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

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

[Out]

-24/35*a^2*cos(d*x+c)*(a+a*sin(d*x+c))^(3/2)/d-2/7*a*cos(d*x+c)*(a+a*sin(d*x+c))^(5/2)/d-256/35*a^4*cos(d*x+c)
/d/(a+a*sin(d*x+c))^(1/2)-64/35*a^3*cos(d*x+c)*(a+a*sin(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} \[ -\frac {256 a^4 \cos (c+d x)}{35 d \sqrt {a \sin (c+d x)+a}}-\frac {64 a^3 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{35 d}-\frac {24 a^2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{35 d}-\frac {2 a \cos (c+d x) (a \sin (c+d x)+a)^{5/2}}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[c + d*x])^(7/2),x]

[Out]

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

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Mathematica [A]  time = 0.72, size = 154, normalized size = 1.29 \[ -\frac {a^3 (\sin (c+d x)+1)^3 \sqrt {a (\sin (c+d x)+1)} \left (-1225 \sin \left (\frac {1}{2} (c+d x)\right )+245 \sin \left (\frac {3}{2} (c+d x)\right )+49 \sin \left (\frac {5}{2} (c+d x)\right )-5 \sin \left (\frac {7}{2} (c+d x)\right )+1225 \cos \left (\frac {1}{2} (c+d x)\right )+245 \cos \left (\frac {3}{2} (c+d x)\right )-49 \cos \left (\frac {5}{2} (c+d x)\right )-5 \cos \left (\frac {7}{2} (c+d x)\right )\right )}{140 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^7} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.43, size = 140, normalized size = 1.18 \[ \frac {2 \, {\left (5 \, a^{3} \cos \left (d x + c\right )^{4} + 27 \, a^{3} \cos \left (d x + c\right )^{3} - 54 \, a^{3} \cos \left (d x + c\right )^{2} - 204 \, a^{3} \cos \left (d x + c\right ) - 128 \, a^{3} + {\left (5 \, a^{3} \cos \left (d x + c\right )^{3} - 22 \, a^{3} \cos \left (d x + c\right )^{2} - 76 \, a^{3} \cos \left (d x + c\right ) + 128 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{35 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/35*(5*a^3*cos(d*x + c)^4 + 27*a^3*cos(d*x + c)^3 - 54*a^3*cos(d*x + c)^2 - 204*a^3*cos(d*x + c) - 128*a^3 +
(5*a^3*cos(d*x + c)^3 - 22*a^3*cos(d*x + c)^2 - 76*a^3*cos(d*x + c) + 128*a^3)*sin(d*x + c))*sqrt(a*sin(d*x +
c) + a)/(d*cos(d*x + c) + d*sin(d*x + c) + d)

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giac [B]  time = 0.98, size = 240, normalized size = 2.02 \[ \frac {1}{140} \, \sqrt {2} {\left (\frac {7 \, a^{3} \cos \left (\frac {1}{4} \, \pi + \frac {5}{2} \, d x + \frac {5}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} - \frac {525 \, a^{3} \cos \left (\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} + \frac {5 \, a^{3} \cos \left (-\frac {1}{4} \, \pi + \frac {7}{2} \, d x + \frac {7}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} - \frac {175 \, a^{3} \cos \left (-\frac {1}{4} \, \pi + \frac {3}{2} \, d x + \frac {3}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} - \frac {70 \, a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{4} \, \pi + \frac {3}{2} \, d x + \frac {3}{2} \, c\right )}{d} - \frac {42 \, a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {5}{2} \, d x + \frac {5}{2} \, c\right )}{d} + \frac {700 \, a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{d}\right )} \sqrt {a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/140*sqrt(2)*(7*a^3*cos(1/4*pi + 5/2*d*x + 5/2*c)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d - 525*a^3*cos(1/4*pi
+ 1/2*d*x + 1/2*c)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d + 5*a^3*cos(-1/4*pi + 7/2*d*x + 7/2*c)*sgn(cos(-1/4*p
i + 1/2*d*x + 1/2*c))/d - 175*a^3*cos(-1/4*pi + 3/2*d*x + 3/2*c)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d - 70*a^
3*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(1/4*pi + 3/2*d*x + 3/2*c)/d - 42*a^3*sgn(cos(-1/4*pi + 1/2*d*x + 1/2
*c))*sin(-1/4*pi + 5/2*d*x + 5/2*c)/d + 700*a^3*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/
2*c)/d)*sqrt(a)

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maple [A]  time = 0.15, size = 75, normalized size = 0.63 \[ \frac {2 \left (1+\sin \left (d x +c \right )\right ) a^{4} \left (\sin \left (d x +c \right )-1\right ) \left (5 \left (\sin ^{3}\left (d x +c \right )\right )+27 \left (\sin ^{2}\left (d x +c \right )\right )+71 \sin \left (d x +c \right )+177\right )}{35 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/35*(1+sin(d*x+c))*a^4*(sin(d*x+c)-1)*(5*sin(d*x+c)^3+27*sin(d*x+c)^2+71*sin(d*x+c)+177)/cos(d*x+c)/(a+a*sin(
d*x+c))^(1/2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*sin(d*x + c) + a)^(7/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(c + d*x))^(7/2),x)

[Out]

int((a + a*sin(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*sin(d*x+c))**(7/2),x)

[Out]

Timed out

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