3.6.38 \(\int \sec ^3(c+d x) (a+b \sec (c+d x))^{3/2} \, dx\) [538]

Optimal. Leaf size=342 \[ \frac {4 a (a-b) \sqrt {a+b} \left (3 a^2-41 b^2\right ) \cot (c+d x) E\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{105 b^3 d}+\frac {2 (a-b) \sqrt {a+b} \left (6 a^2+57 a b-25 b^2\right ) \cot (c+d x) F\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{105 b^2 d}-\frac {2 \left (6 a^2-25 b^2\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{105 b d}-\frac {4 a (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{35 b d}+\frac {2 (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{7 b d} \]

[Out]

4/105*a*(a-b)*(3*a^2-41*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^3/d+2/105*(a-b)*(6*a^2+57*a*b-25*b^2)
*cot(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^2/d-4/35*a*(a+b*sec(d*x+c))^(3/2)*tan(d*x+c)/b/d+2/7*(a+b*sec(d*x
+c))^(5/2)*tan(d*x+c)/b/d-2/105*(6*a^2-25*b^2)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b/d

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Rubi [A]
time = 0.41, antiderivative size = 342, 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 = {3925, 4087, 4090, 3917, 4089} \begin {gather*} \frac {2 (a-b) \sqrt {a+b} \left (6 a^2+57 a b-25 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 )}{105 b^2 d}+\frac {4 a (a-b) \sqrt {a+b} \left (3 a^2-41 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 )}{105 b^3 d}-\frac {2 \left (6 a^2-25 b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{105 b d}+\frac {2 \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}-\frac {4 a \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{35 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a*(a - b)*Sqrt[a + b]*(3*a^2 - 41*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))])/(105*b^3*d) +
(2*(a - b)*Sqrt[a + b]*(6*a^2 + 57*a*b - 25*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))])/(105*b^
2*d) - (2*(6*a^2 - 25*b^2)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(105*b*d) - (4*a*(a + b*Sec[c + d*x])^(3/2)*
Tan[c + d*x])/(35*b*d) + (2*(a + b*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(7*b*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 3925

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-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*
(b*(m + 1) - a*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]

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 ^3(c+d x) (a+b \sec (c+d x))^{3/2} \, dx &=\frac {2 (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{7 b d}+\frac {2 \int \sec (c+d x) \left (\frac {5 b}{2}-a \sec (c+d x)\right ) (a+b \sec (c+d x))^{3/2} \, dx}{7 b}\\ &=-\frac {4 a (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{35 b d}+\frac {2 (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{7 b d}+\frac {4 \int \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (\frac {19 a b}{4}-\frac {1}{4} \left (6 a^2-25 b^2\right ) \sec (c+d x)\right ) \, dx}{35 b}\\ &=-\frac {2 \left (6 a^2-25 b^2\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{105 b d}-\frac {4 a (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{35 b d}+\frac {2 (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{7 b d}+\frac {8 \int \frac {\sec (c+d x) \left (\frac {1}{8} b \left (51 a^2+25 b^2\right )-\frac {1}{4} a \left (3 a^2-41 b^2\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx}{105 b}\\ &=-\frac {2 \left (6 a^2-25 b^2\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{105 b d}-\frac {4 a (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{35 b d}+\frac {2 (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{7 b d}-\frac {\left (2 a \left (3 a^2-41 b^2\right )\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx}{105 b}+\frac {\left ((a-b) \left (6 a^2+57 a b-25 b^2\right )\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{105 b}\\ &=\frac {4 a (a-b) \sqrt {a+b} \left (3 a^2-41 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}}}{105 b^3 d}+\frac {2 (a-b) \sqrt {a+b} \left (6 a^2+57 a b-25 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}}}{105 b^2 d}-\frac {2 \left (6 a^2-25 b^2\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{105 b d}-\frac {4 a (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{35 b d}+\frac {2 (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{7 b d}\\ \end {align*}

