3.154 \(\int \sec ^9(c+d x) (a+a \sin (c+d x))^{7/2} \, dx\)

Optimal. Leaf size=191 \[ -\frac{315 a^4}{2048 d \sqrt{a \sin (c+d x)+a}}+\frac{315 a^{7/2} \tanh ^{-1}\left (\frac{\sqrt{a \sin (c+d x)+a}}{\sqrt{2} \sqrt{a}}\right )}{2048 \sqrt{2} d}+\frac{21 a^2 \sec ^4(c+d x) (a \sin (c+d x)+a)^{3/2}}{256 d}+\frac{105 a^3 \sec ^2(c+d x) \sqrt{a \sin (c+d x)+a}}{1024 d}+\frac{\sec ^8(c+d x) (a \sin (c+d x)+a)^{7/2}}{8 d}+\frac{3 a \sec ^6(c+d x) (a \sin (c+d x)+a)^{5/2}}{32 d} \]

[Out]

(315*a^(7/2)*ArcTanh[Sqrt[a + a*Sin[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(2048*Sqrt[2]*d) - (315*a^4)/(2048*d*Sqrt[a
+ a*Sin[c + d*x]]) + (105*a^3*Sec[c + d*x]^2*Sqrt[a + a*Sin[c + d*x]])/(1024*d) + (21*a^2*Sec[c + d*x]^4*(a +
a*Sin[c + d*x])^(3/2))/(256*d) + (3*a*Sec[c + d*x]^6*(a + a*Sin[c + d*x])^(5/2))/(32*d) + (Sec[c + d*x]^8*(a +
a*Sin[c + d*x])^(7/2))/(8*d)

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Rubi [A]  time = 0.312283, antiderivative size = 191, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {2675, 2667, 51, 63, 206} \[ -\frac{315 a^4}{2048 d \sqrt{a \sin (c+d x)+a}}+\frac{315 a^{7/2} \tanh ^{-1}\left (\frac{\sqrt{a \sin (c+d x)+a}}{\sqrt{2} \sqrt{a}}\right )}{2048 \sqrt{2} d}+\frac{21 a^2 \sec ^4(c+d x) (a \sin (c+d x)+a)^{3/2}}{256 d}+\frac{105 a^3 \sec ^2(c+d x) \sqrt{a \sin (c+d x)+a}}{1024 d}+\frac{\sec ^8(c+d x) (a \sin (c+d x)+a)^{7/2}}{8 d}+\frac{3 a \sec ^6(c+d x) (a \sin (c+d x)+a)^{5/2}}{32 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^9*(a + a*Sin[c + d*x])^(7/2),x]

[Out]

