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

Optimal. Leaf size=200 \[ \frac{a^3 (4 A-9 B) \sin (c+d x)}{4 d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}+\frac{a^2 (4 A+7 B) \sin (c+d x) \sqrt{a \sec (c+d x)+a}}{4 d \sqrt{\cos (c+d x)}}+\frac{a^{5/2} (20 A+19 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{4 d}+\frac{a B \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{2 d \sqrt{\cos (c+d x)}} \]

[Out]

(a^(5/2)*(20*A + 19*B)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c
+ d*x]])/(4*d) + (a^3*(4*A - 9*B)*Sin[c + d*x])/(4*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (a^2*(4*A
+ 7*B)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(4*d*Sqrt[Cos[c + d*x]]) + (a*B*(a + a*Sec[c + d*x])^(3/2)*Sin[c
+ d*x])/(2*d*Sqrt[Cos[c + d*x]])

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Rubi [A]  time = 0.627095, antiderivative size = 200, 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 = {2955, 4018, 4015, 3801, 215} \[ \frac{a^3 (4 A-9 B) \sin (c+d x)}{4 d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}+\frac{a^2 (4 A+7 B) \sin (c+d x) \sqrt{a \sec (c+d x)+a}}{4 d \sqrt{\cos (c+d x)}}+\frac{a^{5/2} (20 A+19 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{4 d}+\frac{a B \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{2 d \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^(5/2)*(20*A + 19*B)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c
+ d*x]])/(4*d) + (a^3*(4*A - 9*B)*Sin[c + d*x])/(4*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (a^2*(4*A
+ 7*B)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(4*d*Sqrt[Cos[c + d*x]]) + (a*B*(a + a*Sec[c + d*x])^(3/2)*Sin[c
+ d*x])/(2*d*Sqrt[Cos[c + d*x]])

Rule 2955

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.
) + (f_.)*(x_)])^(p_.), x_Symbol] :> Dist[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p, Int[((a + b*Csc[e + f*x])^m*(
c + d*Csc[e + f*x])^n)/(g*Csc[e + f*x])^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[b*c - a*d
, 0] &&  !IntegerQ[p] &&  !(IntegerQ[m] && IntegerQ[n])

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 3801

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(-2*a*Sq
rt[(a*d)/b])/(b*f), Subst[Int[1/Sqrt[1 + x^2/a], x], x, (b*Cot[e + f*x])/Sqrt[a + b*Csc[e + f*x]]], x] /; Free
Q[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[(a*d)/b, 0]

Rule 215

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

Rubi steps

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

Mathematica [A]  time = 0.922839, size = 173, normalized size = 0.86 \[ \frac{a^3 \sin (c+d x) \sqrt{\cos (c+d x)} (A+B \sec (c+d x)) \left (\sqrt{1-\sec (c+d x)} \left ((4 A+11 B) \sec (c+d x)+8 A+2 B \sec ^2(c+d x)\right )+20 A \sqrt{\sec (c+d x)} \sin ^{-1}\left (\sqrt{1-\sec (c+d x)}\right )-19 B \sqrt{\sec (c+d x)} \sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right )}{4 d \sqrt{1-\sec (c+d x)} \sqrt{a (\sec (c+d x)+1)} (A \cos (c+d x)+B)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*Sqrt[Cos[c + d*x]]*(A + B*Sec[c + d*x])*(20*A*ArcSin[Sqrt[1 - Sec[c + d*x]]]*Sqrt[Sec[c + d*x]] - 19*B*Ar
cSin[Sqrt[Sec[c + d*x]]]*Sqrt[Sec[c + d*x]] + Sqrt[1 - Sec[c + d*x]]*(8*A + (4*A + 11*B)*Sec[c + d*x] + 2*B*Se
c[c + d*x]^2))*Sin[c + d*x])/(4*d*(B + A*Cos[c + d*x])*Sqrt[1 - Sec[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])

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Maple [B]  time = 0.338, size = 376, normalized size = 1.9 \begin{align*} -{\frac{{a}^{2} \left ( -1+\cos \left ( dx+c \right ) \right ) }{8\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}} \left ( 16\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}-20\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sqrt{2}\arctan \left ( 1/4\,\sqrt{2}\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \left ( \cos \left ( dx+c \right ) +1-\sin \left ( dx+c \right ) \right ) \right ) +20\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sqrt{2}\arctan \left ( 1/4\,\sqrt{2}\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \left ( \cos \left ( dx+c \right ) +1+\sin \left ( dx+c \right ) \right ) \right ) -19\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sqrt{2}\arctan \left ( 1/4\,\sqrt{2}\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \left ( \cos \left ( dx+c \right ) +1-\sin \left ( dx+c \right ) \right ) \right ) +19\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sqrt{2}\arctan \left ( 1/4\,\sqrt{2}\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \left ( \cos \left ( dx+c \right ) +1+\sin \left ( dx+c \right ) \right ) \right ) +8\,A\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}+22\,B\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}+4\,B\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sin \left ( dx+c \right ) \right ) \sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}} \left ( \cos \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}{\frac{1}{\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c))*cos(d*x+c)^(1/2),x)

