3.2.67 \(\int \frac {\sec ^5(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx\) [167]

Optimal. Leaf size=175 \[ \frac {63 \tanh ^{-1}\left (\frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{128 \sqrt {2} \sqrt {a} d}-\frac {21 a}{64 d (a+a \sin (c+d x))^{3/2}}-\frac {9 a \sec ^2(c+d x)}{40 d (a+a \sin (c+d x))^{3/2}}-\frac {63}{128 d \sqrt {a+a \sin (c+d x)}}+\frac {63 \sec ^2(c+d x)}{160 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^4(c+d x)}{4 d \sqrt {a+a \sin (c+d x)}} \]

[Out]

-21/64*a/d/(a+a*sin(d*x+c))^(3/2)-9/40*a*sec(d*x+c)^2/d/(a+a*sin(d*x+c))^(3/2)+63/256*arctanh(1/2*(a+a*sin(d*x
+c))^(1/2)*2^(1/2)/a^(1/2))/d*2^(1/2)/a^(1/2)-63/128/d/(a+a*sin(d*x+c))^(1/2)+63/160*sec(d*x+c)^2/d/(a+a*sin(d
*x+c))^(1/2)+1/4*sec(d*x+c)^4/d/(a+a*sin(d*x+c))^(1/2)

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Rubi [A]
time = 0.18, antiderivative size = 175, 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 = {2766, 2760, 2746, 53, 65, 212} \begin {gather*} -\frac {63}{128 d \sqrt {a \sin (c+d x)+a}}-\frac {21 a}{64 d (a \sin (c+d x)+a)^{3/2}}+\frac {63 \tanh ^{-1}\left (\frac {\sqrt {a \sin (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{128 \sqrt {2} \sqrt {a} d}+\frac {\sec ^4(c+d x)}{4 d \sqrt {a \sin (c+d x)+a}}+\frac {63 \sec ^2(c+d x)}{160 d \sqrt {a \sin (c+d x)+a}}-\frac {9 a \sec ^2(c+d x)}{40 d (a \sin (c+d x)+a)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5/Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(63*ArcTanh[Sqrt[a + a*Sin[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(128*Sqrt[2]*Sqrt[a]*d) - (21*a)/(64*d*(a + a*Sin[c +
d*x])^(3/2)) - (9*a*Sec[c + d*x]^2)/(40*d*(a + a*Sin[c + d*x])^(3/2)) - 63/(128*d*Sqrt[a + a*Sin[c + d*x]]) +
(63*Sec[c + d*x]^2)/(160*d*Sqrt[a + a*Sin[c + d*x]]) + Sec[c + d*x]^4/(4*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 ^5(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx &=\frac {\sec ^4(c+d x)}{4 d \sqrt {a+a \sin (c+d x)}}+\frac {1}{8} (9 a) \int \frac {\sec ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\\ &=-\frac {9 a \sec ^2(c+d x)}{40 d (a+a \sin (c+d x))^{3/2}}+\frac {\sec ^4(c+d x)}{4 d \sqrt {a+a \sin (c+d x)}}+\frac {63}{80} \int \frac {\sec ^3(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx\\ &=-\frac {9 a \sec ^2(c+d x)}{40 d (a+a \sin (c+d x))^{3/2}}+\frac {63 \sec ^2(c+d x)}{160 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^4(c+d x)}{4 d \sqrt {a+a \sin (c+d x)}}+\frac {1}{64} (63 a) \int \frac {\sec (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\\ &=-\frac {9 a \sec ^2(c+d x)}{40 d (a+a \sin (c+d x))^{3/2}}+\frac {63 \sec ^2(c+d x)}{160 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^4(c+d x)}{4 d \sqrt {a+a \sin (c+d x)}}+\frac {\left (63 a^2\right ) \text {Subst}\left (\int \frac {1}{(a-x) (a+x)^{5/2}} \, dx,x,a \sin (c+d x)\right )}{64 d}\\ &=-\frac {21 a}{64 d (a+a \sin (c+d x))^{3/2}}-\frac {9 a \sec ^2(c+d x)}{40 d (a+a \sin (c+d x))^{3/2}}+\frac {63 \sec ^2(c+d x)}{160 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^4(c+d x)}{4 d \sqrt {a+a \sin (c+d x)}}+\frac {(63 a) \text {Subst}\left (\int \frac {1}{(a-x) (a+x)^{3/2}} \, dx,x,a \sin (c+d x)\right )}{128 d}\\ &=-\frac {21 a}{64 d (a+a \sin (c+d x))^{3/2}}-\frac {9 a \sec ^2(c+d x)}{40 d (a+a \sin (c+d x))^{3/2}}-\frac {63}{128 d \sqrt {a+a \sin (c+d x)}}+\frac {63 \sec ^2(c+d x)}{160 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^4(c+d x)}{4 d \sqrt {a+a \sin (c+d x)}}+\frac {63 \text {Subst}\left (\int \frac {1}{(a-x) \sqrt {a+x}} \, dx,x,a \sin (c+d x)\right )}{256 d}\\ &=-\frac {21 a}{64 d (a+a \sin (c+d x))^{3/2}}-\frac {9 a \sec ^2(c+d x)}{40 d (a+a \sin (c+d x))^{3/2}}-\frac {63}{128 d \sqrt {a+a \sin (c+d x)}}+\frac {63 \sec ^2(c+d x)}{160 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^4(c+d x)}{4 d \sqrt {a+a \sin (c+d x)}}+\frac {63 \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+a \sin (c+d x)}\right )}{128 d}\\ &=\frac {63 \tanh ^{-1}\left (\frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{128 \sqrt {2} \sqrt {a} d}-\frac {21 a}{64 d (a+a \sin (c+d x))^{3/2}}-\frac {9 a \sec ^2(c+d x)}{40 d (a+a \sin (c+d x))^{3/2}}-\frac {63}{128 d \sqrt {a+a \sin (c+d x)}}+\frac {63 \sec ^2(c+d x)}{160 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^4(c+d x)}{4 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 = 44, normalized size = 0.25 \begin {gather*} -\frac {a^2 \, _2F_1\left (-\frac {5}{2},3;-\frac {3}{2};\frac {1}{2} (1+\sin (c+d x))\right )}{20 d (a+a \sin (c+d x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5/Sqrt[a + a*Sin[c + d*x]],x]

