3.271 \(\int \sqrt {d \cos (a+b x)} (c \sin (a+b x))^{3/2} \, dx\)

Optimal. Leaf size=320 \[ \frac {c^{3/2} \sqrt {d} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {c} \sqrt {d \cos (a+b x)}}{\sqrt {d} \sqrt {c \sin (a+b x)}}\right )}{4 \sqrt {2} b}-\frac {c^{3/2} \sqrt {d} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {d \cos (a+b x)}}{\sqrt {d} \sqrt {c \sin (a+b x)}}+1\right )}{4 \sqrt {2} b}-\frac {c^{3/2} \sqrt {d} \log \left (-\frac {\sqrt {2} \sqrt {c} \sqrt {d \cos (a+b x)}}{\sqrt {c \sin (a+b x)}}+\sqrt {d} \cot (a+b x)+\sqrt {d}\right )}{8 \sqrt {2} b}+\frac {c^{3/2} \sqrt {d} \log \left (\frac {\sqrt {2} \sqrt {c} \sqrt {d \cos (a+b x)}}{\sqrt {c \sin (a+b x)}}+\sqrt {d} \cot (a+b x)+\sqrt {d}\right )}{8 \sqrt {2} b}-\frac {c \sqrt {c \sin (a+b x)} (d \cos (a+b x))^{3/2}}{2 b d} \]

[Out]

-1/8*c^(3/2)*arctan(-1+2^(1/2)*c^(1/2)*(d*cos(b*x+a))^(1/2)/d^(1/2)/(c*sin(b*x+a))^(1/2))*d^(1/2)/b*2^(1/2)-1/
8*c^(3/2)*arctan(1+2^(1/2)*c^(1/2)*(d*cos(b*x+a))^(1/2)/d^(1/2)/(c*sin(b*x+a))^(1/2))*d^(1/2)/b*2^(1/2)-1/16*c
^(3/2)*ln(d^(1/2)+cot(b*x+a)*d^(1/2)-2^(1/2)*c^(1/2)*(d*cos(b*x+a))^(1/2)/(c*sin(b*x+a))^(1/2))*d^(1/2)/b*2^(1
/2)+1/16*c^(3/2)*ln(d^(1/2)+cot(b*x+a)*d^(1/2)+2^(1/2)*c^(1/2)*(d*cos(b*x+a))^(1/2)/(c*sin(b*x+a))^(1/2))*d^(1
/2)/b*2^(1/2)-1/2*c*(d*cos(b*x+a))^(3/2)*(c*sin(b*x+a))^(1/2)/b/d

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Rubi [A]  time = 0.28, antiderivative size = 320, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {2568, 2575, 297, 1162, 617, 204, 1165, 628} \[ \frac {c^{3/2} \sqrt {d} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {c} \sqrt {d \cos (a+b x)}}{\sqrt {d} \sqrt {c \sin (a+b x)}}\right )}{4 \sqrt {2} b}-\frac {c^{3/2} \sqrt {d} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {d \cos (a+b x)}}{\sqrt {d} \sqrt {c \sin (a+b x)}}+1\right )}{4 \sqrt {2} b}-\frac {c^{3/2} \sqrt {d} \log \left (-\frac {\sqrt {2} \sqrt {c} \sqrt {d \cos (a+b x)}}{\sqrt {c \sin (a+b x)}}+\sqrt {d} \cot (a+b x)+\sqrt {d}\right )}{8 \sqrt {2} b}+\frac {c^{3/2} \sqrt {d} \log \left (\frac {\sqrt {2} \sqrt {c} \sqrt {d \cos (a+b x)}}{\sqrt {c \sin (a+b x)}}+\sqrt {d} \cot (a+b x)+\sqrt {d}\right )}{8 \sqrt {2} b}-\frac {c \sqrt {c \sin (a+b x)} (d \cos (a+b x))^{3/2}}{2 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^(3/2)*Sqrt[d]*ArcTan[1 - (Sqrt[2]*Sqrt[c]*Sqrt[d*Cos[a + b*x]])/(Sqrt[d]*Sqrt[c*Sin[a + b*x]])])/(4*Sqrt[2]
*b) - (c^(3/2)*Sqrt[d]*ArcTan[1 + (Sqrt[2]*Sqrt[c]*Sqrt[d*Cos[a + b*x]])/(Sqrt[d]*Sqrt[c*Sin[a + b*x]])])/(4*S
qrt[2]*b) - (c^(3/2)*Sqrt[d]*Log[Sqrt[d] + Sqrt[d]*Cot[a + b*x] - (Sqrt[2]*Sqrt[c]*Sqrt[d*Cos[a + b*x]])/Sqrt[
c*Sin[a + b*x]]])/(8*Sqrt[2]*b) + (c^(3/2)*Sqrt[d]*Log[Sqrt[d] + Sqrt[d]*Cot[a + b*x] + (Sqrt[2]*Sqrt[c]*Sqrt[
d*Cos[a + b*x]])/Sqrt[c*Sin[a + b*x]]])/(8*Sqrt[2]*b) - (c*(d*Cos[a + b*x])^(3/2)*Sqrt[c*Sin[a + b*x]])/(2*b*d
)

