3.477 \(\int \sec ^3(c+d x) \sqrt {a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=124 \[ -\frac {(2 a-b) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{4 d \sqrt {a-b}}+\frac {(2 a+b) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{4 d \sqrt {a+b}}+\frac {\tan (c+d x) \sec (c+d x) \sqrt {a+b \sin (c+d x)}}{2 d} \]

[Out]

-1/4*(2*a-b)*arctanh((a+b*sin(d*x+c))^(1/2)/(a-b)^(1/2))/d/(a-b)^(1/2)+1/4*(2*a+b)*arctanh((a+b*sin(d*x+c))^(1
/2)/(a+b)^(1/2))/d/(a+b)^(1/2)+1/2*sec(d*x+c)*(a+b*sin(d*x+c))^(1/2)*tan(d*x+c)/d

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Rubi [A]  time = 0.17, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {2668, 737, 827, 1166, 206} \[ -\frac {(2 a-b) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{4 d \sqrt {a-b}}+\frac {(2 a+b) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{4 d \sqrt {a+b}}+\frac {\tan (c+d x) \sec (c+d x) \sqrt {a+b \sin (c+d x)}}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3*Sqrt[a + b*Sin[c + d*x]],x]

[Out]

-((2*a - b)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a - b]])/(4*Sqrt[a - b]*d) + ((2*a + b)*ArcTanh[Sqrt[a + b*S
in[c + d*x]]/Sqrt[a + b]])/(4*Sqrt[a + b]*d) + (Sec[c + d*x]*Sqrt[a + b*Sin[c + d*x]]*Tan[c + d*x])/(2*d)

Rule 206

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

Rule 737

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(x*(d + e*x)^m*(a + c*x^2)^(p + 1)
)/(2*a*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(d + e*x)^(m - 1)*(d*(2*p + 3) + e*(m + 2*p + 3)*x)*(a + c*x^2
)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0] && (LtQ[m, 1]
|| (ILtQ[m + 2*p + 3, 0] && NeQ[m, 2])) && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 827

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2, Subst[Int[(e*f
- d*g + g*x^2)/(c*d^2 + a*e^2 - 2*c*d*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0]

Rule 1166

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

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps

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

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Mathematica [A]  time = 0.71, size = 143, normalized size = 1.15 \[ \frac {(a-b) \left (\sqrt {a+b} (2 a+b) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )+2 (a+b) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sin (c+d x)}\right )-\sqrt {a-b} \left (2 a^2+a b-b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{4 d \left (a^2-b^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3*Sqrt[a + b*Sin[c + d*x]],x]

[Out]

(-(Sqrt[a - b]*(2*a^2 + a*b - b^2)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a - b]]) + (a - b)*(Sqrt[a + b]*(2*a
+ b)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a + b]] + 2*(a + b)*Sec[c + d*x]*Sqrt[a + b*Sin[c + d*x]]*Tan[c + d
*x]))/(4*(a^2 - b^2)*d)

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fricas [B]  time = 1.32, size = 2101, normalized size = 16.94 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/32*((2*a^2 - a*b - b^2)*sqrt(a + b)*cos(d*x + c)^2*log((b^4*cos(d*x + c)^4 + 128*a^4 + 256*a^3*b + 320*a^2*
b^2 + 256*a*b^3 + 72*b^4 - 8*(20*a^2*b^2 + 28*a*b^3 + 9*b^4)*cos(d*x + c)^2 + 8*(16*a^3 + 24*a^2*b + 20*a*b^2
+ 8*b^3 - (10*a*b^2 + 7*b^3)*cos(d*x + c)^2 - (b^3*cos(d*x + c)^2 - 24*a^2*b - 28*a*b^2 - 8*b^3)*sin(d*x + c))
*sqrt(b*sin(d*x + c) + a)*sqrt(a + b) + 4*(64*a^3*b + 112*a^2*b^2 + 64*a*b^3 + 14*b^4 - (8*a*b^3 + 7*b^4)*cos(
d*x + c)^2)*sin(d*x + c))/(cos(d*x + c)^4 - 8*cos(d*x + c)^2 + 4*(cos(d*x + c)^2 - 2)*sin(d*x + c) + 8)) - (2*
a^2 + a*b - b^2)*sqrt(a - b)*cos(d*x + c)^2*log((b^4*cos(d*x + c)^4 + 128*a^4 - 256*a^3*b + 320*a^2*b^2 - 256*
a*b^3 + 72*b^4 - 8*(20*a^2*b^2 - 28*a*b^3 + 9*b^4)*cos(d*x + c)^2 + 8*(16*a^3 - 24*a^2*b + 20*a*b^2 - 8*b^3 -
(10*a*b^2 - 7*b^3)*cos(d*x + c)^2 - (b^3*cos(d*x + c)^2 - 24*a^2*b + 28*a*b^2 - 8*b^3)*sin(d*x + c))*sqrt(b*si
n(d*x + c) + a)*sqrt(a - b) + 4*(64*a^3*b - 112*a^2*b^2 + 64*a*b^3 - 14*b^4 - (8*a*b^3 - 7*b^4)*cos(d*x + c)^2
)*sin(d*x + c))/(cos(d*x + c)^4 - 8*cos(d*x + c)^2 - 4*(cos(d*x + c)^2 - 2)*sin(d*x + c) + 8)) + 16*(a^2 - b^2
)*sqrt(b*sin(d*x + c) + a)*sin(d*x + c))/((a^2 - b^2)*d*cos(d*x + c)^2), -1/32*(2*(2*a^2 - a*b - b^2)*sqrt(-a
- b)*arctan(-1/4*(b^2*cos(d*x + c)^2 - 8*a^2 - 8*a*b - 2*b^2 - 2*(4*a*b + 3*b^2)*sin(d*x + c))*sqrt(b*sin(d*x
+ c) + a)*sqrt(-a - b)/(2*a^3 + 3*a^2*b + 2*a*b^2 + b^3 - (a*b^2 + b^3)*cos(d*x + c)^2 + (3*a^2*b + 4*a*b^2 +
b^3)*sin(d*x + c)))*cos(d*x + c)^2 + (2*a^2 + a*b - b^2)*sqrt(a - b)*cos(d*x + c)^2*log((b^4*cos(d*x + c)^4 +
128*a^4 - 256*a^3*b + 320*a^2*b^2 - 256*a*b^3 + 72*b^4 - 8*(20*a^2*b^2 - 28*a*b^3 + 9*b^4)*cos(d*x + c)^2 + 8*
(16*a^3 - 24*a^2*b + 20*a*b^2 - 8*b^3 - (10*a*b^2 - 7*b^3)*cos(d*x + c)^2 - (b^3*cos(d*x + c)^2 - 24*a^2*b + 2
8*a*b^2 - 8*b^3)*sin(d*x + c))*sqrt(b*sin(d*x + c) + a)*sqrt(a - b) + 4*(64*a^3*b - 112*a^2*b^2 + 64*a*b^3 - 1
4*b^4 - (8*a*b^3 - 7*b^4)*cos(d*x + c)^2)*sin(d*x + c))/(cos(d*x + c)^4 - 8*cos(d*x + c)^2 - 4*(cos(d*x + c)^2
- 2)*sin(d*x + c) + 8)) - 16*(a^2 - b^2)*sqrt(b*sin(d*x + c) + a)*sin(d*x + c))/((a^2 - b^2)*d*cos(d*x + c)^2
), -1/32*(2*(2*a^2 + a*b - b^2)*sqrt(-a + b)*arctan(1/4*(b^2*cos(d*x + c)^2 - 8*a^2 + 8*a*b - 2*b^2 - 2*(4*a*b
- 3*b^2)*sin(d*x + c))*sqrt(b*sin(d*x + c) + a)*sqrt(-a + b)/(2*a^3 - 3*a^2*b + 2*a*b^2 - b^3 - (a*b^2 - b^3)
