3.4.65 \(\int \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [365]

Optimal. Leaf size=234 \[ \frac {4 a^2 (187 B+168 C) \tan (c+d x)}{495 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (187 B+168 C) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (11 B+12 C) \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt {a+a \sec (c+d x)}}-\frac {8 a (187 B+168 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{3465 d}+\frac {2 a C \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{11 d}+\frac {4 (187 B+168 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 d} \]

[Out]

4/1155*(187*B+168*C)*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/d+4/495*a^2*(187*B+168*C)*tan(d*x+c)/d/(a+a*sec(d*x+c))
^(1/2)+2/693*a^2*(187*B+168*C)*sec(d*x+c)^3*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/99*a^2*(11*B+12*C)*sec(d*x+c
)^4*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)-8/3465*a*(187*B+168*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d+2/11*a*C*se
c(d*x+c)^4*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d

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Rubi [A]
time = 0.43, antiderivative size = 234, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4157, 4103, 4101, 3888, 3885, 4086, 3877} \begin {gather*} \frac {2 a^2 (11 B+12 C) \tan (c+d x) \sec ^4(c+d x)}{99 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a^2 (187 B+168 C) \tan (c+d x) \sec ^3(c+d x)}{693 d \sqrt {a \sec (c+d x)+a}}+\frac {4 a^2 (187 B+168 C) \tan (c+d x)}{495 d \sqrt {a \sec (c+d x)+a}}+\frac {4 (187 B+168 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{1155 d}-\frac {8 a (187 B+168 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3465 d}+\frac {2 a C \tan (c+d x) \sec ^4(c+d x) \sqrt {a \sec (c+d x)+a}}{11 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a^2*(187*B + 168*C)*Tan[c + d*x])/(495*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(187*B + 168*C)*Sec[c + d*x]^3*
Tan[c + d*x])/(693*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(11*B + 12*C)*Sec[c + d*x]^4*Tan[c + d*x])/(99*d*Sqrt[
a + a*Sec[c + d*x]]) - (8*a*(187*B + 168*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(3465*d) + (2*a*C*Sec[c + d
*x]^4*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(11*d) + (4*(187*B + 168*C)*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*
x])/(1155*d)

Rule 3877

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]

Rule 3885

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-Cot[e + f*x])*(
(a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[1/(b*(m + 2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*
(b*(m + 1) - a*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)]

