3.6.46 \(\int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} \, dx\) [546]

Optimal. Leaf size=333 \[ -\frac {2 a (a-b) \sqrt {a+b} \left (3 a^2+29 b^2\right ) \cot (c+d x) E\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{21 b^2 d}-\frac {2 (a-b) \sqrt {a+b} \left (3 a^2-24 a b+5 b^2\right ) \cot (c+d x) F\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{21 b d}+\frac {2 \left (3 a^2+5 b^2\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{21 d}+\frac {2 a (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{7 d}+\frac {2 (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{7 d} \]

[Out]

-2/21*a*(a-b)*(3*a^2+29*b^2)*cot(d*x+c)*EllipticE((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(a+b
)^(1/2)*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/b^2/d-2/21*(a-b)*(3*a^2-24*a*b+5*b^2)*c
ot(d*x+c)*EllipticF((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(a+b)^(1/2)*(b*(1-sec(d*x+c))/(a+b
))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/b/d+2/7*a*(a+b*sec(d*x+c))^(3/2)*tan(d*x+c)/d+2/7*(a+b*sec(d*x+c))^(5
/2)*tan(d*x+c)/d+2/21*(3*a^2+5*b^2)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/d

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Rubi [A]
time = 0.39, antiderivative size = 333, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3920, 4087, 4090, 3917, 4089} \begin {gather*} -\frac {2 (a-b) \sqrt {a+b} \left (3 a^2-24 a b+5 b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{21 b d}-\frac {2 a (a-b) \sqrt {a+b} \left (3 a^2+29 b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{21 b^2 d}+\frac {2 \left (3 a^2+5 b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{21 d}+\frac {2 \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}+\frac {2 a \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{7 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*(a - b)*Sqrt[a + b]*(3*a^2 + 29*b^2)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]]
, (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(21*b^2*d) -
(2*(a - b)*Sqrt[a + b]*(3*a^2 - 24*a*b + 5*b^2)*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a
+ b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(21*b*d)
+ (2*(3*a^2 + 5*b^2)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(21*d) + (2*a*(a + b*Sec[c + d*x])^(3/2)*Tan[c +
d*x])/(7*d) + (2*(a + b*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(7*d)

Rule 3917

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(Rt[a + b, 2]/(b*
f*Cot[e + f*x]))*Sqrt[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]*EllipticF[ArcSin
[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 3920

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

Rule 4087

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[1/(m + 1), Int[Csc[e + f
*x]*(a + b*Csc[e + f*x])^(m - 1)*Simp[b*B*m + a*A*(m + 1) + (a*B*m + A*b*(m + 1))*Csc[e + f*x], x], x], x] /;
FreeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0]

Rule 4089

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Simp[-2*(A*b - a*B)*Rt[a + b*(B/A), 2]*Sqrt[b*((1 - Csc[e + f*x])/(a + b))]*(Sqrt[(-b)*((1 + C
sc[e + f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b*(B/A), 2]], (a
*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]

Rule 4090

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

Rubi steps

\begin {align*} \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} \, dx &=\frac {2 (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}+\frac {5}{7} \int \sec (c+d x) (b+a \sec (c+d x)) (a+b \sec (c+d x))^{3/2} \, dx\\ &=\frac {2 a (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{7 d}+\frac {2 (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}+\frac {2}{7} \int \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (4 a b+\frac {1}{2} \left (3 a^2+5 b^2\right ) \sec (c+d x)\right ) \, dx\\ &=\frac {2 \left (3 a^2+5 b^2\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{21 d}+\frac {2 a (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{7 d}+\frac {2 (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}+\frac {4}{21} \int \frac {\sec (c+d x) \left (\frac {1}{4} b \left (27 a^2+5 b^2\right )+\frac {1}{4} a \left (3 a^2+29 b^2\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx\\ &=\frac {2 \left (3 a^2+5 b^2\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{21 d}+\frac {2 a (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{7 d}+\frac {2 (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}-\frac {1}{21} \left ((a-b) \left (3 a^2-24 a b+5 b^2\right )\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx+\frac {1}{21} \left (a \left (3 a^2+29 b^2\right )\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx\\ &=-\frac {2 a (a-b) \sqrt {a+b} \left (3 a^2+29 b^2\right ) \cot (c+d x) E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{21 b^2 d}-\frac {2 (a-b) \sqrt {a+b} \left (3 a^2-24 a b+5 b^2\right ) \cot (c+d x) F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{21 b d}+\frac {2 \left (3 a^2+5 b^2\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{21 d}+\frac {2 a (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{7 d}+\frac {2 (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}\\ \end {align*}

