3.40 \(\int (a+a \sin (c+d x))^4 \, dx\)

Optimal. Leaf size=87 \[ \frac {4 a^4 \cos ^3(c+d x)}{3 d}-\frac {8 a^4 \cos (c+d x)}{d}-\frac {a^4 \sin ^3(c+d x) \cos (c+d x)}{4 d}-\frac {27 a^4 \sin (c+d x) \cos (c+d x)}{8 d}+\frac {35 a^4 x}{8} \]

[Out]

35/8*a^4*x-8*a^4*cos(d*x+c)/d+4/3*a^4*cos(d*x+c)^3/d-27/8*a^4*cos(d*x+c)*sin(d*x+c)/d-1/4*a^4*cos(d*x+c)*sin(d
*x+c)^3/d

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Rubi [A]  time = 0.08, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {2645, 2638, 2635, 8, 2633} \[ \frac {4 a^4 \cos ^3(c+d x)}{3 d}-\frac {8 a^4 \cos (c+d x)}{d}-\frac {a^4 \sin ^3(c+d x) \cos (c+d x)}{4 d}-\frac {27 a^4 \sin (c+d x) \cos (c+d x)}{8 d}+\frac {35 a^4 x}{8} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[c + d*x])^4,x]

[Out]

(35*a^4*x)/8 - (8*a^4*Cos[c + d*x])/d + (4*a^4*Cos[c + d*x]^3)/(3*d) - (27*a^4*Cos[c + d*x]*Sin[c + d*x])/(8*d
) - (a^4*Cos[c + d*x]*Sin[c + d*x]^3)/(4*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2635

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

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2645

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

Rubi steps

\begin {align*} \int (a+a \sin (c+d x))^4 \, dx &=\int \left (a^4+4 a^4 \sin (c+d x)+6 a^4 \sin ^2(c+d x)+4 a^4 \sin ^3(c+d x)+a^4 \sin ^4(c+d x)\right ) \, dx\\ &=a^4 x+a^4 \int \sin ^4(c+d x) \, dx+\left (4 a^4\right ) \int \sin (c+d x) \, dx+\left (4 a^4\right ) \int \sin ^3(c+d x) \, dx+\left (6 a^4\right ) \int \sin ^2(c+d x) \, dx\\ &=a^4 x-\frac {4 a^4 \cos (c+d x)}{d}-\frac {3 a^4 \cos (c+d x) \sin (c+d x)}{d}-\frac {a^4 \cos (c+d x) \sin ^3(c+d x)}{4 d}+\frac {1}{4} \left (3 a^4\right ) \int \sin ^2(c+d x) \, dx+\left (3 a^4\right ) \int 1 \, dx-\frac {\left (4 a^4\right ) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=4 a^4 x-\frac {8 a^4 \cos (c+d x)}{d}+\frac {4 a^4 \cos ^3(c+d x)}{3 d}-\frac {27 a^4 \cos (c+d x) \sin (c+d x)}{8 d}-\frac {a^4 \cos (c+d x) \sin ^3(c+d x)}{4 d}+\frac {1}{8} \left (3 a^4\right ) \int 1 \, dx\\ &=\frac {35 a^4 x}{8}-\frac {8 a^4 \cos (c+d x)}{d}+\frac {4 a^4 \cos ^3(c+d x)}{3 d}-\frac {27 a^4 \cos (c+d x) \sin (c+d x)}{8 d}-\frac {a^4 \cos (c+d x) \sin ^3(c+d x)}{4 d}\\ \end {align*}

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Mathematica [A]  time = 0.40, size = 57, normalized size = 0.66 \[ \frac {a^4 (3 (-56 \sin (2 (c+d x))+\sin (4 (c+d x))+140 c+140 d x)-672 \cos (c+d x)+32 \cos (3 (c+d x)))}{96 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[c + d*x])^4,x]

[Out]

(a^4*(-672*Cos[c + d*x] + 32*Cos[3*(c + d*x)] + 3*(140*c + 140*d*x - 56*Sin[2*(c + d*x)] + Sin[4*(c + d*x)])))
/(96*d)

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fricas [A]  time = 0.42, size = 70, normalized size = 0.80 \[ \frac {32 \, a^{4} \cos \left (d x + c\right )^{3} + 105 \, a^{4} d x - 192 \, a^{4} \cos \left (d x + c\right ) + 3 \, {\left (2 \, a^{4} \cos \left (d x + c\right )^{3} - 29 \, a^{4} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

1/24*(32*a^4*cos(d*x + c)^3 + 105*a^4*d*x - 192*a^4*cos(d*x + c) + 3*(2*a^4*cos(d*x + c)^3 - 29*a^4*cos(d*x +
c))*sin(d*x + c))/d

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giac [A]  time = 1.53, size = 72, normalized size = 0.83 \[ \frac {35}{8} \, a^{4} x + \frac {a^{4} \cos \left (3 \, d x + 3 \, c\right )}{3 \, d} - \frac {7 \, a^{4} \cos \left (d x + c\right )}{d} + \frac {a^{4} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} - \frac {7 \, a^{4} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^4,x, algorithm="giac")

[Out]

