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

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

[Out]

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

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Rubi [A]  time = 0.295161, antiderivative size = 320, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {2569, 2574, 297, 1162, 617, 204, 1165, 628} \[ -\frac{\sqrt{c} d^{3/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d} \sqrt{c \sin (a+b x)}}{\sqrt{c} \sqrt{d \cos (a+b x)}}\right )}{4 \sqrt{2} b}+\frac{\sqrt{c} d^{3/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d} \sqrt{c \sin (a+b x)}}{\sqrt{c} \sqrt{d \cos (a+b x)}}+1\right )}{4 \sqrt{2} b}+\frac{\sqrt{c} d^{3/2} \log \left (-\frac{\sqrt{2} \sqrt{d} \sqrt{c \sin (a+b x)}}{\sqrt{d \cos (a+b x)}}+\sqrt{c} \tan (a+b x)+\sqrt{c}\right )}{8 \sqrt{2} b}-\frac{\sqrt{c} d^{3/2} \log \left (\frac{\sqrt{2} \sqrt{d} \sqrt{c \sin (a+b x)}}{\sqrt{d \cos (a+b x)}}+\sqrt{c} \tan (a+b x)+\sqrt{c}\right )}{8 \sqrt{2} b}+\frac{d (c \sin (a+b x))^{3/2} \sqrt{d \cos (a+b x)}}{2 b c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2569

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(b*Sin[e +
f*x])^(n + 1)*(a*Cos[e + f*x])^(m - 1))/(b*f*(m + n)), x] + Dist[(a^2*(m - 1))/(m + n), Int[(b*Sin[e + f*x])^
n*(a*Cos[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 2574

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), 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*Sin[e + f*x])^(1/k)/(b*Cos
[e + f*x])^(1/k)], x]] /; FreeQ[{a, b, e, f}, x] && EqQ[m + n, 0] && GtQ[m, 0] && LtQ[m, 1]

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 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 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 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 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 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]

Rubi steps

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

Mathematica [C]  time = 0.119344, size = 70, normalized size = 0.22 \[ \frac{2 d^2 \cos ^2(a+b x)^{3/4} \tan (a+b x) \sqrt{c \sin (a+b x)} \, _2F_1\left (-\frac{1}{4},\frac{3}{4};\frac{7}{4};\sin ^2(a+b x)\right )}{3 b \sqrt{d \cos (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*d^2*(Cos[a + b*x]^2)^(3/4)*Hypergeometric2F1[-1/4, 3/4, 7/4, Sin[a + b*x]^2]*Sqrt[c*Sin[a + b*x]]*Tan[a + b
*x])/(3*b*Sqrt[d*Cos[a + b*x]])

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Maple [C]  time = 0.079, size = 514, normalized size = 1.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d \cos \left (b x + a\right )\right )^{\frac{3}{2}} \sqrt{c \sin \left (b x + a\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError