3.217 \(\int \sec ^5(c+b x) \sin (a+b x) \, dx\)

Optimal. Leaf size=59 \[ \frac {\sin (a-c) \tan ^3(b x+c)}{3 b}+\frac {\sin (a-c) \tan (b x+c)}{b}+\frac {\cos (a-c) \sec ^4(b x+c)}{4 b} \]

[Out]

1/4*cos(a-c)*sec(b*x+c)^4/b+sin(a-c)*tan(b*x+c)/b+1/3*sin(a-c)*tan(b*x+c)^3/b

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Rubi [A]  time = 0.05, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {4580, 2606, 30, 3767} \[ \frac {\sin (a-c) \tan ^3(b x+c)}{3 b}+\frac {\sin (a-c) \tan (b x+c)}{b}+\frac {\cos (a-c) \sec ^4(b x+c)}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + b*x]^5*Sin[a + b*x],x]

[Out]

(Cos[a - c]*Sec[c + b*x]^4)/(4*b) + (Sin[a - c]*Tan[c + b*x])/b + (Sin[a - c]*Tan[c + b*x]^3)/(3*b)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 3767

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

Rule 4580

Int[Sec[w_]^(n_.)*Sin[v_], x_Symbol] :> Dist[Cos[v - w], Int[Tan[w]*Sec[w]^(n - 1), x], x] + Dist[Sin[v - w],
Int[Sec[w]^(n - 1), x], x] /; GtQ[n, 0] && FreeQ[v - w, x] && NeQ[w, v]

Rubi steps

\begin {align*} \int \sec ^5(c+b x) \sin (a+b x) \, dx &=\cos (a-c) \int \sec ^4(c+b x) \tan (c+b x) \, dx+\sin (a-c) \int \sec ^4(c+b x) \, dx\\ &=\frac {\cos (a-c) \operatorname {Subst}\left (\int x^3 \, dx,x,\sec (c+b x)\right )}{b}-\frac {\sin (a-c) \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+b x)\right )}{b}\\ &=\frac {\cos (a-c) \sec ^4(c+b x)}{4 b}+\frac {\sin (a-c) \tan (c+b x)}{b}+\frac {\sin (a-c) \tan ^3(c+b x)}{3 b}\\ \end {align*}

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Mathematica [A]  time = 0.34, size = 48, normalized size = 0.81 \[ \frac {\sec (c) \sec ^4(b x+c) (\sin (a-c) (4 \sin (2 b x+c)+\sin (4 b x+3 c))+3 \cos (a))}{12 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + b*x]^5*Sin[a + b*x],x]

[Out]

(Sec[c]*Sec[c + b*x]^4*(3*Cos[a] + Sin[a - c]*(4*Sin[c + 2*b*x] + Sin[3*c + 4*b*x])))/(12*b)

