3.1.72 \(\int \frac {\sec ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx\) [72]

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

[Out]

1/4*arctanh(sin(d*x+c))/a^2/d-1/12*a/d/(a+a*sin(d*x+c))^3-1/8/d/(a+a*sin(d*x+c))^2+1/16/d/(a^2-a^2*sin(d*x+c))
-3/16/d/(a^2+a^2*sin(d*x+c))

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Rubi [A]
time = 0.06, antiderivative size = 104, normalized size of antiderivative = 1.00, 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 = {2746, 46, 212} \begin {gather*} \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} \end {gather*}

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 46

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] && Lt
Q[m + n + 2, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 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 \frac {\sec ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac {a^3 \text {Subst}\left (\int \frac {1}{(a-x)^2 (a+x)^4} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a^3 \text {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 {\text {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*}

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Mathematica [A]
time = 0.08, size = 85, normalized size = 0.82 \begin {gather*} -\frac {\sec ^2(c+d x) \left (4-\sin (c+d x)-6 \sin ^2(c+d x)-3 \sin ^3(c+d x)+3 \tanh ^{-1}(\sin (c+d x)) (-1+\sin (c+d x)) (1+\sin (c+d x))^3\right )}{12 a^2 d (1+\sin (c+d x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*(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))/(a^2*d*(1 + Sin[c + d*x])^2)

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Maple [A]
time = 0.24, size = 79, normalized size = 0.76

method	result	size
derivativedivides	\(\frac {-\frac {1}{16 \left (\sin \left (d x +c \right )-1\right )}-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{8}-\frac {1}{12 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {1}{8 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {3}{16 \left (1+\sin \left (d x +c \right )\right )}+\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{8}}{d \,a^{2}}\)	\(79\)
default	\(\frac {-\frac {1}{16 \left (\sin \left (d x +c \right )-1\right )}-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{8}-\frac {1}{12 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {1}{8 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {3}{16 \left (1+\sin \left (d x +c \right )\right )}+\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{8}}{d \,a^{2}}\)	\(79\)
risch	\(-\frac {i \left (12 i {\mathrm e}^{6 i \left (d x +c \right )}+3 \,{\mathrm e}^{7 i \left (d x +c \right )}+8 i {\mathrm e}^{4 i \left (d x +c \right )}-13 \,{\mathrm e}^{5 i \left (d x +c \right )}+12 i {\mathrm e}^{2 i \left (d x +c \right )}+13 \,{\mathrm e}^{3 i \left (d x +c \right )}-3 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{6 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{6} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{2} a^{2} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{4 a^{2} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{4 a^{2} d}\)	\(162\)
norman	\(\frac {\frac {2 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {2 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d a}+\frac {3 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}-\frac {4 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {7 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d a}+\frac {7 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d a}}{a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4 a^{2} d}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 a^{2} d}\)	\(204\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.30, size = 108, normalized size = 1.04 \begin {gather*} -\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 {gather*}

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 = 0.35, size = 178, normalized size = 1.71 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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 = 3.66, size = 106, normalized size = 1.02 \begin {gather*} \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 {gather*}

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

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Mupad [B]
time = 0.10, size = 93, normalized size = 0.89 \begin {gather*} \frac {\frac {{\sin \left (c+d\,x\right )}^3}{4}+\frac {{\sin \left (c+d\,x\right )}^2}{2}+\frac {\sin \left (c+d\,x\right )}{12}-\frac {1}{3}}{d\,\left (-a^2\,{\sin \left (c+d\,x\right )}^4-2\,a^2\,{\sin \left (c+d\,x\right )}^3+2\,a^2\,\sin \left (c+d\,x\right )+a^2\right )}+\frac {\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )}{4\,a^2\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^3*(a + a*sin(c + d*x))^2),x)

[Out]

(sin(c + d*x)/12 + sin(c + d*x)^2/2 + sin(c + d*x)^3/4 - 1/3)/(d*(2*a^2*sin(c + d*x) + a^2 - 2*a^2*sin(c + d*x
)^3 - a^2*sin(c + d*x)^4)) + atanh(sin(c + d*x))/(4*a^2*d)

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