3.2.94 \(\int \frac {\sec ^3(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx\) [194]

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

[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 = 0.19, 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 = {2760, 2766, 2746, 53, 65, 212} \begin {gather*} \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}} \end {gather*}

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 53

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 65

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2746

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 2760

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*(g*C
os[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 2766

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))), In
t[(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 \text {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 \text {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 \text {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 \text {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] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.05, size = 42, normalized size = 0.23 \begin {gather*} -\frac {a \, _2F_1\left (-\frac {7}{2},2;-\frac {5}{2};\frac {1}{2} (1+\sin (c+d x))\right )}{14 d (a+a \sin (c+d x))^{7/2}} \end {gather*}

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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Maple [A]
time = 0.69, size = 141, normalized size = 0.76

method	result	size
default	\(\frac {2 a^{3} \left (-\frac {\frac {\sqrt {a +a \sin \left (d x +c \right )}}{4 a \sin \left (d x +c \right )-4 a}-\frac {9 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8 \sqrt {a}}}{16 a^{5}}-\frac {1}{8 a^{5} \sqrt {a +a \sin \left (d x +c \right )}}-\frac {1}{16 a^{4} \left (a +a \sin \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {1}{20 a^{3} \left (a +a \sin \left (d x +c \right )\right )^{\frac {5}{2}}}-\frac {1}{28 a^{2} \left (a +a \sin \left (d x +c \right )\right )^{\frac {7}{2}}}\right )}{d}\)	\(141\)

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,method=_RETURNVERBOSE)

[Out]

2*a^3*(-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)))-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))/d

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Maxima [A]
time = 0.52, size = 167, normalized size = 0.90 \begin {gather*} -\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} \end {gather*}

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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Fricas [A]
time = 0.37, size = 225, normalized size = 1.22 \begin {gather*} \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 )}} \end {gather*}

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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sec ^{3}{\left (c + d x \right )}}{\left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \end {gather*}

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]

Integral(sec(c + d*x)**3/(a*(sin(c + d*x) + 1))**(5/2), x)

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Giac [A]
time = 4.88, size = 227, normalized size = 1.23 \begin {gather*} \frac {\sqrt {a} {\left (\frac {315 \, \sqrt {2} \log \left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {315 \, \sqrt {2} \log \left (-\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {70 \, \sqrt {2} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {4 \, \sqrt {2} {\left (140 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 35 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 14 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 5\right )}}{a^{3} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}\right )}}{4480 \, d} \end {gather*}

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]

1/4480*sqrt(a)*(315*sqrt(2)*log(cos(-1/4*pi + 1/2*d*x + 1/2*c) + 1)/(a^3*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)))
- 315*sqrt(2)*log(-cos(-1/4*pi + 1/2*d*x + 1/2*c) + 1)/(a^3*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) - 70*sqrt(2)*
cos(-1/4*pi + 1/2*d*x + 1/2*c)/((cos(-1/4*pi + 1/2*d*x + 1/2*c)^2 - 1)*a^3*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))
) - 4*sqrt(2)*(140*cos(-1/4*pi + 1/2*d*x + 1/2*c)^6 + 35*cos(-1/4*pi + 1/2*d*x + 1/2*c)^4 + 14*cos(-1/4*pi + 1
/2*d*x + 1/2*c)^2 + 5)/(a^3*cos(-1/4*pi + 1/2*d*x + 1/2*c)^7*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))))/d

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

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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