[Out]

-1/8/d*a^2*(-1+cos(d*x+c))*(16*A*cos(d*x+c)^2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2)-20*A*cos(d*x+c)^2*2^(1/2)*a
rctan(1/4*2^(1/2)*(-2/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1-sin(d*x+c)))+20*A*cos(d*x+c)^2*2^(1/2)*arctan(1/4*2^
(1/2)*(-2/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))-19*B*cos(d*x+c)^2*2^(1/2)*arctan(1/4*2^(1/2)*(-2/(c
os(d*x+c)+1))^(1/2)*(cos(d*x+c)+1-sin(d*x+c)))+19*B*cos(d*x+c)^2*2^(1/2)*arctan(1/4*2^(1/2)*(-2/(cos(d*x+c)+1)
)^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))+8*A*cos(d*x+c)*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2)+22*B*cos(d*x+c)*sin(d*x
+c)*(-2/(cos(d*x+c)+1))^(1/2)+4*B*(-2/(cos(d*x+c)+1))^(1/2)*sin(d*x+c))*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/co
s(d*x+c)^(3/2)/sin(d*x+c)^2/(-2/(cos(d*x+c)+1))^(1/2)

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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((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c))*cos(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.692874, size = 1160, normalized size = 5.8 \begin{align*} \left [\frac{4 \,{\left (8 \, A a^{2} \cos \left (d x + c\right )^{2} +{\left (4 \, A + 11 \, B\right )} a^{2} \cos \left (d x + c\right ) + 2 \, B a^{2}\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right ) +{\left ({\left (20 \, A + 19 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} +{\left (20 \, A + 19 \, B\right )} a^{2} \cos \left (d x + c\right )^{2}\right )} \sqrt{a} \log \left (\frac{a \cos \left (d x + c\right )^{3} - 4 \, \sqrt{a} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}{\left (\cos \left (d x + c\right ) - 2\right )} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 7 \, a \cos \left (d x + c\right )^{2} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right )}{16 \,{\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}, \frac{2 \,{\left (8 \, A a^{2} \cos \left (d x + c\right )^{2} +{\left (4 \, A + 11 \, B\right )} a^{2} \cos \left (d x + c\right ) + 2 \, B a^{2}\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right ) +{\left ({\left (20 \, A + 19 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} +{\left (20 \, A + 19 \, B\right )} a^{2} \cos \left (d x + c\right )^{2}\right )} \sqrt{-a} \arctan \left (\frac{2 \, \sqrt{-a} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right )}{8 \,{\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c))*cos(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

[1/16*(4*(8*A*a^2*cos(d*x + c)^2 + (4*A + 11*B)*a^2*cos(d*x + c) + 2*B*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x
+ c))*sqrt(cos(d*x + c))*sin(d*x + c) + ((20*A + 19*B)*a^2*cos(d*x + c)^3 + (20*A + 19*B)*a^2*cos(d*x + c)^2)*
sqrt(a)*log((a*cos(d*x + c)^3 - 4*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*(cos(d*x + c) - 2)*sqrt(cos(
d*x + c))*sin(d*x + c) - 7*a*cos(d*x + c)^2 + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)))/(d*cos(d*x + c)^3 + d*c
os(d*x + c)^2), 1/8*(2*(8*A*a^2*cos(d*x + c)^2 + (4*A + 11*B)*a^2*cos(d*x + c) + 2*B*a^2)*sqrt((a*cos(d*x + c)
+ a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + ((20*A + 19*B)*a^2*cos(d*x + c)^3 + (20*A + 19*B)*a^2*co
s(d*x + c)^2)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x +
c)/(a*cos(d*x + c)^2 - a*cos(d*x + c) - 2*a)))/(d*cos(d*x + c)^3 + 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((a+a*sec(d*x+c))**(5/2)*(A+B*sec(d*x+c))*cos(d*x+c)**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \sec \left (d x + c\right ) + A\right )}{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \sqrt{\cos \left (d x + c\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c))*cos(d*x+c)^(1/2),x, algorithm="giac")

[Out]

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