Rule 3888

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

Rule 4086

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

Rule 4101

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

Rule 4103

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

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 \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \sec ^4(c+d x) (a+a \sec (c+d x))^{3/2} (B+C \sec (c+d x)) \, dx\\ &=\frac {2 a C \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{11 d}+\frac {2}{11} \int \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \left (\frac {1}{2} a (11 B+8 C)+\frac {1}{2} a (11 B+12 C) \sec (c+d x)\right ) \, dx\\ &=\frac {2 a^2 (11 B+12 C) \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a C \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{11 d}+\frac {1}{99} (a (187 B+168 C)) \int \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {2 a^2 (187 B+168 C) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (11 B+12 C) \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a C \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{11 d}+\frac {1}{231} (2 a (187 B+168 C)) \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {2 a^2 (187 B+168 C) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (11 B+12 C) \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a C \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{11 d}+\frac {4 (187 B+168 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 d}+\frac {(4 (187 B+168 C)) \int \sec (c+d x) \left (\frac {3 a}{2}-a \sec (c+d x)\right ) \sqrt {a+a \sec (c+d x)} \, dx}{1155}\\ &=\frac {2 a^2 (187 B+168 C) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (11 B+12 C) \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt {a+a \sec (c+d x)}}-\frac {8 a (187 B+168 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{3465 d}+\frac {2 a C \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{11 d}+\frac {4 (187 B+168 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 d}+\frac {1}{495} (2 a (187 B+168 C)) \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {4 a^2 (187 B+168 C) \tan (c+d x)}{495 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (187 B+168 C) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (11 B+12 C) \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt {a+a \sec (c+d x)}}-\frac {8 a (187 B+168 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{3465 d}+\frac {2 a C \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{11 d}+\frac {4 (187 B+168 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 d}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(487\) vs. \(2(234)=468\).
time = 6.22, size = 487, normalized size = 2.08 \begin {gather*} \frac {544 B (a (1+\sec (c+d x)))^{3/2} \tan (c+d x)}{315 d (1+\sec (c+d x))^2}+\frac {256 C (a (1+\sec (c+d x)))^{3/2} \tan (c+d x)}{165 d (1+\sec (c+d x))^2}+\frac {272 B \sec (c+d x) (a (1+\sec (c+d x)))^{3/2} \tan (c+d x)}{315 d (1+\sec (c+d x))^2}+\frac {128 C \sec (c+d x) (a (1+\sec (c+d x)))^{3/2} \tan (c+d x)}{165 d (1+\sec (c+d x))^2}+\frac {68 B \sec ^2(c+d x) (a (1+\sec (c+d x)))^{3/2} \tan (c+d x)}{105 d (1+\sec (c+d x))^2}+\frac {32 C \sec ^2(c+d x) (a (1+\sec (c+d x)))^{3/2} \tan (c+d x)}{55 d (1+\sec (c+d x))^2}+\frac {34 B \sec ^3(c+d x) (a (1+\sec (c+d x)))^{3/2} \tan (c+d x)}{63 d (1+\sec (c+d x))^2}+\frac {16 C \sec ^3(c+d x) (a (1+\sec (c+d x)))^{3/2} \tan (c+d x)}{33 d (1+\sec (c+d x))^2}+\frac {2 B \sec ^4(c+d x) (a (1+\sec (c+d x)))^{3/2} \tan (c+d x)}{9 d (1+\sec (c+d x))^2}+\frac {14 C \sec ^4(c+d x) (a (1+\sec (c+d x)))^{3/2} \tan (c+d x)}{33 d (1+\sec (c+d x))^2}+\frac {2 C \sec ^5(c+d x) (a (1+\sec (c+d x)))^{3/2} \tan (c+d x)}{11 d (1+\sec (c+d x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(544*B*(a*(1 + Sec[c + d*x]))^(3/2)*Tan[c + d*x])/(315*d*(1 + Sec[c + d*x])^2) + (256*C*(a*(1 + Sec[c + d*x]))
^(3/2)*Tan[c + d*x])/(165*d*(1 + Sec[c + d*x])^2) + (272*B*Sec[c + d*x]*(a*(1 + Sec[c + d*x]))^(3/2)*Tan[c + d
*x])/(315*d*(1 + Sec[c + d*x])^2) + (128*C*Sec[c + d*x]*(a*(1 + Sec[c + d*x]))^(3/2)*Tan[c + d*x])/(165*d*(1 +
Sec[c + d*x])^2) + (68*B*Sec[c + d*x]^2*(a*(1 + Sec[c + d*x]))^(3/2)*Tan[c + d*x])/(105*d*(1 + Sec[c + d*x])^
2) + (32*C*Sec[c + d*x]^2*(a*(1 + Sec[c + d*x]))^(3/2)*Tan[c + d*x])/(55*d*(1 + Sec[c + d*x])^2) + (34*B*Sec[c
+ d*x]^3*(a*(1 + Sec[c + d*x]))^(3/2)*Tan[c + d*x])/(63*d*(1 + Sec[c + d*x])^2) + (16*C*Sec[c + d*x]^3*(a*(1
+ Sec[c + d*x]))^(3/2)*Tan[c + d*x])/(33*d*(1 + Sec[c + d*x])^2) + (2*B*Sec[c + d*x]^4*(a*(1 + Sec[c + d*x]))^
(3/2)*Tan[c + d*x])/(9*d*(1 + Sec[c + d*x])^2) + (14*C*Sec[c + d*x]^4*(a*(1 + Sec[c + d*x]))^(3/2)*Tan[c + d*x
])/(33*d*(1 + Sec[c + d*x])^2) + (2*C*Sec[c + d*x]^5*(a*(1 + Sec[c + d*x]))^(3/2)*Tan[c + d*x])/(11*d*(1 + Sec
[c + d*x])^2)

