3.262 \(\int \frac {\sec ^{\frac {7}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=246 \[ -\frac {(43 A-115 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(2 A-5 B) \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{a^{5/2} d}-\frac {(11 A-35 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{16 a^2 d \sqrt {a \sec (c+d x)+a}}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}+\frac {(7 A-15 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}} \]

[Out]

(2*A-5*B)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(5/2)/d+1/4*(A-B)*sec(d*x+c)^(7/2)*sin(d*x+c)/d
/(a+a*sec(d*x+c))^(5/2)+1/16*(7*A-15*B)*sec(d*x+c)^(5/2)*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^(3/2)-1/32*(43*A-115*
B)*arctanh(1/2*sin(d*x+c)*a^(1/2)*sec(d*x+c)^(1/2)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))/a^(5/2)/d*2^(1/2)-1/16*(11*
A-35*B)*sec(d*x+c)^(3/2)*sin(d*x+c)/a^2/d/(a+a*sec(d*x+c))^(1/2)

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Rubi [A]  time = 0.82, antiderivative size = 246, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4019, 4021, 4023, 3808, 206, 3801, 215} \[ -\frac {(11 A-35 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{16 a^2 d \sqrt {a \sec (c+d x)+a}}-\frac {(43 A-115 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(2 A-5 B) \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{a^{5/2} d}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}+\frac {(7 A-15 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^(7/2)*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^(5/2),x]

[Out]

((2*A - 5*B)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(a^(5/2)*d) - ((43*A - 115*B)*ArcTanh[(
Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) + ((A - B
)*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(4*d*(a + a*Sec[c + d*x])^(5/2)) + ((7*A - 15*B)*Sec[c + d*x]^(5/2)*Sin[c +
d*x])/(16*a*d*(a + a*Sec[c + d*x])^(3/2)) - ((11*A - 35*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(16*a^2*d*Sqrt[a
+ a*Sec[c + d*x]])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 3801

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

Rule 3808

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

Rule 4019

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 4021

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 4023

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

Rubi steps

\begin {align*} \int \frac {\sec ^{\frac {7}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx &=\frac {(A-B) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {\int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (\frac {5}{2} a (A-B)-a (A-5 B) \sec (c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=\frac {(A-B) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {(7 A-15 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {\int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (\frac {3}{4} a^2 (7 A-15 B)-\frac {1}{2} a^2 (11 A-35 B) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{8 a^4}\\ &=\frac {(A-B) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {(7 A-15 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac {(11 A-35 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}}+\frac {\int \frac {\sqrt {\sec (c+d x)} \left (-\frac {1}{4} a^3 (11 A-35 B)+4 a^3 (2 A-5 B) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{8 a^5}\\ &=\frac {(A-B) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {(7 A-15 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac {(11 A-35 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}}-\frac {(43 A-115 B) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+a \sec (c+d x)}} \, dx}{32 a^2}+\frac {(2 A-5 B) \int \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \, dx}{2 a^3}\\ &=\frac {(A-B) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {(7 A-15 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac {(11 A-35 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}}+\frac {(43 A-115 B) \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{16 a^2 d}-\frac {(2 A-5 B) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{a}}} \, dx,x,-\frac {a \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^3 d}\\ &=\frac {(2 A-5 B) \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^{5/2} d}-\frac {(43 A-115 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(A-B) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {(7 A-15 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac {(11 A-35 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}}\\ \end {align*}

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Mathematica [B]  time = 6.17, size = 941, normalized size = 3.83 \[ \frac {7 B (\sec (c+d x)+1) \sin (c+d x) \sec ^{\frac {11}{2}}(c+d x)}{16 d (a (\sec (c+d x)+1))^{5/2}}-\frac {B \sin (c+d x) \sec ^{\frac {11}{2}}(c+d x)}{4 d (a (\sec (c+d x)+1))^{5/2}}-\frac {7 B (\sec (c+d x)+1)^2 \sin (c+d x) \sec ^{\frac {9}{2}}(c+d x)}{16 d (a (\sec (c+d x)+1))^{5/2}}+\frac {3 A (\sec (c+d x)+1) \sin (c+d x) \sec ^{\frac {9}{2}}(c+d x)}{16 d (a (\sec (c+d x)+1))^{5/2}}-\frac {A \sin (c+d x) \sec ^{\frac {9}{2}}(c+d x)}{4 d (a (\sec (c+d x)+1))^{5/2}}-\frac {3 A (\sec (c+d x)+1)^2 \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{16 d (a (\sec (c+d x)+1))^{5/2}}+\frac {11 B (\sec (c+d x)+1)^2 \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{16 d (a (\sec (c+d x)+1))^{5/2}}+\frac {7 A (\sec (c+d x)+1)^2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{16 d (a (\sec (c+d x)+1))^{5/2}}-\frac {15 B (\sec (c+d x)+1)^2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{16 d (a (\sec (c+d x)+1))^{5/2}}-\frac {11 A (\sec (c+d x)+1)^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{16 d (a (\sec (c+d x)+1))^{5/2}}+\frac {35 B (\sec (c+d x)+1)^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{16 d (a (\sec (c+d x)+1))^{5/2}}-\frac {11 A \sin ^{-1}\left (\sqrt {1-\sec (c+d x)}\right ) (\sec (c+d x)+1)^2 \tan (c+d x)}{16 d \sqrt {1-\sec (c+d x)} (a (\sec (c+d x)+1))^{5/2}}+\frac {35 B \sin ^{-1}\left (\sqrt {1-\sec (c+d x)}\right ) (\sec (c+d x)+1)^2 \tan (c+d x)}{16 d \sqrt {1-\sec (c+d x)} (a (\sec (c+d x)+1))^{5/2}}-\frac {43 A \sin ^{-1}\left (\sqrt {\sec (c+d x)}\right ) (\sec (c+d x)+1)^2 \tan (c+d x)}{16 d \sqrt {1-\sec (c+d x)} (a (\sec (c+d x)+1))^{5/2}}+\frac {115 B \sin ^{-1}\left (\sqrt {\sec (c+d x)}\right ) (\sec (c+d x)+1)^2 \tan (c+d x)}{16 d \sqrt {1-\sec (c+d x)} (a (\sec (c+d x)+1))^{5/2}}+\frac {43 A \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) (\sec (c+d x)+1)^2 \tan (c+d x)}{16 \sqrt {2} d \sqrt {1-\sec (c+d x)} (a (\sec (c+d x)+1))^{5/2}}-\frac {115 B \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) (\sec (c+d x)+1)^2 \tan (c+d x)}{16 \sqrt {2} d \sqrt {1-\sec (c+d x)} (a (\sec (c+d x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]^(7/2)*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^(5/2),x]

[Out]

