3.194 \(\int \frac {\sec ^3(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=185 \[ \frac {9 \tanh ^{-1}\left (\frac {\sqrt {a \sin (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{32 \sqrt {2} a^{5/2} d}-\frac {9}{32 a^2 d \sqrt {a \sin (c+d x)+a}}+\frac {9 \sec ^2(c+d x)}{40 a^2 d \sqrt {a \sin (c+d x)+a}}-\frac {3}{16 a d (a \sin (c+d x)+a)^{3/2}}-\frac {9 \sec ^2(c+d x)}{70 a d (a \sin (c+d x)+a)^{3/2}}-\frac {\sec ^2(c+d x)}{7 d (a \sin (c+d x)+a)^{5/2}} \]

[Out]

-1/7*sec(d*x+c)^2/d/(a+a*sin(d*x+c))^(5/2)-3/16/a/d/(a+a*sin(d*x+c))^(3/2)-9/70*sec(d*x+c)^2/a/d/(a+a*sin(d*x+
c))^(3/2)+9/64*arctanh(1/2*(a+a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/a^(5/2)/d*2^(1/2)-9/32/a^2/d/(a+a*sin(d*x+c
))^(1/2)+9/40*sec(d*x+c)^2/a^2/d/(a+a*sin(d*x+c))^(1/2)

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Rubi [A]  time = 0.27, antiderivative size = 185, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2681, 2687, 2667, 51, 63, 206} \[ -\frac {9}{32 a^2 d \sqrt {a \sin (c+d x)+a}}+\frac {9 \tanh ^{-1}\left (\frac {\sqrt {a \sin (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{32 \sqrt {2} a^{5/2} d}+\frac {9 \sec ^2(c+d x)}{40 a^2 d \sqrt {a \sin (c+d x)+a}}-\frac {3}{16 a d (a \sin (c+d x)+a)^{3/2}}-\frac {9 \sec ^2(c+d x)}{70 a d (a \sin (c+d x)+a)^{3/2}}-\frac {\sec ^2(c+d x)}{7 d (a \sin (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

(9*ArcTanh[Sqrt[a + a*Sin[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(32*Sqrt[2]*a^(5/2)*d) - Sec[c + d*x]^2/(7*d*(a + a*Si
n[c + d*x])^(5/2)) - 3/(16*a*d*(a + a*Sin[c + d*x])^(3/2)) - (9*Sec[c + d*x]^2)/(70*a*d*(a + a*Sin[c + d*x])^(
3/2)) - 9/(32*a^2*d*Sqrt[a + a*Sin[c + d*x]]) + (9*Sec[c + d*x]^2)/(40*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])

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]