35/8*a^4*x + 1/3*a^4*cos(3*d*x + 3*c)/d - 7*a^4*cos(d*x + c)/d + 1/32*a^4*sin(4*d*x + 4*c)/d - 7/4*a^4*sin(2*d
*x + 2*c)/d

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maple [A]  time = 0.20, size = 111, normalized size = 1.28 \[ \frac {a^{4} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )-\frac {4 a^{4} \left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}+6 a^{4} \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-4 a^{4} \cos \left (d x +c \right )+a^{4} \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(d*x+c))^4,x)

[Out]

1/d*(a^4*(-1/4*(sin(d*x+c)^3+3/2*sin(d*x+c))*cos(d*x+c)+3/8*d*x+3/8*c)-4/3*a^4*(2+sin(d*x+c)^2)*cos(d*x+c)+6*a
^4*(-1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)-4*a^4*cos(d*x+c)+a^4*(d*x+c))

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maxima [A]  time = 0.30, size = 108, normalized size = 1.24 \[ a^{4} x + \frac {4 \, {\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} a^{4}}{3 \, d} + \frac {{\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) - 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{4}}{32 \, d} + \frac {3 \, {\left (2 \, d x + 2 \, c - \sin \left (2 \, d x + 2 \, c\right )\right )} a^{4}}{2 \, d} - \frac {4 \, a^{4} \cos \left (d x + c\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^4,x, algorithm="maxima")

[Out]

a^4*x + 4/3*(cos(d*x + c)^3 - 3*cos(d*x + c))*a^4/d + 1/32*(12*d*x + 12*c + sin(4*d*x + 4*c) - 8*sin(2*d*x + 2
*c))*a^4/d + 3/2*(2*d*x + 2*c - sin(2*d*x + 2*c))*a^4/d - 4*a^4*cos(d*x + c)/d

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mupad [B]  time = 8.59, size = 237, normalized size = 2.72 \[ \frac {35\,a^4\,x}{8}-\frac {\frac {35\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}-\frac {35\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}-\frac {27\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+\frac {a^4\,\left (105\,c+105\,d\,x\right )}{24}-\frac {a^4\,\left (105\,c+105\,d\,x-320\right )}{24}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {a^4\,\left (105\,c+105\,d\,x\right )}{6}-\frac {a^4\,\left (420\,c+420\,d\,x-192\right )}{24}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {a^4\,\left (105\,c+105\,d\,x\right )}{6}-\frac {a^4\,\left (420\,c+420\,d\,x-1088\right )}{24}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {a^4\,\left (105\,c+105\,d\,x\right )}{4}-\frac {a^4\,\left (630\,c+630\,d\,x-960\right )}{24}\right )+\frac {27\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(c + d*x))^4,x)

[Out]

(35*a^4*x)/8 - ((35*a^4*tan(c/2 + (d*x)/2)^3)/4 - (35*a^4*tan(c/2 + (d*x)/2)^5)/4 - (27*a^4*tan(c/2 + (d*x)/2)
^7)/4 + (a^4*(105*c + 105*d*x))/24 - (a^4*(105*c + 105*d*x - 320))/24 + tan(c/2 + (d*x)/2)^6*((a^4*(105*c + 10
5*d*x))/6 - (a^4*(420*c + 420*d*x - 192))/24) + tan(c/2 + (d*x)/2)^2*((a^4*(105*c + 105*d*x))/6 - (a^4*(420*c
+ 420*d*x - 1088))/24) + tan(c/2 + (d*x)/2)^4*((a^4*(105*c + 105*d*x))/4 - (a^4*(630*c + 630*d*x - 960))/24) +
(27*a^4*tan(c/2 + (d*x)/2))/4)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^4)

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sympy [A]  time = 1.89, size = 224, normalized size = 2.57 \[ \begin {cases} \frac {3 a^{4} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 a^{4} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + 3 a^{4} x \sin ^{2}{\left (c + d x \right )} + \frac {3 a^{4} x \cos ^{4}{\left (c + d x \right )}}{8} + 3 a^{4} x \cos ^{2}{\left (c + d x \right )} + a^{4} x - \frac {5 a^{4} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} - \frac {4 a^{4} \sin ^{2}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{d} - \frac {3 a^{4} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} - \frac {3 a^{4} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{d} - \frac {8 a^{4} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {4 a^{4} \cos {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a \sin {\relax (c )} + a\right )^{4} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))**4,x)

[Out]

Piecewise((3*a**4*x*sin(c + d*x)**4/8 + 3*a**4*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + 3*a**4*x*sin(c + d*x)**2
+ 3*a**4*x*cos(c + d*x)**4/8 + 3*a**4*x*cos(c + d*x)**2 + a**4*x - 5*a**4*sin(c + d*x)**3*cos(c + d*x)/(8*d) -
4*a**4*sin(c + d*x)**2*cos(c + d*x)/d - 3*a**4*sin(c + d*x)*cos(c + d*x)**3/(8*d) - 3*a**4*sin(c + d*x)*cos(c
+ d*x)/d - 8*a**4*cos(c + d*x)**3/(3*d) - 4*a**4*cos(c + d*x)/d, Ne(d, 0)), (x*(a*sin(c) + a)**4, True))

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