\(\int \frac {1}{(a+b \sin (c+d x))^{7/2}} \, dx\) [57]

Optimal result

Integrand size = 14, antiderivative size = 292 \[ \int \frac {1}{(a+b \sin (c+d x))^{7/2}} \, dx=\frac {2 b \cos (c+d x)}{5 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{5/2}}+\frac {16 a b \cos (c+d x)}{15 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^{3/2}}+\frac {2 b \left (23 a^2+9 b^2\right ) \cos (c+d x)}{15 \left (a^2-b^2\right )^3 d \sqrt {a+b \sin (c+d x)}}+\frac {2 \left (23 a^2+9 b^2\right ) E\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )|\frac {2 b}{a+b}\right ) \sqrt {a+b \sin (c+d x)}}{15 \left (a^2-b^2\right )^3 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {16 a \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}{15 \left (a^2-b^2\right )^2 d \sqrt {a+b \sin (c+d x)}} \]

[Out]

2/5*b*cos(d*x+c)/(a^2-b^2)/d/(a+b*sin(d*x+c))^(5/2)+16/15*a*b*cos(d*x+c)/(a^2-b^2)^2/d/(a+b*sin(d*x+c))^(3/2)+
2/15*b*(23*a^2+9*b^2)*cos(d*x+c)/(a^2-b^2)^3/d/(a+b*sin(d*x+c))^(1/2)-2/15*(23*a^2+9*b^2)*(sin(1/2*c+1/4*Pi+1/
2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticE(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2)*(b/(a+b))^(1/2))*(a+b*si
n(d*x+c))^(1/2)/(a^2-b^2)^3/d/((a+b*sin(d*x+c))/(a+b))^(1/2)+16/15*a*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1
/2*c+1/4*Pi+1/2*d*x)*EllipticF(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2)*(b/(a+b))^(1/2))*((a+b*sin(d*x+c))/(a+b))^(1/
2)/(a^2-b^2)^2/d/(a+b*sin(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {2743, 2833, 2831, 2742, 2740, 2734, 2732} \[ \int \frac {1}{(a+b \sin (c+d x))^{7/2}} \, dx=\frac {2 b \left (23 a^2+9 b^2\right ) \cos (c+d x)}{15 d \left (a^2-b^2\right )^3 \sqrt {a+b \sin (c+d x)}}+\frac {16 a b \cos (c+d x)}{15 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^{3/2}}+\frac {2 b \cos (c+d x)}{5 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^{5/2}}-\frac {16 a \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),\frac {2 b}{a+b}\right )}{15 d \left (a^2-b^2\right )^2 \sqrt {a+b \sin (c+d x)}}+\frac {2 \left (23 a^2+9 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{15 d \left (a^2-b^2\right )^3 \sqrt {\frac {a+b \sin (c+d x)}{a+b}}} \]

[In]

Int[(a + b*Sin[c + d*x])^(-7/2),x]

[Out]

(2*b*Cos[c + d*x])/(5*(a^2 - b^2)*d*(a + b*Sin[c + d*x])^(5/2)) + (16*a*b*Cos[c + d*x])/(15*(a^2 - b^2)^2*d*(a
+ b*Sin[c + d*x])^(3/2)) + (2*b*(23*a^2 + 9*b^2)*Cos[c + d*x])/(15*(a^2 - b^2)^3*d*Sqrt[a + b*Sin[c + d*x]])
+ (2*(23*a^2 + 9*b^2)*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(15*(a^2 - b^2)^3
*d*Sqrt[(a + b*Sin[c + d*x])/(a + b)]) - (16*a*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c
+ d*x])/(a + b)])/(15*(a^2 - b^2)^2*d*Sqrt[a + b*Sin[c + d*x]])

Rule 2732

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a + b]/d)*EllipticE[(1/2)*(c - Pi/2
+ d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2734

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2740

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*Sqrt[a + b]))*EllipticF[(1/2)*(c - P
i/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2742

