3.64 \(\int \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=86 \[ \frac {a^3 \sin (c+d x)}{d}-\frac {3 a^2 b \cos (c+d x)}{d}-\frac {3 a b^2 \sin (c+d x)}{d}+\frac {3 a b^2 \tanh ^{-1}(\sin (c+d x))}{d}+\frac {b^3 \cos (c+d x)}{d}+\frac {b^3 \sec (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3090, 2637, 2638, 2592, 321, 206, 2590, 14} \[ -\frac {3 a^2 b \cos (c+d x)}{d}+\frac {a^3 \sin (c+d x)}{d}-\frac {3 a b^2 \sin (c+d x)}{d}+\frac {3 a b^2 \tanh ^{-1}(\sin (c+d x))}{d}+\frac {b^3 \cos (c+d x)}{d}+\frac {b^3 \sec (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^3,x]

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2590

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[f^(-1), Subst[Int[(1 - x^2
)^((m + n - 1)/2)/x^n, x], x, Cos[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/2]

Rule 2592

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, (a*Sin[e + f*x])/ff
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2638

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

Rule 3090

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_.), x_Sym
bol] :> Int[ExpandTrig[cos[c + d*x]^m*(a*cos[c + d*x] + b*sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d}, x] &&
IntegerQ[m] && IGtQ[n, 0]

Rubi steps

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

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Mathematica [A]  time = 1.07, size = 131, normalized size = 1.52 \[ \frac {\sec (c+d x) \left (a^3 \sin (2 (c+d x))+\left (b^3-3 a^2 b\right ) \cos (2 (c+d x))-3 a^2 b-3 a b^2 \sin (2 (c+d x))-6 a b^2 \cos (c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )+3 b^3\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^3,x]

[Out]

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

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fricas [A]  time = 0.65, size = 109, normalized size = 1.27 \[ \frac {3 \, a b^{2} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, a b^{2} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, b^{3} - 2 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.29, size = 150, normalized size = 1.74 \[ \frac {3 \, a b^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, a b^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a^{2} b - 2 \, b^{3}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

(3*a*b^2*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*a*b^2*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(a^3*tan(1/2*d*x
+ 1/2*c)^3 - 3*a*b^2*tan(1/2*d*x + 1/2*c)^3 - 3*a^2*b*tan(1/2*d*x + 1/2*c)^2 - a^3*tan(1/2*d*x + 1/2*c) + 3*a*
b^2*tan(1/2*d*x + 1/2*c) + 3*a^2*b - 2*b^3)/(tan(1/2*d*x + 1/2*c)^4 - 1))/d

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maple [A]  time = 10.96, size = 126, normalized size = 1.47 \[ \frac {a^{3} \sin \left (d x +c \right )}{d}-\frac {3 a^{2} b \cos \left (d x +c \right )}{d}-\frac {3 a \,b^{2} \sin \left (d x +c \right )}{d}+\frac {3 b^{2} a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )}+\frac {b^{3} \cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{d}+\frac {2 b^{3} \cos \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*(a*cos(d*x+c)+b*sin(d*x+c))^3,x)

[Out]

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

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maxima [A]  time = 0.33, size = 84, normalized size = 0.98 \[ \frac {2 \, b^{3} {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )} + 3 \, a b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right ) - 2 \, \sin \left (d x + c\right )\right )} - 6 \, a^{2} b \cos \left (d x + c\right ) + 2 \, a^{3} \sin \left (d x + c\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.05, size = 116, normalized size = 1.35 \[ \frac {6\,a\,b^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (6\,a\,b^2-2\,a^3\right )-6\,a^2\,b-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (6\,a\,b^2-2\,a^3\right )+4\,b^3+6\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*cos(c + d*x) + b*sin(c + d*x))^3/cos(c + d*x)^2,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \cos {\left (c + d x \right )} + b \sin {\left (c + d x \right )}\right )^{3} \sec ^{2}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*(a*cos(d*x+c)+b*sin(d*x+c))**3,x)

[Out]

Integral((a*cos(c + d*x) + b*sin(c + d*x))**3*sec(c + d*x)**2, x)

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