3.113 \(\int \sec ^6(c+d x) \sqrt{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=197 \[ -\frac{63 a^2 \cos (c+d x)}{128 d (a \sin (c+d x)+a)^{3/2}}-\frac{21 a^2 \sec (c+d x)}{80 d (a \sin (c+d x)+a)^{3/2}}+\frac{\sec ^5(c+d x) \sqrt{a \sin (c+d x)+a}}{5 d}+\frac{3 a \sec ^3(c+d x)}{10 d \sqrt{a \sin (c+d x)+a}}+\frac{21 a \sec (c+d x)}{32 d \sqrt{a \sin (c+d x)+a}}-\frac{63 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{128 \sqrt{2} d} \]

[Out]

(-63*Sqrt[a]*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(128*Sqrt[2]*d) - (63*a^2*Cos
[c + d*x])/(128*d*(a + a*Sin[c + d*x])^(3/2)) - (21*a^2*Sec[c + d*x])/(80*d*(a + a*Sin[c + d*x])^(3/2)) + (21*
a*Sec[c + d*x])/(32*d*Sqrt[a + a*Sin[c + d*x]]) + (3*a*Sec[c + d*x]^3)/(10*d*Sqrt[a + a*Sin[c + d*x]]) + (Sec[
c + d*x]^5*Sqrt[a + a*Sin[c + d*x]])/(5*d)

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Rubi [A]  time = 0.293853, antiderivative size = 197, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {2675, 2687, 2681, 2650, 2649, 206} \[ -\frac{63 a^2 \cos (c+d x)}{128 d (a \sin (c+d x)+a)^{3/2}}-\frac{21 a^2 \sec (c+d x)}{80 d (a \sin (c+d x)+a)^{3/2}}+\frac{\sec ^5(c+d x) \sqrt{a \sin (c+d x)+a}}{5 d}+\frac{3 a \sec ^3(c+d x)}{10 d \sqrt{a \sin (c+d x)+a}}+\frac{21 a \sec (c+d x)}{32 d \sqrt{a \sin (c+d x)+a}}-\frac{63 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{128 \sqrt{2} d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^6*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(-63*Sqrt[a]*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(128*Sqrt[2]*d) - (63*a^2*Cos
[c + d*x])/(128*d*(a + a*Sin[c + d*x])^(3/2)) - (21*a^2*Sec[c + d*x])/(80*d*(a + a*Sin[c + d*x])^(3/2)) + (21*
a*Sec[c + d*x])/(32*d*Sqrt[a + a*Sin[c + d*x]]) + (3*a*Sec[c + d*x]^3)/(10*d*Sqrt[a + a*Sin[c + d*x]]) + (Sec[
c + d*x]^5*Sqrt[a + a*Sin[c + d*x]])/(5*d)

Rule 2675

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*(p + 1)), x] + Dist[(a*(m + p + 1))/(g^2*(p + 1)), Int[(
g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0]
&& GtQ[m, 0] && LeQ[p, -2*m] && IntegersQ[m + 1/2, 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]

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 2650

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^n)/(a*
d*(2*n + 1)), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

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 \sec ^6(c+d x) \sqrt{a+a \sin (c+d x)} \, dx &=\frac{\sec ^5(c+d x) \sqrt{a+a \sin (c+d x)}}{5 d}+\frac{1}{10} (9 a) \int \frac{\sec ^4(c+d x)}{\sqrt{a+a \sin (c+d x)}} \, dx\\ &=\frac{3 a \sec ^3(c+d x)}{10 d \sqrt{a+a \sin (c+d x)}}+\frac{\sec ^5(c+d x) \sqrt{a+a \sin (c+d x)}}{5 d}+\frac{1}{20} \left (21 a^2\right ) \int \frac{\sec ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\\ &=-\frac{21 a^2 \sec (c+d x)}{80 d (a+a \sin (c+d x))^{3/2}}+\frac{3 a \sec ^3(c+d x)}{10 d \sqrt{a+a \sin (c+d x)}}+\frac{\sec ^5(c+d x) \sqrt{a+a \sin (c+d x)}}{5 d}+\frac{1}{32} (21 a) \int \frac{\sec ^2(c+d x)}{\sqrt{a+a \sin (c+d x)}} \, dx\\ &=-\frac{21 a^2 \sec (c+d x)}{80 d (a+a \sin (c+d x))^{3/2}}+\frac{21 a \sec (c+d x)}{32 d \sqrt{a+a \sin (c+d x)}}+\frac{3 a \sec ^3(c+d x)}{10 d \sqrt{a+a \sin (c+d x)}}+\frac{\sec ^5(c+d x) \sqrt{a+a \sin (c+d x)}}{5 d}+\frac{1}{64} \left (63 a^2\right ) \int \frac{1}{(a+a \sin (c+d x))^{3/2}} \, dx\\ &=-\frac{63 a^2 \cos (c+d x)}{128 d (a+a \sin (c+d x))^{3/2}}-\frac{21 a^2 \sec (c+d x)}{80 d (a+a \sin (c+d x))^{3/2}}+\frac{21 a \sec (c+d x)}{32 d \sqrt{a+a \sin (c+d x)}}+\frac{3 a \sec ^3(c+d x)}{10 d \sqrt{a+a \sin (c+d x)}}+\frac{\sec ^5(c+d x) \sqrt{a+a \sin (c+d x)}}{5 d}+\frac{1}{256} (63 a) \int \frac{1}{\sqrt{a+a \sin (c+d x)}} \, dx\\ &=-\frac{63 a^2 \cos (c+d x)}{128 d (a+a \sin (c+d x))^{3/2}}-\frac{21 a^2 \sec (c+d x)}{80 d (a+a \sin (c+d x))^{3/2}}+\frac{21 a \sec (c+d x)}{32 d \sqrt{a+a \sin (c+d x)}}+\frac{3 a \sec ^3(c+d x)}{10 d \sqrt{a+a \sin (c+d x)}}+\frac{\sec ^5(c+d x) \sqrt{a+a \sin (c+d x)}}{5 d}-\frac{(63 a) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\frac{a \cos (c+d x)}{\sqrt{a+a \sin (c+d x)}}\right )}{128 d}\\ &=-\frac{63 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a+a \sin (c+d x)}}\right )}{128 \sqrt{2} d}-\frac{63 a^2 \cos (c+d x)}{128 d (a+a \sin (c+d x))^{3/2}}-\frac{21 a^2 \sec (c+d x)}{80 d (a+a \sin (c+d x))^{3/2}}+\frac{21 a \sec (c+d x)}{32 d \sqrt{a+a \sin (c+d x)}}+\frac{3 a \sec ^3(c+d x)}{10 d \sqrt{a+a \sin (c+d x)}}+\frac{\sec ^5(c+d x) \sqrt{a+a \sin (c+d x)}}{5 d}\\ \end{align*}

