3.126 \(\int \sec ^6(c+d x) (a+a \sin (c+d x))^{3/2} \, dx\)

Optimal. Leaf size=169 \[ -\frac{7 a^3 \cos (c+d x)}{16 d (a \sin (c+d x)+a)^{3/2}}+\frac{7 a^2 \sec (c+d x)}{12 d \sqrt{a \sin (c+d x)+a}}-\frac{7 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{16 \sqrt{2} d}+\frac{\sec ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{5 d}+\frac{7 a \sec ^3(c+d x) \sqrt{a \sin (c+d x)+a}}{30 d} \]

[Out]

(-7*a^(3/2)*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(16*Sqrt[2]*d) - (7*a^3*Cos[c
+ d*x])/(16*d*(a + a*Sin[c + d*x])^(3/2)) + (7*a^2*Sec[c + d*x])/(12*d*Sqrt[a + a*Sin[c + d*x]]) + (7*a*Sec[c
+ d*x]^3*Sqrt[a + a*Sin[c + d*x]])/(30*d) + (Sec[c + d*x]^5*(a + a*Sin[c + d*x])^(3/2))/(5*d)

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Rubi [A]  time = 0.221389, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {2675, 2687, 2650, 2649, 206} \[ -\frac{7 a^3 \cos (c+d x)}{16 d (a \sin (c+d x)+a)^{3/2}}+\frac{7 a^2 \sec (c+d x)}{12 d \sqrt{a \sin (c+d x)+a}}-\frac{7 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{16 \sqrt{2} d}+\frac{\sec ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{5 d}+\frac{7 a \sec ^3(c+d x) \sqrt{a \sin (c+d x)+a}}{30 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^6*(a + a*Sin[c + d*x])^(3/2),x]

[Out]

(-7*a^(3/2)*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(16*Sqrt[2]*d) - (7*a^3*Cos[c
+ d*x])/(16*d*(a + a*Sin[c + d*x])^(3/2)) + (7*a^2*Sec[c + d*x])/(12*d*Sqrt[a + a*Sin[c + d*x]]) + (7*a*Sec[c
+ d*x]^3*Sqrt[a + a*Sin[c + d*x]])/(30*d) + (Sec[c + d*x]^5*(a + a*Sin[c + d*x])^(3/2))/(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 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 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) (a+a \sin (c+d x))^{3/2} \, dx &=\frac{\sec ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{5 d}+\frac{1}{10} (7 a) \int \sec ^4(c+d x) \sqrt{a+a \sin (c+d x)} \, dx\\ &=\frac{7 a \sec ^3(c+d x) \sqrt{a+a \sin (c+d x)}}{30 d}+\frac{\sec ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{5 d}+\frac{1}{12} \left (7 a^2\right ) \int \frac{\sec ^2(c+d x)}{\sqrt{a+a \sin (c+d x)}} \, dx\\ &=\frac{7 a^2 \sec (c+d x)}{12 d \sqrt{a+a \sin (c+d x)}}+\frac{7 a \sec ^3(c+d x) \sqrt{a+a \sin (c+d x)}}{30 d}+\frac{\sec ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{5 d}+\frac{1}{8} \left (7 a^3\right ) \int \frac{1}{(a+a \sin (c+d x))^{3/2}} \, dx\\ &=-\frac{7 a^3 \cos (c+d x)}{16 d (a+a \sin (c+d x))^{3/2}}+\frac{7 a^2 \sec (c+d x)}{12 d \sqrt{a+a \sin (c+d x)}}+\frac{7 a \sec ^3(c+d x) \sqrt{a+a \sin (c+d x)}}{30 d}+\frac{\sec ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{5 d}+\frac{1}{32} \left (7 a^2\right ) \int \frac{1}{\sqrt{a+a \sin (c+d x)}} \, dx\\ &=-\frac{7 a^3 \cos (c+d x)}{16 d (a+a \sin (c+d x))^{3/2}}+\frac{7 a^2 \sec (c+d x)}{12 d \sqrt{a+a \sin (c+d x)}}+\frac{7 a \sec ^3(c+d x) \sqrt{a+a \sin (c+d x)}}{30 d}+\frac{\sec ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{5 d}-\frac{\left (7 a^2\right ) \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 )}{16 d}\\ &=-\frac{7 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a+a \sin (c+d x)}}\right )}{16 \sqrt{2} d}-\frac{7 a^3 \cos (c+d x)}{16 d (a+a \sin (c+d x))^{3/2}}+\frac{7 a^2 \sec (c+d x)}{12 d \sqrt{a+a \sin (c+d x)}}+\frac{7 a \sec ^3(c+d x) \sqrt{a+a \sin (c+d x)}}{30 d}+\frac{\sec ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{5 d}\\ \end{align*}

