3.939 \(\int \sin (c+d x) (a+\frac {b}{\sqrt {\sin (c+d x)}}+c \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=148 \[ -\frac {a^2 \cos (c+d x)}{d}+\frac {4 a b E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{d}-\frac {a c \sin (c+d x) \cos (c+d x)}{d}+a c x+b^2 x+\frac {4 b c F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{3 d}-\frac {4 b c \sqrt {\sin (c+d x)} \cos (c+d x)}{3 d}+\frac {c^2 \cos ^3(c+d x)}{3 d}-\frac {c^2 \cos (c+d x)}{d} \]

[Out]

b^2*x+a*c*x-a^2*cos(d*x+c)/d-c^2*cos(d*x+c)/d+1/3*c^2*cos(d*x+c)^3/d-4*a*b*(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))/d-4/3*b*c*(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))/d-a*c*cos(d*x+c)*sin(d*x+c)/d-4
/3*b*c*cos(d*x+c)*sin(d*x+c)^(1/2)/d

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Rubi [A]  time = 0.24, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.258, Rules used = {4395, 4401, 2639, 2638, 2635, 2641, 8, 2633} \[ -\frac {a^2 \cos (c+d x)}{d}+\frac {4 a b E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{d}-\frac {a c \sin (c+d x) \cos (c+d x)}{d}+a c x+b^2 x+\frac {4 b c F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{3 d}-\frac {4 b c \sqrt {\sin (c+d x)} \cos (c+d x)}{3 d}+\frac {c^2 \cos ^3(c+d x)}{3 d}-\frac {c^2 \cos (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]*(a + b/Sqrt[Sin[c + d*x]] + c*Sin[c + d*x])^2,x]

[Out]

b^2*x + a*c*x - (a^2*Cos[c + d*x])/d - (c^2*Cos[c + d*x])/d + (c^2*Cos[c + d*x]^3)/(3*d) + (4*a*b*EllipticE[(c
- Pi/2 + d*x)/2, 2])/d + (4*b*c*EllipticF[(c - Pi/2 + d*x)/2, 2])/(3*d) - (4*b*c*Cos[c + d*x]*Sqrt[Sin[c + d*
x]])/(3*d) - (a*c*Cos[c + d*x]*Sin[c + d*x])/d

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 4395

Int[(u_)*((a_) + (b_.)*(F_)[(d_.) + (e_.)*(x_)]^(p_.) + (c_.)*(F_)[(d_.) + (e_.)*(x_)]^(q_.))^(n_.), x_Symbol]
:> Int[ActivateTrig[u*F[d + e*x]^(n*p)*(b + a/F[d + e*x]^p + c*F[d + e*x]^(q - p))^n], x] /; FreeQ[{a, b, c,
d, e, p, q}, x] && InertTrigQ[F] && IntegerQ[n] && NegQ[p]

Rule 4401

Int[u_, x_Symbol] :> With[{v = ExpandTrig[u, x]}, Int[v, x] /; SumQ[v]] /;  !InertTrigFreeQ[u]

