3.4.91 \(\int \frac {\cos ^3(c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx\) [391]

Optimal. Leaf size=165 \[ \frac {(7 B-4 C) \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 \sqrt {a} d}-\frac {\sqrt {2} (B-C) \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {a} d}-\frac {(B-4 C) \sin (c+d x)}{4 d \sqrt {a+a \sec (c+d x)}}+\frac {B \cos (c+d x) \sin (c+d x)}{2 d \sqrt {a+a \sec (c+d x)}} \]

[Out]

1/4*(7*B-4*C)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d/a^(1/2)-(B-C)*arctan(1/2*a^(1/2)*tan(d*x+c)*
2^(1/2)/(a+a*sec(d*x+c))^(1/2))*2^(1/2)/d/a^(1/2)-1/4*(B-4*C)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/2*B*cos(d*
x+c)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)

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Rubi [A]
time = 0.33, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4157, 4107, 4005, 3859, 209, 3880} \begin {gather*} \frac {(7 B-4 C) \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{4 \sqrt {a} d}-\frac {\sqrt {2} (B-C) \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{\sqrt {a} d}-\frac {(B-4 C) \sin (c+d x)}{4 d \sqrt {a \sec (c+d x)+a}}+\frac {B \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((7*B - 4*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(4*Sqrt[a]*d) - (Sqrt[2]*(B - C)*ArcTan[
(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(Sqrt[a]*d) - ((B - 4*C)*Sin[c + d*x])/(4*d*Sqrt[a
+ a*Sec[c + d*x]]) + (B*Cos[c + d*x]*Sin[c + d*x])/(2*d*Sqrt[a + a*Sec[c + d*x]])

Rule 209

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

Rule 3859

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 3880

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

Rule 4005

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

Rule 4107

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

Rule 4157

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

Rubi steps

\begin {align*} \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx &=\int \frac {\cos ^2(c+d x) (B+C \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx\\ &=\frac {B \cos (c+d x) \sin (c+d x)}{2 d \sqrt {a+a \sec (c+d x)}}+\frac {\int \frac {\cos (c+d x) \left (-\frac {1}{2} a (B-4 C)+\frac {3}{2} a B \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{2 a}\\ &=-\frac {(B-4 C) \sin (c+d x)}{4 d \sqrt {a+a \sec (c+d x)}}+\frac {B \cos (c+d x) \sin (c+d x)}{2 d \sqrt {a+a \sec (c+d x)}}+\frac {\int \frac {\frac {1}{4} a^2 (7 B-4 C)-\frac {1}{4} a^2 (B-4 C) \sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx}{2 a^2}\\ &=-\frac {(B-4 C) \sin (c+d x)}{4 d \sqrt {a+a \sec (c+d x)}}+\frac {B \cos (c+d x) \sin (c+d x)}{2 d \sqrt {a+a \sec (c+d x)}}+\frac {(7 B-4 C) \int \sqrt {a+a \sec (c+d x)} \, dx}{8 a}+(-B+C) \int \frac {\sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx\\ &=-\frac {(B-4 C) \sin (c+d x)}{4 d \sqrt {a+a \sec (c+d x)}}+\frac {B \cos (c+d x) \sin (c+d x)}{2 d \sqrt {a+a \sec (c+d x)}}-\frac {(7 B-4 C) \text {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 {(2 (B-C)) \text {Subst}\left (\int \frac {1}{2 a+x^2} \, dx,x,-\frac {a \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}\\ &=\frac {(7 B-4 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 \sqrt {a} d}-\frac {\sqrt {2} (B-C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {a} d}-\frac {(B-4 C) \sin (c+d x)}{4 d \sqrt {a+a \sec (c+d x)}}+\frac {B \cos (c+d x) \sin (c+d x)}{2 d \sqrt {a+a \sec (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.44, size = 135, normalized size = 0.82 \begin {gather*} \frac {\left ((7 B-4 C) \tanh ^{-1}\left (\sqrt {1-\sec (c+d x)}\right )-4 \sqrt {2} (B-C) \tanh ^{-1}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right )+\cos (c+d x) (-B+4 C+2 B \cos (c+d x)) \sqrt {1-\sec (c+d x)}\right ) \tan (c+d x)}{4 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(((7*B - 4*C)*ArcTanh[Sqrt[1 - Sec[c + d*x]]] - 4*Sqrt[2]*(B - C)*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]] + Co
s[c + d*x]*(-B + 4*C + 2*B*Cos[c + d*x])*Sqrt[1 - Sec[c + d*x]])*Tan[c + d*x])/(4*d*Sqrt[1 - Sec[c + d*x]]*Sqr
t[a*(1 + Sec[c + d*x])])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(716\) vs. \(2(140)=280\).
time = 27.28, size = 717, normalized size = 4.35