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2743

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((a + b*Sin[c + d*x])^(n
+ 1)/(d*(n + 1)*(a^2 - b^2))), x] + Dist[1/((n + 1)*(a^2 - b^2)), Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n +
1) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integ
erQ[2*n]

Rule 2831

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(b*c
- a*d)/b, Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[d/b, Int[Sqrt[a + b*Sin[e + f*x]], x], x] /; FreeQ[{a
, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]

Rule 2833

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(
b*c - a*d))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(f*(m + 1)*(a^2 - b^2))), x] + Dist[1/((m + 1)*(a^2 - b
^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + 2)*Sin[e + f*x], x], x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]

Rubi steps \begin{align*} \text {integral}& = \frac {2 b \cos (c+d x)}{5 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{5/2}}-\frac {2 \int \frac {-\frac {5 a}{2}+\frac {3}{2} b \sin (c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx}{5 \left (a^2-b^2\right )} \\ & = \frac {2 b \cos (c+d x)}{5 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{5/2}}+\frac {16 a b \cos (c+d x)}{15 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^{3/2}}+\frac {4 \int \frac {\frac {3}{4} \left (5 a^2+3 b^2\right )-2 a b \sin (c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx}{15 \left (a^2-b^2\right )^2} \\ & = \frac {2 b \cos (c+d x)}{5 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{5/2}}+\frac {16 a b \cos (c+d x)}{15 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^{3/2}}+\frac {2 b \left (23 a^2+9 b^2\right ) \cos (c+d x)}{15 \left (a^2-b^2\right )^3 d \sqrt {a+b \sin (c+d x)}}-\frac {8 \int \frac {-\frac {1}{8} a \left (15 a^2+17 b^2\right )-\frac {1}{8} b \left (23 a^2+9 b^2\right ) \sin (c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx}{15 \left (a^2-b^2\right )^3} \\ & = \frac {2 b \cos (c+d x)}{5 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{5/2}}+\frac {16 a b \cos (c+d x)}{15 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^{3/2}}+\frac {2 b \left (23 a^2+9 b^2\right ) \cos (c+d x)}{15 \left (a^2-b^2\right )^3 d \sqrt {a+b \sin (c+d x)}}-\frac {(8 a) \int \frac {1}{\sqrt {a+b \sin (c+d x)}} \, dx}{15 \left (a^2-b^2\right )^2}+\frac {\left (23 a^2+9 b^2\right ) \int \sqrt {a+b \sin (c+d x)} \, dx}{15 \left (a^2-b^2\right )^3} \\ & = \frac {2 b \cos (c+d x)}{5 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{5/2}}+\frac {16 a b \cos (c+d x)}{15 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^{3/2}}+\frac {2 b \left (23 a^2+9 b^2\right ) \cos (c+d x)}{15 \left (a^2-b^2\right )^3 d \sqrt {a+b \sin (c+d x)}}+\frac {\left (\left (23 a^2+9 b^2\right ) \sqrt {a+b \sin (c+d x)}\right ) \int \sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}} \, dx}{15 \left (a^2-b^2\right )^3 \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {\left (8 a \sqrt {\frac {a+b \sin (c+d x)}{a+b}}\right ) \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}}} \, dx}{15 \left (a^2-b^2\right )^2 \sqrt {a+b \sin (c+d x)}} \\ & = \frac {2 b \cos (c+d x)}{5 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{5/2}}+\frac {16 a b \cos (c+d x)}{15 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^{3/2}}+\frac {2 b \left (23 a^2+9 b^2\right ) \cos (c+d x)}{15 \left (a^2-b^2\right )^3 d \sqrt {a+b \sin (c+d x)}}+\frac {2 \left (23 a^2+9 b^2\right ) E\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )|\frac {2 b}{a+b}\right ) \sqrt {a+b \sin (c+d x)}}{15 \left (a^2-b^2\right )^3 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {16 a \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}{15 \left (a^2-b^2\right )^2 d \sqrt {a+b \sin (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.97 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.68 \[ \int \frac {1}{(a+b \sin (c+d x))^{7/2}} \, dx=\frac {2 \left (-\frac {\left (\left (23 a^2+9 b^2\right ) E\left (\frac {1}{4} (-2 c+\pi -2 d x)|\frac {2 b}{a+b}\right )+8 a (-a+b) \operatorname {EllipticF}\left (\frac {1}{4} (-2 c+\pi -2 d x),\frac {2 b}{a+b}\right )\right ) \left (\frac {a+b \sin (c+d x)}{a+b}\right )^{5/2}}{(a-b)^3}+\frac {b \cos (c+d x) \left (34 a^4-5 a^2 b^2+3 b^4+2 a b \left (27 a^2+5 b^2\right ) \sin (c+d x)+b^2 \left (23 a^2+9 b^2\right ) \sin ^2(c+d x)\right )}{\left (a^2-b^2\right )^3}\right )}{15 d (a+b \sin (c+d x))^{5/2}} \]