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Maple [A]
time = 13.62, size = 161, normalized size = 0.69

method	result	size
default	\(-\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (2992 B \left (\cos ^{5}\left (d x +c \right )\right )+2688 C \left (\cos ^{5}\left (d x +c \right )\right )+1496 B \left (\cos ^{4}\left (d x +c \right )\right )+1344 C \left (\cos ^{4}\left (d x +c \right )\right )+1122 B \left (\cos ^{3}\left (d x +c \right )\right )+1008 C \left (\cos ^{3}\left (d x +c \right )\right )+935 B \left (\cos ^{2}\left (d x +c \right )\right )+840 C \left (\cos ^{2}\left (d x +c \right )\right )+385 B \cos \left (d x +c \right )+735 C \cos \left (d x +c \right )+315 C \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a}{3465 d \cos \left (d x +c \right )^{5} \sin \left (d x +c \right )}\)	\(161\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3465/d*(-1+cos(d*x+c))*(2992*B*cos(d*x+c)^5+2688*C*cos(d*x+c)^5+1496*B*cos(d*x+c)^4+1344*C*cos(d*x+c)^4+112
2*B*cos(d*x+c)^3+1008*C*cos(d*x+c)^3+935*B*cos(d*x+c)^2+840*C*cos(d*x+c)^2+385*B*cos(d*x+c)+735*C*cos(d*x+c)+3
15*C)*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d*x+c)^5/sin(d*x+c)*a

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Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [A]
time = 2.11, size = 145, normalized size = 0.62 \begin {gather*} \frac {2 \, {\left (16 \, {\left (187 \, B + 168 \, C\right )} a \cos \left (d x + c\right )^{5} + 8 \, {\left (187 \, B + 168 \, C\right )} a \cos \left (d x + c\right )^{4} + 6 \, {\left (187 \, B + 168 \, C\right )} a \cos \left (d x + c\right )^{3} + 5 \, {\left (187 \, B + 168 \, C\right )} a \cos \left (d x + c\right )^{2} + 35 \, {\left (11 \, B + 21 \, C\right )} a \cos \left (d x + c\right ) + 315 \, C a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3465 \, {\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3465*(16*(187*B + 168*C)*a*cos(d*x + c)^5 + 8*(187*B + 168*C)*a*cos(d*x + c)^4 + 6*(187*B + 168*C)*a*cos(d*x
+ c)^3 + 5*(187*B + 168*C)*a*cos(d*x + c)^2 + 35*(11*B + 21*C)*a*cos(d*x + c) + 315*C*a)*sqrt((a*cos(d*x + c)
+ a)/cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c)^6 + d*cos(d*x + c)^5)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}} \left (B + C \sec {\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*(sec(c + d*x) + 1))**(3/2)*(B + C*sec(c + d*x))*sec(c + d*x)**4, x)

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Giac [A]
time = 1.43, size = 305, normalized size = 1.30 \begin {gather*} \frac {4 \, {\left ({\left ({\left ({\left ({\left (2 \, \sqrt {2} {\left (517 \, B a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 483 \, C a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 11 \, \sqrt {2} {\left (517 \, B a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 483 \, C a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 198 \, \sqrt {2} {\left (69 \, B a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 56 \, C a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 462 \, \sqrt {2} {\left (32 \, B a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 33 \, C a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2310 \, \sqrt {2} {\left (4 \, B a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 3 \, C a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3465 \, \sqrt {2} {\left (B a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + C a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{3465 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{5} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/3465*(((((2*sqrt(2)*(517*B*a^7*sgn(cos(d*x + c)) + 483*C*a^7*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2 - 11*
sqrt(2)*(517*B*a^7*sgn(cos(d*x + c)) + 483*C*a^7*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 198*sqrt(2)*(69*
B*a^7*sgn(cos(d*x + c)) + 56*C*a^7*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 - 462*sqrt(2)*(32*B*a^7*sgn(cos(
d*x + c)) + 33*C*a^7*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 2310*sqrt(2)*(4*B*a^7*sgn(cos(d*x + c)) + 3*
C*a^7*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 - 3465*sqrt(2)*(B*a^7*sgn(cos(d*x + c)) + C*a^7*sgn(cos(d*x +
c))))*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^5*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*d)

