3.2.84 \(\int \frac {\sec ^3(c+d x) (A+C \sec ^2(c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx\) [184]

Optimal. Leaf size=193 \[ \frac {\sqrt {2} (A+C) \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {a} d}-\frac {4 (35 A+37 C) \tan (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}-\frac {2 C \sec ^2(c+d x) \tan (c+d x)}{35 d \sqrt {a+a \sec (c+d x)}}+\frac {2 C \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt {a+a \sec (c+d x)}}+\frac {2 (35 A+31 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{105 a d} \]

[Out]

(A+C)*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))*2^(1/2)/d/a^(1/2)-4/105*(35*A+37*C)*tan(d*
x+c)/d/(a+a*sec(d*x+c))^(1/2)-2/35*C*sec(d*x+c)^2*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/7*C*sec(d*x+c)^3*tan(d
*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/105*(35*A+31*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/a/d

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Rubi [A]
time = 0.39, antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {4174, 4106, 4095, 4086, 3880, 209} \begin {gather*} \frac {\sqrt {2} (A+C) \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{\sqrt {a} d}+\frac {2 (35 A+31 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{105 a d}-\frac {4 (35 A+37 C) \tan (c+d x)}{105 d \sqrt {a \sec (c+d x)+a}}+\frac {2 C \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}-\frac {2 C \tan (c+d x) \sec ^2(c+d x)}{35 d \sqrt {a \sec (c+d x)+a}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[2]*(A + C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(Sqrt[a]*d) - (4*(35*A + 3
7*C)*Tan[c + d*x])/(105*d*Sqrt[a + a*Sec[c + d*x]]) - (2*C*Sec[c + d*x]^2*Tan[c + d*x])/(35*d*Sqrt[a + a*Sec[c
+ d*x]]) + (2*C*Sec[c + d*x]^3*Tan[c + d*x])/(7*d*Sqrt[a + a*Sec[c + d*x]]) + (2*(35*A + 31*C)*Sqrt[a + a*Sec
[c + d*x]]*Tan[c + d*x])/(105*a*d)

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 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 4095

Int[csc[(e_.) + (f_.)*(x_)]^2*(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 + 1)/(b*f*(m + 2))), x] + Dist[1/(b*(m + 2)),
Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*B*(m + 1) + (A*b*(m + 2) - a*B)*Csc[e + f*x], x], x], x] /; Fr
eeQ[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] &&  !LtQ[m, -1]

Rule 4106

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

Rule 4174

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

Rubi steps

\begin {align*} \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx &=\frac {2 C \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt {a+a \sec (c+d x)}}+\frac {2 \int \frac {\sec ^3(c+d x) \left (\frac {1}{2} a (7 A+6 C)-\frac {1}{2} a C \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{7 a}\\ &=-\frac {2 C \sec ^2(c+d x) \tan (c+d x)}{35 d \sqrt {a+a \sec (c+d x)}}+\frac {2 C \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt {a+a \sec (c+d x)}}+\frac {4 \int \frac {\sec ^2(c+d x) \left (-a^2 C+\frac {1}{4} a^2 (35 A+31 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{35 a^2}\\ &=-\frac {2 C \sec ^2(c+d x) \tan (c+d x)}{35 d \sqrt {a+a \sec (c+d x)}}+\frac {2 C \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt {a+a \sec (c+d x)}}+\frac {2 (35 A+31 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{105 a d}+\frac {8 \int \frac {\sec (c+d x) \left (\frac {1}{8} a^3 (35 A+31 C)-\frac {1}{4} a^3 (35 A+37 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{105 a^3}\\ &=-\frac {4 (35 A+37 C) \tan (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}-\frac {2 C \sec ^2(c+d x) \tan (c+d x)}{35 d \sqrt {a+a \sec (c+d x)}}+\frac {2 C \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt {a+a \sec (c+d x)}}+\frac {2 (35 A+31 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{105 a d}+(A+C) \int \frac {\sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx\\ &=-\frac {4 (35 A+37 C) \tan (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}-\frac {2 C \sec ^2(c+d x) \tan (c+d x)}{35 d \sqrt {a+a \sec (c+d x)}}+\frac {2 C \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt {a+a \sec (c+d x)}}+\frac {2 (35 A+31 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{105 a d}-\frac {(2 (A+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 )}{d}\\ &=\frac {\sqrt {2} (A+C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {a} d}-\frac {4 (35 A+37 C) \tan (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}-\frac {2 C \sec ^2(c+d x) \tan (c+d x)}{35 d \sqrt {a+a \sec (c+d x)}}+\frac {2 C \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt {a+a \sec (c+d x)}}+\frac {2 (35 A+31 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{105 a d}\\ \end {align*}

