3.4.85 \(\int \frac {\sec ^3(c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx\) [385]

Optimal. Leaf size=202 \[ -\frac {\sqrt {2} (B-C) \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {a} d}+\frac {4 (49 B-37 C) \tan (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {2 (7 B-C) \sec ^2(c+d x) \tan (c+d x)}{35 d \sqrt {a+a \sec (c+d x)}}+\frac {2 C \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt {a+a \sec (c+d x)}}-\frac {2 (7 B-31 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{105 a d} \]

[Out]

-(B-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*(49*B-37*C)*tan(d
*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/35*(7*B-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*(7*B-31*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/a/d

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Rubi [A]
time = 0.48, antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4157, 4106, 4095, 4086, 3880, 209} \begin {gather*} -\frac {\sqrt {2} (B-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 (7 B-C) \tan (c+d x) \sec ^2(c+d x)}{35 d \sqrt {a \sec (c+d x)+a}}-\frac {2 (7 B-31 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{105 a d}+\frac {4 (49 B-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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[2]*(B - C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(Sqrt[a]*d)) + (4*(49*B
- 37*C)*Tan[c + d*x])/(105*d*Sqrt[a + a*Sec[c + d*x]]) + (2*(7*B - 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*(7*B - 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 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 \frac {\sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx &=\int \frac {\sec ^4(c+d x) (B+C \sec (c+d x))}{\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 (3 a C+\frac {1}{2} a (7 B-C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{7 a}\\ &=\frac {2 (7 B-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 (7 B-C)-\frac {1}{4} a^2 (7 B-31 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{35 a^2}\\ &=\frac {2 (7 B-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 (7 B-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 (7 B-31 C)+\frac {1}{4} a^3 (49 B-37 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{105 a^3}\\ &=\frac {4 (49 B-37 C) \tan (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {2 (7 B-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 (7 B-31 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{105 a d}+(-B+C) \int \frac {\sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx\\ &=\frac {4 (49 B-37 C) \tan (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {2 (7 B-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 (7 B-31 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{105 a d}+\frac {(2 (B-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} (B-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 (49 B-37 C) \tan (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {2 (7 B-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 (7 B-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 = 0.54, size = 140, normalized size = 0.69 \begin {gather*} \frac {\left (-105 \sqrt {2} (B-C) \tanh ^{-1}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right )+2 \sqrt {1-\sec (c+d x)} \left (91 B-43 C+(-7 B+31 C) \sec (c+d x)+3 (7 B-C) \sec ^2(c+d x)+15 C \sec ^3(c+d x)\right )\right ) \tan (c+d x)}{105 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-105*Sqrt[2]*(B - C)*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]] + 2*Sqrt[1 - Sec[c + d*x]]*(91*B - 43*C + (-7*B
+ 31*C)*Sec[c + d*x] + 3*(7*B - C)*Sec[c + d*x]^2 + 15*C*Sec[c + d*x]^3))*Tan[c + d*x])/(105*d*Sqrt[1 - Sec[c
+ d*x]]*Sqrt[a*(1 + Sec[c + d*x])])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(784\) vs. \(2(177)=354\).
time = 14.78, size = 785, normalized size = 3.89

method	result	size
default	\(-\frac {\left (-105 B \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 B \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 B \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 B \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{2}} \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 ) \sin \left (d x +c \right )+105 C \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{2}} \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 ) \sin \left (d x +c \right )+1456 B \left (\cos ^{4}\left (d x +c \right )\right )-688 C \left (\cos ^{4}\left (d x +c \right )\right )-1568 B \left (\cos ^{3}\left (d x +c \right )\right )+1184 C \left (\cos ^{3}\left (d x +c \right )\right )+448 B \left (\cos ^{2}\left (d x +c \right )\right )-544 C \left (\cos ^{2}\left (d x +c \right )\right )-336 B \cos \left (d x +c \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}\)	\(785\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/840/d*(-105*B*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)/s
in(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*B*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)/
sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)-315*B*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*B*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*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))*sin(d*x+c)+105*C*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*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))*sin(d*x+c)+1456*B*cos(d*x+c)^4-688*C*cos(d*x+c)^4-1568*B*cos(d*x
+c)^3+1184*C*cos(d*x+c)^3+448*B*cos(d*x+c)^2-544*C*cos(d*x+c)^2-336*B*cos(d*x+c)+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*(B*sec(d*x+c)+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 + B*sec(d*x + c))*sec(d*x + c)^3/sqrt(a*sec(d*x + c) + a), x)

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Fricas [A]
time = 2.69, size = 432, normalized size = 2.14 \begin {gather*} \left [-\frac {105 \, \sqrt {2} {\left ({\left (B - C\right )} a \cos \left (d x + c\right )^{4} + {\left (B - 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 (91 \, B - 43 \, C\right )} \cos \left (d x + c\right )^{3} - {\left (7 \, B - 31 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (7 \, B - C\right )} \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 (91 \, B - 43 \, C\right )} \cos \left (d x + c\right )^{3} - {\left (7 \, B - 31 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (7 \, B - C\right )} \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 (B - C\right )} a \cos \left (d x + c\right )^{4} + {\left (B - 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*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

[-1/210*(105*sqrt(2)*((B - C)*a*cos(d*x + c)^4 + (B - C)*a*cos(d*x + c)^3)*sqrt(-1/a)*log(-(2*sqrt(2)*sqrt((a*
cos(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*((91*B - 43*C)*cos(d*x + c)^3 - (7*B - 31*C)*cos(d*x + c)^2 + 3*(7*
B - C)*cos(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*co
s(d*x + c)^3), 1/105*(2*((91*B - 43*C)*cos(d*x + c)^3 - (7*B - 31*C)*cos(d*x + c)^2 + 3*(7*B - 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)*((B - C)*a*cos(d*x + c)^4 + (B - 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 (B + C \sec {\left (c + d x \right )}\right ) \sec ^{4}{\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*(B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**(1/2),x)

[Out]

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

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Giac [A]
time = 1.57, size = 252, normalized size = 1.25 \begin {gather*} \frac {\frac {105 \, \sqrt {2} {\left (B - 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 {2 \, {\left (\frac {105 \, \sqrt {2} B a^{3}}{\mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - {\left ({\left (\frac {\sqrt {2} {\left (119 \, B a^{3} - 92 \, 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 {7 \, \sqrt {2} {\left (37 \, B a^{3} - 16 \, 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 (7 \, B 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}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\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*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

1/105*(105*sqrt(2)*(B - 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)))/(sqr
t(-a)*sgn(cos(d*x + c))) - 2*(105*sqrt(2)*B*a^3/sgn(cos(d*x + c)) - ((sqrt(2)*(119*B*a^3 - 92*C*a^3)*tan(1/2*d
*x + 1/2*c)^2/sgn(cos(d*x + c)) - 7*sqrt(2)*(37*B*a^3 - 16*C*a^3)/sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2 +
35*sqrt(2)*(7*B*a^3 - 4*C*a^3)/sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)/((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 = -0.00 \begin {gather*} \int \frac {\frac {B}{\cos \left (c+d\,x\right )}+\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((B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^3*(a + a/cos(c + d*x))^(1/2)),x)

[Out]

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

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