-1/4*(A*Sec[c + d*x]^(9/2)*Sin[c + d*x])/(d*(a*(1 + Sec[c + d*x]))^(5/2)) - (B*Sec[c + d*x]^(11/2)*Sin[c + d*x
])/(4*d*(a*(1 + Sec[c + d*x]))^(5/2)) + (3*A*Sec[c + d*x]^(9/2)*(1 + Sec[c + d*x])*Sin[c + d*x])/(16*d*(a*(1 +
Sec[c + d*x]))^(5/2)) + (7*B*Sec[c + d*x]^(11/2)*(1 + Sec[c + d*x])*Sin[c + d*x])/(16*d*(a*(1 + Sec[c + d*x])
)^(5/2)) - (11*A*Sec[c + d*x]^(3/2)*(1 + Sec[c + d*x])^2*Sin[c + d*x])/(16*d*(a*(1 + Sec[c + d*x]))^(5/2)) + (
35*B*Sec[c + d*x]^(3/2)*(1 + Sec[c + d*x])^2*Sin[c + d*x])/(16*d*(a*(1 + Sec[c + d*x]))^(5/2)) + (7*A*Sec[c +
d*x]^(5/2)*(1 + Sec[c + d*x])^2*Sin[c + d*x])/(16*d*(a*(1 + Sec[c + d*x]))^(5/2)) - (15*B*Sec[c + d*x]^(5/2)*(
1 + Sec[c + d*x])^2*Sin[c + d*x])/(16*d*(a*(1 + Sec[c + d*x]))^(5/2)) - (3*A*Sec[c + d*x]^(7/2)*(1 + Sec[c + d
*x])^2*Sin[c + d*x])/(16*d*(a*(1 + Sec[c + d*x]))^(5/2)) + (11*B*Sec[c + d*x]^(7/2)*(1 + Sec[c + d*x])^2*Sin[c
+ d*x])/(16*d*(a*(1 + Sec[c + d*x]))^(5/2)) - (7*B*Sec[c + d*x]^(9/2)*(1 + Sec[c + d*x])^2*Sin[c + d*x])/(16*
d*(a*(1 + Sec[c + d*x]))^(5/2)) - (11*A*ArcSin[Sqrt[1 - Sec[c + d*x]]]*(1 + Sec[c + d*x])^2*Tan[c + d*x])/(16*
d*Sqrt[1 - Sec[c + d*x]]*(a*(1 + Sec[c + d*x]))^(5/2)) + (35*B*ArcSin[Sqrt[1 - Sec[c + d*x]]]*(1 + Sec[c + d*x
])^2*Tan[c + d*x])/(16*d*Sqrt[1 - Sec[c + d*x]]*(a*(1 + Sec[c + d*x]))^(5/2)) - (43*A*ArcSin[Sqrt[Sec[c + d*x]
]]*(1 + Sec[c + d*x])^2*Tan[c + d*x])/(16*d*Sqrt[1 - Sec[c + d*x]]*(a*(1 + Sec[c + d*x]))^(5/2)) + (115*B*ArcS
in[Sqrt[Sec[c + d*x]]]*(1 + Sec[c + d*x])^2*Tan[c + d*x])/(16*d*Sqrt[1 - Sec[c + d*x]]*(a*(1 + Sec[c + d*x]))^
(5/2)) + (43*A*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*(1 + Sec[c + d*x])^2*Tan[c + d*x])/
(16*Sqrt[2]*d*Sqrt[1 - Sec[c + d*x]]*(a*(1 + Sec[c + d*x]))^(5/2)) - (115*B*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]]
)/Sqrt[1 - Sec[c + d*x]]]*(1 + Sec[c + d*x])^2*Tan[c + d*x])/(16*Sqrt[2]*d*Sqrt[1 - Sec[c + d*x]]*(a*(1 + Sec[
c + d*x]))^(5/2))

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fricas [A]  time = 0.72, size = 803, normalized size = 3.26 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(7/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

