3.3.62 \(\int (d \cos (a+b x))^{3/2} \sqrt {c \sin (a+b x)} \, dx\) [262]

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

[Out]

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

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Rubi [A]
time = 0.18, 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 = {2649, 2654, 303, 1176, 631, 210, 1179, 642} \begin {gather*} -\frac {\sqrt {c} d^{3/2} \text {ArcTan}\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} \text {ArcTan}\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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(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]])])/(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]] + Sq
rt[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]])/Sqr
t[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 210

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

Rule 303

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 631

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 642

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 1176

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 1179

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 2649

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 2654

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]

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 ) \text {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 ) \text {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 ) \text {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) \text {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) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.08, size = 70, normalized size = 0.22 \begin {gather*} \frac {2 d^2 \cos ^2(a+b x)^{3/4} \, _2F_1\left (-\frac {1}{4},\frac {3}{4};\frac {7}{4};\sin ^2(a+b x)\right ) \sqrt {c \sin (a+b x)} \tan (a+b x)}{3 b \sqrt {d \cos (a+b x)}} \end {gather*}

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] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.10, size = 514, normalized size = 1.61

method	result	size
default	\(\frac {\left (-i \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 \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 )+\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 )+\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 ^{2}\left (b x +a \right )\right ) \sqrt {2}-2 \cos \left (b x +a \right ) \sqrt {2}\right ) \left (d \cos \left (b x +a \right )\right )^{\frac {3}{2}} \sqrt {c \sin \left (b x +a \right )}\, \sin \left (b x +a \right ) \sqrt {2}}{8 b \left (-1+\cos \left (b x +a \right )\right ) \cos \left (b x +a \right )^{2}}\)	\(514\)

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,method=_RETURNVERBOSE)

[Out]

1/8/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))+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)*El
lipticPi(((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-co
s(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*2^(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((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 [B] Leaf count of result is larger than twice the leaf count of optimal. 2205 vs. \(2 (238) = 476\).
time = 28.54, size = 2205, normalized size = 6.89 \begin {gather*} \text {Too large to display} \end {gather*}

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]

