3.137 \(\int \sec ^6(c+d x) (a+a \sin (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=139 \[ \frac{a^2 \sec (c+d x) \sqrt{a \sin (c+d x)+a}}{4 d}-\frac{a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{4 \sqrt{2} d}+\frac{\sec ^5(c+d x) (a \sin (c+d x)+a)^{5/2}}{5 d}+\frac{a \sec ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{6 d} \]

[Out]

-(a^(5/2)*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(4*Sqrt[2]*d) + (a^2*Sec[c + d*x
]*Sqrt[a + a*Sin[c + d*x]])/(4*d) + (a*Sec[c + d*x]^3*(a + a*Sin[c + d*x])^(3/2))/(6*d) + (Sec[c + d*x]^5*(a +
a*Sin[c + d*x])^(5/2))/(5*d)

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Rubi [A]  time = 0.19727, antiderivative size = 139, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {2675, 2649, 206} \[ \frac{a^2 \sec (c+d x) \sqrt{a \sin (c+d x)+a}}{4 d}-\frac{a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{4 \sqrt{2} d}+\frac{\sec ^5(c+d x) (a \sin (c+d x)+a)^{5/2}}{5 d}+\frac{a \sec ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{6 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^6*(a + a*Sin[c + d*x])^(5/2),x]

[Out]

-(a^(5/2)*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(4*Sqrt[2]*d) + (a^2*Sec[c + d*x
]*Sqrt[a + a*Sin[c + d*x]])/(4*d) + (a*Sec[c + d*x]^3*(a + a*Sin[c + d*x])^(3/2))/(6*d) + (Sec[c + d*x]^5*(a +
a*Sin[c + d*x])^(5/2))/(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 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) (a+a \sin (c+d x))^{5/2} \, dx &=\frac{\sec ^5(c+d x) (a+a \sin (c+d x))^{5/2}}{5 d}+\frac{1}{2} a \int \sec ^4(c+d x) (a+a \sin (c+d x))^{3/2} \, dx\\ &=\frac{a \sec ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{6 d}+\frac{\sec ^5(c+d x) (a+a \sin (c+d x))^{5/2}}{5 d}+\frac{1}{4} a^2 \int \sec ^2(c+d x) \sqrt{a+a \sin (c+d x)} \, dx\\ &=\frac{a^2 \sec (c+d x) \sqrt{a+a \sin (c+d x)}}{4 d}+\frac{a \sec ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{6 d}+\frac{\sec ^5(c+d x) (a+a \sin (c+d x))^{5/2}}{5 d}+\frac{1}{8} a^3 \int \frac{1}{\sqrt{a+a \sin (c+d x)}} \, dx\\ &=\frac{a^2 \sec (c+d x) \sqrt{a+a \sin (c+d x)}}{4 d}+\frac{a \sec ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{6 d}+\frac{\sec ^5(c+d x) (a+a \sin (c+d x))^{5/2}}{5 d}-\frac{a^3 \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 )}{4 d}\\ &=-\frac{a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a+a \sin (c+d x)}}\right )}{4 \sqrt{2} d}+\frac{a^2 \sec (c+d x) \sqrt{a+a \sin (c+d x)}}{4 d}+\frac{a \sec ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{6 d}+\frac{\sec ^5(c+d x) (a+a \sin (c+d x))^{5/2}}{5 d}\\ \end{align*}

Mathematica [C]  time = 5.2959, size = 129, normalized size = 0.93 \[ \frac{(a (\sin (c+d x)+1))^{5/2} \left (\frac{-80 \sin (c+d x)-15 \cos (2 (c+d x))+89}{2 \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^5}+(15+15 i) (-1)^{3/4} \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac{1}{4} (c+d x)\right )-1\right )\right )\right )}{60 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^6*(a + a*Sin[c + d*x])^(5/2),x]

[Out]

(((15 + 15*I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(c + d*x)/4])] + (89 - 15*Cos[2*(c + d*x)] -
80*Sin[c + d*x])/(2*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^5))*(a*(1 + Sin[c + d*x]))^(5/2))/(60*d*(Cos[(c + d
*x)/2] + Sin[(c + d*x)/2])^5)

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Maple [A]  time = 0.126, size = 120, normalized size = 0.9 \begin{align*} -{\frac{1+\sin \left ( dx+c \right ) }{120\, \left ( \sin \left ( dx+c \right ) -1 \right ) ^{2}\cos \left ( dx+c \right ) d} \left ( 30\,{a}^{11/2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}+80\,{a}^{11/2}\sin \left ( dx+c \right ) -104\,{a}^{11/2}+15\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{3} \left ( a-a\sin \left ( dx+c \right ) \right ) ^{5/2} \right ){a}^{-{\frac{5}{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))^(5/2),x)

[Out]

-1/120*(1+sin(d*x+c))*(30*a^(11/2)*cos(d*x+c)^2+80*a^(11/2)*sin(d*x+c)-104*a^(11/2)+15*2^(1/2)*arctanh(1/2*(a-
a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^3*(a-a*sin(d*x+c))^(5/2))/a^(5/2)/(sin(d*x+c)-1)^2/cos(d*x+c)/(a+a*sin(
d*x+c))^(1/2)/d

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Maxima [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))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 1.80312, size = 709, normalized size = 5.1 \begin{align*} \frac{15 \,{\left (\sqrt{2} a^{2} \cos \left (d x + c\right )^{3} + 2 \, \sqrt{2} a^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, \sqrt{2} a^{2} \cos \left (d x + c\right )\right )} \sqrt{a} \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 (15 \, a^{2} \cos \left (d x + c\right )^{2} + 40 \, a^{2} \sin \left (d x + c\right ) - 52 \, a^{2}\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{240 \,{\left (d \cos \left (d x + c\right )^{3} + 2 \, d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, d \cos \left (d x + c\right )\right )}} \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))^(5/2),x, algorithm="fricas")

[Out]

1/240*(15*(sqrt(2)*a^2*cos(d*x + c)^3 + 2*sqrt(2)*a^2*cos(d*x + c)*sin(d*x + c) - 2*sqrt(2)*a^2*cos(d*x + c))*
sqrt(a)*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) + sqr
t(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*(15*a^2*cos(d*x + c)^2 + 40*a^2*sin(d*x + c) - 52*a^2)*sqrt(a*sin(d
*x + c) + a))/(d*cos(d*x + c)^3 + 2*d*cos(d*x + c)*sin(d*x + c) - 2*d*cos(d*x + c))

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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))**(5/2),x)

[Out]

Timed out

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Giac [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))^(5/2),x, algorithm="giac")

[Out]

Timed out