3.804 \(\int (a+a \sin (c+d x))^2 \tan ^4(c+d x) \, dx\)

Optimal. Leaf size=101 \[ -\frac {2 a^2 \cos (c+d x)}{d}-\frac {a^2 \sin (c+d x) \cos (c+d x)}{2 d}-\frac {11 a^2 \cos (c+d x)}{3 d (1-\sin (c+d x))}+\frac {a^2 \cos (c+d x)}{3 d (1-\sin (c+d x))^2}+\frac {7 a^2 x}{2} \]

[Out]

7/2*a^2*x-2*a^2*cos(d*x+c)/d+1/3*a^2*cos(d*x+c)/d/(1-sin(d*x+c))^2-11/3*a^2*cos(d*x+c)/d/(1-sin(d*x+c))-1/2*a^
2*cos(d*x+c)*sin(d*x+c)/d

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Rubi [A]  time = 0.21, antiderivative size = 120, normalized size of antiderivative = 1.19, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2708, 2765, 2977, 2734} \[ -\frac {16 a^2 \cos (c+d x)}{3 d}+\frac {a^4 \sin ^3(c+d x) \cos (c+d x)}{3 d (a-a \sin (c+d x))^2}-\frac {8 a^2 \sin ^2(c+d x) \cos (c+d x)}{3 d (1-\sin (c+d x))}-\frac {7 a^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {7 a^2 x}{2} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[c + d*x])^2*Tan[c + d*x]^4,x]

[Out]

(7*a^2*x)/2 - (16*a^2*Cos[c + d*x])/(3*d) - (7*a^2*Cos[c + d*x]*Sin[c + d*x])/(2*d) - (8*a^2*Cos[c + d*x]*Sin[
c + d*x]^2)/(3*d*(1 - Sin[c + d*x])) + (a^4*Cos[c + d*x]*Sin[c + d*x]^3)/(3*d*(a - a*Sin[c + d*x])^2)

Rule 2708

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_), x_Symbol] :> Dist[a^p, Int[Sin[
e + f*x]^p/(a - b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p] &&
EqQ[p, 2*m]

Rule 2734

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

Rule 2765

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[((b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n - 1))/(a*f*(2*m + 1)), x] + Dist[1/
(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 2)*Simp[b*(c^2*(m + 1) + d^2*(n -
1)) + a*c*d*(m - n + 1) + d*(a*d*(m - n + 1) + b*c*(m + n))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e,
f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ
[2*m, 2*n] || (IntegerQ[m] && EqQ[c, 0]))

Rule 2977

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x]
)^n)/(a*f*(2*m + 1)), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n -
1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], x], x],
x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ
[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])

Rubi steps

\begin {align*} \int (a+a \sin (c+d x))^2 \tan ^4(c+d x) \, dx &=a^4 \int \frac {\sin ^4(c+d x)}{(a-a \sin (c+d x))^2} \, dx\\ &=\frac {a^4 \cos (c+d x) \sin ^3(c+d x)}{3 d (a-a \sin (c+d x))^2}+\frac {1}{3} a^2 \int \frac {\sin ^2(c+d x) (-3 a-5 a \sin (c+d x))}{a-a \sin (c+d x)} \, dx\\ &=-\frac {8 a^2 \cos (c+d x) \sin ^2(c+d x)}{3 d (1-\sin (c+d x))}+\frac {a^4 \cos (c+d x) \sin ^3(c+d x)}{3 d (a-a \sin (c+d x))^2}-\frac {1}{3} \int \sin (c+d x) \left (-16 a^2-21 a^2 \sin (c+d x)\right ) \, dx\\ &=\frac {7 a^2 x}{2}-\frac {16 a^2 \cos (c+d x)}{3 d}-\frac {7 a^2 \cos (c+d x) \sin (c+d x)}{2 d}-\frac {8 a^2 \cos (c+d x) \sin ^2(c+d x)}{3 d (1-\sin (c+d x))}+\frac {a^4 \cos (c+d x) \sin ^3(c+d x)}{3 d (a-a \sin (c+d x))^2}\\ \end {align*}