Rubi steps

\begin {align*} \int \frac {\sec ^3(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx &=-\frac {\sec ^2(c+d x)}{7 d (a+a \sin (c+d x))^{5/2}}+\frac {9 \int \frac {\sec ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx}{14 a}\\ &=-\frac {\sec ^2(c+d x)}{7 d (a+a \sin (c+d x))^{5/2}}-\frac {9 \sec ^2(c+d x)}{70 a d (a+a \sin (c+d x))^{3/2}}+\frac {9 \int \frac {\sec ^3(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx}{20 a^2}\\ &=-\frac {\sec ^2(c+d x)}{7 d (a+a \sin (c+d x))^{5/2}}-\frac {9 \sec ^2(c+d x)}{70 a d (a+a \sin (c+d x))^{3/2}}+\frac {9 \sec ^2(c+d x)}{40 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {9 \int \frac {\sec (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx}{16 a}\\ &=-\frac {\sec ^2(c+d x)}{7 d (a+a \sin (c+d x))^{5/2}}-\frac {9 \sec ^2(c+d x)}{70 a d (a+a \sin (c+d x))^{3/2}}+\frac {9 \sec ^2(c+d x)}{40 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {9 \operatorname {Subst}\left (\int \frac {1}{(a-x) (a+x)^{5/2}} \, dx,x,a \sin (c+d x)\right )}{16 d}\\ &=-\frac {\sec ^2(c+d x)}{7 d (a+a \sin (c+d x))^{5/2}}-\frac {3}{16 a d (a+a \sin (c+d x))^{3/2}}-\frac {9 \sec ^2(c+d x)}{70 a d (a+a \sin (c+d x))^{3/2}}+\frac {9 \sec ^2(c+d x)}{40 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {9 \operatorname {Subst}\left (\int \frac {1}{(a-x) (a+x)^{3/2}} \, dx,x,a \sin (c+d x)\right )}{32 a d}\\ &=-\frac {\sec ^2(c+d x)}{7 d (a+a \sin (c+d x))^{5/2}}-\frac {3}{16 a d (a+a \sin (c+d x))^{3/2}}-\frac {9 \sec ^2(c+d x)}{70 a d (a+a \sin (c+d x))^{3/2}}-\frac {9}{32 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {9 \sec ^2(c+d x)}{40 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {9 \operatorname {Subst}\left (\int \frac {1}{(a-x) \sqrt {a+x}} \, dx,x,a \sin (c+d x)\right )}{64 a^2 d}\\ &=-\frac {\sec ^2(c+d x)}{7 d (a+a \sin (c+d x))^{5/2}}-\frac {3}{16 a d (a+a \sin (c+d x))^{3/2}}-\frac {9 \sec ^2(c+d x)}{70 a d (a+a \sin (c+d x))^{3/2}}-\frac {9}{32 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {9 \sec ^2(c+d x)}{40 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {9 \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+a \sin (c+d x)}\right )}{32 a^2 d}\\ &=\frac {9 \tanh ^{-1}\left (\frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{32 \sqrt {2} a^{5/2} d}-\frac {\sec ^2(c+d x)}{7 d (a+a \sin (c+d x))^{5/2}}-\frac {3}{16 a d (a+a \sin (c+d x))^{3/2}}-\frac {9 \sec ^2(c+d x)}{70 a d (a+a \sin (c+d x))^{3/2}}-\frac {9}{32 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {9 \sec ^2(c+d x)}{40 a^2 d \sqrt {a+a \sin (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 0.09, size = 42, normalized size = 0.23 \[ -\frac {a \, _2F_1\left (-\frac {7}{2},2;-\frac {5}{2};\frac {1}{2} (\sin (c+d x)+1)\right )}{14 d (a \sin (c+d x)+a)^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

-1/14*(a*Hypergeometric2F1[-7/2, 2, -5/2, (1 + Sin[c + d*x])/2])/(d*(a + a*Sin[c + d*x])^(7/2))

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fricas [A]  time = 0.57, size = 225, normalized size = 1.22 \[ \frac {315 \, \sqrt {2} {\left (3 \, \cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {a} \log \left (-\frac {a \sin \left (d x + c\right ) + 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} + 3 \, a}{\sin \left (d x + c\right ) - 1}\right ) - 4 \, {\left (315 \, \cos \left (d x + c\right )^{4} - 1092 \, \cos \left (d x + c\right )^{2} - 120 \, {\left (7 \, \cos \left (d x + c\right )^{2} - 3\right )} \sin \left (d x + c\right ) + 200\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{4480 \, {\left (3 \, a^{3} d \cos \left (d x + c\right )^{4} - 4 \, a^{3} d \cos \left (d x + c\right )^{2} + {\left (a^{3} d \cos \left (d x + c\right )^{4} - 4 \, a^{3} d \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/4480*(315*sqrt(2)*(3*cos(d*x + c)^4 - 4*cos(d*x + c)^2 + (cos(d*x + c)^4 - 4*cos(d*x + c)^2)*sin(d*x + c))*s
qrt(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*(315*c
os(d*x + c)^4 - 1092*cos(d*x + c)^2 - 120*(7*cos(d*x + c)^2 - 3)*sin(d*x + c) + 200)*sqrt(a*sin(d*x + c) + a))
/(3*a^3*d*cos(d*x + c)^4 - 4*a^3*d*cos(d*x + c)^2 + (a^3*d*cos(d*x + c)^4 - 4*a^3*d*cos(d*x + c)^2)*sin(d*x +
c))

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giac [B]  time = 11.92, size = 916, normalized size = 4.95 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a+a*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