(315*a^(7/2)*ArcTanh[Sqrt[a + a*Sin[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(2048*Sqrt[2]*d) - (315*a^4)/(2048*d*Sqrt[a
+ a*Sin[c + d*x]]) + (105*a^3*Sec[c + d*x]^2*Sqrt[a + a*Sin[c + d*x]])/(1024*d) + (21*a^2*Sec[c + d*x]^4*(a +
a*Sin[c + d*x])^(3/2))/(256*d) + (3*a*Sec[c + d*x]^6*(a + a*Sin[c + d*x])^(5/2))/(32*d) + (Sec[c + d*x]^8*(a +
a*Sin[c + d*x])^(7/2))/(8*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 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 \sec ^9(c+d x) (a+a \sin (c+d x))^{7/2} \, dx &=\frac{\sec ^8(c+d x) (a+a \sin (c+d x))^{7/2}}{8 d}+\frac{1}{16} (9 a) \int \sec ^7(c+d x) (a+a \sin (c+d x))^{5/2} \, dx\\ &=\frac{3 a \sec ^6(c+d x) (a+a \sin (c+d x))^{5/2}}{32 d}+\frac{\sec ^8(c+d x) (a+a \sin (c+d x))^{7/2}}{8 d}+\frac{1}{64} \left (21 a^2\right ) \int \sec ^5(c+d x) (a+a \sin (c+d x))^{3/2} \, dx\\ &=\frac{21 a^2 \sec ^4(c+d x) (a+a \sin (c+d x))^{3/2}}{256 d}+\frac{3 a \sec ^6(c+d x) (a+a \sin (c+d x))^{5/2}}{32 d}+\frac{\sec ^8(c+d x) (a+a \sin (c+d x))^{7/2}}{8 d}+\frac{1}{512} \left (105 a^3\right ) \int \sec ^3(c+d x) \sqrt{a+a \sin (c+d x)} \, dx\\ &=\frac{105 a^3 \sec ^2(c+d x) \sqrt{a+a \sin (c+d x)}}{1024 d}+\frac{21 a^2 \sec ^4(c+d x) (a+a \sin (c+d x))^{3/2}}{256 d}+\frac{3 a \sec ^6(c+d x) (a+a \sin (c+d x))^{5/2}}{32 d}+\frac{\sec ^8(c+d x) (a+a \sin (c+d x))^{7/2}}{8 d}+\frac{\left (315 a^4\right ) \int \frac{\sec (c+d x)}{\sqrt{a+a \sin (c+d x)}} \, dx}{2048}\\ &=\frac{105 a^3 \sec ^2(c+d x) \sqrt{a+a \sin (c+d x)}}{1024 d}+\frac{21 a^2 \sec ^4(c+d x) (a+a \sin (c+d x))^{3/2}}{256 d}+\frac{3 a \sec ^6(c+d x) (a+a \sin (c+d x))^{5/2}}{32 d}+\frac{\sec ^8(c+d x) (a+a \sin (c+d x))^{7/2}}{8 d}+\frac{\left (315 a^5\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x) (a+x)^{3/2}} \, dx,x,a \sin (c+d x)\right )}{2048 d}\\ &=-\frac{315 a^4}{2048 d \sqrt{a+a \sin (c+d x)}}+\frac{105 a^3 \sec ^2(c+d x) \sqrt{a+a \sin (c+d x)}}{1024 d}+\frac{21 a^2 \sec ^4(c+d x) (a+a \sin (c+d x))^{3/2}}{256 d}+\frac{3 a \sec ^6(c+d x) (a+a \sin (c+d x))^{5/2}}{32 d}+\frac{\sec ^8(c+d x) (a+a \sin (c+d x))^{7/2}}{8 d}+\frac{\left (315 a^4\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x) \sqrt{a+x}} \, dx,x,a \sin (c+d x)\right )}{4096 d}\\ &=-\frac{315 a^4}{2048 d \sqrt{a+a \sin (c+d x)}}+\frac{105 a^3 \sec ^2(c+d x) \sqrt{a+a \sin (c+d x)}}{1024 d}+\frac{21 a^2 \sec ^4(c+d x) (a+a \sin (c+d x))^{3/2}}{256 d}+\frac{3 a \sec ^6(c+d x) (a+a \sin (c+d x))^{5/2}}{32 d}+\frac{\sec ^8(c+d x) (a+a \sin (c+d x))^{7/2}}{8 d}+\frac{\left (315 a^4\right ) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\sqrt{a+a \sin (c+d x)}\right )}{2048 d}\\ &=\frac{315 a^{7/2} \tanh ^{-1}\left (\frac{\sqrt{a+a \sin (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{2048 \sqrt{2} d}-\frac{315 a^4}{2048 d \sqrt{a+a \sin (c+d x)}}+\frac{105 a^3 \sec ^2(c+d x) \sqrt{a+a \sin (c+d x)}}{1024 d}+\frac{21 a^2 \sec ^4(c+d x) (a+a \sin (c+d x))^{3/2}}{256 d}+\frac{3 a \sec ^6(c+d x) (a+a \sin (c+d x))^{5/2}}{32 d}+\frac{\sec ^8(c+d x) (a+a \sin (c+d x))^{7/2}}{8 d}\\ \end{align*}

