3.3.90 \(\int (a+a \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx\) [290]

Optimal. Leaf size=137 \[ \frac {256 a c^5 \cos ^3(e+f x)}{315 f (c-c \sin (e+f x))^{3/2}}+\frac {64 a c^4 \cos ^3(e+f x)}{105 f \sqrt {c-c \sin (e+f x)}}+\frac {8 a c^3 \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{21 f}+\frac {2 a c^2 \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{9 f} \]

[Out]

256/315*a*c^5*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(3/2)+2/9*a*c^2*cos(f*x+e)^3*(c-c*sin(f*x+e))^(3/2)/f+64/105*a*c
^4*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(1/2)+8/21*a*c^3*cos(f*x+e)^3*(c-c*sin(f*x+e))^(1/2)/f

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Rubi [A]
time = 0.20, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {2815, 2753, 2752} \begin {gather*} \frac {256 a c^5 \cos ^3(e+f x)}{315 f (c-c \sin (e+f x))^{3/2}}+\frac {64 a c^4 \cos ^3(e+f x)}{105 f \sqrt {c-c \sin (e+f x)}}+\frac {8 a c^3 \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{21 f}+\frac {2 a c^2 \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{9 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(256*a*c^5*Cos[e + f*x]^3)/(315*f*(c - c*Sin[e + f*x])^(3/2)) + (64*a*c^4*Cos[e + f*x]^3)/(105*f*Sqrt[c - c*Si
n[e + f*x]]) + (8*a*c^3*Cos[e + f*x]^3*Sqrt[c - c*Sin[e + f*x]])/(21*f) + (2*a*c^2*Cos[e + f*x]^3*(c - c*Sin[e
+ f*x])^(3/2))/(9*f)

Rule 2752

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*(g*C
os[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^(m - 1)/(f*g*(m - 1))), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && Eq
Q[a^2 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]

Rule 2753

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(-b)*(
g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^(m - 1)/(f*g*(m + p))), x] + Dist[a*((2*m + p - 1)/(m + p)), Int
[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2,
0] && IGtQ[Simplify[(2*m + p - 1)/2], 0] && NeQ[m + p, 0]

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 (a+a \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx &=(a c) \int \cos ^2(e+f x) (c-c \sin (e+f x))^{5/2} \, dx\\ &=\frac {2 a c^2 \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{9 f}+\frac {1}{3} \left (4 a c^2\right ) \int \cos ^2(e+f x) (c-c \sin (e+f x))^{3/2} \, dx\\ &=\frac {8 a c^3 \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{21 f}+\frac {2 a c^2 \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{9 f}+\frac {1}{21} \left (32 a c^3\right ) \int \cos ^2(e+f x) \sqrt {c-c \sin (e+f x)} \, dx\\ &=\frac {64 a c^4 \cos ^3(e+f x)}{105 f \sqrt {c-c \sin (e+f x)}}+\frac {8 a c^3 \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{21 f}+\frac {2 a c^2 \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{9 f}+\frac {1}{105} \left (128 a c^4\right ) \int \frac {\cos ^2(e+f x)}{\sqrt {c-c \sin (e+f x)}} \, dx\\ &=\frac {256 a c^5 \cos ^3(e+f x)}{315 f (c-c \sin (e+f x))^{3/2}}+\frac {64 a c^4 \cos ^3(e+f x)}{105 f \sqrt {c-c \sin (e+f x)}}+\frac {8 a c^3 \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{21 f}+\frac {2 a c^2 \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{9 f}\\ \end {align*}

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Mathematica [A]
time = 0.52, size = 104, normalized size = 0.76 \begin {gather*} \frac {a c^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 \sqrt {c-c \sin (e+f x)} (1606-330 \cos (2 (e+f x))-1389 \sin (e+f x)+35 \sin (3 (e+f x)))}{630 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*c^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*Sqrt[c - c*Sin[e + f*x]]*(1606 - 330*Cos[2*(e + f*x)] - 1389*Si
n[e + f*x] + 35*Sin[3*(e + f*x)]))/(630*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]))

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Maple [A]
time = 2.00, size = 79, normalized size = 0.58

method	result	size
default	\(\frac {2 \left (\sin \left (f x +e \right )-1\right ) c^{4} \left (1+\sin \left (f x +e \right )\right )^{2} a \left (35 \left (\sin ^{3}\left (f x +e \right )\right )-165 \left (\sin ^{2}\left (f x +e \right )\right )+321 \sin \left (f x +e \right )-319\right )}{315 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(79\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(sin(f*x+e)-1)*c^4*(1+sin(f*x+e))^2*a*(35*sin(f*x+e)^3-165*sin(f*x+e)^2+321*sin(f*x+e)-319)/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 \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))*(c-c*sin(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.36, size = 192, normalized size = 1.40 \begin {gather*} -\frac {2 \, {\left (35 \, a c^{3} \cos \left (f x + e\right )^{5} - 95 \, a c^{3} \cos \left (f x + e\right )^{4} - 226 \, a c^{3} \cos \left (f x + e\right )^{3} + 32 \, a c^{3} \cos \left (f x + e\right )^{2} - 128 \, a c^{3} \cos \left (f x + e\right ) - 256 \, a c^{3} + {\left (35 \, a c^{3} \cos \left (f x + e\right )^{4} + 130 \, a c^{3} \cos \left (f x + e\right )^{3} - 96 \, a c^{3} \cos \left (f x + e\right )^{2} - 128 \, a c^{3} \cos \left (f x + e\right ) - 256 \, a c^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{315 \, {\left (f \cos \left (f x + e\right ) - f \sin \left (f x + e\right ) + f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/315*(35*a*c^3*cos(f*x + e)^5 - 95*a*c^3*cos(f*x + e)^4 - 226*a*c^3*cos(f*x + e)^3 + 32*a*c^3*cos(f*x + e)^2
- 128*a*c^3*cos(f*x + e) - 256*a*c^3 + (35*a*c^3*cos(f*x + e)^4 + 130*a*c^3*cos(f*x + e)^3 - 96*a*c^3*cos(f*x
+ e)^2 - 128*a*c^3*cos(f*x + e) - 256*a*c^3)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c)/(f*cos(f*x + e) - f*sin(
f*x + e) + f)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3876 deep

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Giac [A]
time = 0.53, size = 144, normalized size = 1.05 \begin {gather*} -\frac {\sqrt {2} {\left (4410 \, a c^{3} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 504 \, a c^{3} \cos \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, f x + \frac {5}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 225 \, a c^{3} \cos \left (-\frac {7}{4} \, \pi + \frac {7}{2} \, f x + \frac {7}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 35 \, a c^{3} \cos \left (-\frac {9}{4} \, \pi + \frac {9}{2} \, f x + \frac {9}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {c}}{2520 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2520*sqrt(2)*(4410*a*c^3*cos(-1/4*pi + 1/2*f*x + 1/2*e)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 504*a*c^3*cos
(-5/4*pi + 5/2*f*x + 5/2*e)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) + 225*a*c^3*cos(-7/4*pi + 7/2*f*x + 7/2*e)*sgn
(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 35*a*c^3*cos(-9/4*pi + 9/2*f*x + 9/2*e)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))
)*sqrt(c)/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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