3.87 \(\int \frac{(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=126 \[ -\frac{a (A-7 B) \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{8 \sqrt{2} c^{5/2} f}-\frac{a (A+9 B) \cos (e+f x)}{8 c f (c-c \sin (e+f x))^{3/2}}+\frac{a (A+B) \cos (e+f x)}{2 f (c-c \sin (e+f x))^{5/2}} \]

[Out]

-(a*(A - 7*B)*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(8*Sqrt[2]*c^(5/2)*f) + (a*(
A + B)*Cos[e + f*x])/(2*f*(c - c*Sin[e + f*x])^(5/2)) - (a*(A + 9*B)*Cos[e + f*x])/(8*c*f*(c - c*Sin[e + f*x])
^(3/2))

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Rubi [A]  time = 0.334508, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.139, Rules used = {2967, 2857, 2750, 2649, 206} \[ -\frac{a (A-7 B) \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{8 \sqrt{2} c^{5/2} f}-\frac{a (A+9 B) \cos (e+f x)}{8 c f (c-c \sin (e+f x))^{3/2}}+\frac{a (A+B) \cos (e+f x)}{2 f (c-c \sin (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[((a + a*Sin[e + f*x])*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(5/2),x]

[Out]

-(a*(A - 7*B)*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(8*Sqrt[2]*c^(5/2)*f) + (a*(
A + B)*Cos[e + f*x])/(2*f*(c - c*Sin[e + f*x])^(5/2)) - (a*(A + 9*B)*Cos[e + f*x])/(8*c*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 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))^{5/2}} \, dx &=(a c) \int \frac{\cos ^2(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx\\ &=\frac{a (A+B) \cos (e+f x)}{2 f (c-c \sin (e+f x))^{5/2}}+\frac{a \int \frac{-A c-5 B c-4 B c \sin (e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx}{4 c^2}\\ &=\frac{a (A+B) \cos (e+f x)}{2 f (c-c \sin (e+f x))^{5/2}}-\frac{a (A+9 B) \cos (e+f x)}{8 c f (c-c \sin (e+f x))^{3/2}}-\frac{(a (A-7 B)) \int \frac{1}{\sqrt{c-c \sin (e+f x)}} \, dx}{16 c^2}\\ &=\frac{a (A+B) \cos (e+f x)}{2 f (c-c \sin (e+f x))^{5/2}}-\frac{a (A+9 B) \cos (e+f x)}{8 c f (c-c \sin (e+f x))^{3/2}}+\frac{(a (A-7 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 )}{8 c^2 f}\\ &=-\frac{a (A-7 B) \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{8 \sqrt{2} c^{5/2} f}+\frac{a (A+B) \cos (e+f x)}{2 f (c-c \sin (e+f x))^{5/2}}-\frac{a (A+9 B) \cos (e+f x)}{8 c f (c-c \sin (e+f x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 2.17522, size = 199, normalized size = 1.58 \[ -\frac{a (\sin (e+f x)-1) (\sin (e+f x)+1) \left (\frac{2 \sqrt{c} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) ((A+9 B) \sin (e+f x)+3 A-5 B)}{\left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^5}+\sqrt{2} (A-7 B) \sec (e+f x) \sqrt{-c (\sin (e+f x)+1)} \tan ^{-1}\left (\frac{\sqrt{-c (\sin (e+f x)+1)}}{\sqrt{2} \sqrt{c}}\right )\right )}{16 c^{5/2} f \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + a*Sin[e + f*x])*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(5/2),x]

[Out]

