3.1.78 \(\int \frac {\sin (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx\) [78]

Optimal. Leaf size=53 \[ \frac {\sin (a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}-\frac {2 \cos (a+b x)}{3 b \sqrt {\sin (2 a+2 b x)}} \]

[Out]

1/3*sin(b*x+a)/b/sin(2*b*x+2*a)^(3/2)-2/3*cos(b*x+a)/b/sin(2*b*x+2*a)^(1/2)

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Rubi [A]
time = 0.03, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {4389, 4376} \begin {gather*} \frac {\sin (a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}-\frac {2 \cos (a+b x)}{3 b \sqrt {\sin (2 a+2 b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4376

Int[(cos[(a_.) + (b_.)*(x_)]*(e_.))^(m_.)*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[(-(e*Cos[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] &&
EqQ[d/b, 2] &&  !IntegerQ[p] && EqQ[m + 2*p + 2, 0]

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 {5}{2}}(2 a+2 b x)} \, dx &=\frac {\sin (a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}+\frac {2}{3} \int \frac {\cos (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx\\ &=\frac {\sin (a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}-\frac {2 \cos (a+b x)}{3 b \sqrt {\sin (2 a+2 b x)}}\\ \end {align*}

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Mathematica [A]
time = 0.13, size = 43, normalized size = 0.81 \begin {gather*} \frac {\sqrt {\sin (2 (a+b x))} \left (-\frac {1}{4} \csc (a+b x)+\frac {1}{12} \sec (a+b x) \tan (a+b x)\right )}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Sin[2*(a + b*x)]]*(-1/4*Csc[a + b*x] + (Sec[a + b*x]*Tan[a + b*x])/12))/b

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 59.40, size = 597, normalized size = 11.26

method	result	size
default	\(\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 (6 \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 )}\, \EllipticE \left (\sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )+1}, \frac {\sqrt {2}}{2}\right ) \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 ^{2}\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 ) \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 ^{2}\left (\frac {a}{2}+\frac {x b}{2}\right )\right )+6 \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 )}\, \EllipticE \left (\sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )+1}, \frac {\sqrt {2}}{2}\right ) \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 )}-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 ) \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 )}+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 ^{4}\left (\frac {a}{2}+\frac {x b}{2}\right )\right )+2 \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 ^{4}\left (\frac {a}{2}+\frac {x b}{2}\right )\right )-2 \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 ^{2}\left (\frac {a}{2}+\frac {x b}{2}\right )\right )-2 \sqrt {\tan ^{3}\left (\frac {a}{2}+\frac {x b}{2}\right )-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}\right )}{8 b \tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \sqrt {\tan ^{3}\left (\frac {a}{2}+\frac {x b}{2}\right )-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}\, \left (1+\tan ^{2}\left (\frac {a}{2}+\frac {x b}{2}\right )\right )}\)	\(597\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/b*(-tan(1/2*a+1/2*x*b)/(tan(1/2*a+1/2*x*b)^2-1))^(1/2)*(6*(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)*EllipticE((tan(1/2*a+1/2*x*b)+1)^(1/2),1/2*2^(1/2))*(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)^2-3*(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)*(
tan(1/2*a+1/2*x*b)^2-1))^(1/2)*tan(1/2*a+1/2*x*b)^2+6*(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)*EllipticE((tan(1/2*a+1/2*x*b)+1)^(1/2),1/2*2^(1/2))*(tan(1/2*a+1/2*x*b)*(tan(
1/2*a+1/2*x*b)^2-1))^(1/2)-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)*(tan(1/2*a+1/2*x*b)^2-1))^(1/2
)+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)^4+2*(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)^4-2*(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)
^2-2*(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)/(tan(1/2*a+1/2*x*b)^3-tan(1/2*a+1/2*x
*b))^(1/2)/(1+tan(1/2*a+1/2*x*b)^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(sin(b*x+a)/sin(2*b*x+2*a)^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 2.69, size = 69, normalized size = 1.30 \begin {gather*} -\frac {4 \, \cos \left (b x + a\right )^{2} \sin \left (b x + a\right ) + \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{2} - 1\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )}}{12 \, b \cos \left (b x + a\right )^{2} \sin \left (b x + a\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)^(5/2),x, algorithm="fricas")

[Out]

-1/12*(4*cos(b*x + a)^2*sin(b*x + a) + sqrt(2)*(4*cos(b*x + a)^2 - 1)*sqrt(cos(b*x + a)*sin(b*x + a)))/(b*cos(
b*x + a)^2*sin(b*x + a))

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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)**(5/2),x)

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 7875 vs. \(2 (45) = 90\).
time = 48.01, size = 7875, normalized size = 148.58 \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)^(5/2),x, algorithm="giac")

[Out]