Rubi steps

\begin {align*} \int \sin (c+d x) \left (a+\frac {b}{\sqrt {\sin (c+d x)}}+c \sin (c+d x)\right )^2 \, dx &=\int \left (b+a \sqrt {\sin (c+d x)}+c \sin ^{\frac {3}{2}}(c+d x)\right )^2 \, dx\\ &=\int \left (b^2+2 a b \sqrt {\sin (c+d x)}+a^2 \sin (c+d x)+2 b c \sin ^{\frac {3}{2}}(c+d x)+2 a c \sin ^2(c+d x)+c^2 \sin ^3(c+d x)\right ) \, dx\\ &=b^2 x+a^2 \int \sin (c+d x) \, dx+(2 a b) \int \sqrt {\sin (c+d x)} \, dx+(2 a c) \int \sin ^2(c+d x) \, dx+(2 b c) \int \sin ^{\frac {3}{2}}(c+d x) \, dx+c^2 \int \sin ^3(c+d x) \, dx\\ &=b^2 x-\frac {a^2 \cos (c+d x)}{d}+\frac {4 a b E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right )}{d}-\frac {4 b c \cos (c+d x) \sqrt {\sin (c+d x)}}{3 d}-\frac {a c \cos (c+d x) \sin (c+d x)}{d}+(a c) \int 1 \, dx+\frac {1}{3} (2 b c) \int \frac {1}{\sqrt {\sin (c+d x)}} \, dx-\frac {c^2 \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=b^2 x+a c x-\frac {a^2 \cos (c+d x)}{d}-\frac {c^2 \cos (c+d x)}{d}+\frac {c^2 \cos ^3(c+d x)}{3 d}+\frac {4 a b E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right )}{d}+\frac {4 b c F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right )}{3 d}-\frac {4 b c \cos (c+d x) \sqrt {\sin (c+d x)}}{3 d}-\frac {a c \cos (c+d x) \sin (c+d x)}{d}\\ \end {align*}

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Mathematica [A]  time = 0.28, size = 137, normalized size = 0.93 \[ \frac {-12 a^2 \cos (c+d x)-48 a b E\left (\left .\frac {1}{4} (-2 c-2 d x+\pi )\right |2\right )+12 a c^2+12 a c d x-6 a c \sin (2 (c+d x))+12 b^2 c+12 b^2 d x-16 b c F\left (\left .\frac {1}{4} (-2 c-2 d x+\pi )\right |2\right )-16 b c \sqrt {\sin (c+d x)} \cos (c+d x)-9 c^2 \cos (c+d x)+c^2 \cos (3 (c+d x))}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*b^2*c + 12*a*c^2 + 12*b^2*d*x + 12*a*c*d*x - 12*a^2*Cos[c + d*x] - 9*c^2*Cos[c + d*x] + c^2*Cos[3*(c + d*x
)] - 48*a*b*EllipticE[(-2*c + Pi - 2*d*x)/4, 2] - 16*b*c*EllipticF[(-2*c + Pi - 2*d*x)/4, 2] - 16*b*c*Cos[c +
d*x]*Sqrt[Sin[c + d*x]] - 6*a*c*Sin[2*(c + d*x)])/(12*d)

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fricas [F]  time = 1.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-2 \, a c \cos \left (d x + c\right )^{2} + b^{2} + 2 \, a c - {\left (c^{2} \cos \left (d x + c\right )^{2} - a^{2} - c^{2}\right )} \sin \left (d x + c\right ) + 2 \, {\left (b c \sin \left (d x + c\right ) + a b\right )} \sqrt {\sin \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-2*a*c*cos(d*x + c)^2 + b^2 + 2*a*c - (c^2*cos(d*x + c)^2 - a^2 - c^2)*sin(d*x + c) + 2*(b*c*sin(d*x
+ c) + a*b)*sqrt(sin(d*x + c)), x)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*p
i/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Warning, choosing ro
ot of [1,0,%%%{-2,[2]%%%}+%%%{-2,[1]%%%}+%%%{-2,[0]%%%},0,%%%{1,[4]%%%}+%%%{-2,[3]%%%}+%%%{3,[2]%%%}+%%%{-2,[1
]%%%}+%%%{1,[0]%%%}] at parameters values [93.1017843988]Warning, choosing root of [1,0,%%%{-2,[2]%%%}+%%%{-2,
[1]%%%}+%%%{-2,[0]%%%},0,%%%{1,[4]%%%}+%%%{-2,[3]%%%}+%%%{3,[2]%%%}+%%%{-2,[1]%%%}+%%%{1,[0]%%%}] at parameter
s values [2.14118046779]Warning, choosing root of [1,0,%%%{-2,[2]%%%}+%%%{-2,[1]%%%}+%%%{-2,[0]%%%},0,%%%{1,[4
]%%%}+%%%{-2,[3]%%%}+%%%{3,[2]%%%}+%%%{-2,[1]%%%}+%%%{1,[0]%%%}] at parameters values [9.72821606882]int()  Er
ror: Bad Argument Value

