3.391 \(\int \frac{\sqrt{\cos (c+d x)}}{(a+a \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=155 \[ -\frac{13 \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{6 a^3 d}+\frac{49 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac{13 \sin (c+d x)}{6 d \sqrt{\cos (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}-\frac{8 \sin (c+d x)}{15 a d \sqrt{\cos (c+d x)} (a \sec (c+d x)+a)^2}-\frac{\sin (c+d x)}{5 d \sqrt{\cos (c+d x)} (a \sec (c+d x)+a)^3} \]

[Out]

(49*EllipticE[(c + d*x)/2, 2])/(10*a^3*d) - (13*EllipticF[(c + d*x)/2, 2])/(6*a^3*d) - Sin[c + d*x]/(5*d*Sqrt[
Cos[c + d*x]]*(a + a*Sec[c + d*x])^3) - (8*Sin[c + d*x])/(15*a*d*Sqrt[Cos[c + d*x]]*(a + a*Sec[c + d*x])^2) -
(13*Sin[c + d*x])/(6*d*Sqrt[Cos[c + d*x]]*(a^3 + a^3*Sec[c + d*x]))

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Rubi [A]  time = 0.36957, antiderivative size = 155, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {4264, 3817, 4020, 3787, 3771, 2639, 2641} \[ -\frac{13 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{6 a^3 d}+\frac{49 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac{13 \sin (c+d x)}{6 d \sqrt{\cos (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}-\frac{8 \sin (c+d x)}{15 a d \sqrt{\cos (c+d x)} (a \sec (c+d x)+a)^2}-\frac{\sin (c+d x)}{5 d \sqrt{\cos (c+d x)} (a \sec (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Cos[c + d*x]]/(a + a*Sec[c + d*x])^3,x]

[Out]

(49*EllipticE[(c + d*x)/2, 2])/(10*a^3*d) - (13*EllipticF[(c + d*x)/2, 2])/(6*a^3*d) - Sin[c + d*x]/(5*d*Sqrt[
Cos[c + d*x]]*(a + a*Sec[c + d*x])^3) - (8*Sin[c + d*x])/(15*a*d*Sqrt[Cos[c + d*x]]*(a + a*Sec[c + d*x])^2) -
(13*Sin[c + d*x])/(6*d*Sqrt[Cos[c + d*x]]*(a^3 + a^3*Sec[c + d*x]))

Rule 4264

Int[(u_)*((c_.)*sin[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Csc[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[
u, x]

Rule 3817

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(Cot[
e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(f*(2*m + 1)), x] + Dist[1/(a^2*(2*m + 1)), Int[(a + b*Csc
[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*(a*(2*m + n + 1) - b*(m + n + 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d
, e, f, n}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -1] && (IntegersQ[2*m, 2*n] || IntegerQ[m])

Rule 4020

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> -Simp[((A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(b*f*(2
*m + 1)), x] - Dist[1/(a^2*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[b*B*n - a*A*(2
*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*
b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0]

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{\sqrt{\cos (c+d x)}}{(a+a \sec (c+d x))^3} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\sec (c+d x)} (a+a \sec (c+d x))^3} \, dx\\ &=-\frac{\sin (c+d x)}{5 d \sqrt{\cos (c+d x)} (a+a \sec (c+d x))^3}-\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{11 a}{2}+\frac{5}{2} a \sec (c+d x)}{\sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac{\sin (c+d x)}{5 d \sqrt{\cos (c+d x)} (a+a \sec (c+d x))^3}-\frac{8 \sin (c+d x)}{15 a d \sqrt{\cos (c+d x)} (a+a \sec (c+d x))^2}-\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{41 a^2}{2}+12 a^2 \sec (c+d x)}{\sqrt{\sec (c+d x)} (a+a \sec (c+d x))} \, dx}{15 a^4}\\ &=-\frac{\sin (c+d x)}{5 d \sqrt{\cos (c+d x)} (a+a \sec (c+d x))^3}-\frac{8 \sin (c+d x)}{15 a d \sqrt{\cos (c+d x)} (a+a \sec (c+d x))^2}-\frac{13 \sin (c+d x)}{6 d \sqrt{\cos (c+d x)} \left (a^3+a^3 \sec (c+d x)\right )}-\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{147 a^3}{4}+\frac{65}{4} a^3 \sec (c+d x)}{\sqrt{\sec (c+d x)}} \, dx}{15 a^6}\\ &=-\frac{\sin (c+d x)}{5 d \sqrt{\cos (c+d x)} (a+a \sec (c+d x))^3}-\frac{8 \sin (c+d x)}{15 a d \sqrt{\cos (c+d x)} (a+a \sec (c+d x))^2}-\frac{13 \sin (c+d x)}{6 d \sqrt{\cos (c+d x)} \left (a^3+a^3 \sec (c+d x)\right )}-\frac{\left (13 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\sec (c+d x)} \, dx}{12 a^3}+\frac{\left (49 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{20 a^3}\\ &=-\frac{\sin (c+d x)}{5 d \sqrt{\cos (c+d x)} (a+a \sec (c+d x))^3}-\frac{8 \sin (c+d x)}{15 a d \sqrt{\cos (c+d x)} (a+a \sec (c+d x))^2}-\frac{13 \sin (c+d x)}{6 d \sqrt{\cos (c+d x)} \left (a^3+a^3 \sec (c+d x)\right )}-\frac{13 \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{12 a^3}+\frac{49 \int \sqrt{\cos (c+d x)} \, dx}{20 a^3}\\ &=\frac{49 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac{13 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{6 a^3 d}-\frac{\sin (c+d x)}{5 d \sqrt{\cos (c+d x)} (a+a \sec (c+d x))^3}-\frac{8 \sin (c+d x)}{15 a d \sqrt{\cos (c+d x)} (a+a \sec (c+d x))^2}-\frac{13 \sin (c+d x)}{6 d \sqrt{\cos (c+d x)} \left (a^3+a^3 \sec (c+d x)\right )}\\ \end{align*}

