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

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

[Out]

-1/5/d/(a+a*sin(d*x+c))^(5/2)-1/6/a/d/(a+a*sin(d*x+c))^(3/2)+1/8*arctanh(1/2*(a+a*sin(d*x+c))^(1/2)*2^(1/2)/a^
(1/2))/a^(5/2)/d*2^(1/2)-1/4/a^2/d/(a+a*sin(d*x+c))^(1/2)

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Rubi [A]  time = 0.09, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, 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 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])

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])

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*}

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Mathematica [C]  time = 0.09, 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]

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

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fricas [A]  time = 0.74, size = 169, normalized size = 1.50 \[ \frac {15 \, \sqrt {2} {\left (3 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right ) - 4\right )} \sqrt {a} \log \left (-\frac {a \sin \left (d x + c\right ) + 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} + 3 \, a}{\sin \left (d x + c\right ) - 1}\right ) - 4 \, {\left (15 \, \cos \left (d x + c\right )^{2} - 40 \, \sin \left (d x + c\right ) - 52\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{240 \, {\left (3 \, a^{3} d \cos \left (d x + c\right )^{2} - 4 \, a^{3} d + {\left (a^{3} d \cos \left (d x + c\right )^{2} - 4 \, a^{3} d\right )} \sin \left (d x + c\right )\right )}} \]

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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*p
i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Warning, integration of abs or sign assumes cons
tant sign by intervals (correct if the argument is real):Check [abs(cos((d*t_nostep+c)/2-pi/4))]Discontinuitie
s at zeroes of cos((d*t_nostep+c)/2-pi/4) were not checkedWarning, integration of abs or sign assumes constant
sign by intervals (correct if the argument is real):Check [abs(t_nostep+1)]Evaluation time: 0.46Not invertibl
e Error: Bad Argument Value

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maple [A]  time = 0.17, size = 88, normalized size = 0.78 \[ -\frac {2 a \left (\frac {1}{8 a^{3} \sqrt {a +a \sin \left (d x +c \right )}}+\frac {1}{12 a^{2} \left (a +a \sin \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {1}{10 a \left (a +a \sin \left (d x +c \right )\right )^{\frac {5}{2}}}-\frac {\sqrt {2}\, \arctanh \left (\frac {\sqrt {a +a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16 a^{\frac {7}{2}}}\right )}{d} \]

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*(1/8/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)-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)))/d

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maxima [A]  time = 1.43, size = 114, normalized size = 1.01 \[ -\frac {\frac {15 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {a \sin \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {a \sin \left (d x + c\right ) + a}}\right )}{a^{\frac {3}{2}}} + \frac {4 \, {\left (15 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{2} + 10 \, {\left (a \sin \left (d x + c\right ) + a\right )} a + 12 \, a^{2}\right )}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a}}{240 \, a d} \]

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]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\cos \left (c+d\,x\right )\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)*(a + a*sin(c + d*x))^(5/2)),x)

[Out]

int(1/(cos(c + d*x)*(a + a*sin(c + d*x))^(5/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec {\left (c + d x \right )}}{\left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

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]

Integral(sec(c + d*x)/(a*(sin(c + d*x) + 1))**(5/2), x)

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