[In]

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

[Out]

(2*(-((((23*a^2 + 9*b^2)*EllipticE[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)] + 8*a*(-a + b)*EllipticF[(-2*c + Pi -
2*d*x)/4, (2*b)/(a + b)])*((a + b*Sin[c + d*x])/(a + b))^(5/2))/(a - b)^3) + (b*Cos[c + d*x]*(34*a^4 - 5*a^2*
b^2 + 3*b^4 + 2*a*b*(27*a^2 + 5*b^2)*Sin[c + d*x] + b^2*(23*a^2 + 9*b^2)*Sin[c + d*x]^2))/(a^2 - b^2)^3))/(15*
d*(a + b*Sin[c + d*x])^(5/2))

Maple [A] (verified)

Time = 0.87 (sec) , antiderivative size = 584, normalized size of antiderivative = 2.00

method	result	size
default	\(\frac {\sqrt {-\left (-b \sin \left (d x +c \right )-a \right ) \left (\cos ^{2}\left (d x +c \right )\right )}\, \left (\frac {2 \sqrt {-\left (-b \sin \left (d x +c \right )-a \right ) \left (\cos ^{2}\left (d x +c \right )\right )}}{5 b^{2} \left (a^{2}-b^{2}\right ) \left (\sin \left (d x +c \right )+\frac {a}{b}\right )^{3}}+\frac {16 a \sqrt {-\left (-b \sin \left (d x +c \right )-a \right ) \left (\cos ^{2}\left (d x +c \right )\right )}}{15 \left (a^{2}-b^{2}\right )^{2} b \left (\sin \left (d x +c \right )+\frac {a}{b}\right )^{2}}+\frac {2 b \left (\cos ^{2}\left (d x +c \right )\right ) \left (23 a^{2}+9 b^{2}\right )}{15 \left (a^{2}-b^{2}\right )^{3} \sqrt {-\left (-b \sin \left (d x +c \right )-a \right ) \left (\cos ^{2}\left (d x +c \right )\right )}}+\frac {2 \left (15 a^{3}+17 a \,b^{2}\right ) \left (\frac {a}{b}-1\right ) \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {\frac {b \left (1-\sin \left (d x +c \right )\right )}{a +b}}\, \sqrt {\frac {\left (-1-\sin \left (d x +c \right )\right ) b}{a -b}}\, F\left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right )}{\left (15 a^{6}-45 a^{4} b^{2}+45 a^{2} b^{4}-15 b^{6}\right ) \sqrt {-\left (-b \sin \left (d x +c \right )-a \right ) \left (\cos ^{2}\left (d x +c \right )\right )}}+\frac {2 b \left (23 a^{2}+9 b^{2}\right ) \left (\frac {a}{b}-1\right ) \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {\frac {b \left (1-\sin \left (d x +c \right )\right )}{a +b}}\, \sqrt {\frac {\left (-1-\sin \left (d x +c \right )\right ) b}{a -b}}\, \left (\left (-\frac {a}{b}-1\right ) E\left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right )+F\left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right )\right )}{15 \left (a^{2}-b^{2}\right )^{3} \sqrt {-\left (-b \sin \left (d x +c \right )-a \right ) \left (\cos ^{2}\left (d x +c \right )\right )}}\right )}{\cos \left (d x +c \right ) \sqrt {a +b \sin \left (d x +c \right )}\, d}\)	\(584\)