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Mathematica [A]
time = 6.42, size = 173, normalized size = 0.90 \begin {gather*} \frac {2 \cos ^2(c+d x) \sqrt {a (1+\sec (c+d x))} \left (A+C \sec ^2(c+d x)\right ) \left (105 \sqrt {2} (A+C) \text {ArcTan}\left (\frac {\sqrt {-1+\sec (c+d x)}}{\sqrt {2}}\right ) \cot (c+d x) \sqrt {-1+\sec (c+d x)}+2 (35 A+73 C+24 C \cos (c+d x)+(35 A+43 C) \cos (2 (c+d x))) \sec ^3(c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{105 a d (A+2 C+A \cos (2 (c+d x)))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Cos[c + d*x]^2*Sqrt[a*(1 + Sec[c + d*x])]*(A + C*Sec[c + d*x]^2)*(105*Sqrt[2]*(A + C)*ArcTan[Sqrt[-1 + Sec[
c + d*x]]/Sqrt[2]]*Cot[c + d*x]*Sqrt[-1 + Sec[c + d*x]] + 2*(35*A + 73*C + 24*C*Cos[c + d*x] + (35*A + 43*C)*C
os[2*(c + d*x)])*Sec[c + d*x]^3*Sin[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(105*a*d*(A + 2*C + A*Cos[2*(c + d*x)]))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(775\) vs. \(2(168)=336\).
time = 13.07, size = 776, normalized size = 4.02

method	result	size
default	\(\frac {\left (-105 A \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \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 ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{2}}-105 C \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \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 ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{2}}-315 A \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \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 ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{2}}-315 C \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \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 ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{2}}-315 A \cos \left (d x +c \right ) \sin \left (d x +c \right ) \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 ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{2}}-315 C \cos \left (d x +c \right ) \sin \left (d x +c \right ) \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 ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{2}}-105 A \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 ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{2}} \sin \left (d x +c \right )-105 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 ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{2}} \sin \left (d x +c \right )+560 A \left (\cos ^{4}\left (d x +c \right )\right )+688 C \left (\cos ^{4}\left (d x +c \right )\right )-1120 A \left (\cos ^{3}\left (d x +c \right )\right )-1184 C \left (\cos ^{3}\left (d x +c \right )\right )+560 A \left (\cos ^{2}\left (d x +c \right )\right )+544 C \left (\cos ^{2}\left (d x +c \right )\right )-288 C \cos \left (d x +c \right )+240 C \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{840 d \cos \left (d x +c \right )^{3} \sin \left (d x +c \right ) a}\)	\(776\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/840/d*(-105*A*cos(d*x+c)^3*sin(d*x+c)*ln(-(-(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/si
n(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)-105*C*cos(d*x+c)^3*sin(d*x+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)))^(7/2)-315*A*cos(d*x+c)^2*sin(d*x
+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)))^(7/2)-315*C*cos(d*x+c)^2*sin(d*x+c)*ln(-(-(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/s
in(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)-315*A*cos(d*x+c)*sin(d*x+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)))^(7/2)-315*C*cos(d*x+c)*sin(d*x+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))
)^(7/2)-105*A*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)))^(7/2)*sin(d*x+c)-105*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)))^(7/2)*sin(d*x+c)+560*A*cos(d*x+c)^4+688*C*cos(d*x+c)^4-1120*A*cos(d*x+c
)^3-1184*C*cos(d*x+c)^3+560*A*cos(d*x+c)^2+544*C*cos(d*x+c)^2-288*C*cos(d*x+c)+240*C)*(a*(1+cos(d*x+c))/cos(d*
x+c))^(1/2)/cos(d*x+c)^3/sin(d*x+c)/a

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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(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate((C*sec(d*x + c)^2 + A)*sec(d*x + c)^3/sqrt(a*sec(d*x + c) + a), x)

