\(\int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx\) [152]

Optimal result

Integrand size = 33, antiderivative size = 216 \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {(11 A-15 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(A-B) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac {(65 A-93 B) \tan (c+d x)}{15 a d \sqrt {a+a \sec (c+d x)}}-\frac {(5 A-9 B) \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}+\frac {(35 A-39 B) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{30 a^2 d} \]

[Out]

1/4*(11*A-15*B)*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))/a^(3/2)/d*2^(1/2)+1/2*(A-B)*sec(
d*x+c)^3*tan(d*x+c)/d/(a+a*sec(d*x+c))^(3/2)-1/15*(65*A-93*B)*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^(1/2)-1/10*(5*A-
9*B)*sec(d*x+c)^2*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^(1/2)+1/30*(35*A-39*B)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/a^2
/d

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {4104, 4106, 4095, 4086, 3880, 209} \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {(11 A-15 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(35 A-39 B) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{30 a^2 d}+\frac {(A-B) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}-\frac {(5 A-9 B) \tan (c+d x) \sec ^2(c+d x)}{10 a d \sqrt {a \sec (c+d x)+a}}-\frac {(65 A-93 B) \tan (c+d x)}{15 a d \sqrt {a \sec (c+d x)+a}} \]

[In]

Int[(Sec[c + d*x]^4*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^(3/2),x]

[Out]

((11*A - 15*B)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) + ((A
- B)*Sec[c + d*x]^3*Tan[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2)) - ((65*A - 93*B)*Tan[c + d*x])/(15*a*d*Sqrt
[a + a*Sec[c + d*x]]) - ((5*A - 9*B)*Sec[c + d*x]^2*Tan[c + d*x])/(10*a*d*Sqrt[a + a*Sec[c + d*x]]) + ((35*A -
39*B)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(30*a^2*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 4104

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

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]

Rubi steps \begin{align*} \text {integral}& = \frac {(A-B) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {\int \frac {\sec ^3(c+d x) \left (3 a (A-B)-\frac {1}{2} a (5 A-9 B) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{2 a^2} \\ & = \frac {(A-B) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac {(5 A-9 B) \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}+\frac {\int \frac {\sec ^2(c+d x) \left (-a^2 (5 A-9 B)+\frac {1}{4} a^2 (35 A-39 B) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{5 a^3} \\ & = \frac {(A-B) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac {(5 A-9 B) \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}+\frac {(35 A-39 B) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{30 a^2 d}+\frac {2 \int \frac {\sec (c+d x) \left (\frac {1}{8} a^3 (35 A-39 B)-\frac {1}{4} a^3 (65 A-93 B) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{15 a^4} \\ & = \frac {(A-B) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac {(65 A-93 B) \tan (c+d x)}{15 a d \sqrt {a+a \sec (c+d x)}}-\frac {(5 A-9 B) \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}+\frac {(35 A-39 B) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{30 a^2 d}+\frac {(11 A-15 B) \int \frac {\sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx}{4 a} \\ & = \frac {(A-B) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac {(65 A-93 B) \tan (c+d x)}{15 a d \sqrt {a+a \sec (c+d x)}}-\frac {(5 A-9 B) \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}+\frac {(35 A-39 B) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{30 a^2 d}-\frac {(11 A-15 B) \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 )}{2 a d} \\ & = \frac {(11 A-15 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(A-B) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac {(65 A-93 B) \tan (c+d x)}{15 a d \sqrt {a+a \sec (c+d x)}}-\frac {(5 A-9 B) \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}+\frac {(35 A-39 B) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{30 a^2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.33 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.74 \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {\left (15 \sqrt {2} (11 A-15 B) \text {arctanh}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)+\sqrt {1-\sec (c+d x)} \left (-95 A+147 B-12 (5 A-9 B) \sec (c+d x)+4 (5 A-3 B) \sec ^2(c+d x)+12 B \sec ^3(c+d x)\right )\right ) \tan (c+d x)}{30 d \sqrt {1-\sec (c+d x)} (a (1+\sec (c+d x)))^{3/2}} \]

