3.4.97 \(\int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx\) [397]

Optimal. Leaf size=87 \[ \frac {(B+3 C) \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(B-C) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}} \]

[Out]

1/4*(B+3*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 = 0.07, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {4137, 12, 3880, 209} \begin {gather*} \frac {(B+3 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 {(B-C) \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((B + 3*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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 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 4137

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

Rubi steps

\begin {align*} \int \frac {B \sec (c+d x)+C \sec ^2(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 {a (B+3 C) \sec (c+d x)}{2 \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+3 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 {(B+3 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 {(B+3 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 = 0.53, size = 127, normalized size = 1.46 \begin {gather*} \frac {2 (B-C) \sqrt {1-\sec (c+d x)} \sin (c+d x)+2 \sqrt {2} (B+3 C) \tanh ^{-1}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right ) \tan (c+d x)}{4 a d (1+\cos (c+d x)) \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(B - C)*Sqrt[1 - Sec[c + d*x]]*Sin[c + d*x] + 2*Sqrt[2]*(B + 3*C)*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]]*C
os[(c + d*x)/2]^2*Tan[c + d*x])/(4*a*d*(1 + Cos[c + d*x])*Sqrt[1 - Sec[c + d*x]]*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. \(401\) vs. \(2(72)=144\).
time = 13.57, size = 402, normalized size = 4.62

method	result	size
default	\(\frac {\sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (B \ln \left (-\frac {-\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+\cos \left (d x +c \right )-1}{\sin \left (d x +c \right )}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \cos \left (d x +c \right )+3 C \ln \left (-\frac {-\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+\cos \left (d x +c \right )-1}{\sin \left (d x +c \right )}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \cos \left (d x +c \right )+B \ln \left (-\frac {-\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+\cos \left (d x +c \right )-1}{\sin \left (d x +c \right )}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+3 C \ln \left (-\frac {-\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+\cos \left (d x +c \right )-1}{\sin \left (d x +c \right )}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-2 B \left (\cos ^{2}\left (d x +c \right )\right )+2 C \left (\cos ^{2}\left (d x +c \right )\right )+2 B \cos \left (d x +c \right )-2 C \cos \left (d x +c \right )\right )}{4 d \left (1+\cos \left (d x +c \right )\right ) \sin \left (d x +c \right ) a^{2}}\)	\(402\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((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)*(B*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)))^(1/2)*sin(d*x+c)*cos(d*x+c)+3*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)))^(1/2)*sin(d*x+c)*cos(d*x+c)+B*l
n(-(-(-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)))^
(1/2)*sin(d*x+c)+3*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)))^(1/2)*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((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))/(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. 159 vs. \(2 (72) = 144\).
time = 2.23, size = 367, normalized size = 4.22 \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 (B + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (B + 3 \, C\right )} \cos \left (d x + c\right ) + B + 3 \, 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 (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 (B + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (B + 3 \, C\right )} \cos \left (d x + c\right ) + B + 3 \, 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 )}{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((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)*((B + 3*C)*cos(d*x
+ c)^2 + 2*(B + 3*C)*cos(d*x + c) + B + 3*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)))/(a^2*d*cos(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)*((B + 3*C)*cos(d*x + c)^2 + 2*(B + 3*C)*cos(d*x + c) + B + 3*C)*
sqrt(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))))/(a^2*d*co
s(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 ) \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((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))*sec(c + d*x)/(a*(sec(c + d*x) + 1))**(3/2), x)

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Giac [A]
time = 1.53, size = 133, normalized size = 1.53 \begin {gather*} -\frac {\frac {{\left (\sqrt {2} B + 3 \, \sqrt {2} 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 |}\right )}{\sqrt {-a} a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {{\left (\sqrt {2} B a \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - \sqrt {2} C a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{3}}}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*((sqrt(2)*B + 3*sqrt(2)*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)))
/(sqrt(-a)*a*sgn(cos(d*x + c))) - (sqrt(2)*B*a*sgn(cos(d*x + c)) - sqrt(2)*C*a*sgn(cos(d*x + c)))*sqrt(-a*tan(
1/2*d*x + 1/2*c)^2 + a)*tan(1/2*d*x + 1/2*c)/a^3)/d

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

[Out]

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

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