3.338 \(\int \frac {1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=191 \[ \frac {7 \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{16 \sqrt {2} a^3 c^{3/2} f}-\frac {\sec ^5(e+f x) (c-c \sin (e+f x))^{3/2}}{5 a^3 c^3 f}-\frac {7 \sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{30 a^3 c^2 f}+\frac {7 \cos (e+f x)}{16 a^3 f (c-c \sin (e+f x))^{3/2}}-\frac {7 \sec (e+f x)}{12 a^3 c f \sqrt {c-c \sin (e+f x)}} \]

[Out]

7/16*cos(f*x+e)/a^3/f/(c-c*sin(f*x+e))^(3/2)-1/5*sec(f*x+e)^5*(c-c*sin(f*x+e))^(3/2)/a^3/c^3/f+7/32*arctanh(1/
2*cos(f*x+e)*c^(1/2)*2^(1/2)/(c-c*sin(f*x+e))^(1/2))/a^3/c^(3/2)/f*2^(1/2)-7/12*sec(f*x+e)/a^3/c/f/(c-c*sin(f*
x+e))^(1/2)-7/30*sec(f*x+e)^3*(c-c*sin(f*x+e))^(1/2)/a^3/c^2/f

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Rubi [A]  time = 0.33, antiderivative size = 191, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {2736, 2675, 2687, 2650, 2649, 206} \[ -\frac {\sec ^5(e+f x) (c-c \sin (e+f x))^{3/2}}{5 a^3 c^3 f}-\frac {7 \sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{30 a^3 c^2 f}+\frac {7 \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{16 \sqrt {2} a^3 c^{3/2} f}+\frac {7 \cos (e+f x)}{16 a^3 f (c-c \sin (e+f x))^{3/2}}-\frac {7 \sec (e+f x)}{12 a^3 c f \sqrt {c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^(3/2)),x]

[Out]

(7*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(16*Sqrt[2]*a^3*c^(3/2)*f) + (7*Cos[e +
f*x])/(16*a^3*f*(c - c*Sin[e + f*x])^(3/2)) - (7*Sec[e + f*x])/(12*a^3*c*f*Sqrt[c - c*Sin[e + f*x]]) - (7*Sec
[e + f*x]^3*Sqrt[c - c*Sin[e + f*x]])/(30*a^3*c^2*f) - (Sec[e + f*x]^5*(c - c*Sin[e + f*x])^(3/2))/(5*a^3*c^3*
f)

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 2675

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

Rule 2687

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> -Simp[(b*(g*
Cos[e + f*x])^(p + 1))/(a*f*g*(p + 1)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(a*(2*p + 1))/(2*g^2*(p + 1)), Int[
(g*Cos[e + f*x])^(p + 2)/(a + b*Sin[e + f*x])^(3/2), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0]
&& LtQ[p, -1] && IntegerQ[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 {1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{3/2}} \, dx &=\frac {\int \sec ^6(e+f x) (c-c \sin (e+f x))^{3/2} \, dx}{a^3 c^3}\\ &=-\frac {\sec ^5(e+f x) (c-c \sin (e+f x))^{3/2}}{5 a^3 c^3 f}+\frac {7 \int \sec ^4(e+f x) \sqrt {c-c \sin (e+f x)} \, dx}{10 a^3 c^2}\\ &=-\frac {7 \sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{30 a^3 c^2 f}-\frac {\sec ^5(e+f x) (c-c \sin (e+f x))^{3/2}}{5 a^3 c^3 f}+\frac {7 \int \frac {\sec ^2(e+f x)}{\sqrt {c-c \sin (e+f x)}} \, dx}{12 a^3 c}\\ &=-\frac {7 \sec (e+f x)}{12 a^3 c f \sqrt {c-c \sin (e+f x)}}-\frac {7 \sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{30 a^3 c^2 f}-\frac {\sec ^5(e+f x) (c-c \sin (e+f x))^{3/2}}{5 a^3 c^3 f}+\frac {7 \int \frac {1}{(c-c \sin (e+f x))^{3/2}} \, dx}{8 a^3}\\ &=\frac {7 \cos (e+f x)}{16 a^3 f (c-c \sin (e+f x))^{3/2}}-\frac {7 \sec (e+f x)}{12 a^3 c f \sqrt {c-c \sin (e+f x)}}-\frac {7 \sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{30 a^3 c^2 f}-\frac {\sec ^5(e+f x) (c-c \sin (e+f x))^{3/2}}{5 a^3 c^3 f}+\frac {7 \int \frac {1}{\sqrt {c-c \sin (e+f x)}} \, dx}{32 a^3 c}\\ &=\frac {7 \cos (e+f x)}{16 a^3 f (c-c \sin (e+f x))^{3/2}}-\frac {7 \sec (e+f x)}{12 a^3 c f \sqrt {c-c \sin (e+f x)}}-\frac {7 \sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{30 a^3 c^2 f}-\frac {\sec ^5(e+f x) (c-c \sin (e+f x))^{3/2}}{5 a^3 c^3 f}-\frac {7 \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 )}{16 a^3 c f}\\ &=\frac {7 \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{16 \sqrt {2} a^3 c^{3/2} f}+\frac {7 \cos (e+f x)}{16 a^3 f (c-c \sin (e+f x))^{3/2}}-\frac {7 \sec (e+f x)}{12 a^3 c f \sqrt {c-c \sin (e+f x)}}-\frac {7 \sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{30 a^3 c^2 f}-\frac {\sec ^5(e+f x) (c-c \sin (e+f x))^{3/2}}{5 a^3 c^3 f}\\ \end {align*}

