3.494 \(\int (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=142 \[ \frac{2 a^2 (15 A+20 B+12 C) \tan (c+d x)}{15 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a^{3/2} A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{d}+\frac{2 a (5 B+3 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{15 d}+\frac{2 C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d} \]

[Out]

(2*a^(3/2)*A*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (2*a^2*(15*A + 20*B + 12*C)*Tan[c +
d*x])/(15*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(5*B + 3*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(15*d) + (2*C*
(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d)

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Rubi [A]  time = 0.23624, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {4054, 3917, 3915, 3774, 203, 3792} \[ \frac{2 a^2 (15 A+20 B+12 C) \tan (c+d x)}{15 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a^{3/2} A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{d}+\frac{2 a (5 B+3 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{15 d}+\frac{2 C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(2*a^(3/2)*A*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (2*a^2*(15*A + 20*B + 12*C)*Tan[c +
d*x])/(15*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(5*B + 3*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(15*d) + (2*C*
(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d)

Rule 4054

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) +
(a_))^(m_.), x_Symbol] :> -Simp[(C*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[1/(b*(m + 1)),
Int[(a + b*Csc[e + f*x])^m*Simp[A*b*(m + 1) + (a*C*m + b*B*(m + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b,
e, f, A, B, C, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)]

Rule 3917

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> -Simp[(b*
d*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1))/(f*m), x] + Dist[1/m, Int[(a + b*Csc[e + f*x])^(m - 1)*Simp[a*c*m
+ (b*c*m + a*d*(2*m - 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && Gt
Q[m, 1] && EqQ[a^2 - b^2, 0] && IntegerQ[2*m]

Rule 3915

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Dist[c, In
t[Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[d, Int[Sqrt[a + b*Csc[e + f*x]]*Csc[e + f*x], x], x] /; FreeQ[{a, b,
c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0]

Rule 3774

Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(-2*b)/d, Subst[Int[1/(a + x^2), x], x, (b*C
ot[c + d*x])/Sqrt[a + b*Csc[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 203

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

Rule 3792

Int[csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*b*Cot[e + f*x])/
(f*Sqrt[a + b*Csc[e + f*x]]), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin{align*} \int (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac{2 C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac{2 \int (a+a \sec (c+d x))^{3/2} \left (\frac{5 a A}{2}+\frac{1}{2} a (5 B+3 C) \sec (c+d x)\right ) \, dx}{5 a}\\ &=\frac{2 a (5 B+3 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac{2 C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac{4 \int \sqrt{a+a \sec (c+d x)} \left (\frac{15 a^2 A}{4}+\frac{1}{4} a^2 (15 A+20 B+12 C) \sec (c+d x)\right ) \, dx}{15 a}\\ &=\frac{2 a (5 B+3 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac{2 C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+(a A) \int \sqrt{a+a \sec (c+d x)} \, dx+\frac{1}{15} (a (15 A+20 B+12 C)) \int \sec (c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{2 a^2 (15 A+20 B+12 C) \tan (c+d x)}{15 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a (5 B+3 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac{2 C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}-\frac{\left (2 a^2 A\right ) \operatorname{Subst}\left (\int \frac{1}{a+x^2} \, dx,x,-\frac{a \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{d}\\ &=\frac{2 a^{3/2} A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{d}+\frac{2 a^2 (15 A+20 B+12 C) \tan (c+d x)}{15 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a (5 B+3 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac{2 C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 1.35286, size = 132, normalized size = 0.93 \[ \frac{a \sec \left (\frac{1}{2} (c+d x)\right ) \sec ^2(c+d x) \sqrt{a (\sec (c+d x)+1)} \left (2 \sin \left (\frac{1}{2} (c+d x)\right ) ((15 A+25 B+18 C) \cos (2 (c+d x))+15 A+2 (5 B+9 C) \cos (c+d x)+25 B+24 C)+30 \sqrt{2} A \sin ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right ) \cos ^{\frac{5}{2}}(c+d x)\right )}{30 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(a*Sec[(c + d*x)/2]*Sec[c + d*x]^2*Sqrt[a*(1 + Sec[c + d*x])]*(30*Sqrt[2]*A*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]]*C
os[c + d*x]^(5/2) + 2*(15*A + 25*B + 24*C + 2*(5*B + 9*C)*Cos[c + d*x] + (15*A + 25*B + 18*C)*Cos[2*(c + d*x)]
)*Sin[(c + d*x)/2]))/(30*d)