[In]

Integrate[(Sec[c + d*x]^4*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^(3/2),x]

[Out]

((15*Sqrt[2]*(11*A - 15*B)*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]]*Cos[(c + d*x)/2]^2*Sec[c + d*x] + Sqrt[1 -
Sec[c + d*x]]*(-95*A + 147*B - 12*(5*A - 9*B)*Sec[c + d*x] + 4*(5*A - 3*B)*Sec[c + d*x]^2 + 12*B*Sec[c + d*x]^
3))*Tan[c + d*x])/(30*d*Sqrt[1 - Sec[c + d*x]]*(a*(1 + Sec[c + d*x]))^(3/2))

Maple [A] (warning: unable to verify)

Time = 4.47 (sec) , antiderivative size = 377, normalized size of antiderivative = 1.75

method	result	size
default	\(\frac {\sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (15 A \left (1-\cos \left (d x +c \right )\right )^{7} \csc \left (d x +c \right )^{7}-15 B \left (1-\cos \left (d x +c \right )\right )^{7} \csc \left (d x +c \right )^{7}+165 A \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {5}{2}} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )-225 B \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {5}{2}} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )-245 A \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}+381 B \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}+365 A \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}-525 B \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}-135 A \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+255 B \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )\right )}{60 a^{2} d \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{2}}\)	\(377\)
parts	\(\frac {A \sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (3 \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}+33 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right ) \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {3}{2}}-46 \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+27 \csc \left (d x +c \right )-27 \cot \left (d x +c \right )\right )}{12 d \,a^{2} \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )}-\frac {B \sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (5 \left (1-\cos \left (d x +c \right )\right )^{7} \csc \left (d x +c \right )^{7}+75 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right ) \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {5}{2}}-127 \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}+175 \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}-85 \csc \left (d x +c \right )+85 \cot \left (d x +c \right )\right )}{20 d \,a^{2} \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{2}}\)	\(410\)

[In]

int(sec(d*x+c)^4*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/60/a^2/d*(-2*a/((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(1/2)*(15*A*(1-cos(d*x+c))^7*csc(d*x+c)^7-15*B*(1-cos(d*x+
c))^7*csc(d*x+c)^7+165*A*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(5/2)*ln(csc(d*x+c)-cot(d*x+c)+((1-cos(d*x+c))^2*cs
c(d*x+c)^2-1)^(1/2))-225*B*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(5/2)*ln(csc(d*x+c)-cot(d*x+c)+((1-cos(d*x+c))^2*
csc(d*x+c)^2-1)^(1/2))-245*A*(1-cos(d*x+c))^5*csc(d*x+c)^5+381*B*(1-cos(d*x+c))^5*csc(d*x+c)^5+365*A*(1-cos(d*
x+c))^3*csc(d*x+c)^3-525*B*(1-cos(d*x+c))^3*csc(d*x+c)^3-135*A*(-cot(d*x+c)+csc(d*x+c))+255*B*(-cot(d*x+c)+csc
(d*x+c)))/((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^2

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 504, normalized size of antiderivative = 2.33 \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx=\left [\frac {15 \, \sqrt {2} {\left ({\left (11 \, A - 15 \, B\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (11 \, A - 15 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (11 \, A - 15 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {-a} \log \left (-\frac {2 \, \sqrt {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 ) - 3 \, a \cos \left (d x + c\right )^{2} - 2 \, a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \, {\left ({\left (95 \, A - 147 \, B\right )} \cos \left (d x + c\right )^{3} + 12 \, {\left (5 \, A - 9 \, B\right )} \cos \left (d x + c\right )^{2} - 4 \, {\left (5 \, A - 3 \, B\right )} \cos \left (d x + c\right ) - 12 \, B\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{120 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\right )}}, -\frac {15 \, \sqrt {2} {\left ({\left (11 \, A - 15 \, B\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (11 \, A - 15 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (11 \, A - 15 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \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 ) + 2 \, {\left ({\left (95 \, A - 147 \, B\right )} \cos \left (d x + c\right )^{3} + 12 \, {\left (5 \, A - 9 \, B\right )} \cos \left (d x + c\right )^{2} - 4 \, {\left (5 \, A - 3 \, B\right )} \cos \left (d x + c\right ) - 12 \, B\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{60 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\right )}}\right ] \]

