3.125 \(\int \sec ^5(c+d x) (a+a \sin (c+d x))^{3/2} \, dx\)

Optimal. Leaf size=127 \[ -\frac{15 a^2}{32 d \sqrt{a \sin (c+d x)+a}}+\frac{15 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a \sin (c+d x)+a}}{\sqrt{2} \sqrt{a}}\right )}{32 \sqrt{2} d}+\frac{\sec ^4(c+d x) (a \sin (c+d x)+a)^{3/2}}{4 d}+\frac{5 a \sec ^2(c+d x) \sqrt{a \sin (c+d x)+a}}{16 d} \]

[Out]

(15*a^(3/2)*ArcTanh[Sqrt[a + a*Sin[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(32*Sqrt[2]*d) - (15*a^2)/(32*d*Sqrt[a + a*Si
n[c + d*x]]) + (5*a*Sec[c + d*x]^2*Sqrt[a + a*Sin[c + d*x]])/(16*d) + (Sec[c + d*x]^4*(a + a*Sin[c + d*x])^(3/
2))/(4*d)

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Rubi [A]  time = 0.181212, antiderivative size = 127, 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, 2667, 51, 63, 206} \[ -\frac{15 a^2}{32 d \sqrt{a \sin (c+d x)+a}}+\frac{15 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a \sin (c+d x)+a}}{\sqrt{2} \sqrt{a}}\right )}{32 \sqrt{2} d}+\frac{\sec ^4(c+d x) (a \sin (c+d x)+a)^{3/2}}{4 d}+\frac{5 a \sec ^2(c+d x) \sqrt{a \sin (c+d x)+a}}{16 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5*(a + a*Sin[c + d*x])^(3/2),x]

[Out]

(15*a^(3/2)*ArcTanh[Sqrt[a + a*Sin[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(32*Sqrt[2]*d) - (15*a^2)/(32*d*Sqrt[a + a*Si
n[c + d*x]]) + (5*a*Sec[c + d*x]^2*Sqrt[a + a*Sin[c + d*x]])/(16*d) + (Sec[c + d*x]^4*(a + a*Sin[c + d*x])^(3/
2))/(4*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 ^5(c+d x) (a+a \sin (c+d x))^{3/2} \, dx &=\frac{\sec ^4(c+d x) (a+a \sin (c+d x))^{3/2}}{4 d}+\frac{1}{8} (5 a) \int \sec ^3(c+d x) \sqrt{a+a \sin (c+d x)} \, dx\\ &=\frac{5 a \sec ^2(c+d x) \sqrt{a+a \sin (c+d x)}}{16 d}+\frac{\sec ^4(c+d x) (a+a \sin (c+d x))^{3/2}}{4 d}+\frac{1}{32} \left (15 a^2\right ) \int \frac{\sec (c+d x)}{\sqrt{a+a \sin (c+d x)}} \, dx\\ &=\frac{5 a \sec ^2(c+d x) \sqrt{a+a \sin (c+d x)}}{16 d}+\frac{\sec ^4(c+d x) (a+a \sin (c+d x))^{3/2}}{4 d}+\frac{\left (15 a^3\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x) (a+x)^{3/2}} \, dx,x,a \sin (c+d x)\right )}{32 d}\\ &=-\frac{15 a^2}{32 d \sqrt{a+a \sin (c+d x)}}+\frac{5 a \sec ^2(c+d x) \sqrt{a+a \sin (c+d x)}}{16 d}+\frac{\sec ^4(c+d x) (a+a \sin (c+d x))^{3/2}}{4 d}+\frac{\left (15 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x) \sqrt{a+x}} \, dx,x,a \sin (c+d x)\right )}{64 d}\\ &=-\frac{15 a^2}{32 d \sqrt{a+a \sin (c+d x)}}+\frac{5 a \sec ^2(c+d x) \sqrt{a+a \sin (c+d x)}}{16 d}+\frac{\sec ^4(c+d x) (a+a \sin (c+d x))^{3/2}}{4 d}+\frac{\left (15 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\sqrt{a+a \sin (c+d x)}\right )}{32 d}\\ &=\frac{15 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+a \sin (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{32 \sqrt{2} d}-\frac{15 a^2}{32 d \sqrt{a+a \sin (c+d x)}}+\frac{5 a \sec ^2(c+d x) \sqrt{a+a \sin (c+d x)}}{16 d}+\frac{\sec ^4(c+d x) (a+a \sin (c+d x))^{3/2}}{4 d}\\ \end{align*}

Mathematica [C]  time = 0.0714919, size = 44, normalized size = 0.35 \[ -\frac{a^2 \, _2F_1\left (-\frac{1}{2},3;\frac{1}{2};\frac{1}{2} (\sin (c+d x)+1)\right )}{4 d \sqrt{a \sin (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5*(a + a*Sin[c + d*x])^(3/2),x]

[Out]

-(a^2*Hypergeometric2F1[-1/2, 3, 1/2, (1 + Sin[c + d*x])/2])/(4*d*Sqrt[a + a*Sin[c + d*x]])

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Maple [A]  time = 0.188, size = 101, normalized size = 0.8 \begin{align*} -2\,{\frac{{a}^{5}}{d} \left ( 1/8\,{\frac{1}{{a}^{3}} \left ( 1/8\,{\frac{\sqrt{a+a\sin \left ( dx+c \right ) }a \left ( 7\,\sin \left ( dx+c \right ) -11 \right ) }{ \left ( a\sin \left ( dx+c \right ) -a \right ) ^{2}}}-{\frac{15\,\sqrt{2}}{16\,\sqrt{a}}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a+a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ) } \right ) }+1/8\,{\frac{1}{{a}^{3}\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)^5*(a+a*sin(d*x+c))^(3/2),x)

[Out]

-2*a^5*(1/8/a^3*(1/8*(a+a*sin(d*x+c))^(1/2)*a*(7*sin(d*x+c)-11)/(a*sin(d*x+c)-a)^2-15/16*2^(1/2)/a^(1/2)*arcta
nh(1/2*(a+a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2)))+1/8/a^3/(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)^5*(a+a*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.76802, size = 417, normalized size = 3.28 \begin{align*} \frac{15 \,{\left (\sqrt{2} a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - \sqrt{2} a \cos \left (d x + c\right )^{2}\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 (15 \, a \cos \left (d x + c\right )^{2} + 20 \, a \sin \left (d x + c\right ) - 12 \, a\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{128 \,{\left (d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - d \cos \left (d x + c\right )^{2}\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))^(3/2),x, algorithm="fricas")

[Out]

1/128*(15*(sqrt(2)*a*cos(d*x + c)^2*sin(d*x + c) - sqrt(2)*a*cos(d*x + c)^2)*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*(15*a*cos(d*x + c)^2 + 20*a*sin(d*x +
c) - 12*a)*sqrt(a*sin(d*x + c) + a))/(d*cos(d*x + c)^2*sin(d*x + c) - d*cos(d*x + c)^2)

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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))**(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)^5*(a+a*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

Timed out