3.168 \(\int \frac {(a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx\)
Optimal. Leaf size=264 \[ \frac {a^4 (A+7 B) \cos (e+f x) \log (1-\sin (e+f x))}{c^3 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {a^3 (A+7 B) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{2 c^3 f \sqrt {c-c \sin (e+f x)}}+\frac {a^2 (A+7 B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{4 c^2 f (c-c \sin (e+f x))^{3/2}}-\frac {a (A+7 B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{12 c f (c-c \sin (e+f x))^{5/2}}+\frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{6 f (c-c \sin (e+f x))^{7/2}} \]
[Out]
1/6*(A+B)*cos(f*x+e)*(a+a*sin(f*x+e))^(7/2)/f/(c-c*sin(f*x+e))^(7/2)-1/12*a*(A+7*B)*cos(f*x+e)*(a+a*sin(f*x+e)
)^(5/2)/c/f/(c-c*sin(f*x+e))^(5/2)+1/4*a^2*(A+7*B)*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)/c^2/f/(c-c*sin(f*x+e))^(3
/2)+a^4*(A+7*B)*cos(f*x+e)*ln(1-sin(f*x+e))/c^3/f/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(1/2)+1/2*a^3*(A+7*B
)*cos(f*x+e)*(a+a*sin(f*x+e))^(1/2)/c^3/f/(c-c*sin(f*x+e))^(1/2)
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Rubi [A] time = 0.61, antiderivative size = 264, 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 =
{2972, 2739, 2740, 2737, 2667, 31} \[ \frac {a^3 (A+7 B) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{2 c^3 f \sqrt {c-c \sin (e+f x)}}+\frac {a^2 (A+7 B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{4 c^2 f (c-c \sin (e+f x))^{3/2}}+\frac {a^4 (A+7 B) \cos (e+f x) \log (1-\sin (e+f x))}{c^3 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {a (A+7 B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{12 c f (c-c \sin (e+f x))^{5/2}}+\frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{6 f (c-c \sin (e+f x))^{7/2}} \]
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])^(7/2),x]
[Out]
((A + B)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2))/(6*f*(c - c*Sin[e + f*x])^(7/2)) - (a*(A + 7*B)*Cos[e + f*x]
*(a + a*Sin[e + f*x])^(5/2))/(12*c*f*(c - c*Sin[e + f*x])^(5/2)) + (a^2*(A + 7*B)*Cos[e + f*x]*(a + a*Sin[e +
f*x])^(3/2))/(4*c^2*f*(c - c*Sin[e + f*x])^(3/2)) + (a^4*(A + 7*B)*Cos[e + f*x]*Log[1 - Sin[e + f*x]])/(c^3*f*
Sqrt[a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]]) + (a^3*(A + 7*B)*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(2*
c^3*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 2667
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 2737
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 2739
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 2740
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Sim
p[(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] && E
qQ[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 2972
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))^{7/2}} \, dx &=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{6 f (c-c \sin (e+f x))^{7/2}}-\frac {(A+7 B) \int \frac {(a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{5/2}} \, dx}{6 c}\\ &=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{6 f (c-c \sin (e+f x))^{7/2}}-\frac {a (A+7 B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{12 c f (c-c \sin (e+f x))^{5/2}}+\frac {(a (A+7 B)) \int \frac {(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{3/2}} \, dx}{4 c^2}\\ &=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{6 f (c-c \sin (e+f x))^{7/2}}-\frac {a (A+7 B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{12 c f (c-c \sin (e+f x))^{5/2}}+\frac {a^2 (A+7 B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{4 c^2 f (c-c \sin (e+f x))^{3/2}}-\frac {\left (a^2 (A+7 B)\right ) \int \frac {(a+a \sin (e+f x))^{3/2}}{\sqrt {c-c \sin (e+f x)}} \, dx}{2 c^3}\\ &=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{6 f (c-c \sin (e+f x))^{7/2}}-\frac {a (A+7 B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{12 c f (c-c \sin (e+f x))^{5/2}}+\frac {a^2 (A+7 B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{4 c^2 f (c-c \sin (e+f x))^{3/2}}+\frac {a^3 (A+7 B) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{2 c^3 f \sqrt {c-c \sin (e+f x)}}-\frac {\left (a^3 (A+7 B)\right ) \int \frac {\sqrt {a+a \sin (e+f x)}}{\sqrt {c-c \sin (e+f x)}} \, dx}{c^3}\\ &=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{6 f (c-c \sin (e+f x))^{7/2}}-\frac {a (A+7 B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{12 c f (c-c \sin (e+f x))^{5/2}}+\frac {a^2 (A+7 B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{4 c^2 f (c-c \sin (e+f x))^{3/2}}+\frac {a^3 (A+7 B) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{2 c^3 f \sqrt {c-c \sin (e+f x)}}-\frac {\left (a^4 (A+7 B) \cos (e+f x)\right ) \int \frac {\cos (e+f x)}{c-c \sin (e+f x)} \, dx}{c^2 \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}}{6 f (c-c \sin (e+f x))^{7/2}}-\frac {a (A+7 B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{12 c f (c-c \sin (e+f x))^{5/2}}+\frac {a^2 (A+7 B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{4 c^2 f (c-c \sin (e+f x))^{3/2}}+\frac {a^3 (A+7 B) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{2 c^3 f \sqrt {c-c \sin (e+f x)}}+\frac {\left (a^4 (A+7 B) \cos (e+f x)\right ) \operatorname {Subst}\left (\int \frac {1}{c+x} \, dx,x,-c \sin (e+f x)\right )}{c^3 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}}{6 f (c-c \sin (e+f x))^{7/2}}-\frac {a (A+7 B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{12 c f (c-c \sin (e+f x))^{5/2}}+\frac {a^2 (A+7 B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{4 c^2 f (c-c \sin (e+f x))^{3/2}}+\frac {a^4 (A+7 B) \cos (e+f x) \log (1-\sin (e+f x))}{c^3 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}+\frac {a^3 (A+7 B) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{2 c^3 f \sqrt {c-c \sin (e+f x)}}\\ \end {align*}
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Mathematica [A] time = 2.90, size = 244, normalized size = 0.92 \[ \frac {(a (\sin (e+f x)+1))^{7/2} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (18 (A+3 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4-6 (3 A+5 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+6 (A+7 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^6 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+8 (A+B)+3 B \sin (e+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^6\right )}{3 f (c-c \sin (e+f x))^{7/2} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^7} \]
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])^(7/2),x]
[Out]
((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(a*(1 + Sin[e + f*x]))^(7/2)*(8*(A + B) - 6*(3*A + 5*B)*(Cos[(e + f*x)/
2] - Sin[(e + f*x)/2])^2 + 18*(A + 3*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4 + 6*(A + 7*B)*Log[Cos[(e + f*x
)/2] - Sin[(e + f*x)/2]]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^6 + 3*B*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^6
*Sin[e + f*x]))/(3*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^7*(c - c*Sin[e + f*x])^(7/2))
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fricas [F] time = 44.32, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B a^{3} \cos \left (f x + e\right )^{4} - {\left (3 \, A + 5 \, B\right )} a^{3} \cos \left (f x + e\right )^{2} + 4 \, {\left (A + B\right )} a^{3} - {\left ({\left (A + 3 \, B\right )} a^{3} \cos \left (f x + e\right )^{2} - 4 \, {\left (A + B\right )} a^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{c^{4} \cos \left (f x + e\right )^{4} - 8 \, c^{4} \cos \left (f x + e\right )^{2} + 8 \, c^{4} + 4 \, {\left (c^{4} \cos \left (f x + e\right )^{2} - 2 \, c^{4}\right )} \sin \left (f x + e\right )}, x\right ) \]
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))^(7/2),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^4*cos(f*x + e)^4 - 8*c^
4*cos(f*x + e)^2 + 8*c^4 + 4*(c^4*cos(f*x + e)^2 - 2*c^4)*sin(f*x + e)), x)
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giac [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((a+a*sin(f*x+e))^(7/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(7/2),x, algorithm="giac")
[Out]
Timed out
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maple [B] time = 0.66, size = 1455, normalized size = 5.51 \[ \text {result too large to display} \]
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))^(7/2),x)
[Out]
1/3/f*(336*B*cos(f*x+e)^2*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-41*B*cos(f*x+e)^3*sin(f*x+e)-20*A*sin(f*x
+e)+14*A*sin(f*x+e)*cos(f*x+e)-8*A*cos(f*x+e)^3*sin(f*x+e)-3*B*sin(f*x+e)*cos(f*x+e)^4+48*A*sin(f*x+e)*ln(-(-1
+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-168*B*cos(f*x+e)^2*ln(2/(cos(f*x+e)+1))+14*A*cos(f*x+e)^2*sin(f*x+e)+98*B*
cos(f*x+e)^2*sin(f*x+e)+44*B*cos(f*x+e)^4+20*A+116*B+6*A*cos(f*x+e)^3-126*B*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin
(f*x+e))*cos(f*x+e)^3+63*B*ln(2/(cos(f*x+e)+1))*cos(f*x+e)^3-24*A*cos(f*x+e)^2*ln(2/(cos(f*x+e)+1))+48*A*cos(f
*x+e)^2*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-24*A*sin(f*x+e)*ln(2/(cos(f*x+e)+1))+24*A*cos(f*x+e)*ln(-(-