[In]

integrate(sec(d*x+c)^4*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

[1/120*(15*sqrt(2)*((11*A - 15*B)*cos(d*x + c)^4 + 2*(11*A - 15*B)*cos(d*x + c)^3 + (11*A - 15*B)*cos(d*x + c)
^2)*sqrt(-a)*log(-(2*sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) - 3*a*
cos(d*x + c)^2 - 2*a*cos(d*x + c) + a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) - 4*((95*A - 147*B)*cos(d*x + c)
^3 + 12*(5*A - 9*B)*cos(d*x + c)^2 - 4*(5*A - 3*B)*cos(d*x + c) - 12*B)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)
)*sin(d*x + c))/(a^2*d*cos(d*x + c)^4 + 2*a^2*d*cos(d*x + c)^3 + a^2*d*cos(d*x + c)^2), -1/60*(15*sqrt(2)*((11
*A - 15*B)*cos(d*x + c)^4 + 2*(11*A - 15*B)*cos(d*x + c)^3 + (11*A - 15*B)*cos(d*x + c)^2)*sqrt(a)*arctan(sqrt
(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) + 2*((95*A - 147*B)*cos(d*x +
c)^3 + 12*(5*A - 9*B)*cos(d*x + c)^2 - 4*(5*A - 3*B)*cos(d*x + c) - 12*B)*sqrt((a*cos(d*x + c) + a)/cos(d*x +
c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^4 + 2*a^2*d*cos(d*x + c)^3 + a^2*d*cos(d*x + c)^2)]

Sympy [F]

\[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(sec(d*x+c)**4*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))**(3/2),x)

[Out]

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

Maxima [F]

\[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{4}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(sec(d*x+c)^4*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate((B*sec(d*x + c) + A)*sec(d*x + c)^4/(a*sec(d*x + c) + a)^(3/2), x)

Giac [A] (verification not implemented)

none

Time = 1.65 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.28 \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx=-\frac {\frac {15 \, \sqrt {2} {\left (11 \, A - 15 \, B\right )} \log \left ({\left | -\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {-a} a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {{\left ({\left ({\left (\frac {15 \, \sqrt {2} {\left (A a^{3} - B a^{3}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {\sqrt {2} {\left (245 \, A a^{3} - 381 \, B a^{3}\right )}}{a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {5 \, \sqrt {2} {\left (73 \, A a^{3} - 105 \, B a^{3}\right )}}{a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {15 \, \sqrt {2} {\left (9 \, A a^{3} - 17 \, B a^{3}\right )}}{a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}}{60 \, d} \]

[In]

integrate(sec(d*x+c)^4*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

-1/60*(15*sqrt(2)*(11*A - 15*B)*log(abs(-sqrt(-a)*tan(1/2*d*x + 1/2*c) + sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))
/(sqrt(-a)*a*sgn(cos(d*x + c))) - (((15*sqrt(2)*(A*a^3 - B*a^3)*tan(1/2*d*x + 1/2*c)^2/(a^2*sgn(cos(d*x + c)))
- sqrt(2)*(245*A*a^3 - 381*B*a^3)/(a^2*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 5*sqrt(2)*(73*A*a^3 - 105
*B*a^3)/(a^2*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 - 15*sqrt(2)*(9*A*a^3 - 17*B*a^3)/(a^2*sgn(cos(d*x + c
))))*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^2*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/d

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^4\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]

[In]

int((A + B/cos(c + d*x))/(cos(c + d*x)^4*(a + a/cos(c + d*x))^(3/2)),x)

[Out]

int((A + B/cos(c + d*x))/(cos(c + d*x)^4*(a + a/cos(c + d*x))^(3/2)), x)