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Mathematica [C]  time = 1.19, size = 174, normalized size = 0.91 \[ \frac {\left (\frac {1}{1920}+\frac {i}{1920}\right ) \cos (e+f x) \left ((1-i) (-231 \sin (e+f x)+105 \sin (3 (e+f x))+350 \cos (2 (e+f x))+206)+840 \sqrt [4]{-1} \tan ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (\tan \left (\frac {1}{4} (e+f x)\right )+1\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^5\right )}{a^3 c f (\sin (e+f x)-1) (\sin (e+f x)+1)^3 \sqrt {c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^(3/2)),x]

[Out]

((1/1920 + I/1920)*Cos[e + f*x]*(840*(-1)^(1/4)*ArcTan[(1/2 + I/2)*(-1)^(1/4)*(1 + Tan[(e + f*x)/4])]*(Cos[(e
+ f*x)/2] - Sin[(e + f*x)/2])^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5 + (1 - I)*(206 + 350*Cos[2*(e + f*x)]
- 231*Sin[e + f*x] + 105*Sin[3*(e + f*x)])))/(a^3*c*f*(-1 + Sin[e + f*x])*(1 + Sin[e + f*x])^3*Sqrt[c - c*Sin[
e + f*x]])

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fricas [A]  time = 0.47, size = 238, normalized size = 1.25 \[ \frac {105 \, \sqrt {2} {\left (\cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) + \cos \left (f x + e\right )^{3}\right )} \sqrt {c} \log \left (-\frac {c \cos \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} \sqrt {c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) + {\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \, {\left (175 \, \cos \left (f x + e\right )^{2} + 21 \, {\left (5 \, \cos \left (f x + e\right )^{2} - 4\right )} \sin \left (f x + e\right ) - 36\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{960 \, {\left (a^{3} c^{2} f \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) + a^{3} c^{2} f \cos \left (f x + e\right )^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

1/960*(105*sqrt(2)*(cos(f*x + e)^3*sin(f*x + e) + cos(f*x + e)^3)*sqrt(c)*log(-(c*cos(f*x + e)^2 + 2*sqrt(2)*s
qrt(-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*(175*cos(f*x +
e)^2 + 21*(5*cos(f*x + e)^2 - 4)*sin(f*x + e) - 36)*sqrt(-c*sin(f*x + e) + c))/(a^3*c^2*f*cos(f*x + e)^3*sin(
f*x + e) + a^3*c^2*f*cos(f*x + e)^3)

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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(1/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(3/2),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: (4*pi/t_nostep/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*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to chec
k sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Warning, integration of abs or sign assumes constant sign by inter
vals (correct if the argument is real):Check [abs(sin((f*t_nostep+exp(1))/2-pi/4))]Unable to check sign: (8*pi
/t_nostep/2)>(-8*pi/t_nostep/2)Discontinuities at zeroes of sin((f*t_nostep+exp(1))/2-pi/4) were not checkedUn
able to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_noste
p/2)Unable 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)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/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 constant sign by intervals (correct if the argument is r
eal):Check [abs(t_nostep-1)]Not invertible Error: Bad Argument Value

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maple [A]  time = 1.25, size = 170, normalized size = 0.89 \[ -\frac {105 \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sin \left (f x +e \right ) c -105 \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c +42 c^{\frac {7}{2}} \sin \left (f x +e \right )-350 c^{\frac {7}{2}} \left (\sin ^{2}\left (f x +e \right )\right )-210 c^{\frac {7}{2}} \left (\sin ^{3}\left (f x +e \right )\right )+278 c^{\frac {7}{2}}}{480 c^{\frac {9}{2}} a^{3} \left (1+\sin \left (f x +e \right )\right )^{2} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(3/2),x)

[Out]

-1/480/c^(9/2)/a^3*(105*(c*(1+sin(f*x+e)))^(5/2)*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-105*(c*(1+sin(f*x+e)))^(5/2)*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*c+42*
c^(7/2)*sin(f*x+e)-350*c^(7/2)*sin(f*x+e)^2-210*c^(7/2)*sin(f*x+e)^3+278*c^(7/2))/(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(-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(1/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^3\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + a*sin(e + f*x))^3*(c - c*sin(e + f*x))^(3/2)),x)

[Out]

int(1/((a + a*sin(e + f*x))^3*(c - c*sin(e + f*x))^(3/2)), 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(1/(a+a*sin(f*x+e))**3/(c-c*sin(f*x+e))**(3/2),x)

[Out]

Timed out

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