3.88 \(\int \frac{(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx\)

Optimal. Leaf size=163 \[ -\frac{a (A-3 B) \cos (e+f x)}{32 c^2 f (c-c \sin (e+f x))^{3/2}}-\frac{a (A-3 B) \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{32 \sqrt{2} c^{7/2} f}-\frac{a (A+13 B) \cos (e+f x)}{24 c f (c-c \sin (e+f x))^{5/2}}+\frac{a (A+B) \cos (e+f x)}{3 f (c-c \sin (e+f x))^{7/2}} \]

[Out]

-(a*(A - 3*B)*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(32*Sqrt[2]*c^(7/2)*f) + (a*
(A + B)*Cos[e + f*x])/(3*f*(c - c*Sin[e + f*x])^(7/2)) - (a*(A + 13*B)*Cos[e + f*x])/(24*c*f*(c - c*Sin[e + f*
x])^(5/2)) - (a*(A - 3*B)*Cos[e + f*x])/(32*c^2*f*(c - c*Sin[e + f*x])^(3/2))

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Rubi [A]  time = 0.372946, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {2967, 2857, 2750, 2650, 2649, 206} \[ -\frac{a (A-3 B) \cos (e+f x)}{32 c^2 f (c-c \sin (e+f x))^{3/2}}-\frac{a (A-3 B) \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{32 \sqrt{2} c^{7/2} f}-\frac{a (A+13 B) \cos (e+f x)}{24 c f (c-c \sin (e+f x))^{5/2}}+\frac{a (A+B) \cos (e+f x)}{3 f (c-c \sin (e+f x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[((a + a*Sin[e + f*x])*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(7/2),x]

[Out]

-(a*(A - 3*B)*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(32*Sqrt[2]*c^(7/2)*f) + (a*
(A + B)*Cos[e + f*x])/(3*f*(c - c*Sin[e + f*x])^(7/2)) - (a*(A + 13*B)*Cos[e + f*x])/(24*c*f*(c - c*Sin[e + f*
x])^(5/2)) - (a*(A - 3*B)*Cos[e + f*x])/(32*c^2*f*(c - c*Sin[e + f*x])^(3/2))

Rule 2967

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m)*(A + B
*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && I
ntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))

Rule 2857

Int[cos[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_
)]), x_Symbol] :> Simp[(2*(b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b^2*f*(2*m + 3)), x] + Dist[
1/(b^3*(2*m + 3)), Int[(a + b*Sin[e + f*x])^(m + 2)*(b*c + 2*a*d*(m + 1) - b*d*(2*m + 3)*Sin[e + f*x]), x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -3/2]

Rule 2750

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)/(a*f*(2*m + 1)), x] + Dist[(a*d*m + b*c*(m + 1))/(a*b*(2*m + 1)
), Int[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 -
b^2, 0] && LtQ[m, -2^(-1)]

Rule 2650

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

Rule 2649

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx &=(a c) \int \frac{\cos ^2(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}} \, dx\\ &=\frac{a (A+B) \cos (e+f x)}{3 f (c-c \sin (e+f x))^{7/2}}+\frac{a \int \frac{-A c-7 B c-6 B c \sin (e+f x)}{(c-c \sin (e+f x))^{5/2}} \, dx}{6 c^2}\\ &=\frac{a (A+B) \cos (e+f x)}{3 f (c-c \sin (e+f x))^{7/2}}-\frac{a (A+13 B) \cos (e+f x)}{24 c f (c-c \sin (e+f x))^{5/2}}-\frac{(a (A-3 B)) \int \frac{1}{(c-c \sin (e+f x))^{3/2}} \, dx}{16 c^2}\\ &=\frac{a (A+B) \cos (e+f x)}{3 f (c-c \sin (e+f x))^{7/2}}-\frac{a (A+13 B) \cos (e+f x)}{24 c f (c-c \sin (e+f x))^{5/2}}-\frac{a (A-3 B) \cos (e+f x)}{32 c^2 f (c-c \sin (e+f x))^{3/2}}-\frac{(a (A-3 B)) \int \frac{1}{\sqrt{c-c \sin (e+f x)}} \, dx}{64 c^3}\\ &=\frac{a (A+B) \cos (e+f x)}{3 f (c-c \sin (e+f x))^{7/2}}-\frac{a (A+13 B) \cos (e+f x)}{24 c f (c-c \sin (e+f x))^{5/2}}-\frac{a (A-3 B) \cos (e+f x)}{32 c^2 f (c-c \sin (e+f x))^{3/2}}+\frac{(a (A-3 B)) \operatorname{Subst}\left (\int \frac{1}{2 c-x^2} \, dx,x,-\frac{c \cos (e+f x)}{\sqrt{c-c \sin (e+f x)}}\right )}{32 c^3 f}\\ &=-\frac{a (A-3 B) \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{32 \sqrt{2} c^{7/2} f}+\frac{a (A+B) \cos (e+f x)}{3 f (c-c \sin (e+f x))^{7/2}}-\frac{a (A+13 B) \cos (e+f x)}{24 c f (c-c \sin (e+f x))^{5/2}}-\frac{a (A-3 B) \cos (e+f x)}{32 c^2 f (c-c \sin (e+f x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 3.31844, size = 217, normalized size = 1.33 \[ -\frac{a (\sin (e+f x)-1) (\sin (e+f x)+1) \left (\frac{\sqrt{c} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) (4 (5 A+17 B) \sin (e+f x)+3 (A-3 B) \cos (2 (e+f x))+47 A-13 B)}{\left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^7}+3 \sqrt{2} (A-3 B) \sec (e+f x) \sqrt{-c (\sin (e+f x)+1)} \tan ^{-1}\left (\frac{\sqrt{-c (\sin (e+f x)+1)}}{\sqrt{2} \sqrt{c}}\right )\right )}{192 c^{7/2} f \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((a + a*Sin[e + f*x])*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(7/2),x]

