3.112 \(\int \sec ^5(c+d x) \sqrt{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=149 \[ -\frac{35 a^2}{96 d (a \sin (c+d x)+a)^{3/2}}-\frac{35 a}{64 d \sqrt{a \sin (c+d x)+a}}+\frac{35 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a \sin (c+d x)+a}}{\sqrt{2} \sqrt{a}}\right )}{64 \sqrt{2} d}+\frac{\sec ^4(c+d x) \sqrt{a \sin (c+d x)+a}}{4 d}+\frac{7 a \sec ^2(c+d x)}{16 d \sqrt{a \sin (c+d x)+a}} \]

[Out]

(35*Sqrt[a]*ArcTanh[Sqrt[a + a*Sin[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(64*Sqrt[2]*d) - (35*a^2)/(96*d*(a + a*Sin[c
+ d*x])^(3/2)) - (35*a)/(64*d*Sqrt[a + a*Sin[c + d*x]]) + (7*a*Sec[c + d*x]^2)/(16*d*Sqrt[a + a*Sin[c + d*x]])
+ (Sec[c + d*x]^4*Sqrt[a + a*Sin[c + d*x]])/(4*d)

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Rubi [A]  time = 0.203373, antiderivative size = 149, 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, 2667, 51, 63, 206} \[ -\frac{35 a^2}{96 d (a \sin (c+d x)+a)^{3/2}}-\frac{35 a}{64 d \sqrt{a \sin (c+d x)+a}}+\frac{35 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a \sin (c+d x)+a}}{\sqrt{2} \sqrt{a}}\right )}{64 \sqrt{2} d}+\frac{\sec ^4(c+d x) \sqrt{a \sin (c+d x)+a}}{4 d}+\frac{7 a \sec ^2(c+d x)}{16 d \sqrt{a \sin (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(35*Sqrt[a]*ArcTanh[Sqrt[a + a*Sin[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(64*Sqrt[2]*d) - (35*a^2)/(96*d*(a + a*Sin[c
+ d*x])^(3/2)) - (35*a)/(64*d*Sqrt[a + a*Sin[c + d*x]]) + (7*a*Sec[c + d*x]^2)/(16*d*Sqrt[a + a*Sin[c + d*x]])
+ (Sec[c + d*x]^4*Sqrt[a + a*Sin[c + d*x]])/(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 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 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) \sqrt{a+a \sin (c+d x)} \, dx &=\frac{\sec ^4(c+d x) \sqrt{a+a \sin (c+d x)}}{4 d}+\frac{1}{8} (7 a) \int \frac{\sec ^3(c+d x)}{\sqrt{a+a \sin (c+d x)}} \, dx\\ &=\frac{7 a \sec ^2(c+d x)}{16 d \sqrt{a+a \sin (c+d x)}}+\frac{\sec ^4(c+d x) \sqrt{a+a \sin (c+d x)}}{4 d}+\frac{1}{32} \left (35 a^2\right ) \int \frac{\sec (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\\ &=\frac{7 a \sec ^2(c+d x)}{16 d \sqrt{a+a \sin (c+d x)}}+\frac{\sec ^4(c+d x) \sqrt{a+a \sin (c+d x)}}{4 d}+\frac{\left (35 a^3\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x) (a+x)^{5/2}} \, dx,x,a \sin (c+d x)\right )}{32 d}\\ &=-\frac{35 a^2}{96 d (a+a \sin (c+d x))^{3/2}}+\frac{7 a \sec ^2(c+d x)}{16 d \sqrt{a+a \sin (c+d x)}}+\frac{\sec ^4(c+d x) \sqrt{a+a \sin (c+d x)}}{4 d}+\frac{\left (35 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x) (a+x)^{3/2}} \, dx,x,a \sin (c+d x)\right )}{64 d}\\ &=-\frac{35 a^2}{96 d (a+a \sin (c+d x))^{3/2}}-\frac{35 a}{64 d \sqrt{a+a \sin (c+d x)}}+\frac{7 a \sec ^2(c+d x)}{16 d \sqrt{a+a \sin (c+d x)}}+\frac{\sec ^4(c+d x) \sqrt{a+a \sin (c+d x)}}{4 d}+\frac{(35 a) \operatorname{Subst}\left (\int \frac{1}{(a-x) \sqrt{a+x}} \, dx,x,a \sin (c+d x)\right )}{128 d}\\ &=-\frac{35 a^2}{96 d (a+a \sin (c+d x))^{3/2}}-\frac{35 a}{64 d \sqrt{a+a \sin (c+d x)}}+\frac{7 a \sec ^2(c+d x)}{16 d \sqrt{a+a \sin (c+d x)}}+\frac{\sec ^4(c+d x) \sqrt{a+a \sin (c+d x)}}{4 d}+\frac{(35 a) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\sqrt{a+a \sin (c+d x)}\right )}{64 d}\\ &=\frac{35 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+a \sin (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{64 \sqrt{2} d}-\frac{35 a^2}{96 d (a+a \sin (c+d x))^{3/2}}-\frac{35 a}{64 d \sqrt{a+a \sin (c+d x)}}+\frac{7 a \sec ^2(c+d x)}{16 d \sqrt{a+a \sin (c+d x)}}+\frac{\sec ^4(c+d x) \sqrt{a+a \sin (c+d x)}}{4 d}\\ \end{align*}