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fricas [A]  time = 0.48, size = 53, normalized size = 0.90 \[ -\frac {4 \, {\left (2 \, \cos \left (b x + c\right )^{3} + \cos \left (b x + c\right )\right )} \sin \left (b x + c\right ) \sin \left (-a + c\right ) - 3 \, \cos \left (-a + c\right )}{12 \, b \cos \left (b x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+c)^5*sin(b*x+a),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.23, size = 327, normalized size = 5.54 \[ \frac {3 \, \tan \left (b x + c\right )^{4} \tan \left (\frac {1}{2} \, a\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} - 3 \, \tan \left (b x + c\right )^{4} \tan \left (\frac {1}{2} \, a\right )^{2} + 12 \, \tan \left (b x + c\right )^{4} \tan \left (\frac {1}{2} \, a\right ) \tan \left (\frac {1}{2} \, c\right ) + 8 \, \tan \left (b x + c\right )^{3} \tan \left (\frac {1}{2} \, a\right )^{2} \tan \left (\frac {1}{2} \, c\right ) - 3 \, \tan \left (b x + c\right )^{4} \tan \left (\frac {1}{2} \, c\right )^{2} - 8 \, \tan \left (b x + c\right )^{3} \tan \left (\frac {1}{2} \, a\right ) \tan \left (\frac {1}{2} \, c\right )^{2} + 6 \, \tan \left (b x + c\right )^{2} \tan \left (\frac {1}{2} \, a\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} + 3 \, \tan \left (b x + c\right )^{4} + 8 \, \tan \left (b x + c\right )^{3} \tan \left (\frac {1}{2} \, a\right ) - 6 \, \tan \left (b x + c\right )^{2} \tan \left (\frac {1}{2} \, a\right )^{2} - 8 \, \tan \left (b x + c\right )^{3} \tan \left (\frac {1}{2} \, c\right ) + 24 \, \tan \left (b x + c\right )^{2} \tan \left (\frac {1}{2} \, a\right ) \tan \left (\frac {1}{2} \, c\right ) + 24 \, \tan \left (b x + c\right ) \tan \left (\frac {1}{2} \, a\right )^{2} \tan \left (\frac {1}{2} \, c\right ) - 6 \, \tan \left (b x + c\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} - 24 \, \tan \left (b x + c\right ) \tan \left (\frac {1}{2} \, a\right ) \tan \left (\frac {1}{2} \, c\right )^{2} + 6 \, \tan \left (b x + c\right )^{2} + 24 \, \tan \left (b x + c\right ) \tan \left (\frac {1}{2} \, a\right ) - 24 \, \tan \left (b x + c\right ) \tan \left (\frac {1}{2} \, c\right )}{12 \, {\left (\tan \left (\frac {1}{2} \, a\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} + \tan \left (\frac {1}{2} \, a\right )^{2} + \tan \left (\frac {1}{2} \, c\right )^{2} + 1\right )} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+c)^5*sin(b*x+a),x, algorithm="giac")

[Out]

1/12*(3*tan(b*x + c)^4*tan(1/2*a)^2*tan(1/2*c)^2 - 3*tan(b*x + c)^4*tan(1/2*a)^2 + 12*tan(b*x + c)^4*tan(1/2*a
)*tan(1/2*c) + 8*tan(b*x + c)^3*tan(1/2*a)^2*tan(1/2*c) - 3*tan(b*x + c)^4*tan(1/2*c)^2 - 8*tan(b*x + c)^3*tan
(1/2*a)*tan(1/2*c)^2 + 6*tan(b*x + c)^2*tan(1/2*a)^2*tan(1/2*c)^2 + 3*tan(b*x + c)^4 + 8*tan(b*x + c)^3*tan(1/
2*a) - 6*tan(b*x + c)^2*tan(1/2*a)^2 - 8*tan(b*x + c)^3*tan(1/2*c) + 24*tan(b*x + c)^2*tan(1/2*a)*tan(1/2*c) +
24*tan(b*x + c)*tan(1/2*a)^2*tan(1/2*c) - 6*tan(b*x + c)^2*tan(1/2*c)^2 - 24*tan(b*x + c)*tan(1/2*a)*tan(1/2*
c)^2 + 6*tan(b*x + c)^2 + 24*tan(b*x + c)*tan(1/2*a) - 24*tan(b*x + c)*tan(1/2*c))/((tan(1/2*a)^2*tan(1/2*c)^2
+ tan(1/2*a)^2 + tan(1/2*c)^2 + 1)*b)

