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

Optimal. Leaf size=182 \[ \frac{2 a^3 (245 A+224 B+160 C) \tan (c+d x)}{105 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a^2 (35 A+56 B+40 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{105 d}+\frac{2 a^{5/2} A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{d}+\frac{2 a (7 B+5 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{35 d}+\frac{2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d} \]

[Out]

(2*a^(5/2)*A*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (2*a^3*(245*A + 224*B + 160*C)*Tan[c
+ d*x])/(105*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(35*A + 56*B + 40*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])
/(105*d) + (2*a*(7*B + 5*C)*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(35*d) + (2*C*(a + a*Sec[c + d*x])^(5/2)*
Tan[c + d*x])/(7*d)

________________________________________________________________________________________

Rubi [A]  time = 0.324399, antiderivative size = 182, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {4054, 3917, 3915, 3774, 203, 3792} \[ \frac{2 a^3 (245 A+224 B+160 C) \tan (c+d x)}{105 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a^2 (35 A+56 B+40 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{105 d}+\frac{2 a^{5/2} A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{d}+\frac{2 a (7 B+5 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{35 d}+\frac{2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^(5/2)*A*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (2*a^3*(245*A + 224*B + 160*C)*Tan[c
+ d*x])/(105*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(35*A + 56*B + 40*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])
/(105*d) + (2*a*(7*B + 5*C)*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(35*d) + (2*C*(a + a*Sec[c + d*x])^(5/2)*
Tan[c + d*x])/(7*d)

Rule 4054

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

Rule 3917

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

Rule 3915

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Dist[c, In
t[Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[d, Int[Sqrt[a + b*Csc[e + f*x]]*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 3774

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 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])

Rule 3792

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

Rubi steps

\begin{align*} \int (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac{2 C (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}+\frac{2 \int (a+a \sec (c+d x))^{5/2} \left (\frac{7 a A}{2}+\frac{1}{2} a (7 B+5 C) \sec (c+d x)\right ) \, dx}{7 a}\\ &=\frac{2 a (7 B+5 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac{2 C (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}+\frac{4 \int (a+a \sec (c+d x))^{3/2} \left (\frac{35 a^2 A}{4}+\frac{1}{4} a^2 (35 A+56 B+40 C) \sec (c+d x)\right ) \, dx}{35 a}\\ &=\frac{2 a^2 (35 A+56 B+40 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{105 d}+\frac{2 a (7 B+5 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac{2 C (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}+\frac{8 \int \sqrt{a+a \sec (c+d x)} \left (\frac{105 a^3 A}{8}+\frac{1}{8} a^3 (245 A+224 B+160 C) \sec (c+d x)\right ) \, dx}{105 a}\\ &=\frac{2 a^2 (35 A+56 B+40 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{105 d}+\frac{2 a (7 B+5 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac{2 C (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}+\left (a^2 A\right ) \int \sqrt{a+a \sec (c+d x)} \, dx+\frac{1}{105} \left (a^2 (245 A+224 B+160 C)\right ) \int \sec (c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{2 a^3 (245 A+224 B+160 C) \tan (c+d x)}{105 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 (35 A+56 B+40 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{105 d}+\frac{2 a (7 B+5 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac{2 C (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}-\frac{\left (2 a^3 A\right ) \operatorname{Subst}\left (\int \frac{1}{a+x^2} \, dx,x,-\frac{a \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{d}\\ &=\frac{2 a^{5/2} A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{d}+\frac{2 a^3 (245 A+224 B+160 C) \tan (c+d x)}{105 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 (35 A+56 B+40 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{105 d}+\frac{2 a (7 B+5 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac{2 C (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}\\ \end{align*}