Mathematica [C]  time = 0.463036, size = 179, normalized size = 1.2 \[ \frac{\sqrt{a (\sin (c+d x)+1)} \left (\frac{329 \sin (c+d x)+105 \sin (3 (c+d x))-70 \cos (2 (c+d x))-102}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^4}-(420-420 i) \sqrt [4]{-1} \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3 \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \sec \left (\frac{d x}{4}\right ) \left (\sin \left (\frac{1}{4} (2 c+d x)\right )+\cos \left (\frac{1}{4} (2 c+d x)\right )\right )\right )\right )}{768 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sec[c + d*x]^5*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(Sqrt[a*(1 + Sin[c + d*x])]*((-420 + 420*I)*(-1)^(1/4)*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])^3 + (-102 - 70*Cos[2*(c + d*x)] + 329*Si
n[c + d*x] + 105*Sin[3*(c + d*x)])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^4))/(768*d*(Cos[(c + d*x)/2] + Sin[(c
+ d*x)/2])^4)

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Maple [A]  time = 0.211, size = 118, normalized size = 0.8 \begin{align*} -2\,{\frac{{a}^{5}}{d} \left ( 1/16\,{\frac{1}{{a}^{4}} \left ( 1/8\,{\frac{\sqrt{a+a\sin \left ( dx+c \right ) }a \left ( 11\,\sin \left ( dx+c \right ) -15 \right ) }{ \left ( a\sin \left ( dx+c \right ) -a \right ) ^{2}}}-{\frac{35\,\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 ) }+3/16\,{\frac{1}{{a}^{4}\sqrt{a+a\sin \left ( dx+c \right ) }}}+1/24\,{\frac{1}{{a}^{3} \left ( a+a\sin \left ( dx+c \right ) \right ) ^{3/2}}} \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))^(1/2),x)

[Out]

-2*a^5*(1/16/a^4*(1/8*(a+a*sin(d*x+c))^(1/2)*a*(11*sin(d*x+c)-15)/(a*sin(d*x+c)-a)^2-35/16*2^(1/2)/a^(1/2)*arc
tanh(1/2*(a+a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2)))+3/16/a^4/(a+a*sin(d*x+c))^(1/2)+1/24/a^3/(a+a*sin(d*x+c))^(3
/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))^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.86235, size = 338, normalized size = 2.27 \begin{align*} \frac{105 \, \sqrt{2} \sqrt{a} \cos \left (d x + c\right )^{4} \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 (35 \, \cos \left (d x + c\right )^{2} - 7 \,{\left (15 \, \cos \left (d x + c\right )^{2} + 8\right )} \sin \left (d x + c\right ) + 8\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{768 \, d \cos \left (d x + c\right )^{4}} \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))^(1/2),x, algorithm="fricas")

[Out]

1/768*(105*sqrt(2)*sqrt(a)*cos(d*x + c)^4*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*(35*cos(d*x + c)^2 - 7*(15*cos(d*x + c)^2 + 8)*sin(d*x + c) + 8)*sqrt(a*sin(d*x +
c) + a))/(d*cos(d*x + c)^4)

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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))**(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)^5*(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

sage2