*cos(d*x + c)^2 + (3*a^2*b - 4*a*b^2 + b^3)*sin(d*x + c)))*cos(d*x + c)^2 - (2*a^2 - a*b - b^2)*sqrt(a + b)*co
s(d*x + c)^2*log((b^4*cos(d*x + c)^4 + 128*a^4 + 256*a^3*b + 320*a^2*b^2 + 256*a*b^3 + 72*b^4 - 8*(20*a^2*b^2
+ 28*a*b^3 + 9*b^4)*cos(d*x + c)^2 + 8*(16*a^3 + 24*a^2*b + 20*a*b^2 + 8*b^3 - (10*a*b^2 + 7*b^3)*cos(d*x + c)
^2 - (b^3*cos(d*x + c)^2 - 24*a^2*b - 28*a*b^2 - 8*b^3)*sin(d*x + c))*sqrt(b*sin(d*x + c) + a)*sqrt(a + b) + 4
*(64*a^3*b + 112*a^2*b^2 + 64*a*b^3 + 14*b^4 - (8*a*b^3 + 7*b^4)*cos(d*x + c)^2)*sin(d*x + c))/(cos(d*x + c)^4
- 8*cos(d*x + c)^2 + 4*(cos(d*x + c)^2 - 2)*sin(d*x + c) + 8)) - 16*(a^2 - b^2)*sqrt(b*sin(d*x + c) + a)*sin(
d*x + c))/((a^2 - b^2)*d*cos(d*x + c)^2), -1/16*((2*a^2 + a*b - b^2)*sqrt(-a + b)*arctan(1/4*(b^2*cos(d*x + c)
^2 - 8*a^2 + 8*a*b - 2*b^2 - 2*(4*a*b - 3*b^2)*sin(d*x + c))*sqrt(b*sin(d*x + c) + a)*sqrt(-a + b)/(2*a^3 - 3*
a^2*b + 2*a*b^2 - b^3 - (a*b^2 - b^3)*cos(d*x + c)^2 + (3*a^2*b - 4*a*b^2 + b^3)*sin(d*x + c)))*cos(d*x + c)^2
+ (2*a^2 - a*b - b^2)*sqrt(-a - b)*arctan(-1/4*(b^2*cos(d*x + c)^2 - 8*a^2 - 8*a*b - 2*b^2 - 2*(4*a*b + 3*b^2
)*sin(d*x + c))*sqrt(b*sin(d*x + c) + a)*sqrt(-a - b)/(2*a^3 + 3*a^2*b + 2*a*b^2 + b^3 - (a*b^2 + b^3)*cos(d*x
+ c)^2 + (3*a^2*b + 4*a*b^2 + b^3)*sin(d*x + c)))*cos(d*x + c)^2 - 8*(a^2 - b^2)*sqrt(b*sin(d*x + c) + a)*sin
(d*x + c))/((a^2 - b^2)*d*cos(d*x + c)^2)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*sin(d*x + c) + a)*sec(d*x + c)^3, x)

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maple [A]  time = 0.61, size = 185, normalized size = 1.49 \[ \frac {2 \sqrt {a +b \sin \left (d x +c \right )}\, \sqrt {-a +b}\, \sqrt {a +b}\, \sin \left (d x +c \right )-\left (-2 \arctanh \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right ) a \sqrt {-a +b}-\arctanh \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right ) b \sqrt {-a +b}-2 \arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right ) a \sqrt {a +b}+\arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right ) b \sqrt {a +b}\right ) \left (\cos ^{2}\left (d x +c \right )\right )}{4 \sqrt {-a +b}\, \sqrt {a +b}\, \cos \left (d x +c \right )^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a-4*b>0)', see `assume?` for
more details)Is 4*a-4*b positive or negative?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {a+b\,\sin \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a + b*sin(c + d*x))*sec(c + d*x)**3, x)

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