128*cos(b*x + a)^6 - 224*cos(b*
x + a)^4 + 84*cos(b*x + a)^2 + 7)*sqrt(cos(b*x + a)*sin(b*x + a)))/(b*cos(b*x + a)^7 - 2*b*cos(b*x + a)^5 + b*
cos(b*x + a)^3)

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

[Out]

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

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Mupad [B]
time = 4.13, size = 350, normalized size = 3.33 \begin {gather*} -\frac {2\,{\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}}}{7\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}^4}+\frac {{\mathrm {e}}^{a\,3{}\mathrm {i}+b\,x\,3{}\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}}\,32{}\mathrm {i}}{35\,b\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}-\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\left (\frac {2}{7\,b}-\frac {16\,{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}}{35\,b}\right )\,\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}}+1\right )}^2\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}^2}+\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\left (\frac {32{}\mathrm {i}}{35\,b}+\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,88{}\mathrm {i}}{35\,b}\right )\,\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}}+1\right )}^3\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(a + b*x)*sin(2*a + 2*b*x)^(7/2)),x)

[Out]

(exp(a*3i + b*x*3i)*((exp(- a*2i - b*x*2i)*1i)/2 - (exp(a*2i + b*x*2i)*1i)/2)^(1/2)*32i)/(35*b*(exp(a*2i + b*x
*2i) + 1)*(exp(a*2i + b*x*2i)*1i - 1i)) - (2*exp(a*1i + b*x*1i)*((exp(- a*2i - b*x*2i)*1i)/2 - (exp(a*2i + b*x
*2i)*1i)/2)^(1/2))/(7*b*(exp(a*2i + b*x*2i)*1i - 1i)^4) - (exp(a*1i + b*x*1i)*(2/(7*b) - (16*exp(a*2i + b*x*2i
))/(35*b))*((exp(- a*2i - b*x*2i)*1i)/2 - (exp(a*2i + b*x*2i)*1i)/2)^(1/2))/((exp(a*2i + b*x*2i) + 1)^2*(exp(a
*2i + b*x*2i)*1i - 1i)^2) + (exp(a*1i + b*x*1i)*(32i/(35*b) + (exp(a*2i + b*x*2i)*88i)/(35*b))*((exp(- a*2i -
b*x*2i)*1i)/2 - (exp(a*2i + b*x*2i)*1i)/2)^(1/2))/((exp(a*2i + b*x*2i) + 1)^3*(exp(a*2i + b*x*2i)*1i - 1i)^3)

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