3.4.98 \(\int \frac {\cos (c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx\) [398]

Optimal. Leaf size=127 \[ \frac {2 B \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^{3/2} d}-\frac {(5 B-C) \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(B-C) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}} \]

[Out]

2*B*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(3/2)/d-1/4*(5*B-C)*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1
/2)/(a+a*sec(d*x+c))^(1/2))/a^(3/2)/d*2^(1/2)-1/2*(B-C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(3/2)

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Rubi [A]
time = 0.20, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4157, 4007, 4005, 3859, 209, 3880} \begin {gather*} -\frac {(5 B-C) \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {2 B \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{a^{3/2} d}-\frac {(B-C) \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*B*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(a^(3/2)*d) - ((5*B - C)*ArcTan[(Sqrt[a]*Tan[c +
d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) - ((B - C)*Tan[c + d*x])/(2*d*(a + a*Sec[c +
d*x])^(3/2))

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 4007

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

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 (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx &=\int \frac {B+C \sec (c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx\\ &=-\frac {(B-C) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac {\int \frac {-2 a B+\frac {1}{2} a (B-C) \sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx}{2 a^2}\\ &=-\frac {(B-C) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {B \int \sqrt {a+a \sec (c+d x)} \, dx}{a^2}-\frac {(5 B-C) \int \frac {\sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx}{4 a}\\ &=-\frac {(B-C) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac {(2 B) \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 )}{a d}+\frac {(5 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 )}{2 a d}\\ &=\frac {2 B \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^{3/2} d}-\frac {(5 B-C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(B-C) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 1.08, size = 147, normalized size = 1.16 \begin {gather*} \frac {\csc (c+d x) \left (8 B \text {ArcTan}\left (\sqrt {-1+\sec (c+d x)}\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {-1+\sec (c+d x)}-\sqrt {2} (5 B-C) \text {ArcTan}\left (\frac {\sqrt {-1+\sec (c+d x)}}{\sqrt {2}}\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {-1+\sec (c+d x)}+2 (-B+C) \sin ^2\left (\frac {1}{2} (c+d x)\right )\right )}{2 a d \sqrt {a (1+\sec (c+d x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Csc[c + d*x]*(8*B*ArcTan[Sqrt[-1 + Sec[c + d*x]]]*Cos[(c + d*x)/2]^2*Sqrt[-1 + Sec[c + d*x]] - Sqrt[2]*(5*B -
C)*ArcTan[Sqrt[-1 + Sec[c + d*x]]/Sqrt[2]]*Cos[(c + d*x)/2]^2*Sqrt[-1 + Sec[c + d*x]] + 2*(-B + C)*Sin[(c + d
*x)/2]^2))/(2*a*d*Sqrt[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. \(553\) vs. \(2(106)=212\).
time = 12.89, size = 554, normalized size = 4.36

method	result	size
default	\(-\frac {\sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (4 B \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \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 )+4 B \sqrt {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 {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+5 B \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\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 )-C \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\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 ) \sin \left (d x +c \right ) \cos \left (d x +c \right )+5 B \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\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 ) \sin \left (d x +c \right )-C \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\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 ) \sin \left (d x +c \right )-2 B \left (\cos ^{2}\left (d x +c \right )\right )+2 C \left (\cos ^{2}\left (d x +c \right )\right )+2 B \cos \left (d x +c \right )-2 C \cos \left (d x +c \right )\right )}{4 d \left (1+\cos \left (d x +c \right )\right ) \sin \left (d x +c \right ) a^{2}}\)	\(554\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/d*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(4*B*cos(d*x+c)*sin(d*x+c)*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(
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))+4*B*(-2*cos(d*x+c)/(1+cos
(d*x+c)))^(1/2)*2^(1/2)*sin(d*x+c)*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))+5*B*sin(d*x+c)*cos(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*s
in(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))-C*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*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)*cos(d*x+c)+5*B*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*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)-C*(-2*cos(d*x+c)/(1+cos(d
*x+c)))^(1/2)*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)-2*B*cos
(d*x+c)^2+2*C*cos(d*x+c)^2+2*B*cos(d*x+c)-2*C*cos(d*x+c))/(1+cos(d*x+c))/sin(d*x+c)/a^2

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 231 vs. \(2 (106) = 212\).
time = 6.71, size = 548, normalized size = 4.31 \begin {gather*} \left [-\frac {4 \, {\left (B - C\right )} \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 ) - \sqrt {2} {\left ({\left (5 \, B - C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (5 \, B - C\right )} \cos \left (d x + c\right ) + 5 \, B - C\right )} \sqrt {-a} \log \left (\frac {2 \, \sqrt {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 ) + 3 \, a \cos \left (d x + c\right )^{2} + 2 \, a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 8 \, {\left (B \cos \left (d x + c\right )^{2} + 2 \, B \cos \left (d x + c\right ) + B\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 )}{8 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}, -\frac {2 \, {\left (B - C\right )} \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 ) - \sqrt {2} {\left ({\left (5 \, B - C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (5 \, B - C\right )} \cos \left (d x + c\right ) + 5 \, B - C\right )} \sqrt {a} \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 ) + 8 \, {\left (B \cos \left (d x + c\right )^{2} + 2 \, B \cos \left (d x + c\right ) + B\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 )}{4 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*(4*(B - C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) - sqrt(2)*((5*B - C)*cos(d*
x + c)^2 + 2*(5*B - C)*cos(d*x + c) + 5*B - C)*sqrt(-a)*log((2*sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(
d*x + c))*cos(d*x + c)*sin(d*x + c) + 3*a*cos(d*x + c)^2 + 2*a*cos(d*x + c) - a)/(cos(d*x + c)^2 + 2*cos(d*x +
c) + 1)) + 8*(B*cos(d*x + c)^2 + 2*B*cos(d*x + c) + B)*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)))/(a^2*d*co
s(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d), -1/4*(2*(B - C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x
+ c)*sin(d*x + c) - sqrt(2)*((5*B - C)*cos(d*x + c)^2 + 2*(5*B - C)*cos(d*x + c) + 5*B - C)*sqrt(a)*arctan(sqr
t(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) + 8*(B*cos(d*x + c)^2 + 2*B*
cos(d*x + c) + B)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))))
/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*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 {\left (c + d x \right )} \sec {\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(co

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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 )\,\left (\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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