3.2.66 \(\int \frac {(a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{3/2}} \, dx\) [166]

Optimal. Leaf size=271 \[ \frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{2 f (c-c \sin (e+f x))^{3/2}}+\frac {4 a^4 (3 A+5 B) \cos (e+f x) \log (1-\sin (e+f x))}{c f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}+\frac {2 a^3 (3 A+5 B) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c f \sqrt {c-c \sin (e+f x)}}+\frac {a^2 (3 A+5 B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c f \sqrt {c-c \sin (e+f x)}}+\frac {a (3 A+5 B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{6 c f \sqrt {c-c \sin (e+f x)}} \]

[Out]

1/2*(A+B)*cos(f*x+e)*(a+a*sin(f*x+e))^(7/2)/f/(c-c*sin(f*x+e))^(3/2)+1/2*a^2*(3*A+5*B)*cos(f*x+e)*(a+a*sin(f*x
+e))^(3/2)/c/f/(c-c*sin(f*x+e))^(1/2)+1/6*a*(3*A+5*B)*cos(f*x+e)*(a+a*sin(f*x+e))^(5/2)/c/f/(c-c*sin(f*x+e))^(
1/2)+4*a^4*(3*A+5*B)*cos(f*x+e)*ln(1-sin(f*x+e))/c/f/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(1/2)+2*a^3*(3*A+
5*B)*cos(f*x+e)*(a+a*sin(f*x+e))^(1/2)/c/f/(c-c*sin(f*x+e))^(1/2)

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Rubi [A]
time = 0.41, antiderivative size = 271, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3051, 2819, 2816, 2746, 31} \begin {gather*} \frac {4 a^4 (3 A+5 B) \cos (e+f x) \log (1-\sin (e+f x))}{c f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {2 a^3 (3 A+5 B) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{c f \sqrt {c-c \sin (e+f x)}}+\frac {a^2 (3 A+5 B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 c f \sqrt {c-c \sin (e+f x)}}+\frac {a (3 A+5 B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{6 c f \sqrt {c-c \sin (e+f x)}}+\frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{2 f (c-c \sin (e+f x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + a*Sin[e + f*x])^(7/2)*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(3/2),x]

[Out]

((A + B)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2))/(2*f*(c - c*Sin[e + f*x])^(3/2)) + (4*a^4*(3*A + 5*B)*Cos[e
+ f*x]*Log[1 - Sin[e + f*x]])/(c*f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]]) + (2*a^3*(3*A + 5*B)*Cos
[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(c*f*Sqrt[c - c*Sin[e + f*x]]) + (a^2*(3*A + 5*B)*Cos[e + f*x]*(a + a*Sin[
e + f*x])^(3/2))/(2*c*f*Sqrt[c - c*Sin[e + f*x]]) + (a*(3*A + 5*B)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(6
*c*f*Sqrt[c - c*Sin[e + f*x]])

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 2746

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rule 2816

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[a
*c*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])), Int[Cos[e + f*x]/(c + d*Sin[e + f*x]),
x], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0]

Rule 2819

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[(-b)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^n/(f*(m + n))), x] + Dist[a*((2*m - 1)/(
m + n)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[m - 1/2, 0] &&  !LtQ[n, -1] &&  !(IGtQ[n - 1/2, 0] && LtQ[n, m
]) &&  !(ILtQ[m + n, 0] && GtQ[2*m + n + 1, 0])

Rule 3051

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x]
)^n/(a*f*(2*m + 1))), x] + Dist[(a*B*(m - n) + A*b*(m + n + 1))/(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2
- b^2, 0] && (LtQ[m, -2^(-1)] || (ILtQ[m + n, 0] &&  !SumSimplerQ[n, 1])) && NeQ[2*m + 1, 0]