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Maple [B]  time = 0.322, size = 361, normalized size = 2.5 \begin{align*} -{\frac{a}{60\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) }\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}} \left ( 15\,A\sqrt{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{5/2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}} \right ) +30\,A\sqrt{2}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{5/2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}} \right ) +15\,A \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{5/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}} \right ) \sin \left ( dx+c \right ) +120\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+200\,B \left ( \cos \left ( dx+c \right ) \right ) ^{3}+144\,C \left ( \cos \left ( dx+c \right ) \right ) ^{3}-120\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}-160\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}-72\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}-40\,B\cos \left ( dx+c \right ) -48\,C\cos \left ( dx+c \right ) -24\,C \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

-1/60/d*a*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(15*A*2^(1/2)*cos(d*x+c)^2*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)
+1))^(5/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))+30*A*2^(1/2)*cos(d*
x+c)*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*
sin(d*x+c)/cos(d*x+c))+15*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*2^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(c
os(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*sin(d*x+c)+120*A*cos(d*x+c)^3+200*B*cos(d*x+c)^3+144*C*cos(d*x+c)^3
-120*A*cos(d*x+c)^2-160*B*cos(d*x+c)^2-72*C*cos(d*x+c)^2-40*B*cos(d*x+c)-48*C*cos(d*x+c)-24*C)/cos(d*x+c)^2/si
n(d*x+c)

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Maxima [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((a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.573072, size = 944, normalized size = 6.65 \begin{align*} \left [\frac{15 \,{\left (A a \cos \left (d x + c\right )^{3} + A a \cos \left (d x + c\right )^{2}\right )} \sqrt{-a} \log \left (\frac{2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt{-a} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \,{\left ({\left (15 \, A + 25 \, B + 18 \, C\right )} a \cos \left (d x + c\right )^{2} +{\left (5 \, B + 9 \, C\right )} a \cos \left (d x + c\right ) + 3 \, C a\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{15 \,{\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}, -\frac{2 \,{\left (15 \,{\left (A a \cos \left (d x + c\right )^{3} + A a \cos \left (d x + c\right )^{2}\right )} \sqrt{a} \arctan \left (\frac{\sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt{a} \sin \left (d x + c\right )}\right ) -{\left ({\left (15 \, A + 25 \, B + 18 \, C\right )} a \cos \left (d x + c\right )^{2} +{\left (5 \, B + 9 \, C\right )} a \cos \left (d x + c\right ) + 3 \, C a\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )\right )}}{15 \,{\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

[1/15*(15*(A*a*cos(d*x + c)^3 + A*a*cos(d*x + c)^2)*sqrt(-a)*log((2*a*cos(d*x + c)^2 - 2*sqrt(-a)*sqrt((a*cos(
d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) + 2*((15*A + 2
5*B + 18*C)*a*cos(d*x + c)^2 + (5*B + 9*C)*a*cos(d*x + c) + 3*C*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin
(d*x + c))/(d*cos(d*x + c)^3 + d*cos(d*x + c)^2), -2/15*(15*(A*a*cos(d*x + c)^3 + A*a*cos(d*x + c)^2)*sqrt(a)*
arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - ((15*A + 25*B + 18*C)*a*
cos(d*x + c)^2 + (5*B + 9*C)*a*cos(d*x + c) + 3*C*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*
cos(d*x + c)^3 + d*cos(d*x + c)^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \left (\sec{\left (c + d x \right )} + 1\right )\right )^{\frac{3}{2}} \left (A + B \sec{\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))**(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Integral((a*(sec(c + d*x) + 1))**(3/2)*(A + B*sec(c + d*x) + C*sec(c + d*x)**2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError