3.192 \(\int \frac{\sec (c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=113 \[ -\frac{1}{4 a^2 d \sqrt{a \sin (c+d x)+a}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a \sin (c+d x)+a}}{\sqrt{2} \sqrt{a}}\right )}{4 \sqrt{2} a^{5/2} d}-\frac{1}{6 a d (a \sin (c+d x)+a)^{3/2}}-\frac{1}{5 d (a \sin (c+d x)+a)^{5/2}} \]

[Out]

ArcTanh[Sqrt[a + a*Sin[c + d*x]]/(Sqrt[2]*Sqrt[a])]/(4*Sqrt[2]*a^(5/2)*d) - 1/(5*d*(a + a*Sin[c + d*x])^(5/2))
- 1/(6*a*d*(a + a*Sin[c + d*x])^(3/2)) - 1/(4*a^2*d*Sqrt[a + a*Sin[c + d*x]])

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Rubi [A]  time = 0.0886034, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2667, 51, 63, 206} \[ -\frac{1}{4 a^2 d \sqrt{a \sin (c+d x)+a}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a \sin (c+d x)+a}}{\sqrt{2} \sqrt{a}}\right )}{4 \sqrt{2} a^{5/2} d}-\frac{1}{6 a d (a \sin (c+d x)+a)^{3/2}}-\frac{1}{5 d (a \sin (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

ArcTanh[Sqrt[a + a*Sin[c + d*x]]/(Sqrt[2]*Sqrt[a])]/(4*Sqrt[2]*a^(5/2)*d) - 1/(5*d*(a + a*Sin[c + d*x])^(5/2))
- 1/(6*a*d*(a + a*Sin[c + d*x])^(3/2)) - 1/(4*a^2*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))^{5/2}} \, dx &=\frac{a \operatorname{Subst}\left (\int \frac{1}{(a-x) (a+x)^{7/2}} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac{1}{5 d (a+a \sin (c+d x))^{5/2}}+\frac{\operatorname{Subst}\left (\int \frac{1}{(a-x) (a+x)^{5/2}} \, dx,x,a \sin (c+d x)\right )}{2 d}\\ &=-\frac{1}{5 d (a+a \sin (c+d x))^{5/2}}-\frac{1}{6 a 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 )}{4 a d}\\ &=-\frac{1}{5 d (a+a \sin (c+d x))^{5/2}}-\frac{1}{6 a d (a+a \sin (c+d x))^{3/2}}-\frac{1}{4 a^2 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 )}{8 a^2 d}\\ &=-\frac{1}{5 d (a+a \sin (c+d x))^{5/2}}-\frac{1}{6 a d (a+a \sin (c+d x))^{3/2}}-\frac{1}{4 a^2 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 )}{4 a^2 d}\\ &=\frac{\tanh ^{-1}\left (\frac{\sqrt{a+a \sin (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{4 \sqrt{2} a^{5/2} d}-\frac{1}{5 d (a+a \sin (c+d x))^{5/2}}-\frac{1}{6 a d (a+a \sin (c+d x))^{3/2}}-\frac{1}{4 a^2 d \sqrt{a+a \sin (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 0.0842239, size = 41, normalized size = 0.36 \[ -\frac{\, _2F_1\left (-\frac{5}{2},1;-\frac{3}{2};\frac{1}{2} (\sin (c+d x)+1)\right )}{5 d (a \sin (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

-Hypergeometric2F1[-5/2, 1, -3/2, (1 + Sin[c + d*x])/2]/(5*d*(a + a*Sin[c + d*x])^(5/2))

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Maple [A]  time = 0.097, size = 88, normalized size = 0.8 \begin{align*} -2\,{\frac{a}{d} \left ( -1/16\,{\frac{\sqrt{2}}{{a}^{7/2}}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a+a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ) }+1/8\,{\frac{1}{{a}^{3}\sqrt{a+a\sin \left ( dx+c \right ) }}}+1/12\,{\frac{1}{{a}^{2} \left ( a+a\sin \left ( dx+c \right ) \right ) ^{3/2}}}+1/10\,{\frac{1}{a \left ( a+a\sin \left ( dx+c \right ) \right ) ^{5/2}}} \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))^(5/2),x)

[Out]

-2*a*(-1/16/a^(7/2)*2^(1/2)*arctanh(1/2*(a+a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))+1/8/a^3/(a+a*sin(d*x+c))^(1/2)
+1/12/a^2/(a+a*sin(d*x+c))^(3/2)+1/10/a/(a+a*sin(d*x+c))^(5/2))/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))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.47456, size = 452, normalized size = 4. \begin{align*} \frac{15 \, \sqrt{2}{\left (3 \, \cos \left (d x + c\right )^{2} +{\left (\cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right ) - 4\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 \,{\left (15 \, \cos \left (d x + c\right )^{2} - 40 \, \sin \left (d x + c\right ) - 52\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{240 \,{\left (3 \, a^{3} d \cos \left (d x + c\right )^{2} - 4 \, a^{3} d +{\left (a^{3} d \cos \left (d x + c\right )^{2} - 4 \, a^{3} d\right )} \sin \left (d x + c\right )\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))^(5/2),x, algorithm="fricas")

[Out]

1/240*(15*sqrt(2)*(3*cos(d*x + c)^2 + (cos(d*x + c)^2 - 4)*sin(d*x + c) - 4)*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*(15*cos(d*x + c)^2 - 40*sin(d*x + c) -
52)*sqrt(a*sin(d*x + c) + a))/(3*a^3*d*cos(d*x + c)^2 - 4*a^3*d + (a^3*d*cos(d*x + c)^2 - 4*a^3*d)*sin(d*x +
c))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)/(a+a*sin(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.12648, size = 130, normalized size = 1.15 \begin{align*} -\frac{1}{120} \, a{\left (\frac{15 \, \sqrt{2} \arctan \left (\frac{\sqrt{2} \sqrt{a \sin \left (d x + c\right ) + a}}{2 \, \sqrt{-a}}\right )}{\sqrt{-a} a^{3} d} + \frac{2 \,{\left (15 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{2} + 10 \,{\left (a \sin \left (d x + c\right ) + a\right )} a + 12 \, a^{2}\right )}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}} a^{3} 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))^(5/2),x, algorithm="giac")

[Out]

-1/120*a*(15*sqrt(2)*arctan(1/2*sqrt(2)*sqrt(a*sin(d*x + c) + a)/sqrt(-a))/(sqrt(-a)*a^3*d) + 2*(15*(a*sin(d*x
+ c) + a)^2 + 10*(a*sin(d*x + c) + a)*a + 12*a^2)/((a*sin(d*x + c) + a)^(5/2)*a^3*d))