3.314 \(\int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^{7/2}} \, dx\)

Optimal. Leaf size=157 \[ -\frac {5 a^3 \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{8 \sqrt {2} c^{7/2} f}+\frac {a^3 c^2 \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^{11/2}}+\frac {5 a^3 \cos (e+f x)}{8 c^2 f (c-c \sin (e+f x))^{3/2}}-\frac {5 a^3 \cos ^3(e+f x)}{12 f (c-c \sin (e+f x))^{7/2}} \]

[Out]

1/3*a^3*c^2*cos(f*x+e)^5/f/(c-c*sin(f*x+e))^(11/2)-5/12*a^3*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(7/2)+5/8*a^3*cos(
f*x+e)/c^2/f/(c-c*sin(f*x+e))^(3/2)-5/16*a^3*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 = 0.32, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2736, 2680, 2649, 206} \[ \frac {a^3 c^2 \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^{11/2}}+\frac {5 a^3 \cos (e+f x)}{8 c^2 f (c-c \sin (e+f x))^{3/2}}-\frac {5 a^3 \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{8 \sqrt {2} c^{7/2} f}-\frac {5 a^3 \cos ^3(e+f x)}{12 f (c-c \sin (e+f x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*a^3*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(8*Sqrt[2]*c^(7/2)*f) + (a^3*c^2*C
os[e + f*x]^5)/(3*f*(c - c*Sin[e + f*x])^(11/2)) - (5*a^3*Cos[e + f*x]^3)/(12*f*(c - c*Sin[e + f*x])^(7/2)) +
(5*a^3*Cos[e + f*x])/(8*c^2*f*(c - c*Sin[e + f*x])^(3/2))

Rule 206

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

Rule 2649

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 2680

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 2736

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))^3}{(c-c \sin (e+f x))^{7/2}} \, dx &=\left (a^3 c^3\right ) \int \frac {\cos ^6(e+f x)}{(c-c \sin (e+f x))^{13/2}} \, dx\\ &=\frac {a^3 c^2 \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^{11/2}}-\frac {1}{6} \left (5 a^3 c\right ) \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^{9/2}} \, dx\\ &=\frac {a^3 c^2 \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^{11/2}}-\frac {5 a^3 \cos ^3(e+f x)}{12 f (c-c \sin (e+f x))^{7/2}}+\frac {\left (5 a^3\right ) \int \frac {\cos ^2(e+f x)}{(c-c \sin (e+f x))^{5/2}} \, dx}{8 c}\\ &=\frac {a^3 c^2 \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^{11/2}}-\frac {5 a^3 \cos ^3(e+f x)}{12 f (c-c \sin (e+f x))^{7/2}}+\frac {5 a^3 \cos (e+f x)}{8 c^2 f (c-c \sin (e+f x))^{3/2}}-\frac {\left (5 a^3\right ) \int \frac {1}{\sqrt {c-c \sin (e+f x)}} \, dx}{16 c^3}\\ &=\frac {a^3 c^2 \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^{11/2}}-\frac {5 a^3 \cos ^3(e+f x)}{12 f (c-c \sin (e+f x))^{7/2}}+\frac {5 a^3 \cos (e+f x)}{8 c^2 f (c-c \sin (e+f x))^{3/2}}+\frac {\left (5 a^3\right ) \operatorname {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 c^3 f}\\ &=-\frac {5 a^3 \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{8 \sqrt {2} c^{7/2} f}+\frac {a^3 c^2 \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^{11/2}}-\frac {5 a^3 \cos ^3(e+f x)}{12 f (c-c \sin (e+f x))^{7/2}}+\frac {5 a^3 \cos (e+f x)}{8 c^2 f (c-c \sin (e+f x))^{3/2}}\\ \end {align*}

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Mathematica [C]  time = 1.57, size = 307, normalized size = 1.96 \[ \frac {a^3 (\sin (e+f x)+1)^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (64 \sin \left (\frac {1}{2} (e+f x)\right )+33 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^5+66 \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4-52 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3-104 \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+32 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+(15+15 i) \sqrt [4]{-1} \tan ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (\tan \left (\frac {1}{4} (e+f x)\right )+1\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^6\right )}{24 f (c-c \sin (e+f x))^{7/2} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(32*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]) - 52*(Cos[(e + f*x)/2] -
Sin[(e + f*x)/2])^3 + 33*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5 + (15 + 15*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] - 104*(Cos[(e
+ f*x)/2] - Sin[(e + f*x)/2])^2*Sin[(e + f*x)/2] + 66*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4*Sin[(e + f*x)/2]
)*(1 + Sin[e + f*x])^3)/(24*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6*(c - c*Sin[e + f*x])^(7/2))

