3.2.77 \(\int \frac {\sec ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\) [177]

Optimal. Leaf size=134 \[ -\frac {15 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{32 \sqrt {2} a^{3/2} d}-\frac {15 \cos (c+d x)}{32 d (a+a \sin (c+d x))^{3/2}}-\frac {\sec (c+d x)}{4 d (a+a \sin (c+d x))^{3/2}}+\frac {5 \sec (c+d x)}{8 a d \sqrt {a+a \sin (c+d x)}} \]

[Out]

-15/32*cos(d*x+c)/d/(a+a*sin(d*x+c))^(3/2)-1/4*sec(d*x+c)/d/(a+a*sin(d*x+c))^(3/2)-15/64*arctanh(1/2*cos(d*x+c
)*a^(1/2)*2^(1/2)/(a+a*sin(d*x+c))^(1/2))*2^(1/2)/a^(3/2)/d+5/8*sec(d*x+c)/a/d/(a+a*sin(d*x+c))^(1/2)

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Rubi [A]
time = 0.11, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {2760, 2766, 2729, 2728, 212} \begin {gather*} -\frac {15 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{32 \sqrt {2} a^{3/2} d}-\frac {15 \cos (c+d x)}{32 d (a \sin (c+d x)+a)^{3/2}}+\frac {5 \sec (c+d x)}{8 a d \sqrt {a \sin (c+d x)+a}}-\frac {\sec (c+d x)}{4 d (a \sin (c+d x)+a)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-15*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(32*Sqrt[2]*a^(3/2)*d) - (15*Cos[c +
d*x])/(32*d*(a + a*Sin[c + d*x])^(3/2)) - Sec[c + d*x]/(4*d*(a + a*Sin[c + d*x])^(3/2)) + (5*Sec[c + d*x])/(8*
a*d*Sqrt[a + a*Sin[c + d*x]])

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 2728

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, b*(C
os[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2729

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*Cos[c + d*x]*((a + b*Sin[c + d*x])^n/(a*d
*(2*n + 1))), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 2760

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*(g*C
os[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(a*f*g*(2*m + p + 1))), x] + Dist[(m + p + 1)/(a*(2*m + p + 1)),
Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^
2, 0] && LtQ[m, -1] && NeQ[2*m + p + 1, 0] && IntegersQ[2*m, 2*p]

Rule 2766

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[(-b)*((
g*Cos[e + f*x])^(p + 1)/(a*f*g*(p + 1)*Sqrt[a + b*Sin[e + f*x]])), x] + Dist[a*((2*p + 1)/(2*g^2*(p + 1))), In
t[(g*Cos[e + f*x])^(p + 2)/(a + b*Sin[e + f*x])^(3/2), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0
] && LtQ[p, -1] && IntegerQ[2*p]

Rubi steps

\begin {align*} \int \frac {\sec ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx &=-\frac {\sec (c+d x)}{4 d (a+a \sin (c+d x))^{3/2}}+\frac {5 \int \frac {\sec ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx}{8 a}\\ &=-\frac {\sec (c+d x)}{4 d (a+a \sin (c+d x))^{3/2}}+\frac {5 \sec (c+d x)}{8 a d \sqrt {a+a \sin (c+d x)}}+\frac {15}{16} \int \frac {1}{(a+a \sin (c+d x))^{3/2}} \, dx\\ &=-\frac {15 \cos (c+d x)}{32 d (a+a \sin (c+d x))^{3/2}}-\frac {\sec (c+d x)}{4 d (a+a \sin (c+d x))^{3/2}}+\frac {5 \sec (c+d x)}{8 a d \sqrt {a+a \sin (c+d x)}}+\frac {15 \int \frac {1}{\sqrt {a+a \sin (c+d x)}} \, dx}{64 a}\\ &=-\frac {15 \cos (c+d x)}{32 d (a+a \sin (c+d x))^{3/2}}-\frac {\sec (c+d x)}{4 d (a+a \sin (c+d x))^{3/2}}+\frac {5 \sec (c+d x)}{8 a d \sqrt {a+a \sin (c+d x)}}-\frac {15 \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\frac {a \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{32 a d}\\ &=-\frac {15 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{32 \sqrt {2} a^{3/2} d}-\frac {15 \cos (c+d x)}{32 d (a+a \sin (c+d x))^{3/2}}-\frac {\sec (c+d x)}{4 d (a+a \sin (c+d x))^{3/2}}+\frac {5 \sec (c+d x)}{8 a d \sqrt {a+a \sin (c+d x)}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.18, size = 224, normalized size = 1.67 \begin {gather*} \frac {-4+\frac {8 \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}+14 \sin \left (\frac {1}{2} (c+d x)\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-7 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2+(15+15 i) (-1)^{3/4} \tanh ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (c+d x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3+\frac {8 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}}{32 d (a (1+\sin (c+d x)))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4 + (8*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + 14*Sin[(c + d*x)/2]*(Cos[(c + d*x)/2] + Sin
[(c + d*x)/2]) - 7*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + (15 + 15*I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(
3/4)*(-1 + Tan[(c + d*x)/4])]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3 + (8*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2
])^3)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]))/(32*d*(a*(1 + Sin[c + d*x]))^(3/2))