Mathematica [A]  time = 2.43599, size = 170, normalized size = 0.93 \[ \frac{a^2 \sec \left (\frac{1}{2} (c+d x)\right ) \sec ^3(c+d x) \sqrt{a (\sec (c+d x)+1)} \left (2 \sin \left (\frac{1}{2} (c+d x)\right ) ((840 A+987 B+930 C) \cos (c+d x)+2 (35 A+98 B+115 C) \cos (2 (c+d x))+280 A \cos (3 (c+d x))+70 A+301 B \cos (3 (c+d x))+196 B+230 C \cos (3 (c+d x))+290 C)+420 \sqrt{2} A \sin ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right ) \cos ^{\frac{7}{2}}(c+d x)\right )}{420 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Sec[(c + d*x)/2]*Sec[c + d*x]^3*Sqrt[a*(1 + Sec[c + d*x])]*(420*Sqrt[2]*A*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]
]*Cos[c + d*x]^(7/2) + 2*(70*A + 196*B + 290*C + (840*A + 987*B + 930*C)*Cos[c + d*x] + 2*(35*A + 98*B + 115*C
)*Cos[2*(c + d*x)] + 280*A*Cos[3*(c + d*x)] + 301*B*Cos[3*(c + d*x)] + 230*C*Cos[3*(c + d*x)])*Sin[(c + d*x)/2
]))/(420*d)

________________________________________________________________________________________

Maple [B]  time = 0.355, size = 476, normalized size = 2.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

-1/840/d*a^2*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(-105*A*sin(d*x+c)*cos(d*x+c)^3*arctanh(1/2*2^(1/2)*(-2*cos(d
*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(7/2)*2^(1/2)-315*A*sin(d*x+
c)*cos(d*x+c)^2*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*(-2*cos(d*x+c)
/(cos(d*x+c)+1))^(7/2)*2^(1/2)-315*A*sin(d*x+c)*cos(d*x+c)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^
(1/2)*sin(d*x+c)/cos(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(7/2)*2^(1/2)-105*A*arctanh(1/2*2^(1/2)*(-2*cos(d*
x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(7/2)*2^(1/2)*sin(d*x+c)+4480
*A*cos(d*x+c)^4+4816*B*cos(d*x+c)^4+3680*C*cos(d*x+c)^4-3920*A*cos(d*x+c)^3-3248*B*cos(d*x+c)^3-1840*C*cos(d*x
+c)^3-560*A*cos(d*x+c)^2-1232*B*cos(d*x+c)^2-880*C*cos(d*x+c)^2-336*B*cos(d*x+c)-720*C*cos(d*x+c)-240*C)/cos(d
*x+c)^3/sin(d*x+c)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Fricas [A]  time = 0.595144, size = 1111, normalized size = 6.1 \begin{align*} \left [\frac{105 \,{\left (A a^{2} \cos \left (d x + c\right )^{4} + A a^{2} \cos \left (d x + c\right )^{3}\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 ({\left (280 \, A + 301 \, B + 230 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} +{\left (35 \, A + 98 \, B + 115 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \,{\left (7 \, B + 20 \, C\right )} a^{2} \cos \left (d x + c\right ) + 15 \, C a^{2}\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{105 \,{\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}}, -\frac{2 \,{\left (105 \,{\left (A a^{2} \cos \left (d x + c\right )^{4} + A a^{2} \cos \left (d x + c\right )^{3}\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 ({\left (280 \, A + 301 \, B + 230 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} +{\left (35 \, A + 98 \, B + 115 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \,{\left (7 \, B + 20 \, C\right )} a^{2} \cos \left (d x + c\right ) + 15 \, C a^{2}\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )\right )}}{105 \,{\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/105*(105*(A*a^2*cos(d*x + c)^4 + A*a^2*cos(d*x + c)^3)*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*((28
0*A + 301*B + 230*C)*a^2*cos(d*x + c)^3 + (35*A + 98*B + 115*C)*a^2*cos(d*x + c)^2 + 3*(7*B + 20*C)*a^2*cos(d*
x + c) + 15*C*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^4 + d*cos(d*x + c)^3)
, -2/105*(105*(A*a^2*cos(d*x + c)^4 + A*a^2*cos(d*x + c)^3)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x +
c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - ((280*A + 301*B + 230*C)*a^2*cos(d*x + c)^3 + (35*A + 98*B + 115*C
)*a^2*cos(d*x + c)^2 + 3*(7*B + 20*C)*a^2*cos(d*x + c) + 15*C*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin
(d*x + c))/(d*cos(d*x + c)^4 + d*cos(d*x + c)^3)]

________________________________________________________________________________________

Sympy [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((a+a*sec(d*x+c))**(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError