3.1.79 \(\int \frac {\sin (a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx\) [79]
Optimal. Leaf size=79 \[ \frac {\sin (a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}-\frac {4 \cos (a+b x)}{15 b \sin ^{\frac {3}{2}}(2 a+2 b x)}+\frac {8 \sin (a+b x)}{15 b \sqrt {\sin (2 a+2 b x)}} \]
[Out]
1/5*sin(b*x+a)/b/sin(2*b*x+2*a)^(5/2)-4/15*cos(b*x+a)/b/sin(2*b*x+2*a)^(3/2)+8/15*sin(b*x+a)/b/sin(2*b*x+2*a)^
(1/2)
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Rubi [A]
time = 0.04, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps
used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4389, 4388,
4377} \begin {gather*} \frac {\sin (a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}+\frac {8 \sin (a+b x)}{15 b \sqrt {\sin (2 a+2 b x)}}-\frac {4 \cos (a+b x)}{15 b \sin ^{\frac {3}{2}}(2 a+2 b x)} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[Sin[a + b*x]/Sin[2*a + 2*b*x]^(7/2),x]
[Out]
Sin[a + b*x]/(5*b*Sin[2*a + 2*b*x]^(5/2)) - (4*Cos[a + b*x])/(15*b*Sin[2*a + 2*b*x]^(3/2)) + (8*Sin[a + b*x])/
(15*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]
Rubi steps
\begin {align*} \int \frac {\sin (a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx &=\frac {\sin (a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}+\frac {4}{5} \int \frac {\cos (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx\\ &=\frac {\sin (a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}-\frac {4 \cos (a+b x)}{15 b \sin ^{\frac {3}{2}}(2 a+2 b x)}+\frac {8}{15} \int \frac {\sin (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx\\ &=\frac {\sin (a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}-\frac {4 \cos (a+b x)}{15 b \sin ^{\frac {3}{2}}(2 a+2 b x)}+\frac {8 \sin (a+b x)}{15 b \sqrt {\sin (2 a+2 b x)}}\\ \end {align*}
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Mathematica [A]
time = 0.21, size = 52, normalized size = 0.66 \begin {gather*} \frac {\left (-5 \cot (a+b x) \csc (a+b x)+3 \sec (a+b x) \left (9+\sec ^2(a+b x)\right )\right ) \sqrt {\sin (2 (a+b x))}}{120 b} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[Sin[a + b*x]/Sin[2*a + 2*b*x]^(7/2),x]
[Out]
((-5*Cot[a + b*x]*Csc[a + b*x] + 3*Sec[a + b*x]*(9 + Sec[a + b*x]^2))*Sqrt[Sin[2*(a + b*x)]])/(120*b)
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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {\sin \left (x b +a \right )}{\sin \left (2 x b +2 a \right )^{\frac {7}{2}}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sin(b*x+a)/sin(2*b*x+2*a)^(7/2),x)
[Out]
int(sin(b*x+a)/sin(2*b*x+2*a)^(7/2),x)
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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(sin(b*x+a)/sin(2*b*x+2*a)^(7/2),x, algorithm="maxima")
[Out]
integrate(sin(b*x + a)/sin(2*b*x + 2*a)^(7/2), x)
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Fricas [A]
time = 4.58, size = 88, normalized size = 1.11 \begin {gather*} \frac {32 \, \cos \left (b x + a\right )^{5} - 32 \, \cos \left (b x + a\right )^{3} + \sqrt {2} {\left (32 \, \cos \left (b x + a\right )^{4} - 24 \, \cos \left (b x + a\right )^{2} - 3\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )}}{120 \, {\left (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(sin(b*x+a)/sin(2*b*x+2*a)^(7/2),x, algorithm="fricas")
[Out]
1/120*(32*cos(b*x + a)^5 - 32*cos(b*x + a)^3 + sqrt(2)*(32*cos(b*x + a)^4 - 24*cos(b*x + a)^2 - 3)*sqrt(cos(b*
x + a)*sin(b*x + a)))/(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(sin(b*x+a)/sin(2*b*x+2*a)**(7/2),x)
[Out]
Timed out
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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 18022 vs.
\(2 (67) = 134\).
time = 162.48, size = 18022, normalized size = 228.13 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sin(b*x+a)/sin(2*b*x+2*a)^(7/2),x, algorithm="giac")
[Out]
1/480*sqrt(2)*sqrt(-tan(1/2*b*x)^4*tan(1/2*a)^3 - tan(1/2*b*x)^3*tan(1/2*a)^4 + tan(1/2*b*x)^4*tan(1/2*a) + 6*
tan(1/2*b*x)^3*tan(1/2*a)^2 + 6*tan(1/2*b*x)^2*tan(1/2*a)^3 + tan(1/2*b*x)*tan(1/2*a)^4 - tan(1/2*b*x)^3 - 6*t
an(1/2*b*x)^2*tan(1/2*a) - 6*tan(1/2*b*x)*tan(1/2*a)^2 - tan(1/2*a)^3 + tan(1/2*b*x) + tan(1/2*a))*((((((((((2
*(sqrt(2)*tan(1/2*a)^87 - 92*sqrt(2)*tan(1/2*a)^85 - 3213*sqrt(2)*tan(1/2*a)^83 - 48008*sqrt(2)*tan(1/2*a)^81
- 443462*sqrt(2)*tan(1/2*a)^79 - 2875040*sqrt(2)*tan(1/2*a)^77 - 13907802*sqrt(2)*tan(1/2*a)^75 - 51781432*sqr
t(2)*tan(1/2*a)^73 - 149943911*sqrt(2)*tan(1/2*a)^71 - 333456564*sqrt(2)*tan(1/2*a)^69 - 536442973*sqrt(2)*tan
(1/2*a)^67 - 482080288*sqrt(2)*tan(1/2*a)^65 + 316221080*sqrt(2)*tan(1/2*a)^63 + 2190937152*sqrt(2)*tan(1/2*a)
^61 + 4607763368*sqrt(2)*tan(1/2*a)^59 + 5742984608*sqrt(2)*tan(1/2*a)^57 + 3316624962*sqrt(2)*tan(1/2*a)^55 -
3241815576*sqrt(2)*tan(1/2*a)^53 - 11030972730*sqrt(2)*tan(1/2*a)^51 - 14712027120*sqrt(2)*tan(1/2*a)^49 - 10
524179460*sqrt(2)*tan(1/2*a)^47 + 10524179460*sqrt(2)*tan(1/2*a)^43 + 14712027120*sqrt(2)*tan(1/2*a)^41 + 1103
0972730*sqrt(2)*tan(1/2*a)^39 + 3241815576*sqrt(2)*tan(1/2*a)^37 - 3316624962*sqrt(2)*tan(1/2*a)^35 - 57429846
08*sqrt(2)*tan(1/2*a)^33 - 4607763368*sqrt(2)*tan(1/2*a)^31 - 2190937152*sqrt(2)*tan(1/2*a)^29 - 316221080*sqr
t(2)*tan(1/2*a)^27 + 482080288*sqrt(2)*tan(1/2*a)^25 + 536442973*sqrt(2)*tan(1/2*a)^23 + 333456564*sqrt(2)*tan
(1/2*a)^21 + 149943911*sqrt(2)*tan(1/2*a)^19 + 51781432*sqrt(2)*tan(1/2*a)^17 + 13907802*sqrt(2)*tan(1/2*a)^15
+ 2875040*sqrt(2)*tan(1/2*a)^13 + 443462*sqrt(2)*tan(1/2*a)^11 + 48008*sqrt(2)*tan(1/2*a)^9 + 3213*sqrt(2)*ta
n(1/2*a)^7 + 92*sqrt(2)*tan(1/2*a)^5 - sqrt(2)*tan(1/2*a)^3)*tan(1/2*b*x)/(tan(1/2*a)^78 + 34*tan(1/2*a)^76 +
559*tan(1/2*a)^74 + 5916*tan(1/2*a)^72 + 45255*tan(1/2*a)^70 + 266322*tan(1/2*a)^68 + 1252713*tan(1/2*a)^66 +
4829088*tan(1/2*a)^64 + 15512772*tan(1/2*a)^62 + 41970280*tan(1/2*a)^60 + 96160636*tan(1/2*a)^58 + 186574864*t
an(1/2*a)^56 + 304253964*tan(1/2*a)^54 + 408239496*tan(1/2*a)^52 + 426395700*tan(1/2*a)^50 + 286097760*tan(1/2
*a)^48 - 31635810*tan(1/2*a)^46 - 450345060*tan(1/2*a)^44 - 811985790*tan(1/2*a)^42 - 955277400*tan(1/2*a)^40
- 811985790*tan(1/2*a)^38 - 450345060*tan(1/2*a)^36 - 31635810*tan(1/2*a)^34 + 286097760*tan(1/2*a)^32 + 42639
5700*tan(1/2*a)^30 + 408239496*tan(1/2*a)^28 + 304253964*tan(1/2*a)^26 + 186574864*tan(1/2*a)^24 + 96160636*ta
n(1/2*a)^22 + 41970280*tan(1/2*a)^20 + 15512772*tan(1/2*a)^18 + 4829088*tan(1/2*a)^16 + 1252713*tan(1/2*a)^14
+ 266322*tan(1/2*a)^12 + 45255*tan(1/2*a)^10 + 5916*tan(1/2*a)^8 + 559*tan(1/2*a)^6 + 34*tan(1/2*a)^4 + tan(1/
2*a)^2) + 5*(sqrt(2)*tan(1/2*a)^88 - 97*sqrt(2)*tan(1/2*a)^86 - 2757*sqrt(2)*tan(1/2*a)^84 - 31579*sqrt(2)*tan
(1/2*a)^82 - 190206*sqrt(2)*tan(1/2*a)^80 - 452482*sqrt(2)*tan(1/2*a)^78 + 2446494*sqrt(2)*tan(1/2*a)^76 + 312
36834*sqrt(2)*tan(1/2*a)^74 + 178073713*sqrt(2)*tan(1/2*a)^72 + 692499183*sqrt(2)*tan(1/2*a)^70 + 2006851683*s
qrt(2)*tan(1/2*a)^68 + 4409972669*sqrt(2)*tan(1/2*a)^66 + 7082232504*sqrt(2)*tan(1/2*a)^64 + 6893762888*sqrt(2
)*tan(1/2*a)^62 - 1327875576*sqrt(2)*tan(1/2*a)^60 - 21040534024*sqrt(2)*tan(1/2*a)^58 - 47574053342*sqrt(2)*t
an(1/2*a)^56 - 64972634082*sqrt(2)*tan(1/2*a)^54 - 53236088394*sqrt(2)*tan(1/2*a)^52 - 5004094710*sqrt(2)*tan(
1/2*a)^50 + 61712915820*sqrt(2)*tan(1/2*a)^48 + 110145965460*sqrt(2)*tan(1/2*a)^46 + 110145965460*sqrt(2)*tan(
1/2*a)^44 + 61712915820*sqrt(2)*tan(1/2*a)^42 - 5004094710*sqrt(2)*tan(1/2*a)^40 - 53236088394*sqrt(2)*tan(1/2
*a)^38 - 64972634082*sqrt(2)*tan(1/2*a)^36 - 47574053342*sqrt(2)*tan(1/2*a)^34 - 21040534024*sqrt(2)*tan(1/2*a
)^32 - 1327875576*sqrt(2)*tan(1/2*a)^30 + 6893762888*sqrt(2)*tan(1/2*a)^28 + 7082232504*sqrt(2)*tan(1/2*a)^26
+ 4409972669*sqrt(2)*tan(1/2*a)^24 + 2006851683*sqrt(2)*tan(1/2*a)^22 + 692499183*sqrt(2)*tan(1/2*a)^20 + 1780
73713*sqrt(2)*tan(1/2*a)^18 + 31236834*sqrt(2)*tan(1/2*a)^16 + 2446494*sqrt(2)*tan(1/2*a)^14 - 452482*sqrt(2)*
tan(1/2*a)^12 - 190206*sqrt(2)*tan(1/2*a)^10 - 31579*sqrt(2)*tan(1/2*a)^8 - 2757*sqrt(2)*tan(1/2*a)^6 - 97*sqr
t(2)*tan(1/2*a)^4 + sqrt(2)*tan(1/2*a)^2)/(tan(1/2*a)^78 + 34*tan(1/2*a)^76 + 559*tan(1/2*a)^74 + 5916*tan(1/2
*a)^72 + 45255*tan(1/2*a)^70 + 266322*tan(1/2*a)^68 + 1252713*tan(1/2*a)^66 + 4829088*tan(1/2*a)^64 + 15512772
*tan(1/2*a)^62 + 41970280*tan(1/2*a)^60 + 96160636*tan(1/2*a)^58 + 186574864*tan(1/2*a)^56 + 304253964*tan(1/2
*a)^54 + 408239496*tan(1/2*a)^52 + 426395700*tan(1/2*a)^50 + 286097760*tan(1/2*a)^48 - 31635810*tan(1/2*a)^46
- 450345060*tan(1/2*a)^44 - 811985790*tan(1/2*a)^42 - 955277400*tan(1/2*a)^40 - 811985790*tan(1/2*a)^38 - 4503
45060*tan(1/2*a)^36 - 31635810*tan(1/2*a)^34 + 286097760*tan(1/2*a)^32 + 426395700*tan(1/2*a)^30 + 408239496*t
an(1/2*a)^28 + 304253964*tan(1/2*a)^26 + 186574864*tan(1/2*a)^24 + 96160636*tan(1/2*a)^22 + 41970280*tan(1/2*a
)^20 + 15512772*tan(1/2*a)^18 + 4829088*tan(1/2*a)^16 + 1252713*tan(1/2*a)^14 + 266322*tan(1/2*a)^12 + 45255*t
an(1/2*a)^10 + 5916*tan(1/2*a)^8 + 559*tan(1/2*...
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Mupad [B]
time = 3.39, size = 131, normalized size = 1.66 \begin {gather*} \frac {4\,{\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}}\,\left (2\,{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}-3\,{\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}+2\,{\mathrm {e}}^{a\,6{}\mathrm {i}+b\,x\,6{}\mathrm {i}}+2\,{\mathrm {e}}^{a\,8{}\mathrm {i}+b\,x\,8{}\mathrm {i}}+2\right )}{15\,b\,{\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\right )}^3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sin(a + b*x)/sin(2*a + 2*b*x)^(7/2),x)
[Out]
(4*exp(a*1i + b*x*1i)*((exp(- a*2i - b*x*2i)*1i)/2 - (exp(a*2i + b*x*2i)*1i)/2)^(1/2)*(2*exp(a*2i + b*x*2i) -
3*exp(a*4i + b*x*4i) + 2*exp(a*6i + b*x*6i) + 2*exp(a*8i + b*x*8i) + 2))/(15*b*(exp(a*2i + b*x*2i) - 1)^2*(exp
(a*2i + b*x*2i) + 1)^3)
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