3.4.11 \(\int \frac {(a+a \sin (e+f x))^3}{\sqrt {c-c \sin (e+f x)}} \, dx\) [311]

Optimal. Leaf size=151 \[ \frac {8 \sqrt {2} a^3 \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{\sqrt {c} f}-\frac {2 a^3 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{5/2}}-\frac {4 a^3 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}-\frac {8 a^3 \cos (e+f x)}{f \sqrt {c-c \sin (e+f x)}} \]

[Out]

-2/5*a^3*c^2*cos(f*x+e)^5/f/(c-c*sin(f*x+e))^(5/2)-4/3*a^3*c*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(3/2)+8*a^3*arcta
nh(1/2*cos(f*x+e)*c^(1/2)*2^(1/2)/(c-c*sin(f*x+e))^(1/2))*2^(1/2)/f/c^(1/2)-8*a^3*cos(f*x+e)/f/(c-c*sin(f*x+e)
)^(1/2)

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Rubi [A]
time = 0.22, antiderivative size = 151, 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 = {2815, 2758, 2728, 212} \begin {gather*} -\frac {2 a^3 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{5/2}}-\frac {4 a^3 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}-\frac {8 a^3 \cos (e+f x)}{f \sqrt {c-c \sin (e+f x)}}+\frac {8 \sqrt {2} a^3 \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{\sqrt {c} f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(8*Sqrt[2]*a^3*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(Sqrt[c]*f) - (2*a^3*c^2*Co
s[e + f*x]^5)/(5*f*(c - c*Sin[e + f*x])^(5/2)) - (4*a^3*c*Cos[e + f*x]^3)/(3*f*(c - c*Sin[e + f*x])^(3/2)) - (
8*a^3*Cos[e + f*x])/(f*Sqrt[c - c*Sin[e + f*x]])

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 2758

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[g*(g*C
os[e + f*x])^(p - 1)*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + p))), x] + Dist[g^2*((p - 1)/(a*(m + p))), Int[(g
*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0]
&& LtQ[m, -1] && GtQ[p, 1] && (GtQ[m, -2] || EqQ[2*m + p + 1, 0] || (EqQ[m, -2] && IntegerQ[p])) && NeQ[m + p,
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))^3}{\sqrt {c-c \sin (e+f x)}} \, dx &=\left (a^3 c^3\right ) \int \frac {\cos ^6(e+f x)}{(c-c \sin (e+f x))^{7/2}} \, dx\\ &=-\frac {2 a^3 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{5/2}}+\left (2 a^3 c^2\right ) \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^{5/2}} \, dx\\ &=-\frac {2 a^3 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{5/2}}-\frac {4 a^3 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}+\left (4 a^3 c\right ) \int \frac {\cos ^2(e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx\\ &=-\frac {2 a^3 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{5/2}}-\frac {4 a^3 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}-\frac {8 a^3 \cos (e+f x)}{f \sqrt {c-c \sin (e+f x)}}+\left (8 a^3\right ) \int \frac {1}{\sqrt {c-c \sin (e+f x)}} \, dx\\ &=-\frac {2 a^3 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{5/2}}-\frac {4 a^3 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}-\frac {8 a^3 \cos (e+f x)}{f \sqrt {c-c \sin (e+f x)}}-\frac {\left (16 a^3\right ) \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 )}{f}\\ &=\frac {8 \sqrt {2} a^3 \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{\sqrt {c} f}-\frac {2 a^3 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{5/2}}-\frac {4 a^3 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}-\frac {8 a^3 \cos (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.46, size = 156, normalized size = 1.03 \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 ) \left ((480+480 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 )+330 \cos \left (\frac {1}{2} (e+f x)\right )-35 \cos \left (\frac {3}{2} (e+f x)\right )-3 \cos \left (\frac {5}{2} (e+f x)\right )+330 \sin \left (\frac {1}{2} (e+f x)\right )+35 \sin \left (\frac {3}{2} (e+f x)\right )-3 \sin \left (\frac {5}{2} (e+f x)\right )\right )}{30 f \sqrt {c-c \sin (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/30*(a^3*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*((480 + 480*I)*(-1)^(1/4)*ArcTan[(1/2 + I/2)*(-1)^(1/4)*(1 +
Tan[(e + f*x)/4])] + 330*Cos[(e + f*x)/2] - 35*Cos[(3*(e + f*x))/2] - 3*Cos[(5*(e + f*x))/2] + 330*Sin[(e + f*
x)/2] + 35*Sin[(3*(e + f*x))/2] - 3*Sin[(5*(e + f*x))/2]))/(f*Sqrt[c - c*Sin[e + f*x]])

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Maple [A]
time = 2.70, size = 129, normalized size = 0.85

method	result	size
default	\(-\frac {2 \left (\sin \left (f x +e \right )-1\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, a^{3} \left (60 c^{\frac {5}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right )-3 \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}}-10 c \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}}-60 c^{2} \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\right )}{15 c^{3} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(129\)