[Out]

-(a*(-1 + Sin[e + f*x])*(1 + Sin[e + f*x])*(3*Sqrt[2]*(A - 3*B)*ArcTan[Sqrt[-(c*(1 + Sin[e + f*x]))]/(Sqrt[2]*
Sqrt[c])]*Sec[e + f*x]*Sqrt[-(c*(1 + Sin[e + f*x]))] + (Sqrt[c]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(47*A -
13*B + 3*(A - 3*B)*Cos[2*(e + f*x)] + 4*(5*A + 17*B)*Sin[e + f*x]))/(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^7))/
(192*c^(7/2)*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2*Sqrt[c - c*Sin[e + f*x]])

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Maple [B]  time = 1.416, size = 352, normalized size = 2.2 \begin{align*}{\frac{a}{192\, \left ( -1+\sin \left ( fx+e \right ) \right ) ^{2}\cos \left ( fx+e \right ) f} \left ( -3\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c+c\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{c}}} \right ){c}^{4} \left ( A-3\,B \right ) \sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+12\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c+c\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{c}}} \right ){c}^{4} \left ( A-3\,B \right ) \sin \left ( fx+e \right ) +9\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c+c\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{c}}} \right ){c}^{4} \left ( A-3\,B \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+24\,A\sqrt{c+c\sin \left ( fx+e \right ) }{c}^{7/2}+32\,A \left ( c+c\sin \left ( fx+e \right ) \right ) ^{3/2}{c}^{5/2}-6\,A \left ( c+c\sin \left ( fx+e \right ) \right ) ^{5/2}{c}^{3/2}-72\,B\sqrt{c+c\sin \left ( fx+e \right ) }{c}^{7/2}+32\,B \left ( c+c\sin \left ( fx+e \right ) \right ) ^{3/2}{c}^{5/2}+18\,B \left ( c+c\sin \left ( fx+e \right ) \right ) ^{5/2}{c}^{3/2}-12\,A\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c+c\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{c}}} \right ){c}^{4}+36\,B\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c+c\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{c}}} \right ){c}^{4} \right ) \sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }{c}^{-{\frac{15}{2}}}{\frac{1}{\sqrt{c-c\sin \left ( fx+e \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(7/2),x)

[Out]

1/192*a*(-3*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c^4*(A-3*B)*sin(f*x+e)*cos(f*x+e)^2+12
*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c^4*(A-3*B)*sin(f*x+e)+9*2^(1/2)*arctanh(1/2*(c+c
*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c^4*(A-3*B)*cos(f*x+e)^2+24*A*(c+c*sin(f*x+e))^(1/2)*c^(7/2)+32*A*(c+c*sin
(f*x+e))^(3/2)*c^(5/2)-6*A*(c+c*sin(f*x+e))^(5/2)*c^(3/2)-72*B*(c+c*sin(f*x+e))^(1/2)*c^(7/2)+32*B*(c+c*sin(f*
x+e))^(3/2)*c^(5/2)+18*B*(c+c*sin(f*x+e))^(5/2)*c^(3/2)-12*A*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2
)/c^(1/2))*c^4+36*B*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c^4)*(c*(1+sin(f*x+e)))^(1/2)/
c^(15/2)/(-1+sin(f*x+e))^2/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \sin \left (f x + e\right ) + A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)/(-c*sin(f*x + e) + c)^(7/2), x)

