3.1.22 \(\int (a+a \sec (c+d x))^2 \sin ^3(c+d x) \, dx\) [22]

Optimal. Leaf size=62 \[ \frac {a^2 \cos ^2(c+d x)}{d}+\frac {a^2 \cos ^3(c+d x)}{3 d}-\frac {2 a^2 \log (\cos (c+d x))}{d}+\frac {a^2 \sec (c+d x)}{d} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3957, 2915, 12, 76} \begin {gather*} \frac {a^2 \cos ^3(c+d x)}{3 d}+\frac {a^2 \cos ^2(c+d x)}{d}+\frac {a^2 \sec (c+d x)}{d}-\frac {2 a^2 \log (\cos (c+d x))}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 2915

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

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))^2 \sin ^3(c+d x) \, dx &=\int (-a-a \cos (c+d x))^2 \sin (c+d x) \tan ^2(c+d x) \, dx\\ &=\frac {\text {Subst}\left (\int \frac {a^2 (-a-x) (-a+x)^3}{x^2} \, dx,x,-a \cos (c+d x)\right )}{a^3 d}\\ &=\frac {\text {Subst}\left (\int \frac {(-a-x) (-a+x)^3}{x^2} \, dx,x,-a \cos (c+d x)\right )}{a d}\\ &=\frac {\text {Subst}\left (\int \left (\frac {a^4}{x^2}-\frac {2 a^3}{x}+2 a x-x^2\right ) \, dx,x,-a \cos (c+d x)\right )}{a d}\\ &=\frac {a^2 \cos ^2(c+d x)}{d}+\frac {a^2 \cos ^3(c+d x)}{3 d}-\frac {2 a^2 \log (\cos (c+d x))}{d}+\frac {a^2 \sec (c+d x)}{d}\\ \end {align*}

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Mathematica [A]
time = 0.14, size = 65, normalized size = 1.05 \begin {gather*} \frac {a^2 (27+4 \cos (2 (c+d x))+6 \cos (3 (c+d x))+\cos (4 (c+d x))-6 \cos (c+d x) (1+8 \log (\cos (c+d x)))) \sec (c+d x)}{24 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(27 + 4*Cos[2*(c + d*x)] + 6*Cos[3*(c + d*x)] + Cos[4*(c + d*x)] - 6*Cos[c + d*x]*(1 + 8*Log[Cos[c + d*x]
]))*Sec[c + d*x])/(24*d)

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Maple [A]
time = 0.09, size = 91, normalized size = 1.47

method	result	size
derivativedivides	\(\frac {a^{2} \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+2 a^{2} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )-\frac {a^{2} \left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}}{d}\)	\(91\)
default	\(\frac {a^{2} \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+2 a^{2} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )-\frac {a^{2} \left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}}{d}\)	\(91\)
risch	\(2 i a^{2} x +\frac {a^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{4 d}+\frac {a^{2} {\mathrm e}^{i \left (d x +c \right )}}{8 d}+\frac {a^{2} {\mathrm e}^{-i \left (d x +c \right )}}{8 d}+\frac {a^{2} {\mathrm e}^{-2 i \left (d x +c \right )}}{4 d}+\frac {4 i a^{2} c}{d}+\frac {2 a^{2} {\mathrm e}^{i \left (d x +c \right )}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {2 a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}+\frac {a^{2} \cos \left (3 d x +3 c \right )}{12 d}\)	\(154\)
norman	\(\frac {-\frac {8 a^{2}}{3 d}-\frac {8 a^{2} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a^{2} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {2 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}-\frac {2 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}+\frac {2 a^{2} \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\)	\(161\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^2*sin(d*x+c)^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.27, size = 56, normalized size = 0.90 \begin {gather*} \frac {a^{2} \cos \left (d x + c\right )^{3} + 3 \, a^{2} \cos \left (d x + c\right )^{2} - 6 \, a^{2} \log \left (\cos \left (d x + c\right )\right ) + \frac {3 \, a^{2}}{\cos \left (d x + c\right )}}{3 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 3.08, size = 76, normalized size = 1.23 \begin {gather*} \frac {2 \, a^{2} \cos \left (d x + c\right )^{4} + 6 \, a^{2} \cos \left (d x + c\right )^{3} - 12 \, a^{2} \cos \left (d x + c\right ) \log \left (-\cos \left (d x + c\right )\right ) - 3 \, a^{2} \cos \left (d x + c\right ) + 6 \, a^{2}}{6 \, d \cos \left (d x + c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Giac [A]
time = 0.67, size = 74, normalized size = 1.19 \begin {gather*} -\frac {2 \, a^{2} \log \left (\frac {{\left | \cos \left (d x + c\right ) \right |}}{{\left | d \right |}}\right )}{d} + \frac {a^{2}}{d \cos \left (d x + c\right )} + \frac {a^{2} d^{5} \cos \left (d x + c\right )^{3} + 3 \, a^{2} d^{5} \cos \left (d x + c\right )^{2}}{3 \, d^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a^2*log(abs(cos(d*x + c))/abs(d))/d + a^2/(d*cos(d*x + c)) + 1/3*(a^2*d^5*cos(d*x + c)^3 + 3*a^2*d^5*cos(d*
x + c)^2)/d^6

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Mupad [B]
time = 0.06, size = 54, normalized size = 0.87 \begin {gather*} \frac {\frac {a^2}{\cos \left (c+d\,x\right )}+a^2\,{\cos \left (c+d\,x\right )}^2+\frac {a^2\,{\cos \left (c+d\,x\right )}^3}{3}-2\,a^2\,\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))^2,x)

[Out]

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

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