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maple [A]  time = 0.32, size = 266, normalized size = 1.80 \[ b^{2} x -\frac {a^{2} \cos \left (d x +c \right )}{d}-\frac {c^{2} \left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3 d}+\frac {2 a c \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 b \left (3 a \sqrt {1+\sin \left (d x +c \right )}\, \sqrt {-2 \sin \left (d x +c \right )+2}\, \sqrt {-\sin \left (d x +c \right )}\, \EllipticF \left (\sqrt {1+\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )+\sqrt {1+\sin \left (d x +c \right )}\, \sqrt {-2 \sin \left (d x +c \right )+2}\, \sqrt {-\sin \left (d x +c \right )}\, \EllipticF \left (\sqrt {1+\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right ) c -6 a \sqrt {1+\sin \left (d x +c \right )}\, \sqrt {-2 \sin \left (d x +c \right )+2}\, \sqrt {-\sin \left (d x +c \right )}\, \EllipticE \left (\sqrt {1+\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-2 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) c \right )}{3 \cos \left (d x +c \right ) \sqrt {\sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^2*x-a^2*cos(d*x+c)/d-1/3*c^2/d*(2+sin(d*x+c)^2)*cos(d*x+c)+2*a*c/d*(-1/2*sin(d*x+c)*cos(d*x+c)+1/2*d*x+1/2*c
)+2/3*b*(3*a*(1+sin(d*x+c))^(1/2)*(-2*sin(d*x+c)+2)^(1/2)*(-sin(d*x+c))^(1/2)*EllipticF((1+sin(d*x+c))^(1/2),1
/2*2^(1/2))+(1+sin(d*x+c))^(1/2)*(-2*sin(d*x+c)+2)^(1/2)*(-sin(d*x+c))^(1/2)*EllipticF((1+sin(d*x+c))^(1/2),1/
2*2^(1/2))*c-6*a*(1+sin(d*x+c))^(1/2)*(-2*sin(d*x+c)+2)^(1/2)*(-sin(d*x+c))^(1/2)*EllipticE((1+sin(d*x+c))^(1/
2),1/2*2^(1/2))-2*cos(d*x+c)^2*sin(d*x+c)*c)/cos(d*x+c)/sin(d*x+c)^(1/2)/d

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

[Out]

Timed out

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mupad [B]  time = 6.68, size = 129, normalized size = 0.87 \[ b^2\,x-\frac {a^2\,\cos \left (c+d\,x\right )}{d}-\frac {a\,c\,\left (\sin \left (2\,c+2\,d\,x\right )-2\,d\,x\right )}{2\,d}+\frac {4\,a\,b\,\mathrm {E}\left (\frac {c}{2}-\frac {\pi }{4}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {c^2\,\cos \left (c+d\,x\right )\,\left ({\cos \left (c+d\,x\right )}^2-3\right )}{3\,d}-\frac {2\,b\,c\,\cos \left (c+d\,x\right )\,{\sin \left (c+d\,x\right )}^{5/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{2};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,{\left ({\sin \left (c+d\,x\right )}^2\right )}^{5/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^2*x - (a^2*cos(c + d*x))/d - (a*c*(sin(2*c + 2*d*x) - 2*d*x))/(2*d) + (4*a*b*ellipticE(c/2 - pi/4 + (d*x)/2,
2))/d + (c^2*cos(c + d*x)*(cos(c + d*x)^2 - 3))/(3*d) - (2*b*c*cos(c + d*x)*sin(c + d*x)^(5/2)*hypergeom([-1/
4, 1/2], 3/2, cos(c + d*x)^2))/(d*(sin(c + d*x)^2)^(5/4))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + \frac {b}{\sqrt {\sin {\left (c + d x \right )}}} + c \sin {\left (c + d x \right )}\right )^{2} \sin {\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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