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Mupad [B]
time = 13.51, size = 720, normalized size = 3.08 \begin {gather*} -\frac {\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (-\frac {B\,a\,16{}\mathrm {i}}{9\,d}+\frac {C\,a\,256{}\mathrm {i}}{33\,d}+\frac {a\,\left (B+2\,C\right )\,16{}\mathrm {i}}{3\,d}\right )-\frac {a\,\left (3\,B+2\,C\right )\,16{}\mathrm {i}}{9\,d}+\frac {C\,a\,64{}\mathrm {i}}{9\,d}+\frac {a\,\left (B+4\,C\right )\,16{}\mathrm {i}}{9\,d}\right )}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^4}+\frac {\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {a\,\left (11\,B-42\,C\right )\,16{}\mathrm {i}}{1155\,d}+\frac {B\,a\,16{}\mathrm {i}}{5\,d}\right )+\frac {a\,\left (3\,B+2\,C\right )\,16{}\mathrm {i}}{5\,d}\right )}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2}-\frac {\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (-\frac {a\,\left (2\,B+3\,C\right )\,32{}\mathrm {i}}{11\,d}+\frac {a\,\left (3\,B+2\,C\right )\,16{}\mathrm {i}}{11\,d}+\frac {B\,a\,16{}\mathrm {i}}{11\,d}\right )-\frac {a\,\left (2\,B+3\,C\right )\,32{}\mathrm {i}}{11\,d}+\frac {a\,\left (3\,B+2\,C\right )\,16{}\mathrm {i}}{11\,d}+\frac {B\,a\,16{}\mathrm {i}}{11\,d}\right )}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^5}-\frac {\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left (-{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {a\,\left (11\,B+39\,C\right )\,32{}\mathrm {i}}{693\,d}-\frac {B\,a\,16{}\mathrm {i}}{7\,d}+\frac {a\,\left (B+3\,C\right )\,32{}\mathrm {i}}{7\,d}\right )+\frac {a\,\left (3\,B+2\,C\right )\,16{}\mathrm {i}}{7\,d}+\frac {a\,\left (B-C\right )\,32{}\mathrm {i}}{7\,d}\right )}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^3}-\frac {a\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left (187\,B+168\,C\right )\,32{}\mathrm {i}}{3465\,d\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )}-\frac {a\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left (187\,B+168\,C\right )\,16{}\mathrm {i}}{3465\,d\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )} \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))/cos(c + d*x)^3,x)

[Out]

((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*((a*(11*B - 42*C)*16i)/(115
5*d) + (B*a*16i)/(5*d)) + (a*(3*B + 2*C)*16i)/(5*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^2) -
((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*((C*a*256i)/(33*d) - (B*a*1
6i)/(9*d) + (a*(B + 2*C)*16i)/(3*d)) - (a*(3*B + 2*C)*16i)/(9*d) + (C*a*64i)/(9*d) + (a*(B + 4*C)*16i)/(9*d)))
/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^4) - ((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/
2))^(1/2)*(exp(c*1i + d*x*1i)*((a*(3*B + 2*C)*16i)/(11*d) - (a*(2*B + 3*C)*32i)/(11*d) + (B*a*16i)/(11*d)) - (
a*(2*B + 3*C)*32i)/(11*d) + (a*(3*B + 2*C)*16i)/(11*d) + (B*a*16i)/(11*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2
i + d*x*2i) + 1)^5) - ((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*((a*(3*B + 2*C)*16i)/(7*d
) - exp(c*1i + d*x*1i)*((a*(11*B + 39*C)*32i)/(693*d) - (B*a*16i)/(7*d) + (a*(B + 3*C)*32i)/(7*d)) + (a*(B - C
)*32i)/(7*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^3) - (a*exp(c*1i + d*x*1i)*(a + a/(exp(- c*1
i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(187*B + 168*C)*32i)/(3465*d*(exp(c*1i + d*x*1i) + 1)) - (a*exp(c
*1i + d*x*1i)*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(187*B + 168*C)*16i)/(3465*d*(exp(
c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1))

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