3.4.5 \(\int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^{7/2}} \, dx\) [305]

Optimal. Leaf size=156 \[ \frac {a^2 \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{16 \sqrt {2} c^{7/2} f}+\frac {a^2 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{9/2}}-\frac {a^2 \cos (e+f x)}{4 c f (c-c \sin (e+f x))^{5/2}}+\frac {a^2 \cos (e+f x)}{16 c^2 f (c-c \sin (e+f x))^{3/2}} \]

[Out]

1/3*a^2*c*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(9/2)-1/4*a^2*cos(f*x+e)/c/f/(c-c*sin(f*x+e))^(5/2)+1/16*a^2*cos(f*x
+e)/c^2/f/(c-c*sin(f*x+e))^(3/2)+1/32*a^2*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 = 156, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {2815, 2759, 2729, 2728, 212} \begin {gather*} \frac {a^2 \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{16 \sqrt {2} c^{7/2} f}+\frac {a^2 \cos (e+f x)}{16 c^2 f (c-c \sin (e+f x))^{3/2}}+\frac {a^2 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{9/2}}-\frac {a^2 \cos (e+f x)}{4 c f (c-c \sin (e+f x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(16*Sqrt[2]*c^(7/2)*f) + (a^2*c*Cos[e
+ f*x]^3)/(3*f*(c - c*Sin[e + f*x])^(9/2)) - (a^2*Cos[e + f*x])/(4*c*f*(c - c*Sin[e + f*x])^(5/2)) + (a^2*Cos
[e + f*x])/(16*c^2*f*(c - c*Sin[e + f*x])^(3/2))

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 2759

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 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 {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^{7/2}} \, dx &=\left (a^2 c^2\right ) \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^{11/2}} \, dx\\ &=\frac {a^2 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{9/2}}-\frac {1}{2} a^2 \int \frac {\cos ^2(e+f x)}{(c-c \sin (e+f x))^{7/2}} \, dx\\ &=\frac {a^2 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{9/2}}-\frac {a^2 \cos (e+f x)}{4 c f (c-c \sin (e+f x))^{5/2}}+\frac {a^2 \int \frac {1}{(c-c \sin (e+f x))^{3/2}} \, dx}{8 c^2}\\ &=\frac {a^2 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{9/2}}-\frac {a^2 \cos (e+f x)}{4 c f (c-c \sin (e+f x))^{5/2}}+\frac {a^2 \cos (e+f x)}{16 c^2 f (c-c \sin (e+f x))^{3/2}}+\frac {a^2 \int \frac {1}{\sqrt {c-c \sin (e+f x)}} \, dx}{32 c^3}\\ &=\frac {a^2 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{9/2}}-\frac {a^2 \cos (e+f x)}{4 c f (c-c \sin (e+f x))^{5/2}}+\frac {a^2 \cos (e+f x)}{16 c^2 f (c-c \sin (e+f x))^{3/2}}-\frac {a^2 \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 )}{16 c^3 f}\\ &=\frac {a^2 \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{16 \sqrt {2} c^{7/2} f}+\frac {a^2 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{9/2}}-\frac {a^2 \cos (e+f x)}{4 c f (c-c \sin (e+f x))^{5/2}}+\frac {a^2 \cos (e+f x)}{16 c^2 f (c-c \sin (e+f x))^{3/2}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.61, size = 307, normalized size = 1.97 \begin {gather*} \frac {a^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (32 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-28 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+3 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^5-(3+3 i) \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 )^6+64 \sin \left (\frac {1}{2} (e+f x)\right )-56 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \sin \left (\frac {1}{2} (e+f x)\right )+6 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^2}{48 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 (c-c \sin (e+f x))^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(32*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]) - 28*(Cos[(e + f*x)/2] -
Sin[(e + f*x)/2])^3 + 3*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5 - (3 + 3*I)*(-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])^6 + 64*Sin[(e + f*x)/2] - 56*(Cos[(e + f*
x)/2] - Sin[(e + f*x)/2])^2*Sin[(e + f*x)/2] + 6*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4*Sin[(e + f*x)/2])*(1
+ Sin[e + f*x])^2)/(48*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4*(c - c*Sin[e + f*x])^(7/2))