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Fricas [A]
time = 2.86, size = 412, normalized size = 2.13 \begin {gather*} \left [\frac {105 \, \sqrt {2} {\left ({\left (A + C\right )} a \cos \left (d x + c\right )^{4} + {\left (A + C\right )} a \cos \left (d x + c\right )^{3}\right )} \sqrt {-\frac {1}{a}} \log \left (-\frac {2 \, \sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {-\frac {1}{a}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 3 \, \cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \, {\left ({\left (35 \, A + 43 \, C\right )} \cos \left (d x + c\right )^{3} - {\left (35 \, A + 31 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, C \cos \left (d x + c\right ) - 15 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{210 \, {\left (a d \cos \left (d x + c\right )^{4} + a d \cos \left (d x + c\right )^{3}\right )}}, -\frac {2 \, {\left ({\left (35 \, A + 43 \, C\right )} \cos \left (d x + c\right )^{3} - {\left (35 \, A + 31 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, C \cos \left (d x + c\right ) - 15 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right ) + \frac {105 \, \sqrt {2} {\left ({\left (A + C\right )} a \cos \left (d x + c\right )^{4} + {\left (A + C\right )} a \cos \left (d x + c\right )^{3}\right )} \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 )}{\sqrt {a}}}{105 \, {\left (a d \cos \left (d x + c\right )^{4} + a d \cos \left (d x + c\right )^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

[1/210*(105*sqrt(2)*((A + C)*a*cos(d*x + c)^4 + (A + C)*a*cos(d*x + c)^3)*sqrt(-1/a)*log(-(2*sqrt(2)*sqrt((a*c
os(d*x + c) + a)/cos(d*x + c))*sqrt(-1/a)*cos(d*x + c)*sin(d*x + c) - 3*cos(d*x + c)^2 - 2*cos(d*x + c) + 1)/(
cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) - 4*((35*A + 43*C)*cos(d*x + c)^3 - (35*A + 31*C)*cos(d*x + c)^2 + 3*C*c
os(d*x + c) - 15*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a*d*cos(d*x + c)^4 + a*d*cos(d*x +
c)^3), -1/105*(2*((35*A + 43*C)*cos(d*x + c)^3 - (35*A + 31*C)*cos(d*x + c)^2 + 3*C*cos(d*x + c) - 15*C)*sqrt(
(a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c) + 105*sqrt(2)*((A + C)*a*cos(d*x + c)^4 + (A + C)*a*cos(d*x +
c)^3)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c)))/sqrt(a))/(a*
d*cos(d*x + c)^4 + a*d*cos(d*x + c)^3)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3*(A+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**(1/2),x)

[Out]

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

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Giac [A]
time = 1.60, size = 219, normalized size = 1.13 \begin {gather*} -\frac {\frac {105 \, \sqrt {2} {\left (A + 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} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} + \frac {4 \, {\left ({\left (\frac {\sqrt {2} {\left (35 \, A a^{3} + 46 \, C a^{3}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{\mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {14 \, \sqrt {2} {\left (5 \, A a^{3} + 4 \, C a^{3}\right )}}{\mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {35 \, \sqrt {2} {\left (A a^{3} + 2 \, C a^{3}\right )}}{\mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{3} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}}{105 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

-1/105*(105*sqrt(2)*(A + 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)))/(sq
rt(-a)*sgn(cos(d*x + c))) + 4*((sqrt(2)*(35*A*a^3 + 46*C*a^3)*tan(1/2*d*x + 1/2*c)^2/sgn(cos(d*x + c)) - 14*sq
rt(2)*(5*A*a^3 + 4*C*a^3)/sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2 + 35*sqrt(2)*(A*a^3 + 2*C*a^3)/sgn(cos(d*x
+ c)))*tan(1/2*d*x + 1/2*c)^3/((a*tan(1/2*d*x + 1/2*c)^2 - a)^3*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + C/cos(c + d*x)^2)/(cos(c + d*x)^3*(a + a/cos(c + d*x))^(1/2)),x)

[Out]

int((A + C/cos(c + d*x)^2)/(cos(c + d*x)^3*(a + a/cos(c + d*x))^(1/2)), x)

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