3.405 \(\int \frac{B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=126 \[ \frac{(3 B+5 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{(3 B+5 C) \tan (c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}}+\frac{(B-C) \tan (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}} \]

[Out]

((3*B + 5*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) + ((B -
C)*Tan[c + d*x])/(4*d*(a + a*Sec[c + d*x])^(5/2)) + ((3*B + 5*C)*Tan[c + d*x])/(16*a*d*(a + a*Sec[c + d*x])^(
3/2))

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Rubi [A]  time = 0.151674, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.147, Rules used = {4052, 12, 3796, 3795, 203} \[ \frac{(3 B+5 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{(3 B+5 C) \tan (c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}}+\frac{(B-C) \tan (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*B + 5*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) + ((B -
C)*Tan[c + d*x])/(4*d*(a + a*Sec[c + d*x])^(5/2)) + ((3*B + 5*C)*Tan[c + d*x])/(16*a*d*(a + a*Sec[c + d*x])^(
3/2))

Rule 4052

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)]

Rule 12

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

Rule 3796

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

Rule 3795

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 203

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

Rubi steps

\begin{align*} \int \frac{B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx &=\frac{(B-C) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{\int -\frac{a (3 B+5 C) \sec (c+d x)}{2 (a+a \sec (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=\frac{(B-C) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac{(3 B+5 C) \int \frac{\sec (c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx}{8 a}\\ &=\frac{(B-C) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac{(3 B+5 C) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac{(3 B+5 C) \int \frac{\sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx}{32 a^2}\\ &=\frac{(B-C) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac{(3 B+5 C) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac{(3 B+5 C) \operatorname{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 )}{16 a^2 d}\\ &=\frac{(3 B+5 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a+a \sec (c+d x)}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{(B-C) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac{(3 B+5 C) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}\\ \end{align*}

Mathematica [C]  time = 1.48807, size = 206, normalized size = 1.63 \[ \frac{64 B \sin \left (\frac{1}{2} (c+d x)\right ) \cos ^5\left (\frac{1}{2} (c+d x)\right ) \sqrt{1-\sec (c+d x)} \sec (c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},3,\frac{3}{2},\frac{1}{2} (1-\sec (c+d x))\right )+C (10 \sin (c+d x)+\sin (2 (c+d x))) \sqrt{1-\sec (c+d x)}+40 \sqrt{2} C \sin \left (\frac{1}{2} (c+d x)\right ) \cos ^5\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) \tanh ^{-1}\left (\frac{\sqrt{1-\sec (c+d x)}}{\sqrt{2}}\right )}{32 a^2 d (\cos (c+d x)+1)^2 \sqrt{1-\sec (c+d x)} \sqrt{a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(40*Sqrt[2]*C*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]]*Cos[(c + d*x)/2]^5*Sec[c + d*x]*Sin[(c + d*x)/2] + 64*B*
Cos[(c + d*x)/2]^5*Hypergeometric2F1[1/2, 3, 3/2, (1 - Sec[c + d*x])/2]*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]*Si
n[(c + d*x)/2] + C*Sqrt[1 - Sec[c + d*x]]*(10*Sin[c + d*x] + Sin[2*(c + d*x)]))/(32*a^2*d*(1 + Cos[c + d*x])^2
*Sqrt[1 - Sec[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])

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Maple [B]  time = 0.207, size = 594, normalized size = 4.7 \begin{align*} \text{result too large to display} \end{align*}

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))^(5/2),x)

[Out]

1/32/d/a^3*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(3*B*cos(d*x+c)^2*sin(d*x+c)*ln(((-2*cos(d*x+c)/(cos(d*x+c)+1))
^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+5*C*cos(d*x+c)^2*sin(d*x+c)*l
n(((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1
/2)+6*B*sin(d*x+c)*cos(d*x+c)*ln(((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-
2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+10*C*sin(d*x+c)*cos(d*x+c)*ln(((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x
+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+3*B*ln(((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2
)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-14*B*cos(d*x+c)^3+5*C*l
n(((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1
/2)*sin(d*x+c)-2*C*cos(d*x+c)^3+8*B*cos(d*x+c)^2-8*C*cos(d*x+c)^2+6*B*cos(d*x+c)+10*C*cos(d*x+c))/(cos(d*x+c)+
1)^2/sin(d*x+c)

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

[Out]

Timed out

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Fricas [A]  time = 0.616432, size = 1219, normalized size = 9.67 \begin{align*} \left [-\frac{\sqrt{2}{\left ({\left (3 \, B + 5 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (3 \, B + 5 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (3 \, B + 5 \, C\right )} \cos \left (d x + c\right ) + 3 \, B + 5 \, 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 ) - 4 \,{\left ({\left (7 \, B + C\right )} \cos \left (d x + c\right )^{2} +{\left (3 \, B + 5 \, 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 )}{64 \,{\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}, -\frac{\sqrt{2}{\left ({\left (3 \, B + 5 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (3 \, B + 5 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (3 \, B + 5 \, C\right )} \cos \left (d x + c\right ) + 3 \, B + 5 \, 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 ) - 2 \,{\left ({\left (7 \, B + C\right )} \cos \left (d x + c\right )^{2} +{\left (3 \, B + 5 \, 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 )}{32 \,{\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}\right ] \end{align*}

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))^(5/2),x, algorithm="fricas")

[Out]

[-1/64*(sqrt(2)*((3*B + 5*C)*cos(d*x + c)^3 + 3*(3*B + 5*C)*cos(d*x + c)^2 + 3*(3*B + 5*C)*cos(d*x + c) + 3*B
+ 5*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)) - 4*((7*B + C)*cos(d*x + c)^2
+ (3*B + 5*C)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^
3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d), -1/32*(sqrt(2)*((3*B + 5*C)*cos(d*x + c)^3 + 3*(3*B + 5*C)
*cos(d*x + c)^2 + 3*(3*B + 5*C)*cos(d*x + c) + 3*B + 5*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))) - 2*((7*B + C)*cos(d*x + c)^2 + (3*B + 5*C)*cos(d*x + c))*sqrt
((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos
(d*x + c) + a^3*d)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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{5}{2}}}\, dx \end{align*}

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))**(5/2),x)

[Out]

Integral((B + C*sec(c + d*x))*sec(c + d*x)/(a*(sec(c + d*x) + 1))**(5/2), x)

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Giac [A]  time = 8.97639, size = 258, normalized size = 2.05 \begin{align*} \frac{\sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}{\left (\frac{2 \, \sqrt{2}{\left (B a^{5} - C a^{5}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}{a^{8} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} - \frac{\sqrt{2}{\left (5 \, B a^{5} + 3 \, C a^{5}\right )}}{a^{8} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \frac{\sqrt{2}{\left (3 \, B + 5 \, 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^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}}{32 \, d} \end{align*}

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))^(5/2),x, algorithm="giac")

[Out]

1/32*(sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*(2*sqrt(2)*(B*a^5 - C*a^5)*tan(1/2*d*x + 1/2*c)^2/(a^8*sgn(tan(1/2*d
*x + 1/2*c)^2 - 1)) - sqrt(2)*(5*B*a^5 + 3*C*a^5)/(a^8*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))*tan(1/2*d*x + 1/2*c)
+ sqrt(2)*(3*B + 5*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^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))/d