3.140 \(\int \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=154 \[ \frac{a^3 (9 A-4 B) \sin (c+d x)}{4 d \sqrt{a \sec (c+d x)+a}}-\frac{a^2 (A-4 B) \sin (c+d x) \sqrt{a \sec (c+d x)+a}}{2 d}+\frac{a^{5/2} (19 A+20 B) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{4 d}+\frac{a A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{3/2}}{2 d} \]

[Out]

(a^(5/2)*(19*A + 20*B)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(4*d) + (a^3*(9*A - 4*B)*Sin[c
+ d*x])/(4*d*Sqrt[a + a*Sec[c + d*x]]) - (a^2*(A - 4*B)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(2*d) + (a*A*C
os[c + d*x]*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(2*d)

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Rubi [A]  time = 0.41877, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {4017, 4018, 4015, 3774, 203} \[ \frac{a^3 (9 A-4 B) \sin (c+d x)}{4 d \sqrt{a \sec (c+d x)+a}}-\frac{a^2 (A-4 B) \sin (c+d x) \sqrt{a \sec (c+d x)+a}}{2 d}+\frac{a^{5/2} (19 A+20 B) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{4 d}+\frac{a A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{3/2}}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x]),x]

[Out]

(a^(5/2)*(19*A + 20*B)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(4*d) + (a^3*(9*A - 4*B)*Sin[c
+ d*x])/(4*d*Sqrt[a + a*Sec[c + d*x]]) - (a^2*(A - 4*B)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(2*d) + (a*A*C
os[c + d*x]*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(2*d)

Rule 4017

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

Rule 4018

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

Rule 4015

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

Rule 3774

Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(-2*b)/d, Subst[Int[1/(a + x^2), x], x, (b*C
ot[c + d*x])/Sqrt[a + b*Csc[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx &=\frac{a A \cos (c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{2 d}+\frac{1}{2} \int \cos (c+d x) (a+a \sec (c+d x))^{3/2} \left (\frac{1}{2} a (7 A+4 B)-\frac{1}{2} a (A-4 B) \sec (c+d x)\right ) \, dx\\ &=-\frac{a^2 (A-4 B) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{2 d}+\frac{a A \cos (c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{2 d}+\int \cos (c+d x) \sqrt{a+a \sec (c+d x)} \left (\frac{1}{4} a^2 (9 A-4 B)+\frac{1}{4} a^2 (5 A+12 B) \sec (c+d x)\right ) \, dx\\ &=\frac{a^3 (9 A-4 B) \sin (c+d x)}{4 d \sqrt{a+a \sec (c+d x)}}-\frac{a^2 (A-4 B) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{2 d}+\frac{a A \cos (c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{2 d}+\frac{1}{8} \left (a^2 (19 A+20 B)\right ) \int \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{a^3 (9 A-4 B) \sin (c+d x)}{4 d \sqrt{a+a \sec (c+d x)}}-\frac{a^2 (A-4 B) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{2 d}+\frac{a A \cos (c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{2 d}-\frac{\left (a^3 (19 A+20 B)\right ) \operatorname{Subst}\left (\int \frac{1}{a+x^2} \, dx,x,-\frac{a \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{4 d}\\ &=\frac{a^{5/2} (19 A+20 B) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{4 d}+\frac{a^3 (9 A-4 B) \sin (c+d x)}{4 d \sqrt{a+a \sec (c+d x)}}-\frac{a^2 (A-4 B) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{2 d}+\frac{a A \cos (c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.80738, size = 116, normalized size = 0.75 \[ \frac{a^2 \sec \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\sec (c+d x)+1)} \left (\sqrt{2} (19 A+20 B) \sin ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right ) \sqrt{\cos (c+d x)}+2 \sin \left (\frac{1}{2} (c+d x)\right ) ((11 A+4 B) \cos (c+d x)+A \cos (2 (c+d x))+A+8 B)\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x]),x]

[Out]

(a^2*Sec[(c + d*x)/2]*Sqrt[a*(1 + Sec[c + d*x])]*(Sqrt[2]*(19*A + 20*B)*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]]*Sqrt[
Cos[c + d*x]] + 2*(A + 8*B + (11*A + 4*B)*Cos[c + d*x] + A*Cos[2*(c + d*x)])*Sin[(c + d*x)/2]))/(8*d)

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Maple [B]  time = 0.325, size = 410, normalized size = 2.7 \begin{align*}{\frac{{a}^{2}}{16\,d\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}} \left ( 19\,A\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{3/2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}} \right ) \sqrt{2}+20\,B\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{3/2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}} \right ) \sqrt{2}+19\,A \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{3/2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}} \right ) \sqrt{2}\sin \left ( dx+c \right ) +20\,B \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{3/2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}} \right ) \sqrt{2}\sin \left ( dx+c \right ) -8\,A \left ( \cos \left ( dx+c \right ) \right ) ^{4}-36\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}-16\,B \left ( \cos \left ( dx+c \right ) \right ) ^{3}+44\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}-16\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}+32\,B\cos \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x)

