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

Optimal. Leaf size=313 \[ -\frac {c^{3/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {c} \sqrt {d \cos (a+b x)}}{\sqrt {d} \sqrt {c \sin (a+b x)}}\right )}{\sqrt {2} b d^{3/2}}+\frac {c^{3/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {d \cos (a+b x)}}{\sqrt {d} \sqrt {c \sin (a+b x)}}+1\right )}{\sqrt {2} b d^{3/2}}+\frac {c^{3/2} \log \left (-\frac {\sqrt {2} \sqrt {c} \sqrt {d \cos (a+b x)}}{\sqrt {c \sin (a+b x)}}+\sqrt {d} \cot (a+b x)+\sqrt {d}\right )}{2 \sqrt {2} b d^{3/2}}-\frac {c^{3/2} \log \left (\frac {\sqrt {2} \sqrt {c} \sqrt {d \cos (a+b x)}}{\sqrt {c \sin (a+b x)}}+\sqrt {d} \cot (a+b x)+\sqrt {d}\right )}{2 \sqrt {2} b d^{3/2}}+\frac {2 c \sqrt {c \sin (a+b x)}}{b d \sqrt {d \cos (a+b x)}} \]

[Out]

1/2*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))/b/d^(3/2)*2^(1/2)+1/2
*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))/b/d^(3/2)*2^(1/2)+1/4*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))/b/d^(3/2)*2^(1/2
)-1/4*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))/b/d^(3/
2)*2^(1/2)+2*c*(c*sin(b*x+a))^(1/2)/b/d/(d*cos(b*x+a))^(1/2)

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Rubi [A]  time = 0.28, antiderivative size = 313, 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 = {2566, 2575, 297, 1162, 617, 204, 1165, 628} \[ -\frac {c^{3/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {c} \sqrt {d \cos (a+b x)}}{\sqrt {d} \sqrt {c \sin (a+b x)}}\right )}{\sqrt {2} b d^{3/2}}+\frac {c^{3/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {d \cos (a+b x)}}{\sqrt {d} \sqrt {c \sin (a+b x)}}+1\right )}{\sqrt {2} b d^{3/2}}+\frac {c^{3/2} \log \left (-\frac {\sqrt {2} \sqrt {c} \sqrt {d \cos (a+b x)}}{\sqrt {c \sin (a+b x)}}+\sqrt {d} \cot (a+b x)+\sqrt {d}\right )}{2 \sqrt {2} b d^{3/2}}-\frac {c^{3/2} \log \left (\frac {\sqrt {2} \sqrt {c} \sqrt {d \cos (a+b x)}}{\sqrt {c \sin (a+b x)}}+\sqrt {d} \cot (a+b x)+\sqrt {d}\right )}{2 \sqrt {2} b d^{3/2}}+\frac {2 c \sqrt {c \sin (a+b x)}}{b d \sqrt {d \cos (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((c^(3/2)*ArcTan[1 - (Sqrt[2]*Sqrt[c]*Sqrt[d*Cos[a + b*x]])/(Sqrt[d]*Sqrt[c*Sin[a + b*x]])])/(Sqrt[2]*b*d^(3/
2))) + (c^(3/2)*ArcTan[1 + (Sqrt[2]*Sqrt[c]*Sqrt[d*Cos[a + b*x]])/(Sqrt[d]*Sqrt[c*Sin[a + b*x]])])/(Sqrt[2]*b*
d^(3/2)) + (c^(3/2)*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]]])/(2*Sqrt[2]*b*d^(3/2)) - (c^(3/2)*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]]])/(2*Sqrt[2]*b*d^(3/2)) + (2*c*Sqrt[c*Sin[a + b*x]])/(b*d*Sqrt[d*Cos[a + b*x]])

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 2566

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

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

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Mathematica [C]  time = 0.15, size = 67, normalized size = 0.21 \[ \frac {2 \sqrt [4]{\cos ^2(a+b x)} (c \sin (a+b x))^{5/2} \, _2F_1\left (\frac {5}{4},\frac {5}{4};\frac {9}{4};\sin ^2(a+b x)\right )}{5 b c d \sqrt {d \cos (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 27.41, size = 1865, normalized size = 5.96 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*sqrt(2)*b*d^2*(c^6/(b^4*d^6))^(1/4)*arctan(1/2*(2*c^10*cos(b*x + a)*sin(b*x + a) + sqrt(4*b^2*c^7*d^3*s
qrt(c^6/(b^4*d^6))*cos(b*x + a)*sin(b*x + a) + c^10 + 2*(sqrt(2)*b^3*c^5*d^4*(c^6/(b^4*d^6))^(3/4)*cos(b*x + a
) + sqrt(2)*b*c^8*d*(c^6/(b^4*d^6))^(1/4)*sin(b*x + a))*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a)))*(sqrt(2)*b^
3*d^4*(c^6/(b^4*d^6))^(3/4)*sin(b*x + a) + sqrt(2)*b*c^3*d*(c^6/(b^4*d^6))^(1/4)*cos(b*x + a))*sqrt(d*cos(b*x
+ a))*sqrt(c*sin(b*x + a)) + (sqrt(2)*b^3*c^5*d^4*(c^6/(b^4*d^6))^(3/4)*sin(b*x + a) + sqrt(2)*b*c^8*d*(c^6/(b
^4*d^6))^(1/4)*cos(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/(b^4*d^6)))/((2*c^10*cos(b*x + a)^3 - c^10*cos(b*x + a))*sin(b*x + a)))*cos(b*
x + a) + 2*sqrt(2)*b*d^2*(c^6/(b^4*d^6))^(1/4)*arctan(-1/2*(2*c^10*cos(b*x + a)*sin(b*x + a) - sqrt(4*b^2*c^7*
d^3*sqrt(c^6/(b^4*d^6))*cos(b*x + a)*sin(b*x + a) + c^10 - 2*(sqrt(2)*b^3*c^5*d^4*(c^6/(b^4*d^6))^(3/4)*cos(b*
x + a) + sqrt(2)*b*c^8*d*(c^6/(b^4*d^6))^(1/4)*sin(b*x + a))*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a)))*(sqrt(
2)*b^3*d^4*(c^6/(b^4*d^6))^(3/4)*sin(b*x + a) + sqrt(2)*b*c^3*d*(c^6/(b^4*d^6))^(1/4)*cos(b*x + a))*sqrt(d*cos
(b*x + a))*sqrt(c*sin(b*x + a)) - (sqrt(2)*b^3*c^5*d^4*(c^6/(b^4*d^6))^(3/4)*sin(b*x + a) + sqrt(2)*b*c^8*d*(c
^6/(b^4*d^6))^(1/4)*cos(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/(b^4*d^6)))/((2*c^10*cos(b*x + a)^3 - c^10*cos(b*x + a))*sin(b*x + a)))*c
os(b*x + a) - 2*sqrt(2)*b*d^2*(c^6/(b^4*d^6))^(1/4)*arctan(1/2*((sqrt(2)*b^3*c^5*d^4*(c^6/(b^4*d^6))^(3/4)*sin
(b*x + a) - sqrt(2)*b*c^8*d*(c^6/(b^4*d^6))^(1/4)*cos(b*x + a))*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a)) + sq
rt(4*b^2*c^7*d^3*sqrt(c^6/(b^4*d^6))*cos(b*x + a)*sin(b*x + a) + c^10 - 2*(sqrt(2)*b^3*c^5*d^4*(c^6/(b^4*d^6))
^(3/4)*cos(b*x + a) + sqrt(2)*b*c^8*d*(c^6/(b^4*d^6))^(1/4)*sin(b*x + a))*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x
+ a)))*(2*c^5*cos(b*x + a)*sin(b*x + a) + (sqrt(2)*b^3*d^4*(c^6/(b^4*d^6))^(3/4)*sin(b*x + a) + sqrt(2)*b*c^3*
d*(c^6/(b^4*d^6))^(1/4)*cos(b*x + a))*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a))))/(c^10*cos(b*x + a)*sin(b*x +
a)))*cos(b*x + a) - 2*sqrt(2)*b*d^2*(c^6/(b^4*d^6))^(1/4)*arctan(1/2*((sqrt(2)*b^3*c^5*d^4*(c^6/(b^4*d^6))^(3
/4)*sin(b*x + a) - sqrt(2)*b*c^8*d*(c^6/(b^4*d^6))^(1/4)*cos(b*x + a))*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a
)) - sqrt(4*b^2*c^7*d^3*sqrt(c^6/(b^4*d^6))*cos(b*x + a)*sin(b*x + a) + c^10 + 2*(sqrt(2)*b^3*c^5*d^4*(c^6/(b^
4*d^6))^(3/4)*cos(b*x + a) + sqrt(2)*b*c^8*d*(c^6/(b^4*d^6))^(1/4)*sin(b*x + a))*sqrt(d*cos(b*x + a))*sqrt(c*s
in(b*x + a)))*(2*c^5*cos(b*x + a)*sin(b*x + a) - (sqrt(2)*b^3*d^4*(c^6/(b^4*d^6))^(3/4)*sin(b*x + a) + sqrt(2)
*b*c^3*d*(c^6/(b^4*d^6))^(1/4)*cos(b*x + a))*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a))))/(c^10*cos(b*x + a)*si
n(b*x + a)))*cos(b*x + a) - sqrt(2)*b*d^2*(c^6/(b^4*d^6))^(1/4)*cos(b*x + a)*log(4*b^2*c^7*d^3*sqrt(c^6/(b^4*d
^6))*cos(b*x + a)*sin(b*x + a) + c^10 + 2*(sqrt(2)*b^3*c^5*d^4*(c^6/(b^4*d^6))^(3/4)*cos(b*x + a) + sqrt(2)*b*
c^8*d*(c^6/(b^4*d^6))^(1/4)*sin(b*x + a))*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a))) + sqrt(2)*b*d^2*(c^6/(b^4
*d^6))^(1/4)*cos(b*x + a)*log(4*b^2*c^7*d^3*sqrt(c^6/(b^4*d^6))*cos(b*x + a)*sin(b*x + a) + c^10 - 2*(sqrt(2)*
b^3*c^5*d^4*(c^6/(b^4*d^6))^(3/4)*cos(b*x + a) + sqrt(2)*b*c^8*d*(c^6/(b^4*d^6))^(1/4)*sin(b*x + a))*sqrt(d*co
s(b*x + a))*sqrt(c*sin(b*x + a))) + 16*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a))*c)/(b*d^2*cos(b*x + a))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.13, size = 642, normalized size = 2.05 \[ \frac {\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 )+\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 )+2 \cos \left (b x +a \right ) \sqrt {2}-2 \sqrt {2}\right ) \left (c \sin \left (b x +a \right )\right )^{\frac {3}{2}} \cos \left (b x +a \right ) \sqrt {2}}{2 b \left (-1+\cos \left (b x +a \right )\right ) \left (d \cos \left (b x +a \right )\right )^{\frac {3}{2}} \sin \left (b x +a \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/b*(I*((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)-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))+s
in(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))+2*cos(b*x+a)*2^(1/
2)-2*2^(1/2))*(c*sin(b*x+a))^(3/2)*cos(b*x+a)/(-1+cos(b*x+a))/(d*cos(b*x+a))^(3/2)/sin(b*x+a)*2^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((c*sin(a + b*x))**(3/2)/(d*cos(a + b*x))**(3/2), x)

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