-1/24*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*(sq
rt(2)*tan(1/2*a)^56 + 23*sqrt(2)*tan(1/2*a)^54 + 251*sqrt(2)*tan(1/2*a)^52 + 1725*sqrt(2)*tan(1/2*a)^50 + 8350
*sqrt(2)*tan(1/2*a)^48 + 30130*sqrt(2)*tan(1/2*a)^46 + 83490*sqrt(2)*tan(1/2*a)^44 + 179630*sqrt(2)*tan(1/2*a)
^42 + 297275*sqrt(2)*tan(1/2*a)^40 + 360525*sqrt(2)*tan(1/2*a)^38 + 264385*sqrt(2)*tan(1/2*a)^36 - 37145*sqrt(
2)*tan(1/2*a)^34 - 445740*sqrt(2)*tan(1/2*a)^32 - 742900*sqrt(2)*tan(1/2*a)^30 - 742900*sqrt(2)*tan(1/2*a)^28
- 445740*sqrt(2)*tan(1/2*a)^26 - 37145*sqrt(2)*tan(1/2*a)^24 + 264385*sqrt(2)*tan(1/2*a)^22 + 360525*sqrt(2)*t
an(1/2*a)^20 + 297275*sqrt(2)*tan(1/2*a)^18 + 179630*sqrt(2)*tan(1/2*a)^16 + 83490*sqrt(2)*tan(1/2*a)^14 + 301
30*sqrt(2)*tan(1/2*a)^12 + 8350*sqrt(2)*tan(1/2*a)^10 + 1725*sqrt(2)*tan(1/2*a)^8 + 251*sqrt(2)*tan(1/2*a)^6 +
23*sqrt(2)*tan(1/2*a)^4 + sqrt(2)*tan(1/2*a)^2)*tan(1/2*b*x)/(tan(1/2*a)^51 + 23*tan(1/2*a)^49 + 252*tan(1/2*
a)^47 + 1748*tan(1/2*a)^45 + 8602*tan(1/2*a)^43 + 31878*tan(1/2*a)^41 + 92092*tan(1/2*a)^39 + 211508*tan(1/2*a
)^37 + 389367*tan(1/2*a)^35 + 572033*tan(1/2*a)^33 + 653752*tan(1/2*a)^31 + 534888*tan(1/2*a)^29 + 208012*tan(
1/2*a)^27 - 208012*tan(1/2*a)^25 - 534888*tan(1/2*a)^23 - 653752*tan(1/2*a)^21 - 572033*tan(1/2*a)^19 - 389367
*tan(1/2*a)^17 - 211508*tan(1/2*a)^15 - 92092*tan(1/2*a)^13 - 31878*tan(1/2*a)^11 - 8602*tan(1/2*a)^9 - 1748*t
an(1/2*a)^7 - 252*tan(1/2*a)^5 - 23*tan(1/2*a)^3 - tan(1/2*a)) + 3*(sqrt(2)*tan(1/2*a)^57 + 18*sqrt(2)*tan(1/2
*a)^55 + 132*sqrt(2)*tan(1/2*a)^53 + 374*sqrt(2)*tan(1/2*a)^51 - 1375*sqrt(2)*tan(1/2*a)^49 - 19620*sqrt(2)*ta
n(1/2*a)^47 - 108560*sqrt(2)*tan(1/2*a)^45 - 399740*sqrt(2)*tan(1/2*a)^43 - 1096755*sqrt(2)*tan(1/2*a)^41 - 23
40250*sqrt(2)*tan(1/2*a)^39 - 3941740*sqrt(2)*tan(1/2*a)^37 - 5204670*sqrt(2)*tan(1/2*a)^35 - 5163155*sqrt(2)*
tan(1/2*a)^33 - 3268760*sqrt(2)*tan(1/2*a)^31 + 3268760*sqrt(2)*tan(1/2*a)^27 + 5163155*sqrt(2)*tan(1/2*a)^25
+ 5204670*sqrt(2)*tan(1/2*a)^23 + 3941740*sqrt(2)*tan(1/2*a)^21 + 2340250*sqrt(2)*tan(1/2*a)^19 + 1096755*sqrt
(2)*tan(1/2*a)^17 + 399740*sqrt(2)*tan(1/2*a)^15 + 108560*sqrt(2)*tan(1/2*a)^13 + 19620*sqrt(2)*tan(1/2*a)^11
+ 1375*sqrt(2)*tan(1/2*a)^9 - 374*sqrt(2)*tan(1/2*a)^7 - 132*sqrt(2)*tan(1/2*a)^5 - 18*sqrt(2)*tan(1/2*a)^3 -
sqrt(2)*tan(1/2*a))/(tan(1/2*a)^51 + 23*tan(1/2*a)^49 + 252*tan(1/2*a)^47 + 1748*tan(1/2*a)^45 + 8602*tan(1/2*
a)^43 + 31878*tan(1/2*a)^41 + 92092*tan(1/2*a)^39 + 211508*tan(1/2*a)^37 + 389367*tan(1/2*a)^35 + 572033*tan(1
/2*a)^33 + 653752*tan(1/2*a)^31 + 534888*tan(1/2*a)^29 + 208012*tan(1/2*a)^27 - 208012*tan(1/2*a)^25 - 534888*
tan(1/2*a)^23 - 653752*tan(1/2*a)^21 - 572033*tan(1/2*a)^19 - 389367*tan(1/2*a)^17 - 211508*tan(1/2*a)^15 - 92