method	result	size
default	\(\frac {\left (7 B \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {2}-4 C \sqrt {2}\, \cos \left (d x +c \right ) \sin \left (d x +c \right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right )+8 B \sin \left (d x +c \right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \cos \left (d x +c \right ) \ln \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-\cos \left (d x +c \right )+1}{\sin \left (d x +c \right )}\right )+7 B \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \sqrt {2}\, \sin \left (d x +c \right )-8 C \sin \left (d x +c \right ) \cos \left (d x +c \right ) \ln \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-\cos \left (d x +c \right )+1}{\sin \left (d x +c \right )}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}}-4 C \sqrt {2}\, \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \sin \left (d x +c \right )+8 B \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \ln \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-\cos \left (d x +c \right )+1}{\sin \left (d x +c \right )}\right ) \sin \left (d x +c \right )-8 C \ln \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-\cos \left (d x +c \right )+1}{\sin \left (d x +c \right )}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \sin \left (d x +c \right )-8 B \left (\cos ^{4}\left (d x +c \right )\right )+12 B \left (\cos ^{3}\left (d x +c \right )\right )-16 C \left (\cos ^{3}\left (d x +c \right )\right )-4 B \left (\cos ^{2}\left (d x +c \right )\right )+16 C \left (\cos ^{2}\left (d x +c \right )\right )\right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{16 d \cos \left (d x +c \right ) \sin \left (d x +c \right ) a}\)	\(717\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/16/d*(7*B*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/c
os(d*x+c)*2^(1/2))*sin(d*x+c)*cos(d*x+c)*2^(1/2)-4*C*2^(1/2)*cos(d*x+c)*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c
)))^(3/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))+8*B*sin(d*x+c)*(-2*c
os(d*x+c)/(1+cos(d*x+c)))^(3/2)*cos(d*x+c)*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/s
in(d*x+c))+7*B*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c
)/cos(d*x+c)*2^(1/2))*2^(1/2)*sin(d*x+c)-8*C*sin(d*x+c)*cos(d*x+c)*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*si
n(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)-4*C*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/
2)*2^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*sin(d*x+c)+8*B*(-2*
cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))
*sin(d*x+c)-8*C*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(
1+cos(d*x+c)))^(3/2)*sin(d*x+c)-8*B*cos(d*x+c)^4+12*B*cos(d*x+c)^3-16*C*cos(d*x+c)^3-4*B*cos(d*x+c)^2+16*C*cos
(d*x+c)^2)*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d*x+c)/sin(d*x+c)/a