-(a*(-1 + Sin[e + f*x])*(1 + Sin[e + f*x])*(Sqrt[2]*(A - 7*B)*ArcTan[Sqrt[-(c*(1 + Sin[e + f*x]))]/(Sqrt[2]*Sq
rt[c])]*Sec[e + f*x]*Sqrt[-(c*(1 + Sin[e + f*x]))] + (2*Sqrt[c]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(3*A - 5
*B + (A + 9*B)*Sin[e + f*x]))/(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5))/(16*c^(5/2)*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 = 1.45, size = 268, normalized size = 2.1 \begin{align*}{\frac{a}{ \left ( -16+16\,\sin \left ( fx+e \right ) \right ) \cos \left ( fx+e \right ) f} \left ( -2\,\sin \left ( fx+e \right ) \sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c+c\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{c}}} \right ){c}^{2} \left ( A-7\,B \right ) -\sqrt{2}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{c+c\sin \left ( fx+e \right ) }{\frac{1}{\sqrt{c}}}} \right ){c}^{2} \left ( A-7\,B \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+2\,A\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c+c\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{c}}} \right ){c}^{2}-4\,A\sqrt{c+c\sin \left ( fx+e \right ) }{c}^{3/2}-2\,A \left ( c+c\sin \left ( fx+e \right ) \right ) ^{3/2}\sqrt{c}-14\,B\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c+c\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{c}}} \right ){c}^{2}+28\,B\sqrt{c+c\sin \left ( fx+e \right ) }{c}^{3/2}-18\,B \left ( c+c\sin \left ( fx+e \right ) \right ) ^{3/2}\sqrt{c} \right ) \sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }{c}^{-{\frac{9}{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))^(5/2),x)

[Out]

1/16*a*(-2*sin(f*x+e)*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c^2*(A-7*B)-2^(1/2)*arctanh(
1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c^2*(A-7*B)*cos(f*x+e)^2+2*A*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^
(1/2)*2^(1/2)/c^(1/2))*c^2-4*A*(c+c*sin(f*x+e))^(1/2)*c^(3/2)-2*A*(c+c*sin(f*x+e))^(3/2)*c^(1/2)-14*B*2^(1/2)*
arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c^2+28*B*(c+c*sin(f*x+e))^(1/2)*c^(3/2)-18*B*(c+c*sin(f*x+
e))^(3/2)*c^(1/2))*(c*(1+sin(f*x+e)))^(1/2)/c^(9/2)/(-1+sin(f*x+e))/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{5}{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))^(5/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)/(-c*sin(f*x + e) + c)^(5/2), x)

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Fricas [B]  time = 1.7798, size = 1035, normalized size = 8.21 \begin{align*} -\frac{\sqrt{2}{\left ({\left (A - 7 \, B\right )} a \cos \left (f x + e\right )^{3} + 3 \,{\left (A - 7 \, B\right )} a \cos \left (f x + e\right )^{2} - 2 \,{\left (A - 7 \, B\right )} a \cos \left (f x + e\right ) - 4 \,{\left (A - 7 \, B\right )} a -{\left ({\left (A - 7 \, B\right )} a \cos \left (f x + e\right )^{2} - 2 \,{\left (A - 7 \, B\right )} a \cos \left (f x + e\right ) - 4 \,{\left (A - 7 \, 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 ({\left (A + 9 \, B\right )} a \cos \left (f x + e\right )^{2} -{\left (3 \, A - 5 \, B\right )} a \cos \left (f x + e\right ) - 4 \,{\left (A + B\right )} a -{\left ({\left (A + 9 \, B\right )} a \cos \left (f x + e\right ) + 4 \,{\left (A + B\right )} a\right )} \sin \left (f x + e\right )\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{32 \,{\left (c^{3} f \cos \left (f x + e\right )^{3} + 3 \, c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f -{\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} 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))^(5/2),x, algorithm="fricas")

[Out]

-1/32*(sqrt(2)*((A - 7*B)*a*cos(f*x + e)^3 + 3*(A - 7*B)*a*cos(f*x + e)^2 - 2*(A - 7*B)*a*cos(f*x + e) - 4*(A
- 7*B)*a - ((A - 7*B)*a*cos(f*x + e)^2 - 2*(A - 7*B)*a*cos(f*x + e) - 4*(A - 7*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*((A + 9*B)*a*cos(f*x + e)^2 - (3*A - 5*B)*a*cos(f*x + e) - 4*(A + B)*a - ((A + 9*B)*a*cos(f
*x + e) + 4*(A + B)*a)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c))/(c^3*f*cos(f*x + e)^3 + 3*c^3*f*cos(f*x + e)^2
- 2*c^3*f*cos(f*x + e) - 4*c^3*f - (c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) - 4*c^3*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))**(5/2),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))^(5/2),x, algorithm="giac")

[Out]

sage2