3.1 \(\int (a+a \cos (c+d x))^{7/2} \, dx\)

Optimal. Leaf size=119 \[ \frac {256 a^4 \sin (c+d x)}{35 d \sqrt {a \cos (c+d x)+a}}+\frac {64 a^3 \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{35 d}+\frac {24 a^2 \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{35 d}+\frac {2 a \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 d} \]

[Out]

24/35*a^2*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/d+2/7*a*(a+a*cos(d*x+c))^(5/2)*sin(d*x+c)/d+256/35*a^4*sin(d*x+c)/
d/(a+a*cos(d*x+c))^(1/2)+64/35*a^3*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.07, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2647, 2646} \[ \frac {256 a^4 \sin (c+d x)}{35 d \sqrt {a \cos (c+d x)+a}}+\frac {64 a^3 \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{35 d}+\frac {24 a^2 \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{35 d}+\frac {2 a \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^(7/2),x]

[Out]

(256*a^4*Sin[c + d*x])/(35*d*Sqrt[a + a*Cos[c + d*x]]) + (64*a^3*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(35*d)
+ (24*a^2*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(35*d) + (2*a*(a + a*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(7*
d)

Rule 2646

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(-2*b*Cos[c + d*x])/(d*Sqrt[a + b*Sin[c + d*
x]]), x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2647

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(n -
1))/(d*n), x] + Dist[(a*(2*n - 1))/n, Int[(a + b*Sin[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && Eq
Q[a^2 - b^2, 0] && IGtQ[n - 1/2, 0]

Rubi steps

\begin {align*} \int (a+a \cos (c+d x))^{7/2} \, dx &=\frac {2 a (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}+\frac {1}{7} (12 a) \int (a+a \cos (c+d x))^{5/2} \, dx\\ &=\frac {24 a^2 (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac {2 a (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}+\frac {1}{35} \left (96 a^2\right ) \int (a+a \cos (c+d x))^{3/2} \, dx\\ &=\frac {64 a^3 \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{35 d}+\frac {24 a^2 (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac {2 a (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}+\frac {1}{35} \left (128 a^3\right ) \int \sqrt {a+a \cos (c+d x)} \, dx\\ &=\frac {256 a^4 \sin (c+d x)}{35 d \sqrt {a+a \cos (c+d x)}}+\frac {64 a^3 \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{35 d}+\frac {24 a^2 (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac {2 a (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}\\ \end {align*}

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Mathematica [A]  time = 0.26, size = 83, normalized size = 0.70 \[ \frac {a^3 \left (1225 \sin \left (\frac {1}{2} (c+d x)\right )+245 \sin \left (\frac {3}{2} (c+d x)\right )+49 \sin \left (\frac {5}{2} (c+d x)\right )+5 \sin \left (\frac {7}{2} (c+d x)\right )\right ) \sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\cos (c+d x)+1)}}{140 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[c + d*x])^(7/2),x]

[Out]

(a^3*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*(1225*Sin[(c + d*x)/2] + 245*Sin[(3*(c + d*x))/2] + 49*Sin[(5
*(c + d*x))/2] + 5*Sin[(7*(c + d*x))/2]))/(140*d)

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fricas [A]  time = 0.68, size = 75, normalized size = 0.63 \[ \frac {2 \, {\left (5 \, a^{3} \cos \left (d x + c\right )^{3} + 27 \, a^{3} \cos \left (d x + c\right )^{2} + 71 \, a^{3} \cos \left (d x + c\right ) + 177 \, a^{3}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{35 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

2/35*(5*a^3*cos(d*x + c)^3 + 27*a^3*cos(d*x + c)^2 + 71*a^3*cos(d*x + c) + 177*a^3)*sqrt(a*cos(d*x + c) + a)*s
in(d*x + c)/(d*cos(d*x + c) + d)

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giac [A]  time = 0.59, size = 117, normalized size = 0.98 \[ \frac {1}{140} \, \sqrt {2} {\left (\frac {5 \, a^{3} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )}{d} + \frac {49 \, a^{3} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )}{d} + \frac {245 \, a^{3} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )}{d} + \frac {1225 \, a^{3} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{d}\right )} \sqrt {a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(7/2),x, algorithm="giac")

[Out]

1/140*sqrt(2)*(5*a^3*sgn(cos(1/2*d*x + 1/2*c))*sin(7/2*d*x + 7/2*c)/d + 49*a^3*sgn(cos(1/2*d*x + 1/2*c))*sin(5
/2*d*x + 5/2*c)/d + 245*a^3*sgn(cos(1/2*d*x + 1/2*c))*sin(3/2*d*x + 3/2*c)/d + 1225*a^3*sgn(cos(1/2*d*x + 1/2*
c))*sin(1/2*d*x + 1/2*c)/d)*sqrt(a)

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maple [A]  time = 0.10, size = 86, normalized size = 0.72 \[ \frac {16 a^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (5 \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+16\right ) \sqrt {2}}{35 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))^(7/2),x)

[Out]

16/35*a^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)*(5*cos(1/2*d*x+1/2*c)^6+6*cos(1/2*d*x+1/2*c)^4+8*cos(1/2*d*x+1
/2*c)^2+16)*2^(1/2)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d

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maxima [A]  time = 1.44, size = 77, normalized size = 0.65 \[ \frac {{\left (5 \, \sqrt {2} a^{3} \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 49 \, \sqrt {2} a^{3} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 245 \, \sqrt {2} a^{3} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 1225 \, \sqrt {2} a^{3} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{140 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

1/140*(5*sqrt(2)*a^3*sin(7/2*d*x + 7/2*c) + 49*sqrt(2)*a^3*sin(5/2*d*x + 5/2*c) + 245*sqrt(2)*a^3*sin(3/2*d*x
+ 3/2*c) + 1225*sqrt(2)*a^3*sin(1/2*d*x + 1/2*c))*sqrt(a)/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*cos(c + d*x))^(7/2),x)

[Out]

int((a + a*cos(c + d*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*cos(d*x+c))**(7/2),x)

[Out]

Timed out

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