Rule 204

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

Rule 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 2568

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(a*(b*Cos[e
+ f*x])^(n + 1)*(a*Sin[e + f*x])^(m - 1))/(b*f*(m + n)), x] + Dist[(a^2*(m - 1))/(m + n), Int[(b*Cos[e + f*x])
^n*(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[2*
m, 2*n]

Rule 2575

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> With[{k = Denomina
tor[m]}, -Dist[(k*a*b)/f, Subst[Int[x^(k*(m + 1) - 1)/(a^2 + b^2*x^(2*k)), x], x, (a*Cos[e + f*x])^(1/k)/(b*Si
n[e + f*x])^(1/k)], x]] /; FreeQ[{a, b, e, f}, x] && EqQ[m + n, 0] && GtQ[m, 0] && LtQ[m, 1]

Rubi steps

\begin {align*} \int \sqrt {d \cos (a+b x)} (c \sin (a+b x))^{3/2} \, dx &=-\frac {c (d \cos (a+b x))^{3/2} \sqrt {c \sin (a+b x)}}{2 b d}+\frac {1}{4} c^2 \int \frac {\sqrt {d \cos (a+b x)}}{\sqrt {c \sin (a+b x)}} \, dx\\ &=-\frac {c (d \cos (a+b x))^{3/2} \sqrt {c \sin (a+b x)}}{2 b d}-\frac {\left (c^3 d\right ) \operatorname {Subst}\left (\int \frac {x^2}{d^2+c^2 x^4} \, dx,x,\frac {\sqrt {d \cos (a+b x)}}{\sqrt {c \sin (a+b x)}}\right )}{2 b}\\ &=-\frac {c (d \cos (a+b x))^{3/2} \sqrt {c \sin (a+b x)}}{2 b d}+\frac {\left (c^2 d\right ) \operatorname {Subst}\left (\int \frac {d-c x^2}{d^2+c^2 x^4} \, dx,x,\frac {\sqrt {d \cos (a+b x)}}{\sqrt {c \sin (a+b x)}}\right )}{4 b}-\frac {\left (c^2 d\right ) \operatorname {Subst}\left (\int \frac {d+c x^2}{d^2+c^2 x^4} \, dx,x,\frac {\sqrt {d \cos (a+b x)}}{\sqrt {c \sin (a+b x)}}\right )}{4 b}\\ &=-\frac {c (d \cos (a+b x))^{3/2} \sqrt {c \sin (a+b x)}}{2 b d}-\frac {\left (c^{3/2} \sqrt {d}\right ) \operatorname {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt {d}}{\sqrt {c}}+2 x}{-\frac {d}{c}-\frac {\sqrt {2} \sqrt {d} x}{\sqrt {c}}-x^2} \, dx,x,\frac {\sqrt {d \cos (a+b x)}}{\sqrt {c \sin (a+b x)}}\right )}{8 \sqrt {2} b}-\frac {\left (c^{3/2} \sqrt {d}\right ) \operatorname {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt {d}}{\sqrt {c}}-2 x}{-\frac {d}{c}+\frac {\sqrt {2} \sqrt {d} x}{\sqrt {c}}-x^2} \, dx,x,\frac {\sqrt {d \cos (a+b x)}}{\sqrt {c \sin (a+b x)}}\right )}{8 \sqrt {2} b}-\frac {(c d) \operatorname {Subst}\left (\int \frac {1}{\frac {d}{c}-\frac {\sqrt {2} \sqrt {d} x}{\sqrt {c}}+x^2} \, dx,x,\frac {\sqrt {d \cos (a+b x)}}{\sqrt {c \sin (a+b x)}}\right )}{8 