3.1.41 \(\int (a+a \sec (c+d x))^3 \sin ^3(c+d x) \, dx\) [41]

Optimal. Leaf size=98 \[ \frac {2 a^3 \cos (c+d x)}{d}+\frac {3 a^3 \cos ^2(c+d x)}{2 d}+\frac {a^3 \cos ^3(c+d x)}{3 d}-\frac {2 a^3 \log (\cos (c+d x))}{d}+\frac {3 a^3 \sec (c+d x)}{d}+\frac {a^3 \sec ^2(c+d x)}{2 d} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 98, 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 = {3957, 2786, 76} \begin {gather*} \frac {a^3 \cos ^3(c+d x)}{3 d}+\frac {3 a^3 \cos ^2(c+d x)}{2 d}+\frac {2 a^3 \cos (c+d x)}{d}+\frac {a^3 \sec ^2(c+d x)}{2 d}+\frac {3 a^3 \sec (c+d x)}{d}-\frac {2 a^3 \log (\cos (c+d x))}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n
+ p + 2, 0] && GtQ[n + 2*p, 0])

Rule 2786

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 - (p + 1)/2)/(a - x)^((p + 1)/2)), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& EqQ[a^2 - b^2, 0] && IntegerQ[(p + 1)/2]

Rule 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[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+a \sec (c+d x))^3 \sin ^3(c+d x) \, dx &=-\int (-a-a \cos (c+d x))^3 \tan ^3(c+d x) \, dx\\ &=\frac {\text {Subst}\left (\int \frac {(-a-x) (-a+x)^4}{x^3} \, dx,x,-a \cos (c+d x)\right )}{d}\\ &=\frac {\text {Subst}\left (\int \left (-2 a^2-\frac {a^5}{x^3}+\frac {3 a^4}{x^2}-\frac {2 a^3}{x}+3 a x-x^2\right ) \, dx,x,-a \cos (c+d x)\right )}{d}\\ &=\frac {2 a^3 \cos (c+d x)}{d}+\frac {3 a^3 \cos ^2(c+d x)}{2 d}+\frac {a^3 \cos ^3(c+d x)}{3 d}-\frac {2 a^3 \log (\cos (c+d x))}{d}+\frac {3 a^3 \sec (c+d x)}{d}+\frac {a^3 \sec ^2(c+d x)}{2 d}\\ \end {align*}

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Mathematica [A]
time = 0.13, size = 86, normalized size = 0.88 \begin {gather*} \frac {a^3 (-41+226 \cos (c+d x)+29 \cos (3 (c+d x))+9 \cos (4 (c+d x))+\cos (5 (c+d x))-48 \log (\cos (c+d x))-8 \cos (2 (c+d x)) (7+6 \log (\cos (c+d x)))) \sec ^2(c+d x)}{48 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(-41 + 226*Cos[c + d*x] + 29*Cos[3*(c + d*x)] + 9*Cos[4*(c + d*x)] + Cos[5*(c + d*x)] - 48*Log[Cos[c + d*
x]] - 8*Cos[2*(c + d*x)]*(7 + 6*Log[Cos[c + d*x]]))*Sec[c + d*x]^2)/(48*d)

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Maple [A]
time = 0.12, size = 114, normalized size = 1.16