Rubi steps

\begin {align*} \int \frac {(a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{3/2}} \, dx &=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{2 f (c-c \sin (e+f x))^{3/2}}-\frac {(3 A+5 B) \int \frac {(a+a \sin (e+f x))^{7/2}}{\sqrt {c-c \sin (e+f x)}} \, dx}{2 c}\\ &=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{2 f (c-c \sin (e+f x))^{3/2}}+\frac {a (3 A+5 B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{6 c f \sqrt {c-c \sin (e+f x)}}-\frac {(a (3 A+5 B)) \int \frac {(a+a \sin (e+f x))^{5/2}}{\sqrt {c-c \sin (e+f x)}} \, dx}{c}\\ &=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{2 f (c-c \sin (e+f x))^{3/2}}+\frac {a^2 (3 A+5 B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c f \sqrt {c-c \sin (e+f x)}}+\frac {a (3 A+5 B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{6 c f \sqrt {c-c \sin (e+f x)}}-\frac {\left (2 a^2 (3 A+5 B)\right ) \int \frac {(a+a \sin (e+f x))^{3/2}}{\sqrt {c-c \sin (e+f x)}} \, dx}{c}\\ &=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{2 f (c-c \sin (e+f x))^{3/2}}+\frac {2 a^3 (3 A+5 B) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c f \sqrt {c-c \sin (e+f x)}}+\frac {a^2 (3 A+5 B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c f \sqrt {c-c \sin (e+f x)}}+\frac {a (3 A+5 B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{6 c f \sqrt {c-c \sin (e+f x)}}-\frac {\left (4 a^3 (3 A+5 B)\right ) \int \frac {\sqrt {a+a \sin (e+f x)}}{\sqrt {c-c \sin (e+f x)}} \, dx}{c}\\ &=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{2 f (c-c \sin (e+f x))^{3/2}}+\frac {2 a^3 (3 A+5 B) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c f \sqrt {c-c \sin (e+f x)}}+\frac {a^2 (3 A+5 B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c f \sqrt {c-c \sin (e+f x)}}+\frac {a (3 A+5 B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{6 c f \sqrt {c-c \sin (e+f x)}}-\frac {\left (4 a^4 (3 A+5 B) \cos (e+f x)\right ) \int \frac {\cos (e+f x)}{c-c \sin (e+f x)} \, dx}{\sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ &=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{2 f (c-c \sin (e+f x))^{3/2}}+\frac {2 a^3 (3 A+5 B) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c f \sqrt {c-c \sin (e+f x)}}+\frac {a^2 (3 A+5 B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c f \sqrt {c-c \sin (e+f x)}}+\frac {a (3 A+5 B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{6 c f \sqrt {c-c \sin (e+f x)}}+\frac {\left (4 a^4 (3 A+5 B) \cos (e+f x)\right ) \text {Subst}\left (\int \frac {1}{c+x} \, dx,x,-c \sin (e+f x)\right )}{c f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ &=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{2 f (c-c \sin (e+f x))^{3/2}}+\frac {4 a^4 (3 A+5 B) \cos (e+f x) \log (1-\sin (e+f x))}{c f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}+\frac {2 a^3 (3 A+5 B) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c f \sqrt {c-c \sin (e+f x)}}+\frac {a^2 (3 A+5 B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c f \sqrt {c-c \sin (e+f x)}}+\frac {a (3 A+5 B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{6 c f \sqrt {c-c \sin (e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 2.32, size = 292, normalized size = 1.08 \begin {gather*} \frac {a^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a (1+\sin (e+f x))} \left (-132 A-45 B-2 (27 A+59 B) \cos (2 (e+f x))+B \cos (4 (e+f x))-576 A \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-960 B \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-117 A \sin (e+f x)-279 B \sin (e+f x)+576 A \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sin (e+f x)+960 B \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sin (e+f x)-3 A \sin (3 (e+f x))-13 B \sin (3 (e+f x))\right )}{24 c f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (-1+\sin (e+f x)) \sqrt {c-c \sin (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + a*Sin[e + f*x])^(7/2)*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(3/2),x]

[Out]

(a^3*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*(-132*A - 45*B - 2*(27*A + 59*B)*Cos[2*(
e + f*x)] + B*Cos[4*(e + f*x)] - 576*A*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] - 960*B*Log[Cos[(e + f*x)/2] -
Sin[(e + f*x)/2]] - 117*A*Sin[e + f*x] - 279*B*Sin[e + f*x] + 576*A*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]]*
Sin[e + f*x] + 960*B*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]]*Sin[e + f*x] - 3*A*Sin[3*(e + f*x)] - 13*B*Sin[3
*(e + f*x)]))/(24*c*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-1 + Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(926\) vs. \(2(245)=490\).
time = 0.31, size = 927, normalized size = 3.42