1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-12*A*cos(f*x+e)*ln(2/(cos(f*x+e)+1))+62*B*sin(f*x+e)*cos(f*x+e)+336*B*sin
(f*x+e)*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-168*B*sin(f*x+e)*ln(2/(cos(f*x+e)+1))+168*B*cos(f*x+e)*ln(-
(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-84*B*cos(f*x+e)*ln(2/(cos(f*x+e)+1))+12*A*cos(f*x+e)*sin(f*x+e)*ln(2/(c
os(f*x+e)+1))-24*A*cos(f*x+e)*sin(f*x+e)*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-3*B*cos(f*x+e)^5-168*B*ln(
-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))*cos(f*x+e)^2*sin(f*x+e)+84*B*ln(2/(cos(f*x+e)+1))*cos(f*x+e)^2*sin(f*x
+e)+84*B*ln(2/(cos(f*x+e)+1))*sin(f*x+e)*cos(f*x+e)-168*B*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))*sin(f*x+e
)*cos(f*x+e)-6*A*cos(f*x+e)^4*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+3*A*cos(f*x+e)^4*ln(2/(cos(f*x+e)+1))
+21*B*cos(f*x+e)^4*ln(2/(cos(f*x+e)+1))-42*B*cos(f*x+e)^4*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-160*B*cos
(f*x+e)^2-18*A*cos(f*x+e)^3*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+9*A*cos(f*x+e)^3*ln(2/(cos(f*x+e)+1))+8
*A*cos(f*x+e)^4+12*A*cos(f*x+e)^2*sin(f*x+e)*ln(2/(cos(f*x+e)+1))-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))+57*B*cos(f*x+e)^3-116*B*sin(f*x+e)-3*A*sin(f*x+e)*cos(f*x+e)^3*ln(2/(cos(f*x+e)+1)
)+6*A*sin(f*x+e)*cos(f*x+e)^3*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+42*B*cos(f*x+e)^3*sin(f*x+e)*ln(-(-1+
cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-21*B*cos(f*x+e)^3*sin(f*x+e)*ln(2/(cos(f*x+e)+1))-6*A*cos(f*x+e)-54*B*cos(f
*x+e)-28*A*cos(f*x+e)^2-48*A*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+24*A*ln(2/(cos(f*x+e)+1))-336*B*ln(-(-
1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+168*B*ln(2/(cos(f*x+e)+1)))*(a*(1+sin(f*x+e)))^(7/2)/(cos(f*x+e)^4+sin(f*
x+e)*cos(f*x+e)^3+3*cos(f*x+e)^3-4*cos(f*x+e)^2*sin(f*x+e)-8*cos(f*x+e)^2-4*sin(f*x+e)*cos(f*x+e)-4*cos(f*x+e)
+8*sin(f*x+e)+8)/(-c*(sin(f*x+e)-1))^(7/2)
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maxima [B] time = 1.19, size = 749, normalized size = 2.84 \[ \text {result too large to display} \]
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))^(7/2),x, algorithm="maxima")
[Out]
-1/3*(B*(42*a^(7/2)*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/c^(7/2) - 21*a^(7/2)*log(sin(f*x + e)^2/(cos(f*x
+ e) + 1)^2 + 1)/c^(7/2) + 2*(21*a^(7/2)*sin(f*x + e)/(cos(f*x + e) + 1) - 102*a^(7/2)*sin(f*x + e)^2/(cos(f*x
+ e) + 1)^2 + 227*a^(7/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 228*a^(7/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)
^4 + 227*a^(7/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 102*a^(7/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 21*a^
(7/2)*sin(f*x + e)^7/(cos(f*x + e) + 1)^7)/(c^(7/2) - 6*c^(7/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 16*c^(7/2)*s
in(f*x + e)^2/(cos(f*x + e) + 1)^2 - 26*c^(7/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 30*c^(7/2)*sin(f*x + e)^
4/(cos(f*x + e) + 1)^4 - 26*c^(7/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 16*c^(7/2)*sin(f*x + e)^6/(cos(f*x +
e) + 1)^6 - 6*c^(7/2)*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + c^(7/2)*sin(f*x + e)^8/(cos(f*x + e) + 1)^8)) + A
*(6*a^(7/2)*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/c^(7/2) - 3*a^(7/2)*log(sin(f*x + e)^2/(cos(f*x + e) + 1)
^2 + 1)/c^(7/2) + 4*(3*a^(7/2)*sqrt(c)*sin(f*x + e)/(cos(f*x + e) + 1) - 6*a^(7/2)*sqrt(c)*sin(f*x + e)^2/(cos
(f*x + e) + 1)^2 + 22*a^(7/2)*sqrt(c)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 6*a^(7/2)*sqrt(c)*sin(f*x + e)^4/(
cos(f*x + e) + 1)^4 + 3*a^(7/2)*sqrt(c)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)/(c^4 - 6*c^4*sin(f*x + e)/(cos(f*
x + e) + 1) + 15*c^4*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 20*c^4*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 15*c^4
*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 6*c^4*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + c^4*sin(f*x + e)^6/(cos(f*x
+ e) + 1)^6)))/f
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{7/2}}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{7/2}} \,d x \]
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))^(7/2),x)
[Out]
int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(7/2))/(c - c*sin(e + f*x))^(7/2), 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((a+a*sin(f*x+e))**(7/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**(7/2),x)
[Out]
Timed out
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