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 4.43, size = 506, normalized size = 3.07 \begin {gather*} \left [-\frac {4 \, \sqrt {2} {\left ({\left (B - C\right )} a \cos \left (d x + c\right ) + {\left (B - C\right )} a\right )} \sqrt {-\frac {1}{a}} \log \left (-\frac {2 \, \sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {-\frac {1}{a}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 3 \, \cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - {\left ({\left (7 \, B - 4 \, C\right )} \cos \left (d x + c\right ) + 7 \, B - 4 \, C\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 \, B \cos \left (d x + c\right )^{2} - {\left (B - 4 \, C\right )} \cos \left (d x + c\right )\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 (a d \cos \left (d x + c\right ) + a d\right )}}, -\frac {{\left ({\left (7 \, B - 4 \, C\right )} \cos \left (d x + c\right ) + 7 \, B - 4 \, C\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 \, B \cos \left (d x + c\right )^{2} - {\left (B - 4 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right ) - \frac {4 \, \sqrt {2} {\left ({\left (B - C\right )} a \cos \left (d x + c\right ) + {\left (B - C\right )} a\right )} \arctan \left (\frac {\sqrt {2} \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 )}{\sqrt {a}}}{4 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*(4*sqrt(2)*((B - C)*a*cos(d*x + c) + (B - C)*a)*sqrt(-1/a)*log(-(2*sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos
(d*x + c))*sqrt(-1/a)*cos(d*x + c)*sin(d*x + c) - 3*cos(d*x + c)^2 - 2*cos(d*x + c) + 1)/(cos(d*x + c)^2 + 2*c
os(d*x + c) + 1)) - ((7*B - 4*C)*cos(d*x + c) + 7*B - 4*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*(2*
B*cos(d*x + c)^2 - (B - 4*C)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a*d*cos(d*x
+ c) + a*d), -1/4*(((7*B - 4*C)*cos(d*x + c) + 7*B - 4*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))) - (2*B*cos(d*x + c)^2 - (B - 4*C)*cos(d*x + c))*sqrt((a*cos(d*x + c) +
a)/cos(d*x + c))*sin(d*x + c) - 4*sqrt(2)*((B - C)*a*cos(d*x + c) + (B - C)*a)*arctan(sqrt(2)*sqrt((a*cos(d*x
+ c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c)))/sqrt(a))/(a*d*cos(d*x + c) + a*d)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (B + C \sec {\left (c + d x \right )}\right ) \cos ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{\sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 621 vs. \(2 (140) = 280\).
time = 2.02, size = 621, normalized size = 3.76 \begin {gather*} \frac {\frac {4 \, \sqrt {2} {\left (B - C\right )} \log \left ({\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2}\right )}{\sqrt {-a} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} + \frac {{\left (7 \, B - 4 \, C\right )} \log \left ({\left | {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - a {\left (2 \, \sqrt {2} + 3\right )} \right |}\right )}{\sqrt {-a} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {{\left (7 \, B - 4 \, C\right )} \log \left ({\left | {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} + a {\left (2 \, \sqrt {2} - 3\right )} \right |}\right )}{\sqrt {-a} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} + \frac {4 \, \sqrt {2} {\left (17 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{6} B \sqrt {-a} - 12 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{6} C \sqrt {-a} - 57 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{4} B \sqrt {-a} a + 76 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{4} C \sqrt {-a} a + 19 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} B \sqrt {-a} a^{2} - 36 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} C \sqrt {-a} a^{2} - 3 \, B \sqrt {-a} a^{3} + 4 \, C \sqrt {-a} a^{3}\right )}}{{\left ({\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{4} - 6 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} a + a^{2}\right )}^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}}{8 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(4*sqrt(2)*(B - C)*log((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2)/(sqrt(-a)*
sgn(cos(d*x + c))) + (7*B - 4*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)))/(sqrt(-a)*sgn(cos(d*x + c))) - (7*B - 4*C)*log(abs((sqrt(-a)*tan(1/2*d*x + 1/2*c) - s
qrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + a*(2*sqrt(2) - 3)))/(sqrt(-a)*sgn(cos(d*x + c))) + 4*sqrt(2)*(17*(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) - 12*(sqrt(-a)*tan(1/2*d*x + 1/2
*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*C*sqrt(-a) - 57*(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 + 76*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))
^4*C*sqrt(-a)*a + 19*(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^2 -
36*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*C*sqrt(-a)*a^2 - 3*B*sqrt(-a)*a^3 +
4*C*sqrt(-a)*a^3)/(((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*sgn(cos(d*x + c))))/d

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\cos \left (c+d\,x\right )}^3\,\left (\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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