[In]

int(1/(a+b*sin(d*x+c))^(7/2),x,method=_RETURNVERBOSE)

[Out]

(-(-b*sin(d*x+c)-a)*cos(d*x+c)^2)^(1/2)*(2/5/b^2/(a^2-b^2)*(-(-b*sin(d*x+c)-a)*cos(d*x+c)^2)^(1/2)/(sin(d*x+c)
+a/b)^3+16/15*a/(a^2-b^2)^2/b*(-(-b*sin(d*x+c)-a)*cos(d*x+c)^2)^(1/2)/(sin(d*x+c)+a/b)^2+2/15*b*cos(d*x+c)^2/(
a^2-b^2)^3*(23*a^2+9*b^2)/(-(-b*sin(d*x+c)-a)*cos(d*x+c)^2)^(1/2)+2*(15*a^3+17*a*b^2)/(15*a^6-45*a^4*b^2+45*a^
2*b^4-15*b^6)*(a/b-1)*((a+b*sin(d*x+c))/(a-b))^(1/2)*(b*(1-sin(d*x+c))/(a+b))^(1/2)*((-1-sin(d*x+c))*b/(a-b))^
(1/2)/(-(-b*sin(d*x+c)-a)*cos(d*x+c)^2)^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))+2/
15*b*(23*a^2+9*b^2)/(a^2-b^2)^3*(a/b-1)*((a+b*sin(d*x+c))/(a-b))^(1/2)*(b*(1-sin(d*x+c))/(a+b))^(1/2)*((-1-sin
(d*x+c))*b/(a-b))^(1/2)/(-(-b*sin(d*x+c)-a)*cos(d*x+c)^2)^(1/2)*((-a/b-1)*EllipticE(((a+b*sin(d*x+c))/(a-b))^(
1/2),((a-b)/(a+b))^(1/2))+EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))))/cos(d*x+c)/(a+b*sin(
d*x+c))^(1/2)/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.16 (sec) , antiderivative size = 1051, normalized size of antiderivative = 3.60 \[ \int \frac {1}{(a+b \sin (c+d x))^{7/2}} \, dx=\text {Too large to display} \]

[In]