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))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2/15*(sin(f*x+e)-1)*(c*(1+sin(f*x+e)))^(1/2)*a^3*(60*c^(5/2)*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(
1/2)/c^(1/2))-3*(c*(1+sin(f*x+e)))^(5/2)-10*c*(c*(1+sin(f*x+e)))^(3/2)-60*c^2*(c*(1+sin(f*x+e)))^(1/2))/c^3/co
s(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))^3/(c-c*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 287 vs. \(2 (140) = 280\).
time = 0.37, size = 287, normalized size = 1.90 \begin {gather*} \frac {2 \, {\left (\frac {30 \, \sqrt {2} {\left (a^{3} c \cos \left (f x + e\right ) - a^{3} c \sin \left (f x + e\right ) + a^{3} c\right )} \log \left (-\frac {\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) - 2\right )} \sin \left (f x + e\right ) + \frac {2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )}}{\sqrt {c}} + 3 \, \cos \left (f x + e\right ) + 2}{\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 )}{\sqrt {c}} + {\left (3 \, a^{3} \cos \left (f x + e\right )^{3} + 19 \, a^{3} \cos \left (f x + e\right )^{2} - 76 \, a^{3} \cos \left (f x + e\right ) - 92 \, a^{3} + {\left (3 \, a^{3} \cos \left (f x + e\right )^{2} - 16 \, a^{3} \cos \left (f x + e\right ) - 92 \, a^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}\right )}}{15 \, {\left (c f \cos \left (f x + e\right ) - c f \sin \left (f x + e\right ) + c f\right )}} \end {gather*}

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))^(1/2),x, algorithm="fricas")

[Out]

2/15*(30*sqrt(2)*(a^3*c*cos(f*x + e) - a^3*c*sin(f*x + e) + a^3*c)*log(-(cos(f*x + e)^2 + (cos(f*x + e) - 2)*s
in(f*x + e) + 2*sqrt(2)*sqrt(-c*sin(f*x + e) + c)*(cos(f*x + e) + sin(f*x + e) + 1)/sqrt(c) + 3*cos(f*x + e) +
2)/(cos(f*x + e)^2 + (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2))/sqrt(c) + (3*a^3*cos(f*x + e)^3 + 1
9*a^3*cos(f*x + e)^2 - 76*a^3*cos(f*x + e) - 92*a^3 + (3*a^3*cos(f*x + e)^2 - 16*a^3*cos(f*x + e) - 92*a^3)*si
n(f*x + e))*sqrt(-c*sin(f*x + e) + c))/(c*f*cos(f*x + e) - c*f*sin(f*x + e) + c*f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a^{3} \left (\int \frac {3 \sin {\left (e + f x \right )}}{\sqrt {- c \sin {\left (e + f x \right )} + c}}\, dx + \int \frac {3 \sin ^{2}{\left (e + f x \right )}}{\sqrt {- c \sin {\left (e + f x \right )} + c}}\, dx + \int \frac {\sin ^{3}{\left (e + f x \right )}}{\sqrt {- c \sin {\left (e + f x \right )} + c}}\, dx + \int \frac {1}{\sqrt {- c \sin {\left (e + f x \right )} + c}}\, dx\right ) \end {gather*}

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))**(1/2),x)

[Out]

a**3*(Integral(3*sin(e + f*x)/sqrt(-c*sin(e + f*x) + c), x) + Integral(3*sin(e + f*x)**2/sqrt(-c*sin(e + f*x)
+ c), x) + Integral(sin(e + f*x)**3/sqrt(-c*sin(e + f*x) + c), x) + Integral(1/sqrt(-c*sin(e + f*x) + c), x))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 304 vs. \(2 (140) = 280\).
time = 0.49, size = 304, normalized size = 2.01 \begin {gather*} \frac {4 \, {\left (\frac {15 \, \sqrt {2} a^{3} \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 )}{\sqrt {c} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {4 \, \sqrt {2} {\left (23 \, a^{3} \sqrt {c} - \frac {70 \, a^{3} \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 {140 \, a^{3} \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 {90 \, a^{3} \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}} + \frac {45 \, a^{3} \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{4}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{4}}\right )}}{c {\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} - 1\right )}^{5} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}\right )}}{15 \, f} \end {gather*}

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))^(1/2),x, algorithm="giac")

[Out]

4/15*(15*sqrt(2)*a^3*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))/(sqrt(c)*
sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) - 4*sqrt(2)*(23*a^3*sqrt(c) - 70*a^3*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) + 140*a^3*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 - 90*a^3*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 + 45*a^3*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^4/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^
4)/(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) - 1)^5*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 = -0.01 \begin {gather*} \int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^3}{\sqrt {c-c\,\sin \left (e+f\,x\right )}} \,d x \end {gather*}

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))^(1/2),x)

[Out]

int((a + a*sin(e + f*x))^3/(c - c*sin(e + f*x))^(1/2), x)

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