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fricas [B]  time = 0.46, size = 440, normalized size = 2.80 \[ \frac {15 \, \sqrt {2} {\left (a^{3} \cos \left (f x + e\right )^{4} - 3 \, a^{3} \cos \left (f x + e\right )^{3} - 8 \, a^{3} \cos \left (f x + e\right )^{2} + 4 \, a^{3} \cos \left (f x + e\right ) + 8 \, a^{3} + {\left (a^{3} \cos \left (f x + e\right )^{3} + 4 \, a^{3} \cos \left (f x + e\right )^{2} - 4 \, a^{3} \cos \left (f x + e\right ) - 8 \, a^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {c} \log \left (-\frac {c \cos \left (f x + e\right )^{2} - 2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} \sqrt {c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) + {\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \, {\left (33 \, a^{3} \cos \left (f x + e\right )^{3} + 19 \, a^{3} \cos \left (f x + e\right )^{2} - 46 \, a^{3} \cos \left (f x + e\right ) - 32 \, a^{3} + {\left (33 \, a^{3} \cos \left (f x + e\right )^{2} + 14 \, a^{3} \cos \left (f x + e\right ) - 32 \, a^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{96 \, {\left (c^{4} f \cos \left (f x + e\right )^{4} - 3 \, c^{4} f \cos \left (f x + e\right )^{3} - 8 \, c^{4} f \cos \left (f x + e\right )^{2} + 4 \, c^{4} f \cos \left (f x + e\right ) + 8 \, c^{4} f + {\left (c^{4} f \cos \left (f x + e\right )^{3} + 4 \, c^{4} f \cos \left (f x + e\right )^{2} - 4 \, c^{4} f \cos \left (f x + e\right ) - 8 \, c^{4} f\right )} \sin \left (f x + e\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(15*sqrt(2)*(a^3*cos(f*x + e)^4 - 3*a^3*cos(f*x + e)^3 - 8*a^3*cos(f*x + e)^2 + 4*a^3*cos(f*x + e) + 8*a^
3 + (a^3*cos(f*x + e)^3 + 4*a^3*cos(f*x + e)^2 - 4*a^3*cos(f*x + e) - 8*a^3)*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*(33*a^3*cos(f*x + e)^3 + 19*a^3*cos(f*x + e)^2 - 46*a^3*cos(f*x + e) - 32*a^3 + (33*a^3*cos(f*x + e
)^2 + 14*a^3*cos(f*x + e) - 32*a^3)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c))/(c^4*f*cos(f*x + e)^4 - 3*c^4*f*c
os(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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (4*p
i/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unab
le to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*p
i/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unab
le to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*p
i/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign
: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Warning, integration of abs or sign assumes constant sign by intervals (
correct if the argument is real):Check [abs(sin((f*t_nostep+exp(1))/2-pi/4))]Unable to check sign: (8*pi/t_nos
tep/2)>(-8*pi/t_nostep/2)Discontinuities at zeroes of sin((f*t_nostep+exp(1))/2-pi/4) were not checkedUnable t
o check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Un
able to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to
check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/t_no
step/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (2*p
i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Warning, integration of abs or sign assumes cons
tant sign by intervals (correct if the argument is real):Check [abs(t_nostep-1)]Evaluation time: 0.99Not inver
tible Error: Bad Argument Value

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maple [A]  time = 1.00, size = 245, normalized size = 1.56 \[ \frac {a^{3} \left (15 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (\sin ^{3}\left (f x +e \right )\right ) c^{3}-45 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (\sin ^{2}\left (f x +e \right )\right ) c^{3}+66 \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}} \sqrt {c}+45 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sin \left (f x +e \right ) c^{3}-160 \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} c^{\frac {3}{2}}-15 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3}+120 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, c^{\frac {5}{2}}\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}}{48 c^{\frac {13}{2}} \left (\sin \left (f x +e \right )-1\right )^{2} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48/c^(13/2)*a^3*(15*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^3-45*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^3+66*(c*(1+sin(f*x+e)))^(5/2)*c^(1/2)+4
5*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^3-160*(c*(1+sin(f*x+e)))^(3/2)*c^
(3/2)-15*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*c^3+120*(c*(1+sin(f*x+e)))^(1/2)*c^(5/2
))*(c*(1+sin(f*x+e)))^(1/2)/(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 \[ \int \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{3}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^3}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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