3.974 \(\int \sec ^3(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=43 \[ \frac{a^3 (A+B)}{d (a-a \sin (c+d x))}+\frac{a^2 B \log (1-\sin (c+d x))}{d} \]

[Out]

(a^2*B*Log[1 - Sin[c + d*x]])/d + (a^3*(A + B))/(d*(a - a*Sin[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 0.0903995, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.065, Rules used = {2836, 43} \[ \frac{a^3 (A+B)}{d (a-a \sin (c+d x))}+\frac{a^2 B \log (1-\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*B*Log[1 - Sin[c + d*x]])/d + (a^3*(A + B))/(d*(a - a*Sin[c + d*x]))

Rule 2836

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*x)/b
)^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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0742413, size = 41, normalized size = 0.95 \[ \frac{a^3 \left (\frac{A+B}{a-a \sin (c+d x)}+\frac{B \log (1-\sin (c+d x))}{a}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*((B*Log[1 - Sin[c + d*x]])/a + (A + B)/(a - a*Sin[c + d*x])))/d

________________________________________________________________________________________

Maple [B]  time = 0.103, size = 189, normalized size = 4.4 \begin{align*}{\frac{{a}^{2}A \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{a}^{2}A\sin \left ( dx+c \right ) }{2\,d}}+{\frac{B{a}^{2} \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,d}}+{\frac{B{a}^{2}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+{\frac{{a}^{2}A}{d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{B{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{B{a}^{2}\sin \left ( dx+c \right ) }{d}}-{\frac{B{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{{a}^{2}A\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{B{a}^{2}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.08152, size = 50, normalized size = 1.16 \begin{align*} \frac{B a^{2} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac{{\left (A + B\right )} a^{2}}{\sin \left (d x + c\right ) - 1}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(B*a^2*log(sin(d*x + c) - 1) - (A + B)*a^2/(sin(d*x + c) - 1))/d

________________________________________________________________________________________

Fricas [A]  time = 1.73785, size = 123, normalized size = 2.86 \begin{align*} -\frac{{\left (A + B\right )} a^{2} -{\left (B a^{2} \sin \left (d x + c\right ) - B a^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{d \sin \left (d x + c\right ) - d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((A + B)*a^2 - (B*a^2*sin(d*x + c) - B*a^2)*log(-sin(d*x + c) + 1))/(d*sin(d*x + c) - d)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3*(a+a*sin(d*x+c))**2*(A+B*sin(d*x+c)),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.33581, size = 151, normalized size = 3.51 \begin{align*} -\frac{B a^{2} \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right ) - 2 \, B a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{3 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 2 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 8 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, B a^{2}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}^{2}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(B*a^2*log(tan(1/2*d*x + 1/2*c)^2 + 1) - 2*B*a^2*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + (3*B*a^2*tan(1/2*d*x +
1/2*c)^2 - 2*A*a^2*tan(1/2*d*x + 1/2*c) - 8*B*a^2*tan(1/2*d*x + 1/2*c) + 3*B*a^2)/(tan(1/2*d*x + 1/2*c) - 1)^2
)/d