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

Optimal. Leaf size=145 \[ -\frac{a \cos (e+f x)}{32 c^2 f (c-c \sin (e+f x))^{3/2}}-\frac{a \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{32 \sqrt{2} c^{7/2} f}-\frac{a \cos (e+f x)}{24 c f (c-c \sin (e+f x))^{5/2}}+\frac{a \cos (e+f x)}{3 f (c-c \sin (e+f x))^{7/2}} \]

[Out]

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

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Rubi [A]  time = 0.193465, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {2736, 2680, 2650, 2649, 206} \[ -\frac{a \cos (e+f x)}{32 c^2 f (c-c \sin (e+f x))^{3/2}}-\frac{a \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{32 \sqrt{2} c^{7/2} f}-\frac{a \cos (e+f x)}{24 c f (c-c \sin (e+f x))^{5/2}}+\frac{a \cos (e+f x)}{3 f (c-c \sin (e+f x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2736

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

Rule 2680

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(2*g*(
g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(2*m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(2*m +
p + 1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && Eq
Q[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] && NeQ[2*m + p + 1, 0] &&  !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*
p]

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

Mathematica [A]  time = 1.1603, size = 189, normalized size = 1.3 \[ \frac{a \left (2 \sqrt{c} (131 \sin (e+f x)+3 (\sin (3 (e+f x))+38)-14 \cos (2 (e+f x)))+12 \sqrt{2} \sqrt{-c (\sin (e+f x)+1)} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^6 \tan ^{-1}\left (\frac{\sqrt{-c (\sin (e+f x)+1)}}{\sqrt{2} \sqrt{c}}\right )\right )}{768 c^{7/2} f \sqrt{c-c \sin (e+f x)} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^5 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(12*Sqrt[2]*ArcTan[Sqrt[-(c*(1 + Sin[e + f*x]))]/(Sqrt[2]*Sqrt[c])]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^6
*Sqrt[-(c*(1 + Sin[e + f*x]))] + 2*Sqrt[c]*(-14*Cos[2*(e + f*x)] + 131*Sin[e + f*x] + 3*(38 + Sin[3*(e + f*x)]
))))/(768*c^(7/2)*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c - c*S
in[e + f*x]])

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Maple [A]  time = 0.719, size = 243, normalized size = 1.7 \begin{align*}{\frac{a}{192\, \left ( -1+\sin \left ( fx+e \right ) \right ) ^{2}\cos \left ( fx+e \right ) f} \left ( 24\,\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }{c}^{9/2}+32\, \left ( c \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{3/2}{c}^{7/2}-6\, \left ( c \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{5/2}{c}^{5/2}+3\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }\sqrt{2}}{\sqrt{c}}} \right ) \left ( \sin \left ( fx+e \right ) \right ) ^{3}{c}^{5}-9\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }\sqrt{2}}{\sqrt{c}}} \right ) \left ( \sin \left ( fx+e \right ) \right ) ^{2}{c}^{5}+9\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }\sqrt{2}}{\sqrt{c}}} \right ) \sin \left ( fx+e \right ){c}^{5}-3\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }\sqrt{2}}{\sqrt{c}}} \right ){c}^{5} \right ) \sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }{c}^{-{\frac{17}{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))/(c-c*sin(f*x+e))^(7/2),x)

[Out]

1/192*a*(24*(c*(1+sin(f*x+e)))^(1/2)*c^(9/2)+32*(c*(1+sin(f*x+e)))^(3/2)*c^(7/2)-6*(c*(1+sin(f*x+e)))^(5/2)*c^
(5/2)+3*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*sin(f*x+e)^3*c^5-9*2^(1/2)*arctanh(1/2*(
c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*sin(f*x+e)^2*c^5+9*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/
2)/c^(1/2))*sin(f*x+e)*c^5-3*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*c^5)*(c*(1+sin(f*x+
e)))^(1/2)/c^(17/2)/(-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{a \sin \left (f x + e\right ) + a}{{\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))/(c-c*sin(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.19135, size = 1073, normalized size = 7.4 \begin{align*} \frac{3 \, \sqrt{2}{\left (a \cos \left (f x + e\right )^{4} - 3 \, a \cos \left (f x + e\right )^{3} - 8 \, a \cos \left (f x + e\right )^{2} + 4 \, a \cos \left (f x + e\right ) +{\left (a \cos \left (f x + e\right )^{3} + 4 \, a \cos \left (f x + e\right )^{2} - 4 \, a \cos \left (f x + e\right ) - 8 \, a\right )} \sin \left (f x + e\right ) + 8 \, a\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 \, a \cos \left (f x + e\right )^{3} - 7 \, a \cos \left (f x + e\right )^{2} + 22 \, a \cos \left (f x + e\right ) +{\left (3 \, a \cos \left (f x + e\right )^{2} + 10 \, a \cos \left (f x + e\right ) + 32 \, a\right )} \sin \left (f x + e\right ) + 32 \, a\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))/(c-c*sin(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

1/384*(3*sqrt(2)*(a*cos(f*x + e)^4 - 3*a*cos(f*x + e)^3 - 8*a*cos(f*x + e)^2 + 4*a*cos(f*x + e) + (a*cos(f*x +
e)^3 + 4*a*cos(f*x + e)^2 - 4*a*cos(f*x + e) - 8*a)*sin(f*x + e) + 8*a)*sqrt(c)*log(-(c*cos(f*x + e)^2 - 2*sq
rt(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*co
s(f*x + e)^3 - 7*a*cos(f*x + e)^2 + 22*a*cos(f*x + e) + (3*a*cos(f*x + e)^2 + 10*a*cos(f*x + e) + 32*a)*sin(f*
x + e) + 32*a)*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))/(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))/(c-c*sin(f*x+e))^(7/2),x, algorithm="giac")

[Out]

sage2