-1/1120*(315*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a) -
sqrt(a))/sqrt(-a))/(sqrt(-a)*a^2*sgn(tan(1/2*d*x + 1/2*c) + 1)) + 70*(3*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a
*tan(1/2*d*x + 1/2*c)^2 + a))^3 - (sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*sqrt(a
) - (sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))*a - a^(3/2))/(((sqrt(a)*tan(1/2*d*x +
1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - 2*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)
^2 + a))*sqrt(a) - a)^2*a^2*sgn(tan(1/2*d*x + 1/2*c) + 1)) + 8*(455*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan
(1/2*d*x + 1/2*c)^2 + a))^13 + 3395*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^12*sqr
t(a) + 10290*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^11*a + 8750*(sqrt(a)*tan(1/2*
d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^10*a^(3/2) - 16807*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*t
an(1/2*d*x + 1/2*c)^2 + a))^9*a^2 - 31423*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^
8*a^(5/2) + 14076*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^7*a^3 + 33908*(sqrt(a)*t
an(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^6*a^(7/2) - 19607*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sq
rt(a*tan(1/2*d*x + 1/2*c)^2 + a))^5*a^4 - 15883*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2
+ a))^4*a^(9/2) + 19698*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^3*a^5 - 8386*(sqrt
(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*a^(11/2) + 1687*(sqrt(a)*tan(1/2*d*x + 1/2*c)
- sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))*a^6 - 153*a^(13/2))/(((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*
x + 1/2*c)^2 + a))^2 + 2*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))*sqrt(a) - a)^7*a^
2*sgn(tan(1/2*d*x + 1/2*c) + 1)))/d

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maple [A]  time = 0.30, size = 141, normalized size = 0.76 \[ \frac {2 a^{3} \left (-\frac {1}{8 a^{5} \sqrt {a +a \sin \left (d x +c \right )}}-\frac {1}{16 a^{4} \left (a +a \sin \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {1}{20 a^{3} \left (a +a \sin \left (d x +c \right )\right )^{\frac {5}{2}}}-\frac {1}{28 a^{2} \left (a +a \sin \left (d x +c \right )\right )^{\frac {7}{2}}}-\frac {\frac {\sqrt {a +a \sin \left (d x +c \right )}}{4 a \sin \left (d x +c \right )-4 a}-\frac {9 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8 \sqrt {a}}}{16 a^{5}}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3/(a+a*sin(d*x+c))^(5/2),x)

[Out]

2*a^3*(-1/8/a^5/(a+a*sin(d*x+c))^(1/2)-1/16/a^4/(a+a*sin(d*x+c))^(3/2)-1/20/a^3/(a+a*sin(d*x+c))^(5/2)-1/28/a^
2/(a+a*sin(d*x+c))^(7/2)-1/16/a^5*(1/4*(a+a*sin(d*x+c))^(1/2)/(a*sin(d*x+c)-a)-9/8*2^(1/2)/a^(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 = 1.23, size = 167, normalized size = 0.90 \[ -\frac {\frac {4 \, {\left (315 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{4} - 420 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{3} a - 168 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{2} a^{2} - 144 \, {\left (a \sin \left (d x + c\right ) + a\right )} a^{3} - 160 \, a^{4}\right )}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {9}{2}} a - 2 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a^{2}} + \frac {315 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {a \sin \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {a \sin \left (d x + c\right ) + a}}\right )}{a^{\frac {3}{2}}}}{4480 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a+a*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

-1/4480*(4*(315*(a*sin(d*x + c) + a)^4 - 420*(a*sin(d*x + c) + a)^3*a - 168*(a*sin(d*x + c) + a)^2*a^2 - 144*(
a*sin(d*x + c) + a)*a^3 - 160*a^4)/((a*sin(d*x + c) + a)^(9/2)*a - 2*(a*sin(d*x + c) + a)^(7/2)*a^2) + 315*sqr
t(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))/
(a*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\cos \left (c+d\,x\right )}^3\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^3*(a + a*sin(c + d*x))^(5/2)),x)

[Out]

int(1/(cos(c + d*x)^3*(a + a*sin(c + d*x))^(5/2)), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3/(a+a*sin(d*x+c))**(5/2),x)

[Out]

Timed out

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