3.186 \(\int (a+b \sec (c+d x))^3 \sin ^3(c+d x) \, dx\)

Optimal. Leaf size=116 \[ \frac {a^3 \cos ^3(c+d x)}{3 d}-\frac {a \left (a^2-3 b^2\right ) \cos (c+d x)}{d}-\frac {b \left (3 a^2-b^2\right ) \log (\cos (c+d x))}{d}+\frac {3 a^2 b \cos ^2(c+d x)}{2 d}+\frac {3 a b^2 \sec (c+d x)}{d}+\frac {b^3 \sec ^2(c+d x)}{2 d} \]

[Out]

-a*(a^2-3*b^2)*cos(d*x+c)/d+3/2*a^2*b*cos(d*x+c)^2/d+1/3*a^3*cos(d*x+c)^3/d-b*(3*a^2-b^2)*ln(cos(d*x+c))/d+3*a
*b^2*sec(d*x+c)/d+1/2*b^3*sec(d*x+c)^2/d

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Rubi [A]  time = 0.13, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3872, 2721, 894} \[ -\frac {a \left (a^2-3 b^2\right ) \cos (c+d x)}{d}-\frac {b \left (3 a^2-b^2\right ) \log (\cos (c+d x))}{d}+\frac {3 a^2 b \cos ^2(c+d x)}{2 d}+\frac {a^3 \cos ^3(c+d x)}{3 d}+\frac {3 a b^2 \sec (c+d x)}{d}+\frac {b^3 \sec ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[c + d*x])^3*Sin[c + d*x]^3,x]

[Out]

-((a*(a^2 - 3*b^2)*Cos[c + d*x])/d) + (3*a^2*b*Cos[c + d*x]^2)/(2*d) + (a^3*Cos[c + d*x]^3)/(3*d) - (b*(3*a^2
- b^2)*Log[Cos[c + d*x]])/d + (3*a*b^2*Sec[c + d*x])/d + (b^3*Sec[c + d*x]^2)/(2*d)

Rule 894

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 2721

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2
- b^2, 0] && IntegerQ[(p + 1)/2]

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rubi steps

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

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Mathematica [A]  time = 0.35, size = 102, normalized size = 0.88 \[ \frac {a^3 \cos (3 (c+d x))-9 a \left (a^2-4 b^2\right ) \cos (c+d x)+9 a^2 b \cos (2 (c+d x))-36 a^2 b \log (\cos (c+d x))+36 a b^2 \sec (c+d x)+6 b^3 \sec ^2(c+d x)+12 b^3 \log (\cos (c+d x))}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sec[c + d*x])^3*Sin[c + d*x]^3,x]

[Out]

(-9*a*(a^2 - 4*b^2)*Cos[c + d*x] + 9*a^2*b*Cos[2*(c + d*x)] + a^3*Cos[3*(c + d*x)] - 36*a^2*b*Log[Cos[c + d*x]
] + 12*b^3*Log[Cos[c + d*x]] + 36*a*b^2*Sec[c + d*x] + 6*b^3*Sec[c + d*x]^2)/(12*d)

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fricas [A]  time = 0.51, size = 123, normalized size = 1.06 \[ \frac {4 \, a^{3} \cos \left (d x + c\right )^{5} + 18 \, a^{2} b \cos \left (d x + c\right )^{4} - 9 \, a^{2} b \cos \left (d x + c\right )^{2} + 36 \, a b^{2} \cos \left (d x + c\right ) - 12 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} - 12 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\cos \left (d x + c\right )\right ) + 6 \, b^{3}}{12 \, d \cos \left (d x + c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^3*sin(d*x+c)^3,x, algorithm="fricas")

[Out]

1/12*(4*a^3*cos(d*x + c)^5 + 18*a^2*b*cos(d*x + c)^4 - 9*a^2*b*cos(d*x + c)^2 + 36*a*b^2*cos(d*x + c) - 12*(a^
3 - 3*a*b^2)*cos(d*x + c)^3 - 12*(3*a^2*b - b^3)*cos(d*x + c)^2*log(-cos(d*x + c)) + 6*b^3)/(d*cos(d*x + c)^2)