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Maple [A]
time = 0.56, size = 202, normalized size = 1.51

method	result	size
default	\(-\frac {\sin \left (d x +c \right ) \left (30 \sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2}-40 a^{\frac {5}{2}}\right )+\left (-15 \sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2}+30 a^{\frac {5}{2}}\right ) \left (\cos ^{2}\left (d x +c \right )\right )+30 \sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2}-24 a^{\frac {5}{2}}}{64 a^{\frac {7}{2}} \left (1+\sin \left (d x +c \right )\right ) \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(202\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2/(a+a*sin(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/64/a^(7/2)*(sin(d*x+c)*(30*(a-a*sin(d*x+c))^(1/2)*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2
))*a^2-40*a^(5/2))+(-15*(a-a*sin(d*x+c))^(1/2)*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^2
+30*a^(5/2))*cos(d*x+c)^2+30*(a-a*sin(d*x+c))^(1/2)*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2)
)*a^2-24*a^(5/2))/(1+sin(d*x+c))/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sec(d*x + c)^2/(a*sin(d*x + c) + a)^(3/2), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 240 vs. \(2 (111) = 222\).
time = 0.37, size = 240, normalized size = 1.79 \begin {gather*} \frac {15 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{3} - 2 \, \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )} + 3 \, a \cos \left (d x + c\right ) - {\left (a \cos \left (d x + c\right ) - 2 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right ) + 4 \, {\left (15 \, \cos \left (d x + c\right )^{2} - 20 \, \sin \left (d x + c\right ) - 12\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{128 \, {\left (a^{2} d \cos \left (d x + c\right )^{3} - 2 \, a^{2} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, a^{2} d \cos \left (d x + c\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/128*(15*sqrt(2)*(cos(d*x + c)^3 - 2*cos(d*x + c)*sin(d*x + c) - 2*cos(d*x + c))*sqrt(a)*log(-(a*cos(d*x + c)
^2 - 2*sqrt(2)*sqrt(a*sin(d*x + c) + a)*sqrt(a)*(cos(d*x + c) - sin(d*x + c) + 1) + 3*a*cos(d*x + c) - (a*cos(
d*x + c) - 2*a)*sin(d*x + c) + 2*a)/(cos(d*x + c)^2 - (cos(d*x + c) + 2)*sin(d*x + c) - cos(d*x + c) - 2)) + 4
*(15*cos(d*x + c)^2 - 20*sin(d*x + c) - 12)*sqrt(a*sin(d*x + c) + a))/(a^2*d*cos(d*x + c)^3 - 2*a^2*d*cos(d*x
+ c)*sin(d*x + c) - 2*a^2*d*cos(d*x + c))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sec ^{2}{\left (c + d x \right )}}{\left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 8.81, size = 199, normalized size = 1.49 \begin {gather*} \frac {\sqrt {a} {\left (\frac {15 \, \sqrt {2} \log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {15 \, \sqrt {2} \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {16 \, \sqrt {2}}{a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )} - \frac {2 \, {\left (7 \, \sqrt {2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, \sqrt {2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}\right )}}{128 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/128*sqrt(a)*(15*sqrt(2)*log(sin(-1/4*pi + 1/2*d*x + 1/2*c) + 1)/(a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) -
15*sqrt(2)*log(-sin(-1/4*pi + 1/2*d*x + 1/2*c) + 1)/(a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) - 16*sqrt(2)/(a^
2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)) - 2*(7*sqrt(2)*sin(-1/4*pi + 1/2*d*x + 1
/2*c)^3 - 9*sqrt(2)*sin(-1/4*pi + 1/2*d*x + 1/2*c))/((sin(-1/4*pi + 1/2*d*x + 1/2*c)^2 - 1)^2*a^2*sgn(cos(-1/4
*pi + 1/2*d*x + 1/2*c))))/d

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\cos \left (c+d\,x\right )}^2\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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