[Out]

-1/20*(a^2*Hypergeometric2F1[-5/2, 3, -3/2, (1 + Sin[c + d*x])/2])/(d*(a + a*Sin[c + d*x])^(5/2))

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Maple [A]
time = 0.82, size = 135, normalized size = 0.77

method	result	size
default	\(-\frac {2 a^{5} \left (\frac {\frac {\sqrt {a +a \sin \left (d x +c \right )}\, a \left (15 \sin \left (d x +c \right )-19\right )}{16 \left (a \sin \left (d x +c \right )-a \right )^{2}}-\frac {63 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{32 \sqrt {a}}}{16 a^{5}}+\frac {3}{16 a^{5} \sqrt {a +a \sin \left (d x +c \right )}}+\frac {1}{16 a^{4} \left (a +a \sin \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {1}{40 a^{3} \left (a +a \sin \left (d x +c \right )\right )^{\frac {5}{2}}}\right )}{d}\)	\(135\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5/(a+a*sin(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2*a^5*(1/16/a^5*(1/16*(a+a*sin(d*x+c))^(1/2)*a*(15*sin(d*x+c)-19)/(a*sin(d*x+c)-a)^2-63/32*2^(1/2)/a^(1/2)*ar
ctanh(1/2*(a+a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2)))+3/16/a^5/(a+a*sin(d*x+c))^(1/2)+1/16/a^4/(a+a*sin(d*x+c))^(
3/2)+1/40/a^3/(a+a*sin(d*x+c))^(5/2))/d

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Maxima [A]
time = 0.50, size = 183, normalized size = 1.05 \begin {gather*} -\frac {315 \, \sqrt {2} \sqrt {a} \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 ) + \frac {4 \, {\left (315 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{4} a - 1050 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{3} a^{2} + 672 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{2} a^{3} + 192 \, {\left (a \sin \left (d x + c\right ) + a\right )} a^{4} + 128 \, a^{5}\right )}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {9}{2}} - 4 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a + 4 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{2}}}{2560 \, a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

-1/2560*(315*sqrt(2)*sqrt(a)*log(-(sqrt(2)*sqrt(a) - sqrt(a*sin(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(a*sin(d
*x + c) + a))) + 4*(315*(a*sin(d*x + c) + a)^4*a - 1050*(a*sin(d*x + c) + a)^3*a^2 + 672*(a*sin(d*x + c) + a)^
2*a^3 + 192*(a*sin(d*x + c) + a)*a^4 + 128*a^5)/((a*sin(d*x + c) + a)^(9/2) - 4*(a*sin(d*x + c) + a)^(7/2)*a +
4*(a*sin(d*x + c) + a)^(5/2)*a^2))/(a*d)

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Fricas [A]
time = 0.37, size = 167, normalized size = 0.95 \begin {gather*} \frac {315 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + \cos \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 (315 \, \cos \left (d x + c\right )^{4} - 42 \, \cos \left (d x + c\right )^{2} - 6 \, {\left (35 \, \cos \left (d x + c\right )^{2} + 24\right )} \sin \left (d x + c\right ) - 16\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{2560 \, {\left (a d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/2560*(315*sqrt(2)*(cos(d*x + c)^4*sin(d*x + c) + cos(d*x + c)^4)*sqrt(a)*log(-(a*sin(d*x + c) + 2*sqrt(2)*sq
rt(a*sin(d*x + c) + a)*sqrt(a) + 3*a)/(sin(d*x + c) - 1)) - 4*(315*cos(d*x + c)^4 - 42*cos(d*x + c)^2 - 6*(35*
cos(d*x + c)^2 + 24)*sin(d*x + c) - 16)*sqrt(a*sin(d*x + c) + a))/(a*d*cos(d*x + c)^4*sin(d*x + c) + a*d*cos(d
*x + c)^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 6.37, size = 233, normalized size = 1.33 \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 \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 \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {10 \, {\left (15 \, \sqrt {2} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 17 \, \sqrt {2} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {16 \, \sqrt {2} {\left (30 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 5 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}}{a \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}\right )}}{2560 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

1/2560*sqrt(a)*(315*sqrt(2)*log(cos(-1/4*pi + 1/2*d*x + 1/2*c) + 1)/(a*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*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) - 10*(15*sqrt(2)*
cos(-1/4*pi + 1/2*d*x + 1/2*c)^3 - 17*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)^2*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) - 16*sqrt(2)*(30*cos(-1/4*pi + 1/2*d*x + 1/2*c)^4 + 5*cos(-1/
4*pi + 1/2*d*x + 1/2*c)^2 + 1)/(a*cos(-1/4*pi + 1/2*d*x + 1/2*c)^5*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 )}^5\,\sqrt {a+a\,\sin \left (c+d\,x\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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