Mathematica [C]  time = 0.421191, size = 288, normalized size = 1.7 \[ \frac{(a (\sin (c+d x)+1))^{3/2} \left (30 \sin \left (\frac{1}{2} (c+d x)\right )+\frac{90 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}+\frac{40 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{24 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^5}-15 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+(105+105 i) (-1)^{3/4} \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2 \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac{1}{4} (c+d x)\right )-1\right )\right )\right )}{240 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^6*(a + a*Sin[c + d*x])^(3/2),x]

[Out]

((30*Sin[(c + d*x)/2] - 15*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + (105 + 105*I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2
)*(-1)^(3/4)*(-1 + Tan[(c + d*x)/4])]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + (24*(Cos[(c + d*x)/2] + Sin[(c
+ d*x)/2])^2)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^5 + (40*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2)/(Cos[(c
+ d*x)/2] - Sin[(c + d*x)/2])^3 + (90*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2)/(Cos[(c + d*x)/2] - Sin[(c + d*
x)/2]))*(a*(1 + Sin[c + d*x]))^(3/2))/(240*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^5)

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Maple [A]  time = 0.132, size = 172, normalized size = 1. \begin{align*} -{\frac{1}{480\, \left ( \sin \left ( dx+c \right ) -1 \right ) ^{2}\cos \left ( dx+c \right ) d} \left ( 210\,{a}^{7/2}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}+ \left ( 105\, \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-168\,{a}^{7/2} \right ) \sin \left ( dx+c \right ) -350\,{a}^{7/2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}+105\, \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+72\,{a}^{7/2} \right ){a}^{-{\frac{3}{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))^(3/2),x)

[Out]

-1/480/a^(3/2)*(210*a^(7/2)*sin(d*x+c)*cos(d*x+c)^2+(105*(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-168*a^(7/2))*sin(d*x+c)-350*a^(7/2)*cos(d*x+c)^2+105*(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+72*a^(7/2))/(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))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.95671, size = 667, normalized size = 3.95 \begin{align*} \frac{105 \,{\left (\sqrt{2} a \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) - \sqrt{2} a \cos \left (d x + c\right )^{3}\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 (175 \, a \cos \left (d x + c\right )^{2} - 21 \,{\left (5 \, a \cos \left (d x + c\right )^{2} - 4 \, a\right )} \sin \left (d x + c\right ) - 36 \, a\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{960 \,{\left (d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) - d \cos \left (d x + c\right )^{3}\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))^(3/2),x, algorithm="fricas")

[Out]

1/960*(105*(sqrt(2)*a*cos(d*x + c)^3*sin(d*x + c) - sqrt(2)*a*cos(d*x + c)^3)*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) + 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*(175*a*cos(d*x + c)^2 - 21*(5*a*cos(d*x + c)^2 - 4*a)*sin(d*x + c) - 36*a)*sqrt(a*sin(d*x + c) + a)
)/(d*cos(d*x + c)^3*sin(d*x + c) - d*cos(d*x + c)^3)

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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))**(3/2),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))^(3/2),x, algorithm="giac")

[Out]

Timed out