b}-\frac {(c d) \operatorname {Subst}\left (\int \frac {1}{\frac {d}{c}+\frac {\sqrt {2} \sqrt {d} x}{\sqrt {c}}+x^2} \, dx,x,\frac {\sqrt {d \cos (a+b x)}}{\sqrt {c \sin (a+b x)}}\right )}{8 b}\\ &=-\frac {c^{3/2} \sqrt {d} \log \left (\sqrt {d}+\sqrt {d} \cot (a+b x)-\frac {\sqrt {2} \sqrt {c} \sqrt {d \cos (a+b x)}}{\sqrt {c \sin (a+b x)}}\right )}{8 \sqrt {2} b}+\frac {c^{3/2} \sqrt {d} \log \left (\sqrt {d}+\sqrt {d} \cot (a+b x)+\frac {\sqrt {2} \sqrt {c} \sqrt {d \cos (a+b x)}}{\sqrt {c \sin (a+b x)}}\right )}{8 \sqrt {2} b}-\frac {c (d \cos (a+b x))^{3/2} \sqrt {c \sin (a+b x)}}{2 b d}-\frac {\left (c^{3/2} \sqrt {d}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {c} \sqrt {d \cos (a+b x)}}{\sqrt {d} \sqrt {c \sin (a+b x)}}\right )}{4 \sqrt {2} b}+\frac {\left (c^{3/2} \sqrt {d}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {c} \sqrt {d \cos (a+b x)}}{\sqrt {d} \sqrt {c \sin (a+b x)}}\right )}{4 \sqrt {2} b}\\ &=\frac {c^{3/2} \sqrt {d} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {c} \sqrt {d \cos (a+b x)}}{\sqrt {d} \sqrt {c \sin (a+b x)}}\right )}{4 \sqrt {2} b}-\frac {c^{3/2} \sqrt {d} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {c} \sqrt {d \cos (a+b x)}}{\sqrt {d} \sqrt {c \sin (a+b x)}}\right )}{4 \sqrt {2} b}-\frac {c^{3/2} \sqrt {d} \log \left (\sqrt {d}+\sqrt {d} \cot (a+b x)-\frac {\sqrt {2} \sqrt {c} \sqrt {d \cos (a+b x)}}{\sqrt {c \sin (a+b x)}}\right )}{8 \sqrt {2} b}+\frac {c^{3/2} \sqrt {d} \log \left (\sqrt {d}+\sqrt {d} \cot (a+b x)+\frac {\sqrt {2} \sqrt {c} \sqrt {d \cos (a+b x)}}{\sqrt {c \sin (a+b x)}}\right )}{8 \sqrt {2} b}-\frac {c (d \cos (a+b x))^{3/2} \sqrt {c \sin (a+b x)}}{2 b d}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 67, normalized size = 0.21 \[ \frac {2 \sqrt [4]{\cos ^2(a+b x)} \tan (a+b x) (c \sin (a+b x))^{3/2} \sqrt {d \cos (a+b x)} \, _2F_1\left (\frac {1}{4},\frac {5}{4};\frac {9}{4};\sin ^2(a+b x)\right )}{5 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[d*Cos[a + b*x]]*(Cos[a + b*x]^2)^(1/4)*Hypergeometric2F1[1/4, 5/4, 9/4, Sin[a + b*x]^2]*(c*Sin[a + b*x
])^(3/2)*Tan[a + b*x])/(5*b)

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fricas [B]  time = 27.20, size = 1868, normalized size = 5.84 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(b*x+a))^(1/2)*(c*sin(b*x+a))^(3/2),x, algorithm="fricas")

[Out]

