3.9.44 \(\int \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2} \, dx\) [844]

Optimal. Leaf size=187 \[ \frac {2 \left (a^2-b^2\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{3 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {8 b \sqrt {\cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{3 d \sqrt {\frac {b+a \cos (c+d x)}{a+b}}}+\frac {2 a \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{3 d} \]

[Out]

2/3*(a^2-b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2)*(a/(a+b))^(
1/2))*((b+a*cos(d*x+c))/(a+b))^(1/2)/d/cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^(1/2)+2/3*a*sin(d*x+c)*cos(d*x+c)^(1/
2)*(a+b*sec(d*x+c))^(1/2)/d+8/3*b*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c)
,2^(1/2)*(a/(a+b))^(1/2))*cos(d*x+c)^(1/2)*(a+b*sec(d*x+c))^(1/2)/d/((b+a*cos(d*x+c))/(a+b))^(1/2)

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Rubi [A]
time = 0.33, antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {4349, 3949, 4120, 3941, 2734, 2732, 3943, 2742, 2740} \begin {gather*} \frac {2 \left (a^2-b^2\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{3 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 a \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}{3 d}+\frac {8 b \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{3 d \sqrt {\frac {a \cos (c+d x)+b}{a+b}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a^2 - b^2)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)])/(3*d*Sqrt[Cos[c + d*x
]]*Sqrt[a + b*Sec[c + d*x]]) + (8*b*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c
+ d*x]])/(3*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]) + (2*a*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d
*x])/(3*d)

Rule 2732

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a + b]/d)*EllipticE[(1/2)*(c - Pi/2
+ d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2734

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2740

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*Sqrt[a + b]))*EllipticF[(1/2)*(c - P
i/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2742

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 3941

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

Rule 3943

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[Sqrt[d*C
sc[e + f*x]]*(Sqrt[b + a*Sin[e + f*x]]/Sqrt[a + b*Csc[e + f*x]]), Int[1/Sqrt[b + a*Sin[e + f*x]], x], x] /; Fr
eeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 3949

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(3/2), x_Symbol] :> Simp[a*Cot
[e + f*x]*Sqrt[a + b*Csc[e + f*x]]*((d*Csc[e + f*x])^n/(f*n)), x] + Dist[1/(2*d*n), Int[((d*Csc[e + f*x])^(n +
1)/Sqrt[a + b*Csc[e + f*x]])*Simp[a*b*(2*n - 1) + 2*(b^2*n + a^2*(n + 1))*Csc[e + f*x] + a*b*(2*n + 3)*Csc[e
+ f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1] && IntegersQ[2*n]

Rule 4120

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

Rule 4349

Int[(u_)*((c_.)*sin[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Csc[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[
u, x]

Rubi steps

\begin {align*} \int \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+b \sec (c+d x))^{3/2}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 a \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{3 d}-\frac {1}{3} \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-4 a b-\left (a^2+3 b^2\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx\\ &=\frac {2 a \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{3 d}+\frac {1}{3} \left (4 b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx-\frac {1}{3} \left (\left (-a^2+b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+b \sec (c+d x)}} \, dx\\ &=\frac {2 a \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{3 d}-\frac {\left (\left (-a^2+b^2\right ) \sqrt {b+a \cos (c+d x)}\right ) \int \frac {1}{\sqrt {b+a \cos (c+d x)}} \, dx}{3 \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {\left (4 b \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}\right ) \int \sqrt {b+a \cos (c+d x)} \, dx}{3 \sqrt {b+a \cos (c+d x)}}\\ &=\frac {2 a \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{3 d}-\frac {\left (\left (-a^2+b^2\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}}\right ) \int \frac {1}{\sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}} \, dx}{3 \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {\left (4 b \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}\right ) \int \sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}} \, dx}{3 \sqrt {\frac {b+a \cos (c+d x)}{a+b}}}\\ &=\frac {2 \left (a^2-b^2\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{3 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {8 b \sqrt {\cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{3 d \sqrt {\frac {b+a \cos (c+d x)}{a+b}}}+\frac {2 a \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{3 d}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 4.05, size = 284, normalized size = 1.52 \begin {gather*} \frac {2 \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^{3/2} \left (\frac {1}{2} a (b+a \cos (c+d x)) \sin (2 (c+d x))+\frac {\sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \left (4 i b (a+b) E\left (i \sinh ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {-a+b}{a+b}\right ) \sqrt {\frac {(b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}-i \left (a^2+4 a b+3 b^2\right ) F\left (i \sinh ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {-a+b}{a+b}\right ) \sqrt {\frac {(b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+4 b (b+a \cos (c+d x)) \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \tan \left (\frac {1}{2} (c+d x)\right )\right )}{\sec ^{\frac {3}{2}}(c+d x)}\right )}{3 d (b+a \cos (c+d x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[Cos[c + d*x]]*(a + b*Sec[c + d*x])^(3/2)*((a*(b + a*Cos[c + d*x])*Sin[2*(c + d*x)])/2 + (Sqrt[Cos[(c +
d*x)/2]^2*Sec[c + d*x]]*((4*I)*b*(a + b)*EllipticE[I*ArcSinh[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sqrt[((b +
a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)] - I*(a^2 + 4*a*b + 3*b^2)*EllipticF[I*ArcSinh[Tan[(c + d*x)/2]],
(-a + b)/(a + b)]*Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)] + 4*b*(b + a*Cos[c + d*x])*Sqrt[Sec[
(c + d*x)/2]^2]*Tan[(c + d*x)/2]))/Sec[c + d*x]^(3/2)))/(3*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. \(1208\) vs. \(2(223)=446\).
time = 0.24, size = 1209, normalized size = 6.47

