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

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

[Out]

1/3*a*cos(f*x+e)/f/(c-c*sin(f*x+e))^(7/2)-1/24*a*cos(f*x+e)/c/f/(c-c*sin(f*x+e))^(5/2)-1/32*a*cos(f*x+e)/c^2/f
/(c-c*sin(f*x+e))^(3/2)-1/64*a*arctanh(1/2*cos(f*x+e)*c^(1/2)*2^(1/2)/(c-c*sin(f*x+e))^(1/2))/c^(7/2)/f*2^(1/2
)

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Rubi [A]  time = 0.19, antiderivative size = 145, normalized size of antiderivative = 1.00, 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 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])

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 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 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 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]))

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*}

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Mathematica [A]  time = 1.16, size = 189, normalized size = 1.30 \[ \frac {a \left (2 \sqrt {c} (131 \sin (e+f x)+3 (\sin (3 (e+f x))+38)-14 \cos (2 (e+f x)))+12 \sqrt {2} \sqrt {-c (\sin (e+f x)+1)} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^6 \tan ^{-1}\left (\frac {\sqrt {-c (\sin (e+f x)+1)}}{\sqrt {2} \sqrt {c}}\right )\right )}{768 c^{7/2} f \sqrt {c-c \sin (e+f x)} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^5 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )} \]

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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fricas [B]  time = 0.46, size = 408, normalized size = 2.81 \[ \frac {3 \, \sqrt {2} {\left (a \cos \left (f x + e\right )^{4} - 3 \, a \cos \left (f x + e\right )^{3} - 8 \, a \cos \left (f x + e\right )^{2} + 4 \, a \cos \left (f x + e\right ) + {\left (a \cos \left (f x + e\right )^{3} + 4 \, a \cos \left (f x + e\right )^{2} - 4 \, a \cos \left (f x + e\right ) - 8 \, a\right )} \sin \left (f x + e\right ) + 8 \, a\right )} \sqrt {c} \log \left (-\frac {c \cos \left (f x + e\right )^{2} - 2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} \sqrt {c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) + {\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) + 4 \, {\left (3 \, a \cos \left (f x + e\right )^{3} - 7 \, a \cos \left (f x + e\right )^{2} + 22 \, a \cos \left (f x + e\right ) + {\left (3 \, a \cos \left (f x + e\right )^{2} + 10 \, a \cos \left (f x + e\right ) + 32 \, a\right )} \sin \left (f x + e\right ) + 32 \, a\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{384 \, {\left (c^{4} f \cos \left (f x + e\right )^{4} - 3 \, c^{4} f \cos \left (f x + e\right )^{3} - 8 \, c^{4} f \cos \left (f x + e\right )^{2} + 4 \, c^{4} f \cos \left (f x + e\right ) + 8 \, c^{4} f + {\left (c^{4} f \cos \left (f x + e\right )^{3} + 4 \, c^{4} f \cos \left (f x + e\right )^{2} - 4 \, c^{4} f \cos \left (f x + e\right ) - 8 \, c^{4} f\right )} \sin \left (f x + e\right )\right )}} \]

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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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((a+a*sin(f*x+e))/(c-c*sin(f*x+e))^(7/2),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: (4*p
i/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unab
le to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*p
i/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unab
le to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*p
i/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign
: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Warning, integration of abs or sign assumes constant sign by intervals (
correct if the argument is real):Check [abs(sin((f*t_nostep+exp(1))/2-pi/4))]Unable to check sign: (8*pi/t_nos
tep/2)>(-8*pi/t_nostep/2)Discontinuities at zeroes of sin((f*t_nostep+exp(1))/2-pi/4) were not checkedUnable t
o check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Un
able to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to
check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/t_no
step/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (2*p
i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Warning, integration of abs or sign assumes cons
tant sign by intervals (correct if the argument is real):Check [abs(t_nostep-1)]Evaluation time: 0.69Not inver
tible Error: Bad Argument Value

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maple [A]  time = 1.06, size = 243, normalized size = 1.68 \[ -\frac {a \left (6 \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}} c^{\frac {5}{2}}-3 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (\sin ^{3}\left (f x +e \right )\right ) c^{5}-32 \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} c^{\frac {7}{2}}+9 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (\sin ^{2}\left (f x +e \right )\right ) c^{5}-24 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, c^{\frac {9}{2}}-9 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sin \left (f x +e \right ) c^{5}+3 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{5}\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}}{192 c^{\frac {17}{2}} \left (\sin \left (f x +e \right )-1\right )^{2} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f} \]

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/c^(17/2)*a*(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^5-32*(c*(1+sin(f*x+e)))^(3/2)*c^(7/2)+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^5-24*(c*(1+sin(f*x+e)))^(1/2)*c^(9/2)-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)/(sin(f*x+e)-1)^2/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a \sin \left (f x + e\right ) + a}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {7}{2}}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+a\,\sin \left (e+f\,x\right )}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [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((a+a*sin(f*x+e))/(c-c*sin(f*x+e))**(7/2),x)

[Out]

Timed out

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