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Mathematica [A]
time = 13.93, size = 474, normalized size = 1.42 \begin {gather*} -\frac {2 \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} (a+b \sec (c+d x))^{5/2} \left (2 a \left (3 a^3+3 a^2 b+29 a b^2+29 b^3\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\text {ArcSin}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )-2 b \left (3 a^3+27 a^2 b+29 a b^2+5 b^3\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} F\left (\text {ArcSin}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )+a \left (3 a^2+29 b^2\right ) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{21 b d (b+a \cos (c+d x))^3 \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \sec ^{\frac {5}{2}}(c+d x)}+\frac {\cos ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (\frac {2 a \left (3 a^2+29 b^2\right ) \sin (c+d x)}{21 b}+\frac {2}{21} \sec (c+d x) \left (9 a^2 \sin (c+d x)+5 b^2 \sin (c+d x)\right )+\frac {6}{7} a b \sec (c+d x) \tan (c+d x)+\frac {2}{7} b^2 \sec ^2(c+d x) \tan (c+d x)\right )}{d (b+a \cos (c+d x))^2} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(a + b*Sec[c + d*x])^(5/2)*(2*a*(3*a^3 + 3*a^2*b + 29*a*b^2 + 29*b^3
)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcS
in[Tan[(c + d*x)/2]], (a - b)/(a + b)] - 2*b*(3*a^3 + 27*a^2*b + 29*a*b^2 + 5*b^3)*Sqrt[Cos[c + d*x]/(1 + Cos[
c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)
/(a + b)] + a*(3*a^2 + 29*b^2)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(21*b*d
*(b + a*Cos[c + d*x])^3*Sqrt[Sec[(c + d*x)/2]^2]*Sec[c + d*x]^(5/2)) + (Cos[c + d*x]^2*(a + b*Sec[c + d*x])^(5
/2)*((2*a*(3*a^2 + 29*b^2)*Sin[c + d*x])/(21*b) + (2*Sec[c + d*x]*(9*a^2*Sin[c + d*x] + 5*b^2*Sin[c + d*x]))/2
1 + (6*a*b*Sec[c + d*x]*Tan[c + d*x])/7 + (2*b^2*Sec[c + d*x]^2*Tan[c + d*x])/7))/(d*(b + a*Cos[c + d*x])^2)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1851\) vs. \(2(299)=598\).
time = 0.35, size = 1852, normalized size = 5.56

method	result	size
default	\(\text {Expression too large to display}\)	\(1852\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*(a+b*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-2/21/d*(1+cos(d*x+c))^2*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))^2*(-3*sin(d*x+c)*cos(d*x+c)^4*(co
s(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x
+c),((a-b)/(a+b))^(1/2))*a^4+5*cos(d*x+c)^4*b^4-2*cos(d*x+c)^2*b^4+3*cos(d*x+c)^5*a^4-3*cos(d*x+c)^4*a^4-3*b^4
+9*cos(d*x+c)^5*a^3*b+29*cos(d*x+c)^5*a^2*b^2+5*cos(d*x+c)^5*a*b^3+3*cos(d*x+c)^4*a^3*b-11*cos(d*x+c)^4*a^2*b^
2+29*cos(d*x+c)^4*a*b^3-12*cos(d*x+c)^3*a^3*b-22*cos(d*x+c)^3*a*b^3-18*cos(d*x+c)^2*a^2*b^2-12*cos(d*x+c)*a*b^
3+5*sin(d*x+c)*cos(d*x+c)^4*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*El
lipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^4-3*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c
)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/
2))*a^4+5*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1
/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^4-3*sin(d*x+c)*cos(d*x+c)^4*(cos(d*x+c)/(1+cos
(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b
))^(1/2))*a^3*b-29*sin(d*x+c)*cos(d*x+c)^4*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/
(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b^2-29*sin(d*x+c)*cos(d*x+c)^4*(cos
(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+
c),((a-b)/(a+b))^(1/2))*a*b^3+3*sin(d*x+c)*cos(d*x+c)^4*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1
+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^3*b+27*sin(d*x+c)*cos(d*
x+c)^4*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c
))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b^2+29*sin(d*x+c)*cos(d*x+c)^4*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*
cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^3-3*sin(
d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE(
(-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^3*b-29*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(
1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a
^2*b^2-29*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1
/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^3+3*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(1+c
os(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a
+b))^(1/2))*a^3*b+27*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c)
)/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b^2+29*sin(d*x+c)*cos(d*x+c)^3*(c
os(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*
x+c),((a-b)/(a+b))^(1/2))*a*b^3)/(b+a*cos(d*x+c))/cos(d*x+c)^3/sin(d*x+c)^5/b

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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)^2*(a+b*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*sec(d*x + c) + a)^(5/2)*sec(d*x + c)^2, x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*sec(d*x + c)^4 + 2*a*b*sec(d*x + c)^3 + a^2*sec(d*x + c)^2)*sqrt(b*sec(d*x + c) + a), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}} \sec ^{2}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*(a+b*sec(d*x+c))**(5/2),x)

[Out]

Integral((a + b*sec(c + d*x))**(5/2)*sec(c + d*x)**2, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*sec(d*x + c) + a)^(5/2)*sec(d*x + c)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/cos(c + d*x))^(5/2)/cos(c + d*x)^2,x)

[Out]

int((a + b/cos(c + d*x))^(5/2)/cos(c + d*x)^2, x)

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