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

Optimal. Leaf size=188 \[ \frac{a^3 (27 A-56 C) \sin (c+d x)}{12 d \sqrt{a \sec (c+d x)+a}}-\frac{a^2 (A-8 C) \sin (c+d x) \sqrt{a \sec (c+d x)+a}}{2 d}+\frac{a^{5/2} (19 A+8 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{4 d}-\frac{a (3 A-4 C) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{6 d}+\frac{A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{5/2}}{2 d} \]

[Out]

(a^(5/2)*(19*A + 8*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(4*d) + (a^3*(27*A - 56*C)*Sin[
c + d*x])/(12*d*Sqrt[a + a*Sec[c + d*x]]) - (a^2*(A - 8*C)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(2*d) - (a*(
3*A - 4*C)*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(6*d) + (A*Cos[c + d*x]*(a + a*Sec[c + d*x])^(5/2)*Sin[c +
d*x])/(2*d)

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Rubi [A]  time = 0.598209, antiderivative size = 188, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {4087, 4018, 4015, 3774, 203} \[ \frac{a^3 (27 A-56 C) \sin (c+d x)}{12 d \sqrt{a \sec (c+d x)+a}}-\frac{a^2 (A-8 C) \sin (c+d x) \sqrt{a \sec (c+d x)+a}}{2 d}+\frac{a^{5/2} (19 A+8 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{4 d}-\frac{a (3 A-4 C) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{6 d}+\frac{A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{5/2}}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^(5/2)*(19*A + 8*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(4*d) + (a^3*(27*A - 56*C)*Sin[
c + d*x])/(12*d*Sqrt[a + a*Sec[c + d*x]]) - (a^2*(A - 8*C)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(2*d) - (a*(
3*A - 4*C)*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(6*d) + (A*Cos[c + d*x]*(a + a*Sec[c + d*x])^(5/2)*Sin[c +
d*x])/(2*d)

Rule 4087

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b
_.) + (a_))^(m_), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(f*n), x] - Dis
t[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*(A*(m + n + 1) + C*n)*Csc[e +
f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -2
^(-1)] || EqQ[m + n + 1, 0])

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} \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{A \cos (c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{2 d}+\frac{\int \cos (c+d x) (a+a \sec (c+d x))^{5/2} \left (\frac{5 a A}{2}-\frac{1}{2} a (3 A-4 C) \sec (c+d x)\right ) \, dx}{2 a}\\ &=-\frac{a (3 A-4 C) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{6 d}+\frac{A \cos (c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{2 d}+\frac{\int \cos (c+d x) (a+a \sec (c+d x))^{3/2} \left (\frac{1}{4} a^2 (21 A-8 C)-\frac{3}{4} a^2 (A-8 C) \sec (c+d x)\right ) \, dx}{3 a}\\ &=-\frac{a^2 (A-8 C) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{2 d}-\frac{a (3 A-4 C) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{6 d}+\frac{A \cos (c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{2 d}+\frac{2 \int \cos (c+d x) \sqrt{a+a \sec (c+d x)} \left (\frac{1}{8} a^3 (27 A-56 C)+\frac{5}{8} a^3 (3 A+8 C) \sec (c+d x)\right ) \, dx}{3 a}\\ &=\frac{a^3 (27 A-56 C) \sin (c+d x)}{12 d \sqrt{a+a \sec (c+d x)}}-\frac{a^2 (A-8 C) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{2 d}-\frac{a (3 A-4 C) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{6 d}+\frac{A \cos (c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{2 d}+\frac{1}{8} \left (a^2 (19 A+8 C)\right ) \int \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{a^3 (27 A-56 C) \sin (c+d x)}{12 d \sqrt{a+a \sec (c+d x)}}-\frac{a^2 (A-8 C) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{2 d}-\frac{a (3 A-4 C) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{6 d}+\frac{A \cos (c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{2 d}-\frac{\left (a^3 (19 A+8 C)\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+8 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{4 d}+\frac{a^3 (27 A-56 C) \sin (c+d x)}{12 d \sqrt{a+a \sec (c+d x)}}-\frac{a^2 (A-8 C) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{2 d}-\frac{a (3 A-4 C) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{6 d}+\frac{A \cos (c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.885886, size = 137, normalized size = 0.73 \[ \frac{a^2 \sec \left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) \sqrt{a (\sec (c+d x)+1)} \left (6 \sqrt{2} (19 A+8 C) \sin ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right ) \cos ^{\frac{3}{2}}(c+d x)+2 \sin \left (\frac{1}{2} (c+d x)\right ) ((9 A+128 C) \cos (c+d x)+33 A \cos (2 (c+d x))+3 A \cos (3 (c+d x))+33 A+16 C)\right )}{48 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Sec[(c + d*x)/2]*Sec[c + d*x]*Sqrt[a*(1 + Sec[c + d*x])]*(6*Sqrt[2]*(19*A + 8*C)*ArcSin[Sqrt[2]*Sin[(c +
d*x)/2]]*Cos[c + d*x]^(3/2) + 2*(33*A + 16*C + (9*A + 128*C)*Cos[c + d*x] + 33*A*Cos[2*(c + d*x)] + 3*A*Cos[3*
(c + d*x)])*Sin[(c + d*x)/2]))/(48*d)