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Fricas [B]  time = 1.71156, size = 1289, normalized size = 7.91 \begin{align*} -\frac{3 \, \sqrt{2}{\left ({\left (A - 3 \, B\right )} a \cos \left (f x + e\right )^{4} - 3 \,{\left (A - 3 \, B\right )} a \cos \left (f x + e\right )^{3} - 8 \,{\left (A - 3 \, B\right )} a \cos \left (f x + e\right )^{2} + 4 \,{\left (A - 3 \, B\right )} a \cos \left (f x + e\right ) + 8 \,{\left (A - 3 \, B\right )} a +{\left ({\left (A - 3 \, B\right )} a \cos \left (f x + e\right )^{3} + 4 \,{\left (A - 3 \, B\right )} a \cos \left (f x + e\right )^{2} - 4 \,{\left (A - 3 \, B\right )} a \cos \left (f x + e\right ) - 8 \,{\left (A - 3 \, B\right )} a\right )} \sin \left (f x + e\right )\right )} \sqrt{c} \log \left (-\frac{c \cos \left (f x + e\right )^{2} + 2 \, \sqrt{2} \sqrt{-c \sin \left (f x + e\right ) + c} \sqrt{c}{\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) +{\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\cos \left (f x + e\right )^{2} +{\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \,{\left (3 \,{\left (A - 3 \, B\right )} a \cos \left (f x + e\right )^{3} -{\left (7 \, A + 43 \, B\right )} a \cos \left (f x + e\right )^{2} + 2 \,{\left (11 \, A - B\right )} a \cos \left (f x + e\right ) + 32 \,{\left (A + B\right )} a +{\left (3 \,{\left (A - 3 \, B\right )} a \cos \left (f x + e\right )^{2} + 2 \,{\left (5 \, A + 17 \, B\right )} a \cos \left (f x + e\right ) + 32 \,{\left (A + B\right )} a\right )} \sin \left (f x + e\right )\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{384 \,{\left (c^{4} f \cos \left (f x + e\right )^{4} - 3 \, c^{4} f \cos \left (f x + e\right )^{3} - 8 \, c^{4} f \cos \left (f x + e\right )^{2} + 4 \, c^{4} f \cos \left (f x + e\right ) + 8 \, c^{4} f +{\left (c^{4} f \cos \left (f x + e\right )^{3} + 4 \, c^{4} f \cos \left (f x + e\right )^{2} - 4 \, c^{4} f \cos \left (f x + e\right ) - 8 \, c^{4} f\right )} \sin \left (f x + e\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

-1/384*(3*sqrt(2)*((A - 3*B)*a*cos(f*x + e)^4 - 3*(A - 3*B)*a*cos(f*x + e)^3 - 8*(A - 3*B)*a*cos(f*x + e)^2 +
4*(A - 3*B)*a*cos(f*x + e) + 8*(A - 3*B)*a + ((A - 3*B)*a*cos(f*x + e)^3 + 4*(A - 3*B)*a*cos(f*x + e)^2 - 4*(A
- 3*B)*a*cos(f*x + e) - 8*(A - 3*B)*a)*sin(f*x + e))*sqrt(c)*log(-(c*cos(f*x + e)^2 + 2*sqrt(2)*sqrt(-c*sin(f
*x + e) + c)*sqrt(c)*(cos(f*x + e) + sin(f*x + e) + 1) + 3*c*cos(f*x + e) + (c*cos(f*x + e) - 2*c)*sin(f*x + e
) + 2*c)/(cos(f*x + e)^2 + (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2)) - 4*(3*(A - 3*B)*a*cos(f*x + e
)^3 - (7*A + 43*B)*a*cos(f*x + e)^2 + 2*(11*A - B)*a*cos(f*x + e) + 32*(A + B)*a + (3*(A - 3*B)*a*cos(f*x + e)
^2 + 2*(5*A + 17*B)*a*cos(f*x + e) + 32*(A + B)*a)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c))/(c^4*f*cos(f*x + e
)^4 - 3*c^4*f*cos(f*x + e)^3 - 8*c^4*f*cos(f*x + e)^2 + 4*c^4*f*cos(f*x + e) + 8*c^4*f + (c^4*f*cos(f*x + e)^3
+ 4*c^4*f*cos(f*x + e)^2 - 4*c^4*f*cos(f*x + e) - 8*c^4*f)*sin(f*x + e))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(7/2),x, algorithm="giac")

[Out]

sage2