092*tan(1/2*a)^13 - 31878*tan(1/2*a)^11 - 8602*tan(1/2*a)^9 - 1748*tan(1/2*a)^7 - 252*tan(1/2*a)^5 - 23*tan(1/
2*a)^3 - tan(1/2*a)))*tan(1/2*b*x) - 30*(sqrt(2)*tan(1/2*a)^56 + 23*sqrt(2)*tan(1/2*a)^54 + 251*sqrt(2)*tan(1/
2*a)^52 + 1725*sqrt(2)*tan(1/2*a)^50 + 8350*sqrt(2)*tan(1/2*a)^48 + 30130*sqrt(2)*tan(1/2*a)^46 + 83490*sqrt(2
)*tan(1/2*a)^44 + 179630*sqrt(2)*tan(1/2*a)^42 + 297275*sqrt(2)*tan(1/2*a)^40 + 360525*sqrt(2)*tan(1/2*a)^38 +
264385*sqrt(2)*tan(1/2*a)^36 - 37145*sqrt(2)*tan(1/2*a)^34 - 445740*sqrt(2)*tan(1/2*a)^32 - 742900*sqrt(2)*ta
n(1/2*a)^30 - 742900*sqrt(2)*tan(1/2*a)^28 - 445740*sqrt(2)*tan(1/2*a)^26 - 37145*sqrt(2)*tan(1/2*a)^24 + 2643
85*sqrt(2)*tan(1/2*a)^22 + 360525*sqrt(2)*tan(1/2*a)^20 + 297275*sqrt(2)*tan(1/2*a)^18 + 179630*sqrt(2)*tan(1/
2*a)^16 + 83490*sqrt(2)*tan(1/2*a)^14 + 30130*sqrt(2)*tan(1/2*a)^12 + 8350*sqrt(2)*tan(1/2*a)^10 + 1725*sqrt(2
)*tan(1/2*a)^8 + 251*sqrt(2)*tan(1/2*a)^6 + 23*sqrt(2)*tan(1/2*a)^4 + sqrt(2)*tan(1/2*a)^2)/(tan(1/2*a)^51 + 2
3*tan(1/2*a)^49 + 252*tan(1/2*a)^47 + 1748*tan(1/2*a)^45 + 8602*tan(1/2*a)^43 + 31878*tan(1/2*a)^41 + 92092*ta
n(1/2*a)^39 + 211508*tan(1/2*a)^37 + 389367*tan(1/2*a)^35 + 572033*tan(1/2*a)^33 + 653752*tan(1/2*a)^31 + 5348
88*tan(1/2*a)^29 + 208012*tan(1/2*a)^27 - 208012*tan(1/2*a)^25 - 534888*tan(1/2*a)^23 - 653752*tan(1/2*a)^21 -
572033*tan(1/2*a)^19 - 389367*tan(1/2*a)^17 - 211508*tan(1/2*a)^15 - 92092*tan(1/2*a)^13 - 31878*tan(1/2*a)^1
1 - 8602*tan(1/2*a)^9 - 1748*tan(1/2*a)^7 - 252*tan(1/2*a)^5 - 23*tan(1/2*a)^3 - tan(1/2*a)))*tan(1/2*b*x) - 1
0*(sqrt(2)*tan(1/2*a)^57 + 18*sqrt(2)*tan(1/2*a)^55 + 132*sqrt(2)*tan(1/2*a)^53 + 374*sqrt(2)*tan(1/2*a)^51 -
1375*sqrt(2)*tan(1/2*a)^49 - 19620*sqrt(2)*tan(1/2*a)^47 - 108560*sqrt(2)*tan(1/2*a)^45 - 399740*sqrt(2)*tan(1
/2*a)^43 - 1096755*sqrt(2)*tan(1/2*a)^41 - 2340250*sqrt(2)*tan(1/2*a)^39 - 3941740*sqrt(2)*tan(1/2*a)^37 - 520
4670*sqrt(2)*tan(1/2*a)^35 - 5163155*sqrt(2)*tan(1/2*a)^33 - 3268760*sqrt(2)*tan(1/2*a)^31 + 3268760*sqrt(2)*t
an(1/2*a)^27 + 5163155*sqrt(2)*tan(1/2*a)^25 + ...

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Mupad [B]
time = 3.08, size = 108, normalized size = 2.04 \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}}\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}+{\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}\,1{}\mathrm {i}+1{}\mathrm {i}\right )}{3\,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\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(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)*(exp(a*2i + b*x*2i)*1i
+ exp(a*4i + b*x*4i)*1i + 1i))/(3*b*(exp(a*2i + b*x*2i) - 1)*(exp(a*2i + b*x*2i) + 1)^2)

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