3.4.30 \(\int \frac {1}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{3/2}} \, dx\) [330]

Optimal. Leaf size=155 \[ \frac {5 \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{8 \sqrt {2} a^2 c^{3/2} f}+\frac {5 \cos (e+f x)}{8 a^2 f (c-c \sin (e+f x))^{3/2}}-\frac {5 \sec (e+f x)}{6 a^2 c f \sqrt {c-c \sin (e+f x)}}-\frac {\sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{3 a^2 c^2 f} \]

[Out]

5/8*cos(f*x+e)/a^2/f/(c-c*sin(f*x+e))^(3/2)+5/16*arctanh(1/2*cos(f*x+e)*c^(1/2)*2^(1/2)/(c-c*sin(f*x+e))^(1/2)
)/a^2/c^(3/2)/f*2^(1/2)-5/6*sec(f*x+e)/a^2/c/f/(c-c*sin(f*x+e))^(1/2)-1/3*sec(f*x+e)^3*(c-c*sin(f*x+e))^(1/2)/
a^2/c^2/f

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Rubi [A]
time = 0.17, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {2815, 2754, 2766, 2729, 2728, 212} \begin {gather*} \frac {5 \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{8 \sqrt {2} a^2 c^{3/2} f}-\frac {\sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{3 a^2 c^2 f}+\frac {5 \cos (e+f x)}{8 a^2 f (c-c \sin (e+f x))^{3/2}}-\frac {5 \sec (e+f x)}{6 a^2 c f \sqrt {c-c \sin (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(8*Sqrt[2]*a^2*c^(3/2)*f) + (5*Cos[e +
f*x])/(8*a^2*f*(c - c*Sin[e + f*x])^(3/2)) - (5*Sec[e + f*x])/(6*a^2*c*f*Sqrt[c - c*Sin[e + f*x]]) - (Sec[e +
f*x]^3*Sqrt[c - c*Sin[e + f*x]])/(3*a^2*c^2*f)

Rule 212

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

Rule 2728

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 2729

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 2754

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 2766

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))), In
t[(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 2815

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))^2 (c-c \sin (e+f x))^{3/2}} \, dx &=\frac {\int \sec ^4(e+f x) \sqrt {c-c \sin (e+f x)} \, dx}{a^2 c^2}\\ &=-\frac {\sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{3 a^2 c^2 f}+\frac {5 \int \frac {\sec ^2(e+f x)}{\sqrt {c-c \sin (e+f x)}} \, dx}{6 a^2 c}\\ &=-\frac {5 \sec (e+f x)}{6 a^2 c f \sqrt {c-c \sin (e+f x)}}-\frac {\sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{3 a^2 c^2 f}+\frac {5 \int \frac {1}{(c-c \sin (e+f x))^{3/2}} \, dx}{4 a^2}\\ &=\frac {5 \cos (e+f x)}{8 a^2 f (c-c \sin (e+f x))^{3/2}}-\frac {5 \sec (e+f x)}{6 a^2 c f \sqrt {c-c \sin (e+f x)}}-\frac {\sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{3 a^2 c^2 f}+\frac {5 \int \frac {1}{\sqrt {c-c \sin (e+f x)}} \, dx}{16 a^2 c}\\ &=\frac {5 \cos (e+f x)}{8 a^2 f (c-c \sin (e+f x))^{3/2}}-\frac {5 \sec (e+f x)}{6 a^2 c f \sqrt {c-c \sin (e+f x)}}-\frac {\sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{3 a^2 c^2 f}-\frac {5 \text {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 a^2 c f}\\ &=\frac {5 \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{8 \sqrt {2} a^2 c^{3/2} f}+\frac {5 \cos (e+f x)}{8 a^2 f (c-c \sin (e+f x))^{3/2}}-\frac {5 \sec (e+f x)}{6 a^2 c f \sqrt {c-c \sin (e+f x)}}-\frac {\sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{3 a^2 c^2 f}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.49, size = 164, normalized size = 1.06 \begin {gather*} \frac {\left (\frac {1}{96}+\frac {i}{96}\right ) \cos (e+f x) \left (60 \sqrt [4]{-1} \tan ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+(1-i) (11+15 \cos (2 (e+f x))-20 \sin (e+f x))\right )}{a^2 c f (-1+\sin (e+f x)) (1+\sin (e+f x))^2 \sqrt {c-c \sin (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((1/96 + I/96)*Cos[e + f*x]*(60*(-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])^3 + (1 - I)*(11 + 15*Cos[2*(e + f*x)] - 20*Si
n[e + f*x])))/(a^2*c*f*(-1 + Sin[e + f*x])*(1 + Sin[e + f*x])^2*Sqrt[c - c*Sin[e + f*x]])

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Maple [A]
time = 2.19, size = 157, normalized size = 1.01

method	result	size
default	\(-\frac {15 \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sin \left (f x +e \right ) c -15 \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c -20 c^{\frac {5}{2}} \sin \left (f x +e \right )-30 c^{\frac {5}{2}} \left (\sin ^{2}\left (f x +e \right )\right )+26 c^{\frac {5}{2}}}{48 c^{\frac {7}{2}} a^{2} \left (1+\sin \left (f x +e \right )\right ) \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(157\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48/c^(7/2)/a^2*(15*(c*(1+sin(f*x+e)))^(3/2)*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*s
in(f*x+e)*c-15*(c*(1+sin(f*x+e)))^(3/2)*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*c-20*c^(
5/2)*sin(f*x+e)-30*c^(5/2)*sin(f*x+e)^2+26*c^(5/2))/(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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.38, size = 202, normalized size = 1.30 \begin {gather*} \frac {15 \, \sqrt {2} \sqrt {c} \cos \left (f x + e\right )^{3} \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 (15 \, \cos \left (f x + e\right )^{2} - 10 \, \sin \left (f x + e\right ) - 2\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{96 \, a^{2} c^{2} f \cos \left (f x + e\right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(15*sqrt(2)*sqrt(c)*cos(f*x + e)^3*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*(15*cos(f*x + e)^2 - 10*sin(f*x + e) - 2)*sqrt(-
c*sin(f*x + e) + c))/(a^2*c^2*f*cos(f*x + e)^3)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 368 vs. \(2 (140) = 280\).
time = 0.54, size = 368, normalized size = 2.37 \begin {gather*} -\frac {\frac {3 \, \sqrt {2} {\left (\frac {10 \, {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} - 1\right )} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}}{a^{2} c^{\frac {3}{2}} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {30 \, \sqrt {2} \log \left (-\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1}\right )}{a^{2} c^{\frac {3}{2}} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {3 \, \sqrt {2} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}}{a^{2} c^{\frac {3}{2}} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {16 \, \sqrt {2} {\left (7 \, \sqrt {c} + \frac {12 \, \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} + \frac {9 \, \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}}\right )}}{a^{2} c^{2} {\left (\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} + 1\right )}^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{192 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/192*(3*sqrt(2)*(10*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) - 1)*(cos(-1/4
*pi + 1/2*f*x + 1/2*e) + 1)/(a^2*c^(3/2)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*
e))) - 30*sqrt(2)*log(-(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1))/(a^2*c^(3/2)
*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) + 3*sqrt(2)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(a^2*c^(3/2)*(cos(-1/4*
pi + 1/2*f*x + 1/2*e) + 1)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) - 16*sqrt(2)*(7*sqrt(c) + 12*sqrt(c)*(cos(-1/4
*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) + 9*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) -
1)^2/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2)/(a^2*c^2*((cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/
2*f*x + 1/2*e) + 1) + 1)^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))))/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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