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Maple [B]  time = 0.361, size = 402, normalized size = 2.1 \begin{align*} -{\frac{{a}^{2}}{48\,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 ( -57\,A\sqrt{2}\sin \left ( dx+c \right ) \cos \left ( dx+c \right ){\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 ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{3/2}-24\,C\cos \left ( dx+c \right ) \sqrt{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 ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{3/2}\sin \left ( dx+c \right ) -57\,A\sqrt{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 ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{3/2}\sin \left ( dx+c \right ) -24\,C{\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} \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{3/2}\sin \left ( dx+c \right ) +24\,A \left ( \cos \left ( dx+c \right ) \right ) ^{4}+108\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}-132\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+256\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}-224\,C\cos \left ( dx+c \right ) -32\,C \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+C*sec(d*x+c)^2),x)

[Out]

-1/48/d*a^2*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(-57*A*2^(1/2)*sin(d*x+c)*cos(d*x+c)*arctanh(1/2*2^(1/2)*(-2*c
os(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)-24*C*cos(d*x+c)*2^
(1/2)*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*cos(d*x+c)/(cos(d*x+
c)+1))^(3/2)*sin(d*x+c)-57*A*2^(1/2)*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*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*sin(d*x+c)-24*C*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)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*sin(d*x+c)+24*A*cos(d*x+c)^4+108*A*
cos(d*x+c)^3-132*A*cos(d*x+c)^2+256*C*cos(d*x+c)^2-224*C*cos(d*x+c)-32*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+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.671674, size = 1022, normalized size = 5.44 \begin{align*} \left [\frac{3 \,{\left ({\left (19 \, A + 8 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} +{\left (19 \, A + 8 \, C\right )} a^{2} \cos \left (d x + c\right )\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 (6 \, A a^{2} \cos \left (d x + c\right )^{3} + 33 \, A a^{2} \cos \left (d x + c\right )^{2} + 64 \, C a^{2} \cos \left (d x + c\right ) + 8 \, C 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 )}{24 \,{\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}, -\frac{3 \,{\left ({\left (19 \, A + 8 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} +{\left (19 \, A + 8 \, C\right )} a^{2} \cos \left (d x + c\right )\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 (6 \, A a^{2} \cos \left (d x + c\right )^{3} + 33 \, A a^{2} \cos \left (d x + c\right )^{2} + 64 \, C a^{2} \cos \left (d x + c\right ) + 8 \, C 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 )}{12 \,{\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\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+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

[1/24*(3*((19*A + 8*C)*a^2*cos(d*x + c)^2 + (19*A + 8*C)*a^2*cos(d*x + c))*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*(6*A*a^2*cos(d*x + c)^3 + 33*A*a^2*cos(d*x + c)^2 + 64*C*a^2*cos(d*x + c) + 8*C*a^2)*sqrt((a*cos(
d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2 + d*cos(d*x + c)), -1/12*(3*((19*A + 8*C)*a^2*cos(
d*x + c)^2 + (19*A + 8*C)*a^2*cos(d*x + c))*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c
)/(sqrt(a)*sin(d*x + c))) - (6*A*a^2*cos(d*x + c)^3 + 33*A*a^2*cos(d*x + c)^2 + 64*C*a^2*cos(d*x + c) + 8*C*a^
2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2 + d*cos(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(cos(d*x+c)**2*(a+a*sec(d*x+c))**(5/2)*(A+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [B]  time = 7.03085, size = 748, normalized size = 3.98 \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+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

-1/24*(3*(19*A*sqrt(-a)*a^2*sgn(cos(d*x + c)) + 8*C*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))) - 3*(19*A*sqrt(-a)*a^2*sgn(cos(d*x
+ c)) + 8*C*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))) - 16*(7*sqrt(2)*C*a^4*sgn(cos(d*x + c))*tan(1/2*d*x + 1/2*c)^2 - 9*sqrt(2
)*C*a^4*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2
+ a)) + 12*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)) - 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)) + 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)) - 9*A*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