[Out]

1/16/d*a^2*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(19*A*sin(d*x+c)*cos(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2
)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*2^(1/2)+20*B*sin(d*x+c)*cos(
d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c
)/cos(d*x+c))*2^(1/2)+19*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)
+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*2^(1/2)*sin(d*x+c)+20*B*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*arctanh(1/2*2^(
1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*2^(1/2)*sin(d*x+c)-8*A*cos(d*x+c)^4-36*A*cos(
d*x+c)^3-16*B*cos(d*x+c)^3+44*A*cos(d*x+c)^2-16*B*cos(d*x+c)^2+32*B*cos(d*x+c))/cos(d*x+c)/sin(d*x+c)

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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(cos(d*x+c)^2*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.631955, size = 890, normalized size = 5.78 \begin{align*} \left [\frac{{\left ({\left (19 \, A + 20 \, B\right )} a^{2} \cos \left (d x + c\right ) +{\left (19 \, A + 20 \, B\right )} a^{2}\right )} \sqrt{-a} \log \left (\frac{2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt{-a} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \,{\left (2 \, A a^{2} \cos \left (d x + c\right )^{2} +{\left (11 \, A + 4 \, B\right )} a^{2} \cos \left (d x + c\right ) + 8 \, B a^{2}\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{8 \,{\left (d \cos \left (d x + c\right ) + d\right )}}, -\frac{{\left ({\left (19 \, A + 20 \, B\right )} a^{2} \cos \left (d x + c\right ) +{\left (19 \, A + 20 \, B\right )} a^{2}\right )} \sqrt{a} \arctan \left (\frac{\sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt{a} \sin \left (d x + c\right )}\right ) -{\left (2 \, A a^{2} \cos \left (d x + c\right )^{2} +{\left (11 \, A + 4 \, B\right )} a^{2} \cos \left (d x + c\right ) + 8 \, B a^{2}\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{4 \,{\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

[1/8*(((19*A + 20*B)*a^2*cos(d*x + c) + (19*A + 20*B)*a^2)*sqrt(-a)*log((2*a*cos(d*x + c)^2 - 2*sqrt(-a)*sqrt(
(a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) + 2*(2*
A*a^2*cos(d*x + c)^2 + (11*A + 4*B)*a^2*cos(d*x + c) + 8*B*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*
x + c))/(d*cos(d*x + c) + d), -1/4*(((19*A + 20*B)*a^2*cos(d*x + c) + (19*A + 20*B)*a^2)*sqrt(a)*arctan(sqrt((
a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - (2*A*a^2*cos(d*x + c)^2 + (11*A + 4*B
)*a^2*cos(d*x + c) + 8*B*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c) + d)]

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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(cos(d*x+c)**2*(a+a*sec(d*x+c))**(5/2)*(A+B*sec(d*x+c)),x)

[Out]

Timed out

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Giac [B]  time = 7.37288, size = 957, normalized size = 6.21 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

-1/8*(16*sqrt(2)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*B*a^3*sgn(cos(d*x + c))*tan(1/2*d*x + 1/2*c)/(a*tan(1/2*d
*x + 1/2*c)^2 - a) + (19*A*sqrt(-a)*a^2*sgn(cos(d*x + c)) + 20*B*sqrt(-a)*a^2*sgn(cos(d*x + c)))*log(abs((sqrt
(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - a*(2*sqrt(2) + 3))) - (19*A*sqrt(-a)*a^2*
sgn(cos(d*x + c)) + 20*B*sqrt(-a)*a^2*sgn(cos(d*x + c)))*log(abs((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(
1/2*d*x + 1/2*c)^2 + a))^2 + a*(2*sqrt(2) - 3))) + 4*sqrt(2)*(19*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(
1/2*d*x + 1/2*c)^2 + a))^6*A*sqrt(-a)*a^3*sgn(cos(d*x + c)) + 12*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(
1/2*d*x + 1/2*c)^2 + a))^6*B*sqrt(-a)*a^3*sgn(cos(d*x + c)) - 171*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan
(1/2*d*x + 1/2*c)^2 + a))^4*A*sqrt(-a)*a^4*sgn(cos(d*x + c)) - 76*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan
(1/2*d*x + 1/2*c)^2 + a))^4*B*sqrt(-a)*a^4*sgn(cos(d*x + c)) + 89*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan
(1/2*d*x + 1/2*c)^2 + a))^2*A*sqrt(-a)*a^5*sgn(cos(d*x + c)) + 36*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan
(1/2*d*x + 1/2*c)^2 + a))^2*B*sqrt(-a)*a^5*sgn(cos(d*x + c)) - 9*A*sqrt(-a)*a^6*sgn(cos(d*x + c)) - 4*B*sqrt(-
a)*a^6*sgn(cos(d*x + c)))/((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4 - 6*(sqrt(-
a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*a + a^2)^2)/d