integrate(1/(a+b*sin(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

-1/45*((3*sqrt(2)*(a^4*b^2 - 33*a^2*b^4)*cos(d*x + c)^2 + (sqrt(2)*(a^3*b^3 - 33*a*b^5)*cos(d*x + c)^2 - sqrt(
2)*(3*a^5*b - 98*a^3*b^3 - 33*a*b^5))*sin(d*x + c) - sqrt(2)*(a^6 - 30*a^4*b^2 - 99*a^2*b^4))*sqrt(I*b)*weiers
trassPInverse(-4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*I*a^3 - 9*I*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) - 3*I*b*sin(d*x
+ c) - 2*I*a)/b) + (3*sqrt(2)*(a^4*b^2 - 33*a^2*b^4)*cos(d*x + c)^2 + (sqrt(2)*(a^3*b^3 - 33*a*b^5)*cos(d*x +
c)^2 - sqrt(2)*(3*a^5*b - 98*a^3*b^3 - 33*a*b^5))*sin(d*x + c) - sqrt(2)*(a^6 - 30*a^4*b^2 - 99*a^2*b^4))*sqr
t(-I*b)*weierstrassPInverse(-4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(-8*I*a^3 + 9*I*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c)
+ 3*I*b*sin(d*x + c) + 2*I*a)/b) - 3*(3*sqrt(2)*(-23*I*a^3*b^3 - 9*I*a*b^5)*cos(d*x + c)^2 + (sqrt(2)*(-23*I*a
^2*b^4 - 9*I*b^6)*cos(d*x + c)^2 + sqrt(2)*(69*I*a^4*b^2 + 50*I*a^2*b^4 + 9*I*b^6))*sin(d*x + c) + sqrt(2)*(23
*I*a^5*b + 78*I*a^3*b^3 + 27*I*a*b^5))*sqrt(I*b)*weierstrassZeta(-4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*I*a^3 - 9*
I*a*b^2)/b^3, weierstrassPInverse(-4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*I*a^3 - 9*I*a*b^2)/b^3, 1/3*(3*b*cos(d*x
+ c) - 3*I*b*sin(d*x + c) - 2*I*a)/b)) - 3*(3*sqrt(2)*(23*I*a^3*b^3 + 9*I*a*b^5)*cos(d*x + c)^2 + (sqrt(2)*(23
*I*a^2*b^4 + 9*I*b^6)*cos(d*x + c)^2 + sqrt(2)*(-69*I*a^4*b^2 - 50*I*a^2*b^4 - 9*I*b^6))*sin(d*x + c) + sqrt(2
)*(-23*I*a^5*b - 78*I*a^3*b^3 - 27*I*a*b^5))*sqrt(-I*b)*weierstrassZeta(-4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(-8*I*
a^3 + 9*I*a*b^2)/b^3, weierstrassPInverse(-4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(-8*I*a^3 + 9*I*a*b^2)/b^3, 1/3*(3*b
*cos(d*x + c) + 3*I*b*sin(d*x + c) + 2*I*a)/b)) - 6*((23*a^2*b^4 + 9*b^6)*cos(d*x + c)^3 - 2*(27*a^3*b^3 + 5*a
*b^5)*cos(d*x + c)*sin(d*x + c) - 2*(17*a^4*b^2 + 9*a^2*b^4 + 6*b^6)*cos(d*x + c))*sqrt(b*sin(d*x + c) + a))/(
3*(a^7*b^3 - 3*a^5*b^5 + 3*a^3*b^7 - a*b^9)*d*cos(d*x + c)^2 - (a^9*b - 6*a^5*b^5 + 8*a^3*b^7 - 3*a*b^9)*d + (
(a^6*b^4 - 3*a^4*b^6 + 3*a^2*b^8 - b^10)*d*cos(d*x + c)^2 - (3*a^8*b^2 - 8*a^6*b^4 + 6*a^4*b^6 - b^10)*d)*sin(
d*x + c))

Sympy [F]

\[ \int \frac {1}{(a+b \sin (c+d x))^{7/2}} \, dx=\int \frac {1}{\left (a + b \sin {\left (c + d x \right )}\right )^{\frac {7}{2}}}\, dx \]

[In]

integrate(1/(a+b*sin(d*x+c))**(7/2),x)

[Out]

Integral((a + b*sin(c + d*x))**(-7/2), x)

Maxima [F]

\[ \int \frac {1}{(a+b \sin (c+d x))^{7/2}} \, dx=\int { \frac {1}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}}} \,d x } \]

[In]

integrate(1/(a+b*sin(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

integrate((b*sin(d*x + c) + a)^(-7/2), x)

Giac [F]

\[ \int \frac {1}{(a+b \sin (c+d x))^{7/2}} \, dx=\int { \frac {1}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}}} \,d x } \]

[In]

integrate(1/(a+b*sin(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate((b*sin(d*x + c) + a)^(-7/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a+b \sin (c+d x))^{7/2}} \, dx=\int \frac {1}{{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{7/2}} \,d x \]

[In]

int(1/(a + b*sin(c + d*x))^(7/2),x)

[Out]

int(1/(a + b*sin(c + d*x))^(7/2), x)