3.528 \(\int \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=144 \[ \frac {a^{3/2} (2 A+3 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a^2 (2 A-B) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \sec (c+d x)+a}}+\frac {a B \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{d \sqrt {\cos (c+d x)}} \]

[Out]

a^(3/2)*(2*A+3*B)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d+a^2*(
2*A-B)*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)+a*B*sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d/cos(d*x+c)
^(1/2)

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Rubi [A]  time = 0.43, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 5, 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} \[ \frac {a^2 (2 A-B) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \sec (c+d x)+a}}+\frac {a^{3/2} (2 A+3 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a B \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{d \sqrt {\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Cos[c + d*x]]*(a + a*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x]),x]

[Out]

(a^(3/2)*(2*A + 3*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]])/d + (a^2*(2*A - B)*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (a*B*Sqrt[a + a*Sec[
c + d*x]]*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]])

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]

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 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 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 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]

Rubi steps

\begin {align*} \int \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{3/2} (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))^{3/2} (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx\\ &=\frac {a B \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \sec (c+d x)} \left (\frac {1}{2} a (2 A-B)+\frac {1}{2} a (2 A+3 B) \sec (c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx\\ &=\frac {a^2 (2 A-B) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {a B \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {1}{2} \left (a (2 A+3 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^2 (2 A-B) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {a B \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {\left (a (2 A+3 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 )}{d}\\ &=\frac {a^{3/2} (2 A+3 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)}}{d}+\frac {a^2 (2 A-B) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {a B \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{d \sqrt {\cos (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.86, size = 133, normalized size = 0.92 \[ \frac {a \sqrt {\cos (c+d x)} \tan \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\sec (c+d x)+1)} \left (\sqrt {1-\sec (c+d x)} (2 A+B \sec (c+d x))+2 A \sqrt {\sec (c+d x)} \sin ^{-1}\left (\sqrt {1-\sec (c+d x)}\right )-3 B \sqrt {\sec (c+d x)} \sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right )}{d \sqrt {1-\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[Cos[c + d*x]]*(a + a*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x]),x]

[Out]

(a*Sqrt[Cos[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])]*(2*A*ArcSin[Sqrt[1 - Sec[c + d*x]]]*Sqrt[Sec[c + d*x]] - 3*B*
ArcSin[Sqrt[Sec[c + d*x]]]*Sqrt[Sec[c + d*x]] + Sqrt[1 - Sec[c + d*x]]*(2*A + B*Sec[c + d*x]))*Tan[(c + d*x)/2
])/(d*Sqrt[1 - Sec[c + d*x]])

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fricas [A]  time = 0.51, size = 389, normalized size = 2.70 \[ \left [\frac {4 \, {\left (2 \, A a \cos \left (d x + c\right ) + B a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left ({\left (2 \, A + 3 \, B\right )} a \cos \left (d x + c\right )^{2} + {\left (2 \, A + 3 \, B\right )} a \cos \left (d x + c\right )\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 4 \, \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} {\left (\cos \left (d x + c\right ) - 2\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 7 \, a \cos \left (d x + c\right )^{2} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right )}{4 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}, \frac {2 \, {\left (2 \, A a \cos \left (d x + c\right ) + B a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left ({\left (2 \, A + 3 \, B\right )} a \cos \left (d x + c\right )^{2} + {\left (2 \, A + 3 \, B\right )} a \cos \left (d x + c\right )\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right )}{2 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(4*(2*A*a*cos(d*x + c) + B*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + (
(2*A + 3*B)*a*cos(d*x + c)^2 + (2*A + 3*B)*a*cos(d*x + c))*sqrt(a)*log((a*cos(d*x + c)^3 - 4*sqrt(a)*sqrt((a*c
os(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)^2 + d*cos(d*x + c)), 1/2*(2*(2*A*a*cos(d*x + c) + B*a)*sqr
t((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + ((2*A + 3*B)*a*cos(d*x + c)^2 + (2*A +
3*B)*a*cos(d*x + c))*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)^2 + d*cos(d*x + c))]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sqrt {\cos \left (d x + c\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [B]  time = 1.88, size = 306, normalized size = 2.12 \[ -\frac {a \left (-1+\cos \left (d x +c \right )\right ) \left (4 A \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}+2 A \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1+\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right ) \cos \left (d x +c \right ) \sqrt {2}-2 A \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1-\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right ) \cos \left (d x +c \right ) \sqrt {2}+3 B \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1+\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right ) \cos \left (d x +c \right ) \sqrt {2}-3 B \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1-\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right ) \cos \left (d x +c \right ) \sqrt {2}+2 B \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )\right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{2 d \sqrt {\cos \left (d x +c \right )}\, \sin \left (d x +c \right )^{2} \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/d*a*(-1+cos(d*x+c))*(4*A*sin(d*x+c)*cos(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2)+2*A*arctan(1/4*(-2/(1+cos(d*x+c)
))^(1/2)*(cos(d*x+c)+1+sin(d*x+c))*2^(1/2))*cos(d*x+c)*2^(1/2)-2*A*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(cos(d
*x+c)+1-sin(d*x+c))*2^(1/2))*cos(d*x+c)*2^(1/2)+3*B*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1+sin(d*x
+c))*2^(1/2))*cos(d*x+c)*2^(1/2)-3*B*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1-sin(d*x+c))*2^(1/2))*c
os(d*x+c)*2^(1/2)+2*B*(-2/(1+cos(d*x+c)))^(1/2)*sin(d*x+c))*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d*x+c)^(1/
2)/sin(d*x+c)^2/(-2/(1+cos(d*x+c)))^(1/2)

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maxima [B]  time = 0.80, size = 1417, normalized size = 9.84 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(sqrt(2)*(sqrt(2)*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*
c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - sqrt(2)*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 +
2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + sqrt(2)*a*log(2*cos(1/2*d*x + 1/2*c)^2
+ 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - sqrt(2)*a
*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*
d*x + 1/2*c) + 2) + 8*a*sin(1/2*d*x + 1/2*c))*A*sqrt(a) + (3*(a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x +
1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - a*log(2*cos(1/2*d*x + 1/2*c
)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + a*log(
2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x +
1/2*c) + 2) - a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*
sqrt(2)*sin(1/2*d*x + 1/2*c) + 2))*cos(2*d*x + 2*c)^2 + 3*(a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/
2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - a*log(2*cos(1/2*d*x + 1/2*c)^2
+ 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + a*log(2*c
os(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/
2*c) + 2) - a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqr
t(2)*sin(1/2*d*x + 1/2*c) + 2))*sin(2*d*x + 2*c)^2 + 4*sqrt(2)*a*sin(3/2*d*x + 3/2*c) - 4*sqrt(2)*a*sin(1/2*d*
x + 1/2*c) + 2*(2*sqrt(2)*a*sin(3/2*d*x + 3/2*c) - 2*sqrt(2)*a*sin(1/2*d*x + 1/2*c) + 3*a*log(2*cos(1/2*d*x +
1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - 3
*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/
2*d*x + 1/2*c) + 2) + 3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/
2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - 3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*s
qrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2))*cos(2*d*x + 2*c) + 3*a*log(2*cos(1/2*d*x +
1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - 3
*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/
2*d*x + 1/2*c) + 2) + 3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/
2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - 3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*s
qrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - 4*(sqrt(2)*a*cos(3/2*d*x + 3/2*c) - sqrt(2
)*a*cos(1/2*d*x + 1/2*c))*sin(2*d*x + 2*c))*B*sqrt(a)/(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x +
2*c) + 1))/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \sqrt {\cos \left (c+d\,x\right )}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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