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

Optimal. Leaf size=89 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{a \sin (c+d x)+a}}{\sqrt{2} \sqrt{a}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{1}{2 a d \sqrt{a \sin (c+d x)+a}}-\frac{1}{3 d (a \sin (c+d x)+a)^{3/2}} \]

[Out]

ArcTanh[Sqrt[a + a*Sin[c + d*x]]/(Sqrt[2]*Sqrt[a])]/(2*Sqrt[2]*a^(3/2)*d) - 1/(3*d*(a + a*Sin[c + d*x])^(3/2))
- 1/(2*a*d*Sqrt[a + a*Sin[c + d*x]])

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Rubi [A]  time = 0.0768569, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2667, 51, 63, 206} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{a \sin (c+d x)+a}}{\sqrt{2} \sqrt{a}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{1}{2 a d \sqrt{a \sin (c+d x)+a}}-\frac{1}{3 d (a \sin (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[Sqrt[a + a*Sin[c + d*x]]/(Sqrt[2]*Sqrt[a])]/(2*Sqrt[2]*a^(3/2)*d) - 1/(3*d*(a + a*Sin[c + d*x])^(3/2))
- 1/(2*a*d*Sqrt[a + a*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 51

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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 (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx &=\frac{a \operatorname{Subst}\left (\int \frac{1}{(a-x) (a+x)^{5/2}} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac{1}{3 d (a+a \sin (c+d x))^{3/2}}+\frac{\operatorname{Subst}\left (\int \frac{1}{(a-x) (a+x)^{3/2}} \, dx,x,a \sin (c+d x)\right )}{2 d}\\ &=-\frac{1}{3 d (a+a \sin (c+d x))^{3/2}}-\frac{1}{2 a d \sqrt{a+a \sin (c+d x)}}+\frac{\operatorname{Subst}\left (\int \frac{1}{(a-x) \sqrt{a+x}} \, dx,x,a \sin (c+d x)\right )}{4 a d}\\ &=-\frac{1}{3 d (a+a \sin (c+d x))^{3/2}}-\frac{1}{2 a d \sqrt{a+a \sin (c+d x)}}+\frac{\operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\sqrt{a+a \sin (c+d x)}\right )}{2 a d}\\ &=\frac{\tanh ^{-1}\left (\frac{\sqrt{a+a \sin (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{1}{3 d (a+a \sin (c+d x))^{3/2}}-\frac{1}{2 a d \sqrt{a+a \sin (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 0.06767, size = 41, normalized size = 0.46 \[ -\frac{\, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{1}{2} (\sin (c+d x)+1)\right )}{3 d (a \sin (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Hypergeometric2F1[-3/2, 1, -1/2, (1 + Sin[c + d*x])/2]/(3*d*(a + a*Sin[c + d*x])^(3/2))

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Maple [A]  time = 0.085, size = 71, normalized size = 0.8 \begin{align*} -{\frac{a}{d} \left ( -{\frac{\sqrt{2}}{4}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{a+a\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{a}}}} \right ){a}^{-{\frac{5}{2}}}}+{\frac{1}{2\,{a}^{2}}{\frac{1}{\sqrt{a+a\sin \left ( dx+c \right ) }}}}+{\frac{1}{3\,a} \left ( a+a\sin \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*(-1/4/a^(5/2)*2^(1/2)*arctanh(1/2*(a+a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))+1/2/a^2/(a+a*sin(d*x+c))^(1/2)+1/
3/a/(a+a*sin(d*x+c))^(3/2))/d

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.34851, size = 355, normalized size = 3.99 \begin{align*} \frac{3 \, \sqrt{2}{\left (\cos \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) - 2\right )} \sqrt{a} \log \left (-\frac{a \sin \left (d x + c\right ) + 2 \, \sqrt{2} \sqrt{a \sin \left (d x + c\right ) + a} \sqrt{a} + 3 \, a}{\sin \left (d x + c\right ) - 1}\right ) + 4 \, \sqrt{a \sin \left (d x + c\right ) + a}{\left (3 \, \sin \left (d x + c\right ) + 5\right )}}{24 \,{\left (a^{2} d \cos \left (d x + c\right )^{2} - 2 \, a^{2} d \sin \left (d x + c\right ) - 2 \, a^{2} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*sqrt(2)*(cos(d*x + c)^2 - 2*sin(d*x + c) - 2)*sqrt(a)*log(-(a*sin(d*x + c) + 2*sqrt(2)*sqrt(a*sin(d*x
+ c) + a)*sqrt(a) + 3*a)/(sin(d*x + c) - 1)) + 4*sqrt(a*sin(d*x + c) + a)*(3*sin(d*x + c) + 5))/(a^2*d*cos(d*x
+ c)^2 - 2*a^2*d*sin(d*x + c) - 2*a^2*d)

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

[Out]

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

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Giac [A]  time = 1.11997, size = 103, normalized size = 1.16 \begin{align*} -\frac{1}{12} \, a{\left (\frac{3 \, \sqrt{2} \arctan \left (\frac{\sqrt{2} \sqrt{a \sin \left (d x + c\right ) + a}}{2 \, \sqrt{-a}}\right )}{\sqrt{-a} a^{2} d} + \frac{2 \,{\left (3 \, a \sin \left (d x + c\right ) + 5 \, a\right )}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}} a^{2} d}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*a*(3*sqrt(2)*arctan(1/2*sqrt(2)*sqrt(a*sin(d*x + c) + a)/sqrt(-a))/(sqrt(-a)*a^2*d) + 2*(3*a*sin(d*x + c
) + 5*a)/((a*sin(d*x + c) + a)^(3/2)*a^2*d))