3.2.67 \(\int \frac {(a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx\) [167]

Optimal. Leaf size=263 \[ \frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{4 f (c-c \sin (e+f x))^{5/2}}-\frac {a (A+3 B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{2 c f (c-c \sin (e+f x))^{3/2}}-\frac {6 a^4 (A+3 B) \cos (e+f x) \log (1-\sin (e+f x))}{c^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}-\frac {3 a^3 (A+3 B) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {3 a^2 (A+3 B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{4 c^2 f \sqrt {c-c \sin (e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.40, antiderivative size = 263, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {3051, 2818, 2819, 2816, 2746, 31} \begin {gather*} -\frac {6 a^4 (A+3 B) \cos (e+f x) \log (1-\sin (e+f x))}{c^2 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {3 a^3 (A+3 B) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {3 a^2 (A+3 B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{4 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {a (A+3 B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{2 c f (c-c \sin (e+f x))^{3/2}}+\frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{4 f (c-c \sin (e+f x))^{5/2}} \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])^(5/2),x]

[Out]

((A + B)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2))/(4*f*(c - c*Sin[e + f*x])^(5/2)) - (a*(A + 3*B)*Cos[e + f*x]
*(a + a*Sin[e + f*x])^(5/2))/(2*c*f*(c - c*Sin[e + f*x])^(3/2)) - (6*a^4*(A + 3*B)*Cos[e + f*x]*Log[1 - Sin[e
+ f*x]])/(c^2*f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]]) - (3*a^3*(A + 3*B)*Cos[e + f*x]*Sqrt[a + a*
Sin[e + f*x]])/(c^2*f*Sqrt[c - c*Sin[e + f*x]]) - (3*a^2*(A + 3*B)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(4
*c^2*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 2818

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[-2*b*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^n/(f*(2*n + 1))), x] - Dist[b*((2*m - 1)
/(d*(2*n + 1))), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e
, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[m - 1/2, 0] && LtQ[n, -1] &&  !(ILtQ[m + n, 0] && G
tQ[2*m + n + 1, 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))^{5/2}} \, dx &=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{4 f (c-c \sin (e+f x))^{5/2}}-\frac {(A+3 B) \int \frac {(a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{3/2}} \, dx}{2 c}\\ &=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{4 f (c-c \sin (e+f x))^{5/2}}-\frac {a (A+3 B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{2 c f (c-c \sin (e+f x))^{3/2}}+\frac {(3 a (A+3 B)) \int \frac {(a+a \sin (e+f x))^{5/2}}{\sqrt {c-c \sin (e+f x)}} \, dx}{2 c^2}\\ &=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{4 f (c-c \sin (e+f x))^{5/2}}-\frac {a (A+3 B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{2 c f (c-c \sin (e+f x))^{3/2}}-\frac {3 a^2 (A+3 B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{4 c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {\left (3 a^2 (A+3 B)\right ) \int \frac {(a+a \sin (e+f x))^{3/2}}{\sqrt {c-c \sin (e+f x)}} \, dx}{c^2}\\ &=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{4 f (c-c \sin (e+f x))^{5/2}}-\frac {a (A+3 B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{2 c f (c-c \sin (e+f x))^{3/2}}-\frac {3 a^3 (A+3 B) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {3 a^2 (A+3 B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{4 c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {\left (6 a^3 (A+3 B)\right ) \int \frac {\sqrt {a+a \sin (e+f x)}}{\sqrt {c-c \sin (e+f x)}} \, dx}{c^2}\\ &=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{4 f (c-c \sin (e+f x))^{5/2}}-\frac {a (A+3 B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{2 c f (c-c \sin (e+f x))^{3/2}}-\frac {3 a^3 (A+3 B) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {3 a^2 (A+3 B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{4 c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {\left (6 a^4 (A+3 B) \cos (e+f x)\right ) \int \frac {\cos (e+f x)}{c-c \sin (e+f x)} \, dx}{c \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}}{4 f (c-c \sin (e+f x))^{5/2}}-\frac {a (A+3 B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{2 c f (c-c \sin (e+f x))^{3/2}}-\frac {3 a^3 (A+3 B) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {3 a^2 (A+3 B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{4 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {\left (6 a^4 (A+3 B) \cos (e+f x)\right ) \text {Subst}\left (\int \frac {1}{c+x} \, dx,x,-c \sin (e+f x)\right )}{c^2 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}}{4 f (c-c \sin (e+f x))^{5/2}}-\frac {a (A+3 B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{2 c f (c-c \sin (e+f x))^{3/2}}-\frac {6 a^4 (A+3 B) \cos (e+f x) \log (1-\sin (e+f x))}{c^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}-\frac {3 a^3 (A+3 B) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {3 a^2 (A+3 B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{4 c^2 f \sqrt {c-c \sin (e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 1.69, size = 251, normalized size = 0.95 \begin {gather*} \frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (a (1+\sin (e+f x)))^{7/2} \left (16 (A+B)-16 (3 A+5 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+B \cos (2 (e+f x)) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4-48 (A+3 B) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4-4 (A+6 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \sin (e+f x)\right )}{4 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^7 (c-c \sin (e+f x))^{5/2}} \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])^(5/2),x]

[Out]

((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(a*(1 + Sin[e + f*x]))^(7/2)*(16*(A + B) - 16*(3*A + 5*B)*(Cos[(e + f*x
)/2] - Sin[(e + f*x)/2])^2 + B*Cos[2*(e + f*x)]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4 - 48*(A + 3*B)*Log[Cos
[(e + f*x)/2] - Sin[(e + f*x)/2]]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4 - 4*(A + 6*B)*(Cos[(e + f*x)/2] - Si
n[(e + f*x)/2])^4*Sin[e + f*x]))/(4*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^7*(c - c*Sin[e + f*x])^(5/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1188\) vs. \(2(237)=474\).
time = 0.31, size = 1189, normalized size = 4.52

method	result	size
default	\(\text {Expression too large to display}\)	\(1189\)

