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

Optimal. Leaf size=170 \[ \frac{2 a^3 (49 A+32 C) \tan (c+d x)}{21 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a^2 (7 A+8 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{21 d}+\frac{2 a^{5/2} A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{d}+\frac{2 a C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{7 d}+\frac{2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d} \]

[Out]

(2*a^(5/2)*A*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (2*a^3*(49*A + 32*C)*Tan[c + d*x])/(
21*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(7*A + 8*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(21*d) + (2*a*C*(a
+ a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(7*d) + (2*C*(a + a*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(7*d)

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Rubi [A]  time = 0.308362, antiderivative size = 170, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {4055, 3917, 3915, 3774, 203, 3792} \[ \frac{2 a^3 (49 A+32 C) \tan (c+d x)}{21 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a^2 (7 A+8 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{21 d}+\frac{2 a^{5/2} A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{d}+\frac{2 a C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{7 d}+\frac{2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^(5/2)*(A + 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*(49*A + 32*C)*Tan[c + d*x])/(
21*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(7*A + 8*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(21*d) + (2*a*C*(a
+ a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(7*d) + (2*C*(a + a*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(7*d)

Rule 4055

Int[((A_.) + 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*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, 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+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{5}{2} a C \sec (c+d x)\right ) \, dx}{7 a}\\ &=\frac{2 a C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{7 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{5}{4} a^2 (7 A+8 C) \sec (c+d x)\right ) \, dx}{35 a}\\ &=\frac{2 a^2 (7 A+8 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{21 d}+\frac{2 a C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{7 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{5}{8} a^3 (49 A+32 C) \sec (c+d x)\right ) \, dx}{105 a}\\ &=\frac{2 a^2 (7 A+8 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{21 d}+\frac{2 a C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{7 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}{21} \left (a^2 (49 A+32 C)\right ) \int \sec (c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{2 a^3 (49 A+32 C) \tan (c+d x)}{21 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 (7 A+8 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{21 d}+\frac{2 a C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{7 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 (49 A+32 C) \tan (c+d x)}{21 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 (7 A+8 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{21 d}+\frac{2 a C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{7 d}+\frac{2 C (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}\\ \end{align*}

Mathematica [A]  time = 1.83551, size = 151, normalized size = 0.89 \[ \frac{a^2 \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^3(c+d x) \sqrt{a (\sec (c+d x)+1)} \left (\sqrt{\sec (c+d x)-1} ((84 A+93 C) \cos (c+d x)+(7 A+23 C) \cos (2 (c+d x))+28 A \cos (3 (c+d x))+7 A+23 C \cos (3 (c+d x))+29 C)+42 A \cos ^3(c+d x) \tan ^{-1}\left (\sqrt{\sec (c+d x)-1}\right )\right )}{21 d \sqrt{\sec (c+d x)-1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(42*A*ArcTan[Sqrt[-1 + Sec[c + d*x]]]*Cos[c + d*x]^3 + (7*A + 29*C + (84*A + 93*C)*Cos[c + d*x] + (7*A +
23*C)*Cos[2*(c + d*x)] + 28*A*Cos[3*(c + d*x)] + 23*C*Cos[3*(c + d*x)])*Sqrt[-1 + Sec[c + d*x]])*Sec[c + d*x]^
3*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d*x)/2])/(21*d*Sqrt[-1 + Sec[c + d*x]])

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Maple [B]  time = 0.314, size = 434, normalized size = 2.6 \begin{align*}{\frac{{a}^{2}}{168\,d \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) }\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}} \left ( 21\,A\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}} \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{7/2}\sqrt{2}+63\,A\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}} \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{7/2}\sqrt{2}+63\,A\sin \left ( dx+c \right ) \cos \left ( dx+c \right ){\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}} \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{7/2}\sqrt{2}+21\,A{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}} \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{7/2}\sqrt{2}\sin \left ( dx+c \right ) -896\,A \left ( \cos \left ( dx+c \right ) \right ) ^{4}-736\,C \left ( \cos \left ( dx+c \right ) \right ) ^{4}+784\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+368\,C \left ( \cos \left ( dx+c \right ) \right ) ^{3}+112\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+176\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}+144\,C\cos \left ( dx+c \right ) +48\,C \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/168/d*a^2*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(21*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)+63*A*sin(d*x+c)*c
os(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)/(co
s(d*x+c)+1))^(7/2)*2^(1/2)+63*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)+21*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)-896*A*cos(
d*x+c)^4-736*C*cos(d*x+c)^4+784*A*cos(d*x+c)^3+368*C*cos(d*x+c)^3+112*A*cos(d*x+c)^2+176*C*cos(d*x+c)^2+144*C*
cos(d*x+c)+48*C)/cos(d*x+c)^3/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((a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.582656, size = 1030, normalized size = 6.06 \begin{align*} \left [\frac{21 \,{\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 (2 \,{\left (28 \, A + 23 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} +{\left (7 \, A + 23 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 12 \, C a^{2} \cos \left (d x + c\right ) + 3 \, 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 )}{21 \,{\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}}, -\frac{2 \,{\left (21 \,{\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 (2 \,{\left (28 \, A + 23 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} +{\left (7 \, A + 23 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 12 \, C a^{2} \cos \left (d x + c\right ) + 3 \, 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 )}}{21 \,{\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+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

[1/21*(21*(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*(2*(28
*A + 23*C)*a^2*cos(d*x + c)^3 + (7*A + 23*C)*a^2*cos(d*x + c)^2 + 12*C*a^2*cos(d*x + c) + 3*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), -2/21*(21*(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))) - (2*(28*A + 23*C)*a^2*cos(d*x + c)^3 + (7*A + 23*C)*a^2*cos(d*x + c)^2 + 12*C*a^2*cos(d*x + c) + 3*
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)]

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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+C*sec(d*x+c)**2),x)

[Out]

Timed out

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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+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError