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

Optimal. Leaf size=61 \[ \frac{(a-b)^2 \log (\sin (c+d x)+1)}{2 d}-\frac{(a+b)^2 \log (1-\sin (c+d x))}{2 d}-\frac{b^2 \sin (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.0818299, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {2668, 702, 633, 31} \[ \frac{(a-b)^2 \log (\sin (c+d x)+1)}{2 d}-\frac{(a+b)^2 \log (1-\sin (c+d x))}{2 d}-\frac{b^2 \sin (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(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] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 702

Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[(d + e*x)^m, a + c*x^2,
x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[m, 1] && (NeQ[d, 0] || GtQ[m, 2])

Rule 633

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[e/2 + (c*d)/(2*q),
Int[1/(-q + c*x), x], x] + Dist[e/2 - (c*d)/(2*q), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[-(a*c)]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

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

Mathematica [A]  time = 0.0625328, size = 54, normalized size = 0.89 \[ \frac{(a-b)^2 \log (\sin (c+d x)+1)-(a+b)^2 \log (1-\sin (c+d x))-2 b^2 \sin (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.043, size = 72, normalized size = 1.2 \begin{align*}{\frac{{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}-{\frac{{b}^{2}\sin \left ( dx+c \right ) }{d}}-2\,{\frac{ab\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.962998, size = 81, normalized size = 1.33 \begin{align*} -\frac{2 \, b^{2} \sin \left (d x + c\right ) -{\left (a^{2} - 2 \, a b + b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) +{\left (a^{2} + 2 \, a b + b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.27886, size = 159, normalized size = 2.61 \begin{align*} -\frac{2 \, b^{2} \sin \left (d x + c\right ) -{\left (a^{2} - 2 \, a b + b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) +{\left (a^{2} + 2 \, a b + b^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.12927, size = 84, normalized size = 1.38 \begin{align*} -\frac{2 \, b^{2} \sin \left (d x + c\right ) -{\left (a^{2} - 2 \, a b + b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) +{\left (a^{2} + 2 \, a b + b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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