3.167 \(\int \frac{\sec ^5(c+d x)}{\sqrt{a+a \sin (c+d x)}} \, dx\)

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

[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]])

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Rubi [A]  time = 0.262906, antiderivative size = 175, normalized size of antiderivative = 1., 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 = {2687, 2681, 2667, 51, 63, 206} \[ -\frac{63}{128 d \sqrt{a \sin (c+d x)+a}}-\frac{21 a}{64 d (a \sin (c+d x)+a)^{3/2}}+\frac{63 \tanh ^{-1}\left (\frac{\sqrt{a \sin (c+d x)+a}}{\sqrt{2} \sqrt{a}}\right )}{128 \sqrt{2} \sqrt{a} d}+\frac{\sec ^4(c+d x)}{4 d \sqrt{a \sin (c+d x)+a}}+\frac{63 \sec ^2(c+d x)}{160 d \sqrt{a \sin (c+d x)+a}}-\frac{9 a \sec ^2(c+d x)}{40 d (a \sin (c+d x)+a)^{3/2}} \]

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 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]

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 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 ^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 ) \operatorname{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) \operatorname{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 \operatorname{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 \operatorname{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*}

Mathematica [C]  time = 0.0796178, size = 44, normalized size = 0.25 \[ -\frac{a^2 \, _2F_1\left (-\frac{5}{2},3;-\frac{3}{2};\frac{1}{2} (\sin (c+d x)+1)\right )}{20 d (a \sin (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.244, size = 135, normalized size = 0.8 \begin{align*} -2\,{\frac{{a}^{5}}{d} \left ( 1/16\,{\frac{1}{{a}^{5}} \left ( 1/16\,{\frac{a\sqrt{a+a\sin \left ( dx+c \right ) } \left ( 15\,\sin \left ( dx+c \right ) -19 \right ) }{ \left ( a\sin \left ( dx+c \right ) -a \right ) ^{2}}}-{\frac{63\,\sqrt{2}}{32\,\sqrt{a}}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a+a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ) } \right ) }+3/16\,{\frac{1}{{a}^{5}\sqrt{a+a\sin \left ( dx+c \right ) }}}+1/16\,{\frac{1}{{a}^{4} \left ( a+a\sin \left ( dx+c \right ) \right ) ^{3/2}}}+1/40\,{\frac{1}{{a}^{3} \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)^5/(a+a*sin(d*x+c))^(1/2),x)

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.56751, size = 462, normalized size = 2.64 \begin{align*} \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{align*}

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

[Out]

Timed out

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Giac [A]  time = 1.11151, size = 194, normalized size = 1.11 \begin{align*} -\frac{a^{5}{\left (\frac{315 \, \sqrt{2} \arctan \left (\frac{\sqrt{2} \sqrt{a \sin \left (d x + c\right ) + a}}{2 \, \sqrt{-a}}\right )}{\sqrt{-a} a^{5}} + \frac{10 \,{\left (15 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}} - 34 \, \sqrt{a \sin \left (d x + c\right ) + a} a\right )}}{{\left (a \sin \left (d x + c\right ) - a\right )}^{2} a^{5}} + \frac{32 \,{\left (15 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{2} + 5 \,{\left (a \sin \left (d x + c\right ) + a\right )} a + 2 \, a^{2}\right )}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}} a^{5}}\right )}}{1280 \, d} \end{align*}

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]

-1/1280*a^5*(315*sqrt(2)*arctan(1/2*sqrt(2)*sqrt(a*sin(d*x + c) + a)/sqrt(-a))/(sqrt(-a)*a^5) + 10*(15*(a*sin(
d*x + c) + a)^(3/2) - 34*sqrt(a*sin(d*x + c) + a)*a)/((a*sin(d*x + c) - a)^2*a^5) + 32*(15*(a*sin(d*x + c) + a
)^2 + 5*(a*sin(d*x + c) + a)*a + 2*a^2)/((a*sin(d*x + c) + a)^(5/2)*a^5))/d