method	result	size
default	\(\text {Expression too large to display}\)	\(1209\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3/d*(4*sin(d*x+c)*cos(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*Ellipti
cE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b-4*sin(d*x+c)*cos(d*x+c)*((b+a*cos(
d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin
(d*x+c),(-(a+b)/(a-b))^(1/2))*b^2+sin(d*x+c)*cos(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+co
s(d*x+c)))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^2-4*sin(d*x+
c)*cos(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticF((-1+cos(d*x+c))
*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b+3*sin(d*x+c)*cos(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x
+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a
-b))^(1/2))*b^2+4*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticE((-1+cos(d*x
+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b*sin(d*x+c)-4*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a
+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1
/2))*b^2*sin(d*x+c)+((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticF((-1+cos(d
*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^2*sin(d*x+c)-4*((b+a*cos(d*x+c))/(1+cos(d*x+c))/
(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^
(1/2))*a*b*sin(d*x+c)+3*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticF((-1+c
os(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*b^2*sin(d*x+c)+cos(d*x+c)^3*((a-b)/(a+b))^(1/2
)*a^2+5*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a*b-((a-b)/(a+b))^(1/2)*a^2*cos(d*x+c)-4*cos(d*x+c)*((a-b)/(a+b))^(1/
2)*a*b+4*cos(d*x+c)*((a-b)/(a+b))^(1/2)*b^2-((a-b)/(a+b))^(1/2)*a*b-4*((a-b)/(a+b))^(1/2)*b^2)*((b+a*cos(d*x+c
))/cos(d*x+c))^(1/2)*cos(d*x+c)^(1/2)/(b+a*cos(d*x+c))/sin(d*x+c)/((a-b)/(a+b))^(1/2)

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

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 1.24, size = 415, normalized size = 2.22 \begin {gather*} \frac {6 \, a^{2} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 12 i \, \sqrt {2} a^{\frac {3}{2}} b {\rm weierstrassZeta}\left (-\frac {4 \, {\left (3 \, a^{2} - 4 \, b^{2}\right )}}{3 \, a^{2}}, \frac {8 \, {\left (9 \, a^{2} b - 8 \, b^{3}\right )}}{27 \, a^{3}}, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (3 \, a^{2} - 4 \, b^{2}\right )}}{3 \, a^{2}}, \frac {8 \, {\left (9 \, a^{2} b - 8 \, b^{3}\right )}}{27 \, a^{3}}, \frac {3 \, a \cos \left (d x + c\right ) + 3 i \, a \sin \left (d x + c\right ) + 2 \, b}{3 \, a}\right )\right ) - 12 i \, \sqrt {2} a^{\frac {3}{2}} b {\rm weierstrassZeta}\left (-\frac {4 \, {\left (3 \, a^{2} - 4 \, b^{2}\right )}}{3 \, a^{2}}, \frac {8 \, {\left (9 \, a^{2} b - 8 \, b^{3}\right )}}{27 \, a^{3}}, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (3 \, a^{2} - 4 \, b^{2}\right )}}{3 \, a^{2}}, \frac {8 \, {\left (9 \, a^{2} b - 8 \, b^{3}\right )}}{27 \, a^{3}}, \frac {3 \, a \cos \left (d x + c\right ) - 3 i \, a \sin \left (d x + c\right ) + 2 \, b}{3 \, a}\right )\right ) + \sqrt {2} {\left (-3 i \, a^{2} - i \, b^{2}\right )} \sqrt {a} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (3 \, a^{2} - 4 \, b^{2}\right )}}{3 \, a^{2}}, \frac {8 \, {\left (9 \, a^{2} b - 8 \, b^{3}\right )}}{27 \, a^{3}}, \frac {3 \, a \cos \left (d x + c\right ) + 3 i \, a \sin \left (d x + c\right ) + 2 \, b}{3 \, a}\right ) + \sqrt {2} {\left (3 i \, a^{2} + i \, b^{2}\right )} \sqrt {a} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (3 \, a^{2} - 4 \, b^{2}\right )}}{3 \, a^{2}}, \frac {8 \, {\left (9 \, a^{2} b - 8 \, b^{3}\right )}}{27 \, a^{3}}, \frac {3 \, a \cos \left (d x + c\right ) - 3 i \, a \sin \left (d x + c\right ) + 2 \, b}{3 \, a}\right )}{9 \, a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*(6*a^2*sqrt((a*cos(d*x + c) + b)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + 12*I*sqrt(2)*a^(3/2)*b*we
ierstrassZeta(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, weierstrassPInverse(-4/3*(3*a^2 - 4*b^2)/a
^2, 8/27*(9*a^2*b - 8*b^3)/a^3, 1/3*(3*a*cos(d*x + c) + 3*I*a*sin(d*x + c) + 2*b)/a)) - 12*I*sqrt(2)*a^(3/2)*b
*weierstrassZeta(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, weierstrassPInverse(-4/3*(3*a^2 - 4*b^2
)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, 1/3*(3*a*cos(d*x + c) - 3*I*a*sin(d*x + c) + 2*b)/a)) + sqrt(2)*(-3*I*a^2 -
I*b^2)*sqrt(a)*weierstrassPInverse(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, 1/3*(3*a*cos(d*x + c
) + 3*I*a*sin(d*x + c) + 2*b)/a) + sqrt(2)*(3*I*a^2 + I*b^2)*sqrt(a)*weierstrassPInverse(-4/3*(3*a^2 - 4*b^2)/
a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, 1/3*(3*a*cos(d*x + c) - 3*I*a*sin(d*x + c) + 2*b)/a))/(a*d)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\cos \left (c+d\,x\right )}^{3/2}\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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