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giac [A]  time = 0.40, size = 128, normalized size = 1.10 \[ -\frac {{\left (3 \, a^{2} b - b^{3}\right )} \log \left (\frac {{\left | \cos \left (d x + c\right ) \right |}}{{\left | d \right |}}\right )}{d} + \frac {6 \, a b^{2} \cos \left (d x + c\right ) + b^{3}}{2 \, d \cos \left (d x + c\right )^{2}} + \frac {2 \, a^{3} d^{8} \cos \left (d x + c\right )^{3} + 9 \, a^{2} b d^{8} \cos \left (d x + c\right )^{2} - 6 \, a^{3} d^{8} \cos \left (d x + c\right ) + 18 \, a b^{2} d^{8} \cos \left (d x + c\right )}{6 \, d^{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^3*sin(d*x+c)^3,x, algorithm="giac")

[Out]

-(3*a^2*b - b^3)*log(abs(cos(d*x + c))/abs(d))/d + 1/2*(6*a*b^2*cos(d*x + c) + b^3)/(d*cos(d*x + c)^2) + 1/6*(
2*a^3*d^8*cos(d*x + c)^3 + 9*a^2*b*d^8*cos(d*x + c)^2 - 6*a^3*d^8*cos(d*x + c) + 18*a*b^2*d^8*cos(d*x + c))/d^
9

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maple [A]  time = 0.65, size = 164, normalized size = 1.41 \[ -\frac {a^{3} \cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{3 d}-\frac {2 a^{3} \cos \left (d x +c \right )}{3 d}-\frac {3 a^{2} b \left (\sin ^{2}\left (d x +c \right )\right )}{2 d}-\frac {3 a^{2} b \ln \left (\cos \left (d x +c \right )\right )}{d}+\frac {3 b^{2} a \left (\sin ^{4}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )}+\frac {3 b^{2} a \cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{d}+\frac {6 a \,b^{2} \cos \left (d x +c \right )}{d}+\frac {b^{3} \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}+\frac {b^{3} \ln \left (\cos \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(d*x+c))^3*sin(d*x+c)^3,x)

[Out]

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

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maxima [A]  time = 0.61, size = 98, normalized size = 0.84 \[ \frac {2 \, a^{3} \cos \left (d x + c\right )^{3} + 9 \, a^{2} b \cos \left (d x + c\right )^{2} - 6 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right ) - 6 \, {\left (3 \, a^{2} b - b^{3}\right )} \log \left (\cos \left (d x + c\right )\right ) + \frac {3 \, {\left (6 \, a b^{2} \cos \left (d x + c\right ) + b^{3}\right )}}{\cos \left (d x + c\right )^{2}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^3*sin(d*x+c)^3,x, algorithm="maxima")

[Out]

1/6*(2*a^3*cos(d*x + c)^3 + 9*a^2*b*cos(d*x + c)^2 - 6*(a^3 - 3*a*b^2)*cos(d*x + c) - 6*(3*a^2*b - b^3)*log(co
s(d*x + c)) + 3*(6*a*b^2*cos(d*x + c) + b^3)/cos(d*x + c)^2)/d

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mupad [B]  time = 0.07, size = 99, normalized size = 0.85 \[ \frac {\frac {\frac {b^3}{2}+3\,a\,\cos \left (c+d\,x\right )\,b^2}{{\cos \left (c+d\,x\right )}^2}-\ln \left (\cos \left (c+d\,x\right )\right )\,\left (3\,a^2\,b-b^3\right )+\cos \left (c+d\,x\right )\,\left (3\,a\,b^2-a^3\right )+\frac {a^3\,{\cos \left (c+d\,x\right )}^3}{3}+\frac {3\,a^2\,b\,{\cos \left (c+d\,x\right )}^2}{2}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^3*(a + b/cos(c + d*x))^3,x)

[Out]

((b^3/2 + 3*a*b^2*cos(c + d*x))/cos(c + d*x)^2 - log(cos(c + d*x))*(3*a^2*b - b^3) + cos(c + d*x)*(3*a*b^2 - a
^3) + (a^3*cos(c + d*x)^3)/3 + (3*a^2*b*cos(c + d*x)^2)/2)/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \sec {\left (c + d x \right )}\right )^{3} \sin ^{3}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))**3*sin(d*x+c)**3,x)

[Out]

Integral((a + b*sec(c + d*x))**3*sin(c + d*x)**3, x)

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