-1/32*(16*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a))*c*cos(b*x + a) + 2*sqrt(2)*(c^6*d^2/b^4)^(1/4)*b*arctan(1/
2*(2*c^10*d^4*cos(b*x + a)*sin(b*x + a) + sqrt(4*sqrt(c^6*d^2/b^4)*b^2*c^7*d^3*cos(b*x + a)*sin(b*x + a) + c^1
0*d^4 + 2*(sqrt(2)*(c^6*d^2/b^4)^(1/4)*b*c^8*d^3*sin(b*x + a) + sqrt(2)*(c^6*d^2/b^4)^(3/4)*b^3*c^5*d^2*cos(b*
x + a))*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a)))*(sqrt(2)*(c^6*d^2/b^4)^(1/4)*b*c^3*d*cos(b*x + a) + sqrt(2)
*(c^6*d^2/b^4)^(3/4)*b^3*sin(b*x + a))*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a)) + (sqrt(2)*(c^6*d^2/b^4)^(1/4
)*b*c^8*d^3*cos(b*x + a) + sqrt(2)*(c^6*d^2/b^4)^(3/4)*b^3*c^5*d^2*sin(b*x + a))*sqrt(d*cos(b*x + a))*sqrt(c*s
in(b*x + a)) - 4*(b^2*c^7*d^3*cos(b*x + a)^4 - b^2*c^7*d^3*cos(b*x + a)^2)*sqrt(c^6*d^2/b^4))/((2*c^10*d^4*cos
(b*x + a)^3 - c^10*d^4*cos(b*x + a))*sin(b*x + a))) + 2*sqrt(2)*(c^6*d^2/b^4)^(1/4)*b*arctan(-1/2*(2*c^10*d^4*
cos(b*x + a)*sin(b*x + a) - sqrt(4*sqrt(c^6*d^2/b^4)*b^2*c^7*d^3*cos(b*x + a)*sin(b*x + a) + c^10*d^4 - 2*(sqr
t(2)*(c^6*d^2/b^4)^(1/4)*b*c^8*d^3*sin(b*x + a) + sqrt(2)*(c^6*d^2/b^4)^(3/4)*b^3*c^5*d^2*cos(b*x + a))*sqrt(d
*cos(b*x + a))*sqrt(c*sin(b*x + a)))*(sqrt(2)*(c^6*d^2/b^4)^(1/4)*b*c^3*d*cos(b*x + a) + sqrt(2)*(c^6*d^2/b^4)
^(3/4)*b^3*sin(b*x + a))*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a)) - (sqrt(2)*(c^6*d^2/b^4)^(1/4)*b*c^8*d^3*co
s(b*x + a) + sqrt(2)*(c^6*d^2/b^4)^(3/4)*b^3*c^5*d^2*sin(b*x + a))*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a)) -
4*(b^2*c^7*d^3*cos(b*x + a)^4 - b^2*c^7*d^3*cos(b*x + a)^2)*sqrt(c^6*d^2/b^4))/((2*c^10*d^4*cos(b*x + a)^3 -
c^10*d^4*cos(b*x + a))*sin(b*x + a))) + 2*sqrt(2)*(c^6*d^2/b^4)^(1/4)*b*arctan(1/2*((sqrt(2)*(c^6*d^2/b^4)^(1/
4)*b*c^8*d^3*cos(b*x + a) - sqrt(2)*(c^6*d^2/b^4)^(3/4)*b^3*c^5*d^2*sin(b*x + a))*sqrt(d*cos(b*x + a))*sqrt(c*
sin(b*x + a)) + sqrt(4*sqrt(c^6*d^2/b^4)*b^2*c^7*d^3*cos(b*x + a)*sin(b*x + a) + c^10*d^4 - 2*(sqrt(2)*(c^6*d^
2/b^4)^(1/4)*b*c^8*d^3*sin(b*x + a) + sqrt(2)*(c^6*d^2/b^4)^(3/4)*b^3*c^5*d^2*cos(b*x + a))*sqrt(d*cos(b*x + a
))*sqrt(c*sin(b*x + a)))*(2*c^5*d^2*cos(b*x + a)*sin(b*x + a) + (sqrt(2)*(c^6*d^2/b^4)^(1/4)*b*c^3*d*cos(b*x +
a) + sqrt(2)*(c^6*d^2/b^4)^(3/4)*b^3*sin(b*x + a))*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a))))/(c^10*d^4*cos(
b*x + a)*sin(b*x + a))) + 2*sqrt(2)*(c^6*d^2/b^4)^(1/4)*b*arctan(1/2*((sqrt(2)*(c^6*d^2/b^4)^(1/4)*b*c^8*d^3*c
os(b*x + a) - sqrt(2)*(c^6*d^2/b^4)^(3/4)*b^3*c^5*d^2*sin(b*x + a))*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a))
- sqrt(4*sqrt(c^6*d^2/b^4)*b^2*c^7*d^3*cos(b*x + a)*sin(b*x + a) + c^10*d^4 + 2*(sqrt(2)*(c^6*d^2/b^4)^(1/4)*b
*c^8*d^3*sin(b*x + a) + sqrt(2)*(c^6*d^2/b^4)^(3/4)*b^3*c^5*d^2*cos(b*x + a))*sqrt(d*cos(b*x + a))*sqrt(c*sin(
b*x + a)))*(2*c^5*d^2*cos(b*x + a)*sin(b*x + a) - (sqrt(2)*(c^6*d^2/b^4)^(1/4)*b*c^3*d*cos(b*x + a) + sqrt(2)*
(c^6*d^2/b^4)^(3/4)*b^3*sin(b*x + a))*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a))))/(c^10*d^4*cos(b*x + a)*sin(b
*x + a))) - sqrt(2)*(c^6*d^2/b^4)^(1/4)*b*log(4*sqrt(c^6*d^2/b^4)*b^2*c^7*d^3*cos(b*x + a)*sin(b*x + a) + c^10
*d^4 + 2*(sqrt(2)*(c^6*d^2/b^4)^(1/4)*b*c^8*d^3*sin(b*x + a) + sqrt(2)*(c^6*d^2/b^4)^(3/4)*b^3*c^5*d^2*cos(b*x
+ a))*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a))) + sqrt(2)*(c^6*d^2/b^4)^(1/4)*b*log(4*sqrt(c^6*d^2/b^4)*b^2*
c^7*d^3*cos(b*x + a)*sin(b*x + a) + c^10*d^4 - 2*(sqrt(2)*(c^6*d^2/b^4)^(1/4)*b*c^8*d^3*sin(b*x + a) + sqrt(2)
*(c^6*d^2/b^4)^(3/4)*b^3*c^5*d^2*cos(b*x + a))*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a))))/b