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Mathematica [A]
time = 14.22, size = 471, normalized size = 1.38 \begin {gather*} \frac {4 \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} (a+b \sec (c+d x))^{3/2} \left (2 a \left (3 a^3+3 a^2 b-41 a b^2-41 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 )+b \left (-6 a^3+51 a^2 b+82 a b^2+25 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^2-41 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 )}{105 b^2 d (b+a \cos (c+d x))^2 \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \sec ^{\frac {3}{2}}(c+d x)}+\frac {\cos (c+d x) (a+b \sec (c+d x))^{3/2} \left (-\frac {4 a \left (3 a^2-41 b^2\right ) \sin (c+d x)}{105 b^2}+\frac {2 \sec (c+d x) \left (3 a^2 \sin (c+d x)+25 b^2 \sin (c+d x)\right )}{105 b}+\frac {16}{35} a \sec (c+d x) \tan (c+d x)+\frac {2}{7} b \sec ^2(c+d x) \tan (c+d x)\right )}{d (b+a \cos (c+d x))} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(4*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(a + b*Sec[c + d*x])^(3/2)*(2*a*(3*a^3 + 3*a^2*b - 41*a*b^2 - 41*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[ArcSi
n[Tan[(c + d*x)/2]], (a - b)/(a + b)] + b*(-6*a^3 + 51*a^2*b + 82*a*b^2 + 25*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 - 41*b^2)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(105*b^2
*d*(b + a*Cos[c + d*x])^2*Sqrt[Sec[(c + d*x)/2]^2]*Sec[c + d*x]^(3/2)) + (Cos[c + d*x]*(a + b*Sec[c + d*x])^(3
/2)*((-4*a*(3*a^2 - 41*b^2)*Sin[c + d*x])/(105*b^2) + (2*Sec[c + d*x]*(3*a^2*Sin[c + d*x] + 25*b^2*Sin[c + d*x
]))/(105*b) + (16*a*Sec[c + d*x]*Tan[c + d*x])/35 + (2*b*Sec[c + d*x]^2*Tan[c + d*x])/7))/(d*(b + a*Cos[c + d*
x]))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1851\) vs. \(2(308)=616\).
time = 0.36, size = 1852, normalized size = 5.42

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)^3*(a+b*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-2/105/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*(6*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+25*cos(d*x+c)^4*b^4-10*cos(d*x+c)^2*b^4-6*cos(d*x+c)^5*a^4+6*cos(d*x+c)^4*a^4-15*
b^4+3*cos(d*x+c)^5*a^3*b+82*cos(d*x+c)^5*a^2*b^2+25*cos(d*x+c)^5*a*b^3-6*cos(d*x+c)^4*a^3*b-55*cos(d*x+c)^4*a^
2*b^2+82*cos(d*x+c)^4*a*b^3+3*cos(d*x+c)^3*a^3*b-68*cos(d*x+c)^3*a*b^3-27*cos(d*x+c)^2*a^2*b^2-39*cos(d*x+c)*a
*b^3+25*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))*b^4+6*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+25*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+6*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-82*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^2-82*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^3-6*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+51*sin(d*x+c)*c
os(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+82*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+6
*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)*Ellip
ticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^3*b-82*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^2-82*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^3-6*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^3*b+51*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+82*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))/s
in(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^2

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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+b*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

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

[Out]

integral((b*sec(d*x + c)^4 + a*sec(d*x + c)^3)*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 {3}{2}} \sec ^{3}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3*(a+b*sec(d*x+c))**(3/2),x)

[Out]

Integral((a + b*sec(c + d*x))**(3/2)*sec(c + d*x)**3, 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)^3*(a+b*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((b*sec(d*x + c) + a)^(3/2)*sec(d*x + c)^3, 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 )}^{3/2}}{{\cos \left (c+d\,x\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/cos(c + d*x))^(3/2)/cos(c + d*x)^3,x)

[Out]

int((a + b/cos(c + d*x))^(3/2)/cos(c + d*x)^3, x)

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