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Mathematica [A]  time = 1.26, size = 159, normalized size = 1.57 \[ -\frac {a^2 \left (-21 (12 c+12 d x+7) \cos \left (\frac {1}{2} (c+d x)\right )+(84 c+84 d x+239) \cos \left (\frac {3}{2} (c+d x)\right )+3 \left (-5 \cos \left (\frac {5}{2} (c+d x)\right )+\cos \left (\frac {7}{2} (c+d x)\right )+2 \sin \left (\frac {1}{2} (c+d x)\right ) ((28 c+28 d x-27) \cos (c+d x)-6 \cos (2 (c+d x))-\cos (3 (c+d x))+56 c+56 d x+50)\right )\right )}{48 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[c + d*x])^2*Tan[c + d*x]^4,x]

[Out]

-1/48*(a^2*(-21*(7 + 12*c + 12*d*x)*Cos[(c + d*x)/2] + (239 + 84*c + 84*d*x)*Cos[(3*(c + d*x))/2] + 3*(-5*Cos[
(5*(c + d*x))/2] + Cos[(7*(c + d*x))/2] + 2*(50 + 56*c + 56*d*x + (-27 + 28*c + 28*d*x)*Cos[c + d*x] - 6*Cos[2
*(c + d*x)] - Cos[3*(c + d*x)])*Sin[(c + d*x)/2])))/(d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3)

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fricas [B]  time = 0.50, size = 196, normalized size = 1.94 \[ \frac {3 \, a^{2} \cos \left (d x + c\right )^{4} - 6 \, a^{2} \cos \left (d x + c\right )^{3} - 42 \, a^{2} d x + {\left (21 \, a^{2} d x + 31 \, a^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - {\left (21 \, a^{2} d x - 38 \, a^{2}\right )} \cos \left (d x + c\right ) - {\left (3 \, a^{2} \cos \left (d x + c\right )^{3} - 42 \, a^{2} d x + 9 \, a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} - {\left (21 \, a^{2} d x - 40 \, a^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left (d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) + {\left (d \cos \left (d x + c\right ) + 2 \, d\right )} \sin \left (d x + c\right ) - 2 \, d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^4*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/6*(3*a^2*cos(d*x + c)^4 - 6*a^2*cos(d*x + c)^3 - 42*a^2*d*x + (21*a^2*d*x + 31*a^2)*cos(d*x + c)^2 - 2*a^2 -
(21*a^2*d*x - 38*a^2)*cos(d*x + c) - (3*a^2*cos(d*x + c)^3 - 42*a^2*d*x + 9*a^2*cos(d*x + c)^2 + 2*a^2 - (21*
a^2*d*x - 40*a^2)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2 - d*cos(d*x + c) + (d*cos(d*x + c) + 2*d)*sin(
d*x + c) - 2*d)

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giac [A]  time = 0.21, size = 135, normalized size = 1.34 \[ \frac {21 \, {\left (d x + c\right )} a^{2} + \frac {6 \, {\left (a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, a^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}} + \frac {4 \, {\left (9 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 21 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 10 \, a^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^4*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/6*(21*(d*x + c)*a^2 + 6*(a^2*tan(1/2*d*x + 1/2*c)^3 - 4*a^2*tan(1/2*d*x + 1/2*c)^2 - a^2*tan(1/2*d*x + 1/2*c
) - 4*a^2)/(tan(1/2*d*x + 1/2*c)^2 + 1)^2 + 4*(9*a^2*tan(1/2*d*x + 1/2*c)^2 - 21*a^2*tan(1/2*d*x + 1/2*c) + 10
*a^2)/(tan(1/2*d*x + 1/2*c) - 1)^3)/d

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maple [A]  time = 0.49, size = 186, normalized size = 1.84 \[ \frac {a^{2} \left (\frac {\sin ^{7}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {4 \left (\sin ^{7}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )}-\frac {4 \left (\sin ^{5}\left (d x +c \right )+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{3}+\frac {5 d x}{2}+\frac {5 c}{2}\right )+2 a^{2} \left (\frac {\sin ^{6}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{6}\left (d x +c \right )}{\cos \left (d x +c \right )}-\left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )\right )+a^{2} \left (\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-\tan \left (d x +c \right )+d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*sin(d*x+c)^4*(a+a*sin(d*x+c))^2,x)

[Out]

