3.72 \(\int \frac{\sec ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=104 \[ \frac{1}{16 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac{3}{16 d \left (a^2 \sin (c+d x)+a^2\right )}+\frac{\tanh ^{-1}(\sin (c+d x))}{4 a^2 d}-\frac{a}{12 d (a \sin (c+d x)+a)^3}-\frac{1}{8 d (a \sin (c+d x)+a)^2} \]

[Out]

ArcTanh[Sin[c + d*x]]/(4*a^2*d) - a/(12*d*(a + a*Sin[c + d*x])^3) - 1/(8*d*(a + a*Sin[c + d*x])^2) + 1/(16*d*(
a^2 - a^2*Sin[c + d*x])) - 3/(16*d*(a^2 + a^2*Sin[c + d*x]))

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Rubi [A]  time = 0.0820373, antiderivative size = 104, normalized size of antiderivative = 1., 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 = {2667, 44, 206} \[ \frac{1}{16 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac{3}{16 d \left (a^2 \sin (c+d x)+a^2\right )}+\frac{\tanh ^{-1}(\sin (c+d x))}{4 a^2 d}-\frac{a}{12 d (a \sin (c+d x)+a)^3}-\frac{1}{8 d (a \sin (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[Sin[c + d*x]]/(4*a^2*d) - a/(12*d*(a + a*Sin[c + d*x])^3) - 1/(8*d*(a + a*Sin[c + d*x])^2) + 1/(16*d*(
a^2 - a^2*Sin[c + d*x])) - 3/(16*d*(a^2 + a^2*Sin[c + d*x]))

Rule 2667

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])

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\sec ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac{a^3 \operatorname{Subst}\left (\int \frac{1}{(a-x)^2 (a+x)^4} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{a^3 \operatorname{Subst}\left (\int \left (\frac{1}{16 a^4 (a-x)^2}+\frac{1}{4 a^2 (a+x)^4}+\frac{1}{4 a^3 (a+x)^3}+\frac{3}{16 a^4 (a+x)^2}+\frac{1}{4 a^4 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac{a}{12 d (a+a \sin (c+d x))^3}-\frac{1}{8 d (a+a \sin (c+d x))^2}+\frac{1}{16 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac{3}{16 d \left (a^2+a^2 \sin (c+d x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{4 a d}\\ &=\frac{\tanh ^{-1}(\sin (c+d x))}{4 a^2 d}-\frac{a}{12 d (a+a \sin (c+d x))^3}-\frac{1}{8 d (a+a \sin (c+d x))^2}+\frac{1}{16 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac{3}{16 d \left (a^2+a^2 \sin (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.119599, size = 85, normalized size = 0.82 \[ -\frac{\sec ^2(c+d x) \left (-3 \sin ^3(c+d x)-6 \sin ^2(c+d x)-\sin (c+d x)+3 (\sin (c+d x)-1) (\sin (c+d x)+1)^3 \tanh ^{-1}(\sin (c+d x))+4\right )}{12 a^2 d (\sin (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sec[c + d*x]^2*(4 - Sin[c + d*x] - 6*Sin[c + d*x]^2 - 3*Sin[c + d*x]^3 + 3*ArcTanh[Sin[c + d*x]]*(-1 + Sin[c
+ d*x])*(1 + Sin[c + d*x])^3))/(12*a^2*d*(1 + Sin[c + d*x])^2)

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Maple [A]  time = 0.087, size = 108, normalized size = 1. \begin{align*} -{\frac{1}{16\,d{a}^{2} \left ( \sin \left ( dx+c \right ) -1 \right ) }}-{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ) }{8\,d{a}^{2}}}-{\frac{1}{12\,d{a}^{2} \left ( 1+\sin \left ( dx+c \right ) \right ) ^{3}}}-{\frac{1}{8\,d{a}^{2} \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{3}{16\,d{a}^{2} \left ( 1+\sin \left ( dx+c \right ) \right ) }}+{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{8\,d{a}^{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,x)