method	result	size
default	\(\text {Expression too large to display}\)	\(927\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^(7/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/6/f*(-24*A*cos(f*x+e)^2*sin(f*x+e)-48*B*cos(f*x+e)^2*sin(f*x+e)-102*A-166*B+240*B*ln(-(-1+cos(f*x+e)+sin(f*
x+e))/sin(f*x+e))*sin(f*x+e)*cos(f*x+e)-120*B*ln(2/(1+cos(f*x+e)))*sin(f*x+e)*cos(f*x+e)+27*A*cos(f*x+e)+166*B
*sin(f*x+e)+102*A*sin(f*x+e)-27*A*cos(f*x+e)^3-61*B*cos(f*x+e)^3+59*B*cos(f*x+e)-13*B*cos(f*x+e)^3*sin(f*x+e)+
11*B*cos(f*x+e)^4+99*A*cos(f*x+e)^2+155*B*cos(f*x+e)^2-3*A*cos(f*x+e)^3*sin(f*x+e)+72*A*cos(f*x+e)^2*ln(2/(1+c
os(f*x+e)))+72*A*cos(f*x+e)*ln(2/(1+cos(f*x+e)))-240*B*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))*cos(f*x+e)^2
+2*B*cos(f*x+e)^5+240*B*ln(2/(1+cos(f*x+e)))*sin(f*x+e)-107*B*sin(f*x+e)*cos(f*x+e)-240*B*cos(f*x+e)*ln(-(-1+c
os(f*x+e)+sin(f*x+e))/sin(f*x+e))-480*B*sin(f*x+e)*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+144*A*ln(2/(1+co
s(f*x+e)))*sin(f*x+e)-144*A*cos(f*x+e)*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-288*A*sin(f*x+e)*ln(-(-1+cos
(f*x+e)+sin(f*x+e))/sin(f*x+e))+120*B*cos(f*x+e)*ln(2/(1+cos(f*x+e)))+3*A*cos(f*x+e)^4+144*A*cos(f*x+e)*sin(f*
x+e)*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-72*A*cos(f*x+e)*sin(f*x+e)*ln(2/(1+cos(f*x+e)))-144*A*ln(2/(1+
cos(f*x+e)))+288*A*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-240*B*ln(2/(1+cos(f*x+e)))+480*B*ln(-(-1+cos(f*x
+e)+sin(f*x+e))/sin(f*x+e))-144*A*cos(f*x+e)^2*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+120*B*ln(2/(1+cos(f*
x+e)))*cos(f*x+e)^2-75*A*sin(f*x+e)*cos(f*x+e)+2*B*sin(f*x+e)*cos(f*x+e)^4)*(a*(1+sin(f*x+e)))^(7/2)/(cos(f*x+
e)^4+cos(f*x+e)^3*sin(f*x+e)+3*cos(f*x+e)^3-4*cos(f*x+e)^2*sin(f*x+e)-8*cos(f*x+e)^2-4*cos(f*x+e)*sin(f*x+e)-4
*cos(f*x+e)+8*sin(f*x+e)+8)/(-c*(sin(f*x+e)-1))^(3/2)

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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))^(7/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^(7/2)/(-c*sin(f*x + e) + c)^(3/2), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(7/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

integral(-(B*a^3*cos(f*x + e)^4 - (3*A + 5*B)*a^3*cos(f*x + e)^2 + 4*(A + B)*a^3 - ((A + 3*B)*a^3*cos(f*x + e)
^2 - 4*(A + B)*a^3)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(c^2*cos(f*x + e)^2 + 2*c
^2*sin(f*x + e) - 2*c^2), x)

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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))**(7/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**(3/2),x)

[Out]

Timed out

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Giac [A]
time = 0.62, size = 411, normalized size = 1.52 \begin {gather*} -\frac {\sqrt {2} \sqrt {a} {\left (\frac {6 \, {\left (3 \, \sqrt {2} A a^{3} \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 5 \, \sqrt {2} B a^{3} \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \log \left (-\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}{c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {6 \, {\left (\sqrt {2} A a^{3} \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + \sqrt {2} B a^{3} \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )} c^{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 (4 \, B a^{3} c^{\frac {9}{2}} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 3 \, A a^{3} c^{\frac {9}{2}} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 9 \, B a^{3} c^{\frac {9}{2}} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 12 \, A a^{3} c^{\frac {9}{2}} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 24 \, B a^{3} c^{\frac {9}{2}} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )}}{c^{6} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}\right )}}{3 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(7/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(3/2),x, algorithm="giac")

[Out]

-1/3*sqrt(2)*sqrt(a)*(6*(3*sqrt(2)*A*a^3*sqrt(c)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 5*sqrt(2)*B*a^3*sqrt(c)
*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*log(-cos(-1/4*pi + 1/2*f*x + 1/2*e)^2 + 1)/(c^2*sgn(sin(-1/4*pi + 1/2*f*
x + 1/2*e))) - 6*(sqrt(2)*A*a^3*sqrt(c)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + sqrt(2)*B*a^3*sqrt(c)*sgn(cos(-1
/4*pi + 1/2*f*x + 1/2*e)))/((cos(-1/4*pi + 1/2*f*x + 1/2*e)^2 - 1)*c^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) +
sqrt(2)*(4*B*a^3*c^(9/2)*cos(-1/4*pi + 1/2*f*x + 1/2*e)^6*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 3*A*a^3*c^(9/2
)*cos(-1/4*pi + 1/2*f*x + 1/2*e)^4*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 9*B*a^3*c^(9/2)*cos(-1/4*pi + 1/2*f*x
+ 1/2*e)^4*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 12*A*a^3*c^(9/2)*cos(-1/4*pi + 1/2*f*x + 1/2*e)^2*sgn(cos(-1
/4*pi + 1/2*f*x + 1/2*e)) + 24*B*a^3*c^(9/2)*cos(-1/4*pi + 1/2*f*x + 1/2*e)^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*
e)))/(c^6*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 = -0.00 \begin {gather*} \int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{7/2}}{{\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(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(7/2))/(c - c*sin(e + f*x))^(3/2),x)

[Out]

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(7/2))/(c - c*sin(e + f*x))^(3/2), x)

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