method	result	size
derivativedivides	\(\frac {a^{3} \left (\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\cos \left (d x +c \right )\right )\right )+3 a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+3 a^{3} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )-\frac {a^{3} \left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}}{d}\)	\(114\)
default	\(\frac {a^{3} \left (\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\cos \left (d x +c \right )\right )\right )+3 a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+3 a^{3} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )-\frac {a^{3} \left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}}{d}\)	\(114\)
norman	\(\frac {\frac {32 a^{3}}{3 d}-\frac {4 a^{3} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a^{3} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {20 a^{3} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {20 a^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {2 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}-\frac {2 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}+\frac {2 a^{3} \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\)	\(180\)
risch	\(2 i a^{3} x +\frac {a^{3} {\mathrm e}^{3 i \left (d x +c \right )}}{24 d}+\frac {3 a^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {9 a^{3} {\mathrm e}^{i \left (d x +c \right )}}{8 d}+\frac {9 a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{8 d}+\frac {3 a^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {a^{3} {\mathrm e}^{-3 i \left (d x +c \right )}}{24 d}+\frac {4 i a^{3} c}{d}+\frac {2 a^{3} \left (3 \,{\mathrm e}^{3 i \left (d x +c \right )}+{\mathrm e}^{2 i \left (d x +c \right )}+3 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {2 a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\)	\(194\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^3*sin(d*x+c)^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.26, size = 80, normalized size = 0.82 \begin {gather*} \frac {2 \, a^{3} \cos \left (d x + c\right )^{3} + 9 \, a^{3} \cos \left (d x + c\right )^{2} + 12 \, a^{3} \cos \left (d x + c\right ) - 12 \, a^{3} \log \left (\cos \left (d x + c\right )\right ) + \frac {3 \, {\left (6 \, a^{3} \cos \left (d x + c\right ) + a^{3}\right )}}{\cos \left (d x + c\right )^{2}}}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*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^3*cos(d*x + c)^2 + 12*a^3*cos(d*x + c) - 12*a^3*log(cos(d*x + c)) + 3*(6*a^3*c
os(d*x + c) + a^3)/cos(d*x + c)^2)/d

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Fricas [A]
time = 2.97, size = 104, normalized size = 1.06 \begin {gather*} \frac {4 \, a^{3} \cos \left (d x + c\right )^{5} + 18 \, a^{3} \cos \left (d x + c\right )^{4} + 24 \, a^{3} \cos \left (d x + c\right )^{3} - 24 \, a^{3} \cos \left (d x + c\right )^{2} \log \left (-\cos \left (d x + c\right )\right ) - 9 \, a^{3} \cos \left (d x + c\right )^{2} + 36 \, a^{3} \cos \left (d x + c\right ) + 6 \, a^{3}}{12 \, d \cos \left (d x + c\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*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^3*cos(d*x + c)^4 + 24*a^3*cos(d*x + c)^3 - 24*a^3*cos(d*x + c)^2*log(-cos(d*
x + c)) - 9*a^3*cos(d*x + c)^2 + 36*a^3*cos(d*x + c) + 6*a^3)/(d*cos(d*x + c)^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a^{3} \left (\int 3 \sin ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 3 \sin ^{3}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \sin ^{3}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int \sin ^{3}{\left (c + d x \right )}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.49, size = 102, normalized size = 1.04 \begin {gather*} -\frac {2 \, a^{3} \log \left (\frac {{\left | \cos \left (d x + c\right ) \right |}}{{\left | d \right |}}\right )}{d} + \frac {6 \, a^{3} \cos \left (d x + c\right ) + a^{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^{3} d^{8} \cos \left (d x + c\right )^{2} + 12 \, a^{3} d^{8} \cos \left (d x + c\right )}{6 \, d^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a^3*log(abs(cos(d*x + c))/abs(d))/d + 1/2*(6*a^3*cos(d*x + c) + a^3)/(d*cos(d*x + c)^2) + 1/6*(2*a^3*d^8*co
s(d*x + c)^3 + 9*a^3*d^8*cos(d*x + c)^2 + 12*a^3*d^8*cos(d*x + c))/d^9

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Mupad [B]
time = 0.88, size = 80, normalized size = 0.82 \begin {gather*} \frac {\frac {3\,a^3\,\cos \left (c+d\,x\right )+\frac {a^3}{2}}{{\cos \left (c+d\,x\right )}^2}+2\,a^3\,\cos \left (c+d\,x\right )+\frac {3\,a^3\,{\cos \left (c+d\,x\right )}^2}{2}+\frac {a^3\,{\cos \left (c+d\,x\right )}^3}{3}-2\,a^3\,\ln \left (\cos \left (c+d\,x\right )\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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