[Out]

-1/16/d/a^2/(sin(d*x+c)-1)-1/8/d/a^2*ln(sin(d*x+c)-1)-1/12/d/a^2/(1+sin(d*x+c))^3-1/8/d/a^2/(1+sin(d*x+c))^2-3
/16/d/a^2/(1+sin(d*x+c))+1/8*ln(1+sin(d*x+c))/a^2/d

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Maxima [A]  time = 1.14414, size = 146, normalized size = 1.4 \begin{align*} -\frac{\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{3} + 6 \, \sin \left (d x + c\right )^{2} + \sin \left (d x + c\right ) - 4\right )}}{a^{2} \sin \left (d x + c\right )^{4} + 2 \, a^{2} \sin \left (d x + c\right )^{3} - 2 \, a^{2} \sin \left (d x + c\right ) - a^{2}} - \frac{3 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}} + \frac{3 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2}}}{24 \, 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,x, algorithm="maxima")

[Out]

-1/24*(2*(3*sin(d*x + c)^3 + 6*sin(d*x + c)^2 + sin(d*x + c) - 4)/(a^2*sin(d*x + c)^4 + 2*a^2*sin(d*x + c)^3 -
2*a^2*sin(d*x + c) - a^2) - 3*log(sin(d*x + c) + 1)/a^2 + 3*log(sin(d*x + c) - 1)/a^2)/d

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Fricas [A]  time = 2.20367, size = 466, normalized size = 4.48 \begin{align*} \frac{12 \, \cos \left (d x + c\right )^{2} + 3 \,{\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (3 \, \cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right ) - 4}{24 \,{\left (a^{2} d \cos \left (d x + c\right )^{4} - 2 \, a^{2} d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \, a^{2} d \cos \left (d x + c\right )^{2}\right )}} \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,x, algorithm="fricas")

[Out]

1/24*(12*cos(d*x + c)^2 + 3*(cos(d*x + c)^4 - 2*cos(d*x + c)^2*sin(d*x + c) - 2*cos(d*x + c)^2)*log(sin(d*x +
c) + 1) - 3*(cos(d*x + c)^4 - 2*cos(d*x + c)^2*sin(d*x + c) - 2*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) + 2*(3*
cos(d*x + c)^2 - 4)*sin(d*x + c) - 4)/(a^2*d*cos(d*x + c)^4 - 2*a^2*d*cos(d*x + c)^2*sin(d*x + c) - 2*a^2*d*co
s(d*x + c)^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\sec ^{3}{\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin{\left (c + d x \right )} + 1}\, dx}{a^{2}} \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,x)

[Out]

Integral(sec(c + d*x)**3/(sin(c + d*x)**2 + 2*sin(c + d*x) + 1), x)/a**2

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Giac [A]  time = 1.19462, size = 143, normalized size = 1.38 \begin{align*} \frac{\frac{6 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2}} - \frac{6 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{2}} + \frac{3 \,{\left (2 \, \sin \left (d x + c\right ) - 3\right )}}{a^{2}{\left (\sin \left (d x + c\right ) - 1\right )}} - \frac{11 \, \sin \left (d x + c\right )^{3} + 42 \, \sin \left (d x + c\right )^{2} + 57 \, \sin \left (d x + c\right ) + 30}{a^{2}{\left (\sin \left (d x + c\right ) + 1\right )}^{3}}}{48 \, 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,x, algorithm="giac")

[Out]

1/48*(6*log(abs(sin(d*x + c) + 1))/a^2 - 6*log(abs(sin(d*x + c) - 1))/a^2 + 3*(2*sin(d*x + c) - 3)/(a^2*(sin(d
*x + c) - 1)) - (11*sin(d*x + c)^3 + 42*sin(d*x + c)^2 + 57*sin(d*x + c) + 30)/(a^2*(sin(d*x + c) + 1)^3))/d