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {d \cos \left (b x + a\right )} \left (c \sin \left (b x + a\right )\right )^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(b*x+a))^(1/2)*(c*sin(b*x+a))^(3/2),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.11, size = 656, normalized size = 2.05 \[ \frac {\left (c \sin \left (b x +a \right )\right )^{\frac {3}{2}} \sqrt {d \cos \left (b x +a \right )}\, \left (i \sin \left (b x +a \right ) \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \EllipticPi \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-i \sin \left (b x +a \right ) \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \EllipticPi \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-\sin \left (b x +a \right ) \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \EllipticPi \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+2 \sin \left (b x +a \right ) \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \EllipticF \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right )-\sin \left (b x +a \right ) \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \EllipticPi \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-2 \left (\cos ^{3}\left (b x +a \right )\right ) \sqrt {2}+2 \left (\cos ^{2}\left (b x +a \right )\right ) \sqrt {2}\right ) \sqrt {2}}{8 b \sin \left (b x +a \right ) \cos \left (b x +a \right ) \left (-1+\cos \left (b x +a \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*cos(b*x+a))^(1/2)*(c*sin(b*x+a))^(3/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {d \cos \left (b x + a\right )} \left (c \sin \left (b x + a\right )\right )^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(b*x+a))^(1/2)*(c*sin(b*x+a))^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \sqrt {d\,\cos \left (a+b\,x\right )}\,{\left (c\,\sin \left (a+b\,x\right )\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*cos(a + b*x))^(1/2)*(c*sin(a + b*x))^(3/2),x)

[Out]

int((d*cos(a + b*x))^(1/2)*(c*sin(a + b*x))^(3/2), 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((d*cos(b*x+a))**(1/2)*(c*sin(b*x+a))**(3/2),x)

[Out]

Timed out

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