Optimal. Leaf size=264 \[ \frac {\left (a^2 (-B)+4 a A b-8 b^2 B\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{4 b^{3/2} d}+\frac {\sqrt {-b+i a} (-B+i A) \tan ^{-1}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {(4 A b-a B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 b d}+\frac {\sqrt {b+i a} (B+i A) \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 b d} \]
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Rubi [A] time = 1.91, antiderivative size = 264, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3607, 3647, 3655, 6725, 63, 217, 206, 93, 205, 208} \[ \frac {\left (a^2 (-B)+4 a A b-8 b^2 B\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{4 b^{3/2} d}+\frac {\sqrt {-b+i a} (-B+i A) \tan ^{-1}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {(4 A b-a B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 b d}+\frac {\sqrt {b+i a} (B+i A) \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 b d} \]
Antiderivative was successfully verified.
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Rule 63
Rule 93
Rule 205
Rule 206
Rule 208
Rule 217
Rule 3607
Rule 3647
Rule 3655
Rule 6725
Rubi steps
\begin {align*} \int \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx &=\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 b d}+\frac {\int \frac {\sqrt {a+b \tan (c+d x)} \left (-\frac {a B}{2}-2 b B \tan (c+d x)+\frac {1}{2} (4 A b-a B) \tan ^2(c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx}{2 b}\\ &=\frac {(4 A b-a B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 b d}+\frac {\int \frac {-\frac {1}{4} a (4 A b+a B)-2 b (A b+a B) \tan (c+d x)+\frac {1}{4} \left (4 a A b-a^2 B-8 b^2 B\right ) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{2 b}\\ &=\frac {(4 A b-a B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 b d}+\frac {\operatorname {Subst}\left (\int \frac {-\frac {1}{4} a (4 A b+a B)-2 b (A b+a B) x+\frac {1}{4} \left (4 a A b-a^2 B-8 b^2 B\right ) x^2}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{2 b d}\\ &=\frac {(4 A b-a B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 b d}+\frac {\operatorname {Subst}\left (\int \left (\frac {4 a A b-a^2 B-8 b^2 B}{4 \sqrt {x} \sqrt {a+b x}}-\frac {2 (b (a A-b B)+b (A b+a B) x)}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{2 b d}\\ &=\frac {(4 A b-a B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 b d}-\frac {\operatorname {Subst}\left (\int \frac {b (a A-b B)+b (A b+a B) x}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{b d}+\frac {\left (4 a A b-a^2 B-8 b^2 B\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{8 b d}\\ &=\frac {(4 A b-a B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 b d}-\frac {\operatorname {Subst}\left (\int \left (\frac {-b (A b+a B)+i b (a A-b B)}{2 (i-x) \sqrt {x} \sqrt {a+b x}}+\frac {b (A b+a B)+i b (a A-b B)}{2 \sqrt {x} (i+x) \sqrt {a+b x}}\right ) \, dx,x,\tan (c+d x)\right )}{b d}+\frac {\left (4 a A b-a^2 B-8 b^2 B\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{4 b d}\\ &=\frac {(4 A b-a B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 b d}-\frac {((i a+b) (A-i B)) \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} (i+x) \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}-\frac {((i a-b) (A+i B)) \operatorname {Subst}\left (\int \frac {1}{(i-x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac {\left (4 a A b-a^2 B-8 b^2 B\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{4 b d}\\ &=\frac {\left (4 a A b-a^2 B-8 b^2 B\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{4 b^{3/2} d}+\frac {(4 A b-a B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 b d}-\frac {((i a+b) (A-i B)) \operatorname {Subst}\left (\int \frac {1}{i-(-a+i b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {((i a-b) (A+i B)) \operatorname {Subst}\left (\int \frac {1}{i-(a+i b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}\\ &=\frac {\sqrt {i a-b} (i A-B) \tan ^{-1}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\left (4 a A b-a^2 B-8 b^2 B\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{4 b^{3/2} d}+\frac {\sqrt {i a+b} (i A+B) \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {(4 A b-a B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 b d}\\ \end {align*}
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Mathematica [A] time = 3.46, size = 304, normalized size = 1.15 \[ \frac {-\sqrt {a} \left (a^2 B-4 a A b+8 b^2 B\right ) \sqrt {\frac {b \tan (c+d x)}{a}+1} \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )+\sqrt {b} \left (-4 \sqrt [4]{-1} b \sqrt {-a+i b} (B+i A) \sqrt {a+b \tan (c+d x)} \tan ^{-1}\left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+4 (-1)^{3/4} b \sqrt {a+i b} (A+i B) \tan ^{-1}\left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {a+b \tan (c+d x)}+\sqrt {\tan (c+d x)} (a+b \tan (c+d x)) (a B+4 A b+2 b B \tan (c+d x))\right )}{4 b^{3/2} d \sqrt {a+b \tan (c+d x)}} \]
Antiderivative was successfully verified.
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fricas [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 1.03, size = 2181075, normalized size = 8261.65 \[ \text {output too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \tan \left (d x + c\right ) + A\right )} \sqrt {b \tan \left (d x + c\right ) + a} \tan \left (d x + c\right )^{\frac {3}{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int {\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (A + B \tan {\left (c + d x \right )}\right ) \sqrt {a + b \tan {\left (c + d x \right )}} \tan ^{\frac {3}{2}}{\left (c + d x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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