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Maple [A]
time = 2.47, size = 245, normalized size = 1.57

method	result	size
default	\(-\frac {a^{2} \left (3 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (\sin ^{3}\left (f x +e \right )\right ) c^{4}-9 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (\sin ^{2}\left (f x +e \right )\right ) c^{4}-6 \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}} c^{\frac {3}{2}}-32 \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} c^{\frac {5}{2}}+24 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, c^{\frac {7}{2}}+9 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sin \left (f x +e \right ) c^{4}-3 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{4}\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}}{96 c^{\frac {15}{2}} \left (\sin \left (f x +e \right )-1\right )^{2} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(245\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96*a^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^4-9*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^4-6*(c*(1+sin(f*x+e)))^(5/2)*c^(3/2)-32*(c*(1+sin
(f*x+e)))^(3/2)*c^(5/2)+24*(c*(1+sin(f*x+e)))^(1/2)*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)*c^4-3*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*c^4)*(c*(1+sin(f*
x+e)))^(1/2)/c^(15/2)/(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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 474 vs. \(2 (141) = 282\).
time = 0.37, size = 474, normalized size = 3.04 \begin {gather*} \frac {3 \, \sqrt {2} {\left (a^{2} \cos \left (f x + e\right )^{4} - 3 \, a^{2} \cos \left (f x + e\right )^{3} - 8 \, a^{2} \cos \left (f x + e\right )^{2} + 4 \, a^{2} \cos \left (f x + e\right ) + 8 \, a^{2} + {\left (a^{2} \cos \left (f x + e\right )^{3} + 4 \, a^{2} \cos \left (f x + e\right )^{2} - 4 \, a^{2} \cos \left (f x + e\right ) - 8 \, a^{2}\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 (3 \, a^{2} \cos \left (f x + e\right )^{3} + 25 \, a^{2} \cos \left (f x + e\right )^{2} - 10 \, a^{2} \cos \left (f x + e\right ) - 32 \, a^{2} + {\left (3 \, a^{2} \cos \left (f x + e\right )^{2} - 22 \, a^{2} \cos \left (f x + e\right ) - 32 \, a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{192 \, {\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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*(3*sqrt(2)*(a^2*cos(f*x + e)^4 - 3*a^2*cos(f*x + e)^3 - 8*a^2*cos(f*x + e)^2 + 4*a^2*cos(f*x + e) + 8*a^
2 + (a^2*cos(f*x + e)^3 + 4*a^2*cos(f*x + e)^2 - 4*a^2*cos(f*x + e) - 8*a^2)*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*(3*a^2*cos(f*x + e)^3 + 25*a^2*cos(f*x + e)^2 - 10*a^2*cos(f*x + e) - 32*a^2 + (3*a^2*cos(f*x + e)^
2 - 22*a^2*cos(f*x + e) - 32*a^2)*sin(f*x + e))*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)] 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((a+a*sin(f*x+e))**2/(c-c*sin(f*x+e))**(7/2),x)

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 405 vs. \(2 (141) = 282\).
time = 0.64, size = 405, normalized size = 2.60 \begin {gather*} \frac {\frac {12 \, \sqrt {2} a^{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 )}{c^{\frac {7}{2}} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {\sqrt {2} {\left (a^{2} \sqrt {c} + \frac {3 \, a^{2} \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 {3 \, a^{2} \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}} - \frac {22 \, a^{2} \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{3}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{3}}\right )} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{3}}{c^{4} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {\sqrt {2} {\left (\frac {3 \, a^{2} c^{\frac {17}{2}} {\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 {3 \, a^{2} c^{\frac {17}{2}} {\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}} - \frac {a^{2} c^{\frac {17}{2}} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{3}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{3}}\right )}}{c^{12} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{768 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/768*(12*sqrt(2)*a^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))/(c^(7/2)
*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) + sqrt(2)*(a^2*sqrt(c) + 3*a^2*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) - 3*a^2*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 - 22*a^2*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^3/(cos(-1/4*pi + 1/2*f*x + 1/2*
e) + 1)^3)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^3/(c^4*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^3*sgn(sin(-1/4*pi
+ 1/2*f*x + 1/2*e))) + sqrt(2)*(3*a^2*c^(17/2)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1
/2*e) + 1) - 3*a^2*c^(17/2)*(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^(17/2)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^3/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^3)/(c^12*sgn(sin(-1/4*p
i + 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 {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^2}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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