Mathematica [C]  time = 0.101034, size = 44, normalized size = 0.23 \[ -\frac{a^4 \, _2F_1\left (-\frac{1}{2},5;\frac{1}{2};\frac{1}{2} (\sin (c+d x)+1)\right )}{16 d \sqrt{a \sin (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^9*(a + a*Sin[c + d*x])^(7/2),x]

[Out]

-(a^4*Hypergeometric2F1[-1/2, 5, 1/2, (1 + Sin[c + d*x])/2])/(16*d*Sqrt[a + a*Sin[c + d*x]])

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Maple [A]  time = 0.26, size = 129, normalized size = 0.7 \begin{align*} -2\,{\frac{{a}^{9}}{d} \left ( 1/32\,{\frac{1}{{a}^{5}} \left ( -{\frac{\sqrt{a+a\sin \left ( dx+c \right ) }{a}^{3} \left ( 187\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) -725\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}-1236\,\sin \left ( dx+c \right ) +1364 \right ) }{128\, \left ( a\sin \left ( dx+c \right ) -a \right ) ^{4}}}-{\frac{315\,\sqrt{2}}{256\,\sqrt{a}}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a+a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ) } \right ) }+1/32\,{\frac{1}{{a}^{5}\sqrt{a+a\sin \left ( dx+c \right ) }}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^9*(a+a*sin(d*x+c))^(7/2),x)

[Out]

-2*a^9*(1/32/a^5*(-1/128*(a+a*sin(d*x+c))^(1/2)*a^3*(187*cos(d*x+c)^2*sin(d*x+c)-725*cos(d*x+c)^2-1236*sin(d*x
+c)+1364)/(a*sin(d*x+c)-a)^4-315/256*2^(1/2)/a^(1/2)*arctanh(1/2*(a+a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2)))+1/32
/a^5/(a+a*sin(d*x+c))^(1/2))/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)^9*(a+a*sin(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.24515, size = 663, normalized size = 3.47 \begin{align*} \frac{315 \,{\left (3 \, \sqrt{2} a^{3} \cos \left (d x + c\right )^{4} - 4 \, \sqrt{2} a^{3} \cos \left (d x + c\right )^{2} -{\left (\sqrt{2} a^{3} \cos \left (d x + c\right )^{4} - 4 \, \sqrt{2} a^{3} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\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 \, a^{3} \cos \left (d x + c\right )^{4} - 1722 \, a^{3} \cos \left (d x + c\right )^{2} + 896 \, a^{3} + 6 \,{\left (175 \, a^{3} \cos \left (d x + c\right )^{2} - 192 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{8192 \,{\left (3 \, d \cos \left (d x + c\right )^{4} - 4 \, d \cos \left (d x + c\right )^{2} -{\left (d \cos \left (d x + c\right )^{4} - 4 \, d \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*(a+a*sin(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

1/8192*(315*(3*sqrt(2)*a^3*cos(d*x + c)^4 - 4*sqrt(2)*a^3*cos(d*x + c)^2 - (sqrt(2)*a^3*cos(d*x + c)^4 - 4*sqr
t(2)*a^3*cos(d*x + c)^2)*sin(d*x + c))*sqrt(a)*log(-(a*sin(d*x + c) + 2*sqrt(2)*sqrt(a*sin(d*x + c) + a)*sqrt(
a) + 3*a)/(sin(d*x + c) - 1)) + 4*(315*a^3*cos(d*x + c)^4 - 1722*a^3*cos(d*x + c)^2 + 896*a^3 + 6*(175*a^3*cos
(d*x + c)^2 - 192*a^3)*sin(d*x + c))*sqrt(a*sin(d*x + c) + a))/(3*d*cos(d*x + c)^4 - 4*d*cos(d*x + c)^2 - (d*c
os(d*x + c)^4 - 4*d*cos(d*x + c)^2)*sin(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)**9*(a+a*sin(d*x+c))**(7/2),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)^9*(a+a*sin(d*x+c))^(7/2),x, algorithm="giac")

[Out]

Timed out