Mathematica [C]  time = 0.620729, size = 191, normalized size = 0.97 \[ \frac{\sqrt{a (\sin (c+d x)+1)} \left (\frac{1572 \sin (c+d x)+420 \sin (3 (c+d x))+1092 \cos (2 (c+d x))+315 \cos (4 (c+d x))+649}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^5}-(2520+2520 i) (-1)^{3/4} \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4 \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \sec \left (\frac{d x}{4}\right ) \left (\cos \left (\frac{1}{4} (2 c+d x)\right )-\sin \left (\frac{1}{4} (2 c+d x)\right )\right )\right )\right )}{5120 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^6*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(Sqrt[a*(1 + Sin[c + d*x])]*((-2520 - 2520*I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*Sec[(d*x)/4]*(Cos[(2*c
+ d*x)/4] - Sin[(2*c + d*x)/4])]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4 + (649 + 1092*Cos[2*(c + d*x)] + 315
*Cos[4*(c + d*x)] + 1572*Sin[c + d*x] + 420*Sin[3*(c + d*x)])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^5))/(5120*
d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^5)

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Maple [A]  time = 0.152, size = 244, normalized size = 1.2 \begin{align*} -{\frac{1}{1280\, \left ( \sin \left ( dx+c \right ) -1 \right ) ^{2} \left ( 1+\sin \left ( dx+c \right ) \right ) \cos \left ( dx+c \right ) d} \left ( -420\,{a}^{9/2}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}+ \left ( 630\, \left ( a-a\sin \left ( dx+c \right ) \right ) ^{5/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{2}-288\,{a}^{9/2} \right ) \sin \left ( dx+c \right ) -630\,{a}^{9/2} \left ( \cos \left ( dx+c \right ) \right ) ^{4}+ \left ( -315\, \left ( a-a\sin \left ( dx+c \right ) \right ) ^{5/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{2}+84\,{a}^{9/2} \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}+630\, \left ( a-a\sin \left ( dx+c \right ) \right ) ^{5/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{2}+32\,{a}^{9/2} \right ){a}^{-{\frac{7}{2}}}{\frac{1}{\sqrt{a+a\sin \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^6*(a+a*sin(d*x+c))^(1/2),x)

[Out]

-1/1280/a^(7/2)*(-420*a^(9/2)*sin(d*x+c)*cos(d*x+c)^2+(630*(a-a*sin(d*x+c))^(5/2)*2^(1/2)*arctanh(1/2*(a-a*sin
(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^2-288*a^(9/2))*sin(d*x+c)-630*a^(9/2)*cos(d*x+c)^4+(-315*(a-a*sin(d*x+c))^(5
/2)*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^2+84*a^(9/2))*cos(d*x+c)^2+630*(a-a*sin(d*x+
c))^(5/2)*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^2+32*a^(9/2))/(sin(d*x+c)-1)^2/(1+sin(
d*x+c))/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a \sin \left (d x + c\right ) + a} \sec \left (d x + c\right )^{6}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(a*sin(d*x + c) + a)*sec(d*x + c)^6, x)

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Fricas [A]  time = 1.89484, size = 583, normalized size = 2.96 \begin{align*} \frac{315 \, \sqrt{2} \sqrt{a} \cos \left (d x + c\right )^{5} \log \left (-\frac{a \cos \left (d x + c\right )^{2} - 2 \, \sqrt{a \sin \left (d x + c\right ) + a}{\left (\sqrt{2} \cos \left (d x + c\right ) - \sqrt{2} \sin \left (d x + c\right ) + \sqrt{2}\right )} \sqrt{a} + 3 \, a \cos \left (d x + c\right ) -{\left (a \cos \left (d x + c\right ) - 2 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{\cos \left (d x + c\right )^{2} -{\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\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 \, d \cos \left (d x + c\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/2560*(315*sqrt(2)*sqrt(a)*cos(d*x + c)^5*log(-(a*cos(d*x + c)^2 - 2*sqrt(a*sin(d*x + c) + a)*(sqrt(2)*cos(d*
x + c) - sqrt(2)*sin(d*x + c) + sqrt(2))*sqrt(a) + 3*a*cos(d*x + c) - (a*cos(d*x + c) - 2*a)*sin(d*x + c) + 2*
a)/(cos(d*x + c)^2 - (cos(d*x + c) + 2)*sin(d*x + c) - cos(d*x + c) - 2)) + 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))/(d*cos(d*x + c)^5)

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

sage2