[-1/64*(sqrt(2)*((43*A - 115*B)*cos(d*x + c)^3 + 3*(43*A - 115*B)*cos(d*x + c)^2 + 3*(43*A - 115*B)*cos(d*x +
c) + 43*A - 115*B)*sqrt(a)*log(-(a*cos(d*x + c)^2 - 2*sqrt(2)*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*
sqrt(cos(d*x + c))*sin(d*x + c) - 2*a*cos(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 16*((2*A -
5*B)*cos(d*x + c)^3 + 3*(2*A - 5*B)*cos(d*x + c)^2 + 3*(2*A - 5*B)*cos(d*x + c) + 2*A - 5*B)*sqrt(a)*log((a*co
s(d*x + c)^3 - 7*a*cos(d*x + c)^2 + 4*(cos(d*x + c)^2 - 2*cos(d*x + c))*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(
d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)) + 4*((11*A - 35*B)*cos(d*x
+ c)^2 + 5*(3*A - 11*B)*cos(d*x + c) - 16*B)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*
x + c)))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d), 1/32*(sqrt(2)*((43*A
- 115*B)*cos(d*x + c)^3 + 3*(43*A - 115*B)*cos(d*x + c)^2 + 3*(43*A - 115*B)*cos(d*x + c) + 43*A - 115*B)*sqrt
(-a)*arctan(sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))/(a*sin(d*x + c))) + 16
*((2*A - 5*B)*cos(d*x + c)^3 + 3*(2*A - 5*B)*cos(d*x + c)^2 + 3*(2*A - 5*B)*cos(d*x + c) + 2*A - 5*B)*sqrt(-a)
*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 -
a*cos(d*x + c) - 2*a)) - 2*((11*A - 35*B)*cos(d*x + c)^2 + 5*(3*A - 11*B)*cos(d*x + c) - 16*B)*sqrt((a*cos(d*
x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*
a^3*d*cos(d*x + c) + a^3*d)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {7}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(7/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((B*sec(d*x + c) + A)*sec(d*x + c)^(7/2)/(a*sec(d*x + c) + a)^(5/2), x)

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maple [B]  time = 2.64, size = 831, normalized size = 3.38 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^(7/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(5/2),x)

[Out]

-1/16/d*(1/cos(d*x+c))^(7/2)*cos(d*x+c)^3*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))^2*(16*A*2^(1/2)*
arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1-sin(d*x+c))*2^(1/2))*sin(d*x+c)*cos(d*x+c)^2-16*A*2^(1/2)*a
rctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1+sin(d*x+c))*2^(1/2))*sin(d*x+c)*cos(d*x+c)^2-40*B*2^(1/2)*ar
ctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1-sin(d*x+c))*2^(1/2))*sin(d*x+c)*cos(d*x+c)^2+40*B*2^(1/2)*arc
tan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1+sin(d*x+c))*2^(1/2))*sin(d*x+c)*cos(d*x+c)^2+16*A*cos(d*x+c)*s
in(d*x+c)*2^(1/2)*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1-sin(d*x+c))*2^(1/2))-16*A*cos(d*x+c)*sin(
d*x+c)*2^(1/2)*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1+sin(d*x+c))*2^(1/2))+43*A*arctan(1/2*sin(d*x
+c)*(-2/(1+cos(d*x+c)))^(1/2))*sin(d*x+c)*cos(d*x+c)^2-11*A*(-2/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^3-40*B*cos(d*
x+c)*sin(d*x+c)*2^(1/2)*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1-sin(d*x+c))*2^(1/2))+40*B*cos(d*x+c
)*sin(d*x+c)*2^(1/2)*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1+sin(d*x+c))*2^(1/2))-115*B*arctan(1/2*
sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2))*sin(d*x+c)*cos(d*x+c)^2+35*B*(-2/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^3+43*A
*cos(d*x+c)*sin(d*x+c)*arctan(1/2*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2))-4*A*(-2/(1+cos(d*x+c)))^(1/2)*cos(d*x+
c)^2-115*B*cos(d*x+c)*sin(d*x+c)*arctan(1/2*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2))+20*B*(-2/(1+cos(d*x+c)))^(1/
2)*cos(d*x+c)^2+15*A*(-2/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)-39*B*(-2/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)-16*B*(-2/(
1+cos(d*x+c)))^(1/2))/(-2/(1+cos(d*x+c)))^(1/2)/sin(d*x+c)^5/a^3

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(7/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B/cos(c + d*x))*(1/cos(c + d*x))^(7/2))/(a + a/cos(c + d*x))^(5/2),x)

[Out]

int(((A + B/cos(c + d*x))*(1/cos(c + d*x))^(7/2))/(a + a/cos(c + d*x))^(5/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**(7/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))**(5/2),x)

[Out]

Timed out

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