Mathematica [C]  time = 2.16161, size = 357, normalized size = 2.3 \[ \frac{\cos ^6\left (\frac{1}{2} (c+d x)\right ) \left (-\frac{\csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) \left (806 \cos \left (\frac{1}{2} (c-d x)\right )+664 \cos \left (\frac{1}{2} (3 c+d x)\right )+470 \cos \left (\frac{1}{2} (c+3 d x)\right )+265 \cos \left (\frac{1}{2} (5 c+3 d x)\right )+117 \cos \left (\frac{1}{2} (3 c+5 d x)\right )+30 \cos \left (\frac{1}{2} (7 c+5 d x)\right )\right ) \sec ^5\left (\frac{1}{2} (c+d x)\right )}{8 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{4 i \sqrt{2} e^{-i (c+d x)} \sec ^3(c+d x) \left (147 \left (-1+e^{2 i c}\right ) \sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (-\frac{1}{4},\frac{1}{2},\frac{3}{4},-e^{2 i (c+d x)}\right )+65 \left (-1+e^{2 i c}\right ) e^{i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{4},\frac{1}{2},\frac{5}{4},-e^{2 i (c+d x)}\right )+147 \left (1+e^{2 i (c+d x)}\right )\right )}{\left (-1+e^{2 i c}\right ) d \sqrt{e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )}}\right )}{15 a^3 (\sec (c+d x)+1)^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sqrt[Cos[c + d*x]]/(a + a*Sec[c + d*x])^3,x]

[Out]

(Cos[(c + d*x)/2]^6*(-((806*Cos[(c - d*x)/2] + 664*Cos[(3*c + d*x)/2] + 470*Cos[(c + 3*d*x)/2] + 265*Cos[(5*c
+ 3*d*x)/2] + 117*Cos[(3*c + 5*d*x)/2] + 30*Cos[(7*c + 5*d*x)/2])*Csc[c/2]*Sec[c/2]*Sec[(c + d*x)/2]^5)/(8*d*C
os[c + d*x]^(5/2)) + ((4*I)*Sqrt[2]*(147*(1 + E^((2*I)*(c + d*x))) + 147*(-1 + E^((2*I)*c))*Sqrt[1 + E^((2*I)*
(c + d*x))]*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x))] + 65*E^(I*(c + d*x))*(-1 + E^((2*I)*c))*Sq
rt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/4, 1/2, 5/4, -E^((2*I)*(c + d*x))])*Sec[c + d*x]^3)/(d*E^(I*(c
+ d*x))*(-1 + E^((2*I)*c))*Sqrt[(1 + E^((2*I)*(c + d*x)))/E^(I*(c + d*x))])))/(15*a^3*(1 + Sec[c + d*x])^3)

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Maple [A]  time = 1.601, size = 270, normalized size = 1.7 \begin{align*}{\frac{1}{60\,{a}^{3}d}\sqrt{ \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( 348\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{8}+130\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}+294\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1} \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -578\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}+264\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}-37\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+3 \right ){\frac{1}{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}}} \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-5} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^3,x)

[Out]

1/60*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(348*cos(1/2*d*x+1/2*c)^8+130*(sin(1/2*d*x+1/2*c)
^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^5+294*(si
n(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*cos(1/2*d*x+1/2*c)^5*EllipticE(cos(1/2*d*x+1/2*c),
2^(1/2))-578*cos(1/2*d*x+1/2*c)^6+264*cos(1/2*d*x+1/2*c)^4-37*cos(1/2*d*x+1/2*c)^2+3)/a^3/(-2*sin(1/2*d*x+1/2*
c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)^5/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\cos \left (d x + c\right )}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

integrate(sqrt(cos(d*x + c))/(a*sec(d*x + c) + a)^3, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{\cos \left (d x + c\right )}}{a^{3} \sec \left (d x + c\right )^{3} + 3 \, a^{3} \sec \left (d x + c\right )^{2} + 3 \, a^{3} \sec \left (d x + c\right ) + a^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

integral(sqrt(cos(d*x + c))/(a^3*sec(d*x + c)^3 + 3*a^3*sec(d*x + c)^2 + 3*a^3*sec(d*x + c) + a^3), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\sqrt{\cos{\left (c + d x \right )}}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec{\left (c + d x \right )} + 1}\, dx}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**(1/2)/(a+a*sec(d*x+c))**3,x)

[Out]

Integral(sqrt(cos(c + d*x))/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x)/a**3

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\cos \left (d x + c\right )}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

integrate(sqrt(cos(d*x + c))/(a*sec(d*x + c) + a)^3, x)