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maple [B]  time = 13.27, size = 333, normalized size = 5.64 \[ \frac {\frac {\left (\cos \relax (a ) \cos \relax (c )+\sin \relax (a ) \sin \relax (c )\right ) \left (\left (\cos ^{2}\relax (a )\right ) \left (\cos ^{2}\relax (c )\right )+\left (\cos ^{2}\relax (a )\right ) \left (\sin ^{2}\relax (c )\right )+\left (\cos ^{2}\relax (c )\right ) \left (\sin ^{2}\relax (a )\right )+\left (\sin ^{2}\relax (a )\right ) \left (\sin ^{2}\relax (c )\right )\right )}{4 \left (\cos \relax (a ) \sin \relax (c )-\sin \relax (a ) \cos \relax (c )\right )^{4} \left (\tan \left (b x +a \right ) \cos \relax (a ) \sin \relax (c )-\tan \left (b x +a \right ) \sin \relax (a ) \cos \relax (c )-\sin \relax (a ) \sin \relax (c )-\cos \relax (a ) \cos \relax (c )\right )^{4}}-\frac {-3 \cos \relax (a ) \cos \relax (c )-3 \sin \relax (a ) \sin \relax (c )}{2 \left (\cos \relax (a ) \sin \relax (c )-\sin \relax (a ) \cos \relax (c )\right )^{4} \left (\tan \left (b x +a \right ) \cos \relax (a ) \sin \relax (c )-\tan \left (b x +a \right ) \sin \relax (a ) \cos \relax (c )-\sin \relax (a ) \sin \relax (c )-\cos \relax (a ) \cos \relax (c )\right )^{2}}+\frac {1}{\left (\cos \relax (a ) \sin \relax (c )-\sin \relax (a ) \cos \relax (c )\right )^{4} \left (\tan \left (b x +a \right ) \cos \relax (a ) \sin \relax (c )-\tan \left (b x +a \right ) \sin \relax (a ) \cos \relax (c )-\sin \relax (a ) \sin \relax (c )-\cos \relax (a ) \cos \relax (c )\right )}-\frac {-\left (\cos ^{2}\relax (a )\right ) \left (\sin ^{2}\relax (c )\right )-3 \left (\sin ^{2}\relax (a )\right ) \left (\sin ^{2}\relax (c )\right )-4 \cos \relax (a ) \cos \relax (c ) \sin \relax (a ) \sin \relax (c )-3 \left (\cos ^{2}\relax (a )\right ) \left (\cos ^{2}\relax (c )\right )-\left (\cos ^{2}\relax (c )\right ) \left (\sin ^{2}\relax (a )\right )}{3 \left (\cos \relax (a ) \sin \relax (c )-\sin \relax (a ) \cos \relax (c )\right )^{4} \left (\tan \left (b x +a \right ) \cos \relax (a ) \sin \relax (c )-\tan \left (b x +a \right ) \sin \relax (a ) \cos \relax (c )-\sin \relax (a ) \sin \relax (c )-\cos \relax (a ) \cos \relax (c )\right )^{3}}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+c)^5*sin(b*x+a),x)

[Out]

1/b*(1/4*(cos(a)*cos(c)+sin(a)*sin(c))*(cos(a)^2*cos(c)^2+cos(a)^2*sin(c)^2+cos(c)^2*sin(a)^2+sin(a)^2*sin(c)^
2)/(cos(a)*sin(c)-sin(a)*cos(c))^4/(tan(b*x+a)*cos(a)*sin(c)-tan(b*x+a)*sin(a)*cos(c)-sin(a)*sin(c)-cos(a)*cos
(c))^4-1/2*(-3*cos(a)*cos(c)-3*sin(a)*sin(c))/(cos(a)*sin(c)-sin(a)*cos(c))^4/(tan(b*x+a)*cos(a)*sin(c)-tan(b*
x+a)*sin(a)*cos(c)-sin(a)*sin(c)-cos(a)*cos(c))^2+1/(cos(a)*sin(c)-sin(a)*cos(c))^4/(tan(b*x+a)*cos(a)*sin(c)-
tan(b*x+a)*sin(a)*cos(c)-sin(a)*sin(c)-cos(a)*cos(c))-1/3*(-cos(a)^2*sin(c)^2-3*sin(a)^2*sin(c)^2-4*cos(a)*cos
(c)*sin(a)*sin(c)-3*cos(a)^2*cos(c)^2-cos(c)^2*sin(a)^2)/(cos(a)*sin(c)-sin(a)*cos(c))^4/(tan(b*x+a)*cos(a)*si
n(c)-tan(b*x+a)*sin(a)*cos(c)-sin(a)*sin(c)-cos(a)*cos(c))^3)

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maxima [B]  time = 0.35, size = 1074, normalized size = 18.20 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+c)^5*sin(b*x+a),x, algorithm="maxima")

[Out]

2/3*((6*cos(4*b*x + 2*a + 4*c) + 4*cos(2*b*x + 2*a + 2*c) - 4*cos(2*b*x + 4*c) + cos(2*a) - cos(2*c))*cos(8*b*
x + a + 9*c) + 4*(6*cos(4*b*x + 2*a + 4*c) + 4*cos(2*b*x + 2*a + 2*c) - 4*cos(2*b*x + 4*c) + cos(2*a) - cos(2*
c))*cos(6*b*x + a + 7*c) + 6*(4*cos(2*b*x + a + 3*c) + cos(a + c))*cos(4*b*x + 2*a + 4*c) + 6*(6*cos(4*b*x + 2
*a + 4*c) + 4*cos(2*b*x + 2*a + 2*c) - 4*cos(2*b*x + 4*c) + cos(2*a) - cos(2*c))*cos(4*b*x + a + 5*c) + 4*(4*c
os(2*b*x + 2*a + 2*c) + cos(2*a) - cos(2*c))*cos(2*b*x + a + 3*c) - 4*(4*cos(2*b*x + a + 3*c) + cos(a + c))*co
s(2*b*x + 4*c) + (cos(2*a) - cos(2*c))*cos(a + c) + 4*cos(2*b*x + 2*a + 2*c)*cos(a + c) + (6*sin(4*b*x + 2*a +
4*c) + 4*sin(2*b*x + 2*a + 2*c) - 4*sin(2*b*x + 4*c) + sin(2*a) - sin(2*c))*sin(8*b*x + a + 9*c) + 4*(6*sin(4
*b*x + 2*a + 4*c) + 4*sin(2*b*x + 2*a + 2*c) - 4*sin(2*b*x + 4*c) + sin(2*a) - sin(2*c))*sin(6*b*x + a + 7*c)
+ 6*(4*sin(2*b*x + a + 3*c) + sin(a + c))*sin(4*b*x + 2*a + 4*c) + 6*(6*sin(4*b*x + 2*a + 4*c) + 4*sin(2*b*x +
2*a + 2*c) - 4*sin(2*b*x + 4*c) + sin(2*a) - sin(2*c))*sin(4*b*x + a + 5*c) + 4*(4*sin(2*b*x + 2*a + 2*c) + s
in(2*a) - sin(2*c))*sin(2*b*x + a + 3*c) - 4*(4*sin(2*b*x + a + 3*c) + sin(a + c))*sin(2*b*x + 4*c) + (sin(2*a
) - sin(2*c))*sin(a + c) + 4*sin(2*b*x + 2*a + 2*c)*sin(a + c))/(b*cos(8*b*x + a + 9*c)^2 + 16*b*cos(6*b*x + a
+ 7*c)^2 + 36*b*cos(4*b*x + a + 5*c)^2 + 16*b*cos(2*b*x + a + 3*c)^2 + 8*b*cos(2*b*x + a + 3*c)*cos(a + c) +
b*cos(a + c)^2 + b*sin(8*b*x + a + 9*c)^2 + 16*b*sin(6*b*x + a + 7*c)^2 + 36*b*sin(4*b*x + a + 5*c)^2 + 16*b*s
in(2*b*x + a + 3*c)^2 + 8*b*sin(2*b*x + a + 3*c)*sin(a + c) + b*sin(a + c)^2 + 2*(4*b*cos(6*b*x + a + 7*c) + 6
*b*cos(4*b*x + a + 5*c) + 4*b*cos(2*b*x + a + 3*c) + b*cos(a + c))*cos(8*b*x + a + 9*c) + 8*(6*b*cos(4*b*x + a
+ 5*c) + 4*b*cos(2*b*x + a + 3*c) + b*cos(a + c))*cos(6*b*x + a + 7*c) + 12*(4*b*cos(2*b*x + a + 3*c) + b*cos
(a + c))*cos(4*b*x + a + 5*c) + 2*(4*b*sin(6*b*x + a + 7*c) + 6*b*sin(4*b*x + a + 5*c) + 4*b*sin(2*b*x + a + 3
*c) + b*sin(a + c))*sin(8*b*x + a + 9*c) + 8*(6*b*sin(4*b*x + a + 5*c) + 4*b*sin(2*b*x + a + 3*c) + b*sin(a +
c))*sin(6*b*x + a + 7*c) + 12*(4*b*sin(2*b*x + a + 3*c) + b*sin(a + c))*sin(4*b*x + a + 5*c))

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mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.02 \[ \text {Hanged} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)/cos(c + b*x)^5,x)

[Out]

\text{Hanged}

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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(b*x+c)**5*sin(b*x+a),x)

[Out]

Timed out

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