3.1.19 \(\int \sec (c+d x) (a+a \sin (c+d x))^2 \, dx\) [19]

Optimal. Leaf size=34 \[ -\frac {2 a^2 \log (1-\sin (c+d x))}{d}-\frac {a^2 \sin (c+d x)}{d} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2746, 45} \begin {gather*} -\frac {a^2 \sin (c+d x)}{d}-\frac {2 a^2 \log (1-\sin (c+d x))}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 45

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])

Rule 2746

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

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 29, normalized size = 0.85 \begin {gather*} \frac {a^2 (-2 \log (1-\sin (c+d x))-\sin (c+d x))}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.12, size = 63, normalized size = 1.85

method	result	size
derivativedivides	\(\frac {a^{2} \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )-2 a^{2} \ln \left (\cos \left (d x +c \right )\right )+a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(63\)
default	\(\frac {a^{2} \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )-2 a^{2} \ln \left (\cos \left (d x +c \right )\right )+a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(63\)
risch	\(2 i a^{2} x +\frac {i a^{2} {\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {i a^{2} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {4 i a^{2} c}{d}-\frac {4 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}\)	\(76\)
norman	\(\frac {-\frac {2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 a^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {4 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}+\frac {2 a^{2} \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\)	\(97\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.33, size = 30, normalized size = 0.88 \begin {gather*} -\frac {2 \, a^{2} \log \left (\sin \left (d x + c\right ) - 1\right ) + a^{2} \sin \left (d x + c\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.39, size = 32, normalized size = 0.94 \begin {gather*} -\frac {2 \, a^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + a^{2} \sin \left (d x + c\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 91 vs. \(2 (34) = 68\).
time = 3.61, size = 91, normalized size = 2.68 \begin {gather*} \frac {2 \, {\left (a^{2} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right ) - 2 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1}\right )}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.05, size = 26, normalized size = 0.76 \begin {gather*} -\frac {a^2\,\left (2\,\ln \left (\sin \left (c+d\,x\right )-1\right )+\sin \left (c+d\,x\right )\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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