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))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/2/f*(36*B*ln(2/(1+cos(f*x+e)))*cos(f*x+e)^3-72*B*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))*cos(f*x+e)^3+22
*A*cos(f*x+e)^2*sin(f*x+e)+63*B*cos(f*x+e)^2*sin(f*x+e)+32*A+100*B-144*B*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*
x+e))*sin(f*x+e)*cos(f*x+e)+72*B*ln(2/(1+cos(f*x+e)))*sin(f*x+e)*cos(f*x+e)-24*A*cos(f*x+e)^3*ln(-(-1+cos(f*x+
e)+sin(f*x+e))/sin(f*x+e))+12*A*cos(f*x+e)^3*ln(2/(1+cos(f*x+e)))-20*A*cos(f*x+e)-100*B*sin(f*x+e)-32*A*sin(f*
x+e)+20*A*cos(f*x+e)^3+53*B*cos(f*x+e)^3-54*B*cos(f*x+e)-10*B*cos(f*x+e)^3*sin(f*x+e)+36*B*ln(2/(1+cos(f*x+e))
)*cos(f*x+e)^2*sin(f*x+e)-72*B*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))*cos(f*x+e)^2*sin(f*x+e)+9*B*cos(f*x+
e)^4-34*A*cos(f*x+e)^2-109*B*cos(f*x+e)^2-2*A*cos(f*x+e)^3*sin(f*x+e)-36*A*cos(f*x+e)^2*ln(2/(1+cos(f*x+e)))-2
4*A*cos(f*x+e)*ln(2/(1+cos(f*x+e)))+216*B*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))*cos(f*x+e)^2+B*cos(f*x+e)
^5-144*B*ln(2/(1+cos(f*x+e)))*sin(f*x+e)+46*B*sin(f*x+e)*cos(f*x+e)+144*B*cos(f*x+e)*ln(-(-1+cos(f*x+e)+sin(f*
x+e))/sin(f*x+e))+288*B*sin(f*x+e)*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-24*A*cos(f*x+e)^2*sin(f*x+e)*ln(
-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+12*A*cos(f*x+e)^2*sin(f*x+e)*ln(2/(1+cos(f*x+e)))-48*A*ln(2/(1+cos(f*x
+e)))*sin(f*x+e)+48*A*cos(f*x+e)*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+96*A*sin(f*x+e)*ln(-(-1+cos(f*x+e)
+sin(f*x+e))/sin(f*x+e))-72*B*cos(f*x+e)*ln(2/(1+cos(f*x+e)))+2*A*cos(f*x+e)^4-48*A*cos(f*x+e)*sin(f*x+e)*ln(-
(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+24*A*cos(f*x+e)*sin(f*x+e)*ln(2/(1+cos(f*x+e)))+48*A*ln(2/(1+cos(f*x+e)
))-96*A*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+144*B*ln(2/(1+cos(f*x+e)))-288*B*ln(-(-1+cos(f*x+e)+sin(f*x
+e))/sin(f*x+e))+72*A*cos(f*x+e)^2*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-108*B*ln(2/(1+cos(f*x+e)))*cos(f
*x+e)^2+12*A*sin(f*x+e)*cos(f*x+e)+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))^(5/2)

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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))^(5/2),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)^(5/2), 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))^(5/2),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)/(3*c^3*cos(f*x + e)^2 - 4
*c^3 - (c^3*cos(f*x + e)^2 - 4*c^3)*sin(f*x + e)), 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))**(5/2),x)

[Out]

Timed out

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Giac [A]
time = 0.59, size = 433, normalized size = 1.65 \begin {gather*} \frac {\sqrt {2} \sqrt {a} {\left (\frac {6 \, \sqrt {2} {\left (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 ) + 3 \, 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^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {2 \, {\left (\sqrt {2} B a^{3} c^{\frac {7}{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 ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + \sqrt {2} A a^{3} c^{\frac {7}{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 ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 5 \, \sqrt {2} B a^{3} c^{\frac {7}{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 ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )}}{c^{6}} + \frac {5 \, \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 ) + 9 \, \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 ) - 2 \, {\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 )} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{2} c^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}\right )}}{2 \, 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))^(5/2),x, algorithm="giac")

[Out]

1/2*sqrt(2)*sqrt(a)*(6*sqrt(2)*(A*a^3*sqrt(c)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 3*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^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))
) + 2*(sqrt(2)*B*a^3*c^(7/2)*cos(-1/4*pi + 1/2*f*x + 1/2*e)^4*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4
*pi + 1/2*f*x + 1/2*e)) + sqrt(2)*A*a^3*c^(7/2)*cos(-1/4*pi + 1/2*f*x + 1/2*e)^2*sgn(cos(-1/4*pi + 1/2*f*x + 1
/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) + 5*sqrt(2)*B*a^3*c^(7/2)*cos(-1/4*pi + 1/2*f*x + 1/2*e)^2*sgn(cos(
-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)))/c^6 + (5*sqrt(2)*A*a^3*sqrt(c)*sgn(cos(-1/4*p
i + 1/2*f*x + 1/2*e)) + 9*sqrt(2)*B*a^3*sqrt(c)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - 2*(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)))*cos(-1/4
*pi + 1/2*f*x + 1/2*e)^2)/((cos(-1/4*pi + 1/2*f*x + 1/2*e)^2 - 1)^2*c^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 = -0.00 \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 )}^{5/2}} \,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))^(5/2),x)

[Out]

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(7/2))/(c - c*sin(e + f*x))^(5/2), x)

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