1/d*(a^2*(1/3*sin(d*x+c)^7/cos(d*x+c)^3-4/3*sin(d*x+c)^7/cos(d*x+c)-4/3*(sin(d*x+c)^5+5/4*sin(d*x+c)^3+15/8*si
n(d*x+c))*cos(d*x+c)+5/2*d*x+5/2*c)+2*a^2*(1/3*sin(d*x+c)^6/cos(d*x+c)^3-sin(d*x+c)^6/cos(d*x+c)-(8/3+sin(d*x+
c)^4+4/3*sin(d*x+c)^2)*cos(d*x+c))+a^2*(1/3*tan(d*x+c)^3-tan(d*x+c)+d*x+c))

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maxima [A]  time = 0.43, size = 120, normalized size = 1.19 \[ \frac {{\left (2 \, \tan \left (d x + c\right )^{3} + 15 \, d x + 15 \, c - \frac {3 \, \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 12 \, \tan \left (d x + c\right )\right )} a^{2} + 2 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} a^{2} - 4 \, a^{2} {\left (\frac {6 \, \cos \left (d x + c\right )^{2} - 1}{\cos \left (d x + c\right )^{3}} + 3 \, \cos \left (d x + c\right )\right )}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^4*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/6*((2*tan(d*x + c)^3 + 15*d*x + 15*c - 3*tan(d*x + c)/(tan(d*x + c)^2 + 1) - 12*tan(d*x + c))*a^2 + 2*(tan(d
*x + c)^3 + 3*d*x + 3*c - 3*tan(d*x + c))*a^2 - 4*a^2*((6*cos(d*x + c)^2 - 1)/cos(d*x + c)^3 + 3*cos(d*x + c))
)/d

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mupad [B]  time = 14.69, size = 287, normalized size = 2.84 \[ \frac {7\,a^2\,x}{2}+\frac {\frac {7\,a^2\,\left (c+d\,x\right )}{2}-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {21\,a^2\,\left (c+d\,x\right )}{2}-\frac {a^2\,\left (63\,c+63\,d\,x-150\right )}{6}\right )-\frac {a^2\,\left (21\,c+21\,d\,x-64\right )}{6}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {21\,a^2\,\left (c+d\,x\right )}{2}-\frac {a^2\,\left (63\,c+63\,d\,x-42\right )}{6}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {35\,a^2\,\left (c+d\,x\right )}{2}-\frac {a^2\,\left (105\,c+105\,d\,x-126\right )}{6}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {35\,a^2\,\left (c+d\,x\right )}{2}-\frac {a^2\,\left (105\,c+105\,d\,x-194\right )}{6}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {49\,a^2\,\left (c+d\,x\right )}{2}-\frac {a^2\,\left (147\,c+147\,d\,x-196\right )}{6}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {49\,a^2\,\left (c+d\,x\right )}{2}-\frac {a^2\,\left (147\,c+147\,d\,x-252\right )}{6}\right )}{d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}^3\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(c + d*x)^4*(a + a*sin(c + d*x))^2)/cos(c + d*x)^4,x)

[Out]

(7*a^2*x)/2 + ((7*a^2*(c + d*x))/2 - tan(c/2 + (d*x)/2)*((21*a^2*(c + d*x))/2 - (a^2*(63*c + 63*d*x - 150))/6)
- (a^2*(21*c + 21*d*x - 64))/6 + tan(c/2 + (d*x)/2)^6*((21*a^2*(c + d*x))/2 - (a^2*(63*c + 63*d*x - 42))/6) -
tan(c/2 + (d*x)/2)^5*((35*a^2*(c + d*x))/2 - (a^2*(105*c + 105*d*x - 126))/6) + tan(c/2 + (d*x)/2)^2*((35*a^2
*(c + d*x))/2 - (a^2*(105*c + 105*d*x - 194))/6) + tan(c/2 + (d*x)/2)^4*((49*a^2*(c + d*x))/2 - (a^2*(147*c +
147*d*x - 196))/6) - tan(c/2 + (d*x)/2)^3*((49*a^2*(c + d*x))/2 - (a^2*(147*c + 147*d*x - 252))/6))/(d*(tan(c/
2 + (d*x)/2) - 1)^3*(tan(c/2 + (d*x)/2)^2 + 1)^2)

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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(sec(d*x+c)**4*sin(d*x+c)**4*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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