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

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

[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]])

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Rubi [A]  time = 0.161535, antiderivative size = 134, normalized size of antiderivative = 1., 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 = {2681, 2687, 2650, 2649, 206} \[ -\frac{15 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{32 \sqrt{2} a^{3/2} d}-\frac{15 \cos (c+d x)}{32 d (a \sin (c+d x)+a)^{3/2}}+\frac{5 \sec (c+d x)}{8 a d \sqrt{a \sin (c+d x)+a}}-\frac{\sec (c+d x)}{4 d (a \sin (c+d x)+a)^{3/2}} \]

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 2681

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*(g*
Cos[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 2687

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)), Int[
(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]

Rule 2650

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 2649

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 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 ^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 \operatorname{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*}

Mathematica [C]  time = 0.264118, size = 224, normalized size = 1.67 \[ \frac{\frac{8 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}-7 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2+14 \sin \left (\frac{1}{2} (c+d x)\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{8 \sin \left (\frac{1}{2} (c+d x)\right )}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}+(15+15 i) (-1)^{3/4} \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3 \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac{1}{4} (c+d x)\right )-1\right )\right )-4}{32 d (a (\sin (c+d x)+1))^{3/2}} \]

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 = 0.148, size = 202, normalized size = 1.5 \begin{align*} -{\frac{1}{ \left ( 64+64\,\sin \left ( dx+c \right ) \right ) \cos \left ( dx+c \right ) d} \left ( \sin \left ( dx+c \right ) \left ( 30\,\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{2}-40\,{a}^{5/2} \right ) + \left ( -15\,\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{2}+30\,{a}^{5/2} \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}+30\,\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{2}-24\,{a}^{5/2} \right ){a}^{-{\frac{7}{2}}}{\frac{1}{\sqrt{a+a\sin \left ( dx+c \right ) }}}} \end{align*}

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)

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

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]  time = 2.33137, size = 652, normalized size = 4.87 \begin{align*} \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{align*}

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., size = 0, normalized size = 0. \begin{align*} \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{align*}

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 [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{2} \end{align*}

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]

sage2