1/64*(32*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a))*d*sin(b*x + a) + 4*sqrt(2)*(c^2*d^6/b^4)^(1/4)*b*arctan(((s
qrt(2)*(c^2*d^6/b^4)^(1/4)*b*c^3*d^8*cos(b*x + a) + sqrt(2)*(c^2*d^6/b^4)^(3/4)*b^3*c^2*d^5*sin(b*x + a))*sqrt
(d*cos(b*x + a))*sqrt(c*sin(b*x + a)) + sqrt(4*sqrt(c^2*d^6/b^4)*b^2*c^3*d^7*cos(b*x + a)*sin(b*x + a) + c^4*d
^10 - 2*(sqrt(2)*(c^2*d^6/b^4)^(1/4)*b*c^3*d^8*cos(b*x + a) + sqrt(2)*(c^2*d^6/b^4)^(3/4)*b^3*c^2*d^5*sin(b*x
+ a))*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a)))*(2*c^2*d^5*cos(b*x + a)*sin(b*x + a) + sqrt(c^2*d^6/b^4)*b^2*
c*d^2 + (sqrt(2)*(c^2*d^6/b^4)^(1/4)*b*c*d^3*sin(b*x + a) + sqrt(2)*(c^2*d^6/b^4)^(3/4)*b^3*cos(b*x + a))*sqrt
(d*cos(b*x + a))*sqrt(c*sin(b*x + a))))/(2*c^4*d^10*cos(b*x + a)^2 - c^4*d^10)) + 4*sqrt(2)*(c^2*d^6/b^4)^(1/4
)*b*arctan(((sqrt(2)*(c^2*d^6/b^4)^(1/4)*b*c^3*d^8*cos(b*x + a) + sqrt(2)*(c^2*d^6/b^4)^(3/4)*b^3*c^2*d^5*sin(
b*x + a))*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a)) - sqrt(4*sqrt(c^2*d^6/b^4)*b^2*c^3*d^7*cos(b*x + a)*sin(b*
x + a) + c^4*d^10 + 2*(sqrt(2)*(c^2*d^6/b^4)^(1/4)*b*c^3*d^8*cos(b*x + a) + sqrt(2)*(c^2*d^6/b^4)^(3/4)*b^3*c^
2*d^5*sin(b*x + a))*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a)))*(2*c^2*d^5*cos(b*x + a)*sin(b*x + a) + sqrt(c^2
*d^6/b^4)*b^2*c*d^2 - (sqrt(2)*(c^2*d^6/b^4)^(1/4)*b*c*d^3*sin(b*x + a) + sqrt(2)*(c^2*d^6/b^4)^(3/4)*b^3*cos(
b*x + a))*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a))))/(2*c^4*d^10*cos(b*x + a)^2 - c^4*d^10)) + 4*sqrt(2)*(c^2
*d^6/b^4)^(1/4)*b*arctan(-1/2*(2*c^4*d^10*cos(b*x + a)*sin(b*x + a) - sqrt(4*sqrt(c^2*d^6/b^4)*b^2*c^3*d^7*cos
(b*x + a)*sin(b*x + a) + c^4*d^10 + 2*(sqrt(2)*(c^2*d^6/b^4)^(1/4)*b*c^3*d^8*cos(b*x + a) + sqrt(2)*(c^2*d^6/b
^4)^(3/4)*b^3*c^2*d^5*sin(b*x + a))*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a)))*(sqrt(2)*(c^2*d^6/b^4)^(1/4)*b*
c*d^3*sin(b*x + a) + sqrt(2)*(c^2*d^6/b^4)^(3/4)*b^3*cos(b*x + a))*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a)) +
(sqrt(2)*(c^2*d^6/b^4)^(1/4)*b*c^3*d^8*sin(b*x + a) + sqrt(2)*(c^2*d^6/b^4)^(3/4)*b^3*c^2*d^5*cos(b*x + a))*s
qrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a)) - 4*(b^2*c^3*d^7*cos(b*x + a)^4 - b^2*c^3*d^7*cos(b*x + a)^2)*sqrt(c^
2*d^6/b^4))/((2*c^4*d^10*cos(b*x + a)^3 - c^4*d^10*cos(b*x + a))*sin(b*x + a))) + 4*sqrt(2)*(c^2*d^6/b^4)^(1/4
)*b*arctan(1/2*(2*c^4*d^10*cos(b*x + a)*sin(b*x + a) + sqrt(4*sqrt(c^2*d^6/b^4)*b^2*c^3*d^7*cos(b*x + a)*sin(b
*x + a) + c^4*d^10 - 2*(sqrt(2)*(c^2*d^6/b^4)^(1/4)*b*c^3*d^8*cos(b*x + a) + sqrt(2)*(c^2*d^6/b^4)^(3/4)*b^3*c
^2*d^5*sin(b*x + a))*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a)))*(sqrt(2)*(c^2*d^6/b^4)^(1/4)*b*c*d^3*sin(b*x +
a) + sqrt(2)*(c^2*d^6/b^4)^(3/4)*b^3*cos(b*x + a))*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a)) - (sqrt(2)*(c^2*
d^6/b^4)^(1/4)*b*c^3*d^8*sin(b*x + a) + sqrt(2)*(c^2*d^6/b^4)^(3/4)*b^3*c^2*d^5*cos(b*x + a))*sqrt(d*cos(b*x +
a))*sqrt(c*sin(b*x + a)) - 4*(b^2*c^3*d^7*cos(b*x + a)^4 - b^2*c^3*d^7*cos(b*x + a)^2)*sqrt(c^2*d^6/b^4))/((2
*c^4*d^10*cos(b*x + a)^3 - c^4*d^10*cos(b*x + a))*sin(b*x + a))) - sqrt(2)*(c^2*d^6/b^4)^(1/4)*b*log(4*sqrt(c^
2*d^6/b^4)*b^2*c^3*d^7*cos(b*x + a)*sin(b*x + a) + c^4*d^10 + 2*(sqrt(2)*(c^2*d^6/b^4)^(1/4)*b*c^3*d^8*cos(b*x
+ a) + sqrt(2)*(c^2*d^6/b^4)^(3/4)*b^3*c^2*d^5*sin(b*x + a))*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a))) + sqr
t(2)*(c^2*d^6/b^4)^(1/4)*b*log(4*sqrt(c^2*d^6/b^4)*b^2*c^3*d^7*cos(b*x + a)*sin(b*x + a) + c^4*d^10 - 2*(sqrt(
2)*(c^2*d^6/b^4)^(1/4)*b*c^3*d^8*cos(b*x + a) + sqrt(2)*(c^2*d^6/b^4)^(3/4)*b^3*c^2*d^5*sin(b*x + a))*sqrt(d*c
os(b*x + a))*sqrt(c*sin(b*x + a))) - sqrt(2)*(c^2*d^6/b^4)^(1/4)*b*log(1/4*sqrt(c^2*d^6/b^4)*b^2*c^3*d^7*cos(b
*x + a)*sin(b*x + a) + 1/16*c^4*d^10 + 1/8*(sqrt(2)*(c^2*d^6/b^4)^(1/4)*b*c^3*d^8*cos(b*x + a) + sqrt(2)*(c^2*
d^6/b^4)^(3/4)*b^3*c^2*d^5*sin(b*x + a))*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a))) + sqrt(2)*(c^2*d^6/b^4)^(1
/4)*b*log(1/4*sqrt(c^2*d^6/b^4)*b^2*c^3*d^7*cos(b*x + a)*sin(b*x + a) + 1/16*c^4*d^10 - 1/8*(sqrt(2)*(c^2*d^6/
b^4)^(1/4)*b*c^3*d^8*cos(b*x + a) + sqrt(2)*(c^2*d^6/b^4)^(3/4)*b^3*c^2*d^5*sin(b*x + a))*sqrt(d